Chem32-33-34SaltsBuffer

Chem32-33-34SaltsBuffer - Module 32 — pH of Salts...

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Unformatted text preview: Module 32 — pH of Salts Calculations In Chemistry ***** Module 32: pH of Salts Module 33: Buffers Module 34: pH During Titration Module 32 – pH of Salts................................................................................................ 979 Lesson 32A: Lesson 32B: Lesson 32C: Lesson 32D: The Acid-Base Behavior of Salts...................................................................... 979 Will A Salt Acid-Base React? .......................................................................... 986 Calculating the pH of a Salt Solution ............................................................ 990 Salts That Contain Amphoteric Ions............................................................... 995 Module 33 – Buffers..................................................................................................... 1000 Lesson 33A: Lesson 33B: Lesson 33C: Lesson 33D: Lesson 33E: Lesson 33F: Acid-Base Common Ions, Buffers ................................................................. 1000 Buffer Example ................................................................................................ 1003 Buffer Components ......................................................................................... 1009 Methodical Buffer Calculations..................................................................... 1013 Buffer Quick Steps........................................................................................... 1017 The Henderson-Hasselbalch Equation ......................................................... 1024 Module 34 – pH During Titration ............................................................................. 1031 Lesson 34A: Lesson 34B: Lesson 34C: Lesson 34D: Lesson 34E: pH In Mixtures................................................................................................. 1031 pH After Neutralization ................................................................................. 1036 Distinguishing Types of Acid-Base Calculations........................................ 1046 pH During Strong-Strong Titration .............................................................. 1051 Titration pH: Weak by Strong ...................................................................... 1063 For additional modules, visit www.ChemReview.Net © 2009 ChemReview.Net v. 7u Page i Module 32 — pH of Salts Module 32 — pH of Salts Prerequisites: Complete Modules 30 and 31 before starting this module. ***** Lesson 32A: The Acid-Base Behavior of Salts Timing: Do this lesson if you are assigned problems that ask you to predict the acidity or basicity of salt solutions. ***** Salts Historically in chemistry, salt is a term that has been used to describe the ionic compounds that are a product in acid-base neutralization. In current usage, “salt” can refer to any ionic compound. Salt = Ionic Compound As with all ionic compounds, salts are solids at room temperature. All salts dissolve to some extent in water. All particles that dissolve will be present as ions that can move about freely in the solution. In water, some salts form pH neutral solutions, but others react with water (hydrolyze) to form acidic or basic solutions. Distinguishing the Types of “Neutral” To describe acidic and basic ions, it is necessary to distinguish between the two uses in chemistry for the word neutral. • Particles that have a zero overall charge are termed electrically neutral. Positive or negative ions are particles that are not electrically neutral. • Whether a particle is electrically neutral or is an ion, in an aqueous solution the particle can also be o o pH neutral, also termed acid-base neutral (such particles include H2O and Na+ and Cl─), or not pH neutral and can be acidic (such as HF or NH4+), basic (such as NH3 or F─), or amphoteric (can react as acids or bases, such as HCO3─). To avoid confusion, the terms electrically neutral and pH neutral, rather than simply neutral, are preferred in situations where the meaning of “neutral”may not be clear. Soluble Salts Some ionic compounds are quite soluble in water, but others are only slightly soluble. Recall from Lesson 13A that if compounds contain these ions, they are nearly always soluble in water: • NH4+ , the row 3-7 alkali metal ions (Na+, K+, Rb+, Cs+, Fr+), NO3─, and CH3COO─ except when combined with Ag+. Other ion combinations are also soluble, but the ions above are the most frequently encountered in problems. © 2011 ChemReview.Net v. 7w Page 979 Module 32 — pH of Salts To solve problems in this module, it is especially important to be able to recall from memory the ion combinations that are soluble: those combinations that dissolve and separate into ions essentially 100% at substance concentrations of 0.10 M or less. When a soluble ionic compound dissolves, if the moles of solid dissolved per liter of solution is known, the concentration of the ions in solution can be found by the REC steps (Lessons 7C and 12B). To briefly review, try this problem. Q. Write the chemical formula and concentration for each particle formed when 0.10 M Ca(NO3)2 dissolves in water. ***** Answer Nitrates are ionic compounds that are soluble in water: in dilute solutions, they dissolve and separate into ions essentially 100%. Write the REC steps (Lesson 12B) by inspection. 1 Ca(NO3)2(aq) ^ 0.10 M 0 M Reaction and Extent: Concentrations: 1 Ca2+(aq) + 2 NO3─(aq) (goes ~100%) ^ ^ 0.10 M and 0.20 M formed In this solution are no Ca(NO3)2 particles, 0.10 M Ca2+ ions, and 0.20 M NO3─ ions. Coefficients represent the mole ratios that react and form. However, if the reaction goes to completion and all of the particles in the same solution, the coefficients will also represent the mole per liter (concentration) ratios of the particles used up and formed. Practice A: (For additional review, see Module 7 and Lessons 12B and 13A.) 1. In dilute aqueous solutions, will these dissolve ~100% and dissociate ~100% into ions? State your reasoning. a. KCl b. CH4 c. Ra(NO3)2 d. Sodium acetate f. Calcium phosphate g. Ammonium bromide j. Silver chloride k. PbCl2 h. HCl e. Cl2 i. RbOH l. Lead nitrate 2. Which compound above is a. A strong base? b. An element? c. Radioactive? 3. Write symbols for the ions formed, then beside or below, write the concentration for each ion formed in solution when these salts dissolve in water. a. 0.20 M NH4Cl b. 0.50 M Ba(NO3)2 c. 0.25 M potassium cyanide d. 0.10 M sodium sulfate © 2011 ChemReview.Net v. 7w Page 980 Module 32 — pH of Salts Ions in Aqueous Solution: Acidic, Basic, or Neutral? When mixed with water, an ion has a characteristic acidic, basic, or neutral behavior. An ion that is acidic, when combined in a salt with an ion that is pH-neutral, forms an acidic aqueous solution. A basic ion, in combination with a pH-neutral ion, forms a basic solution in water. To predict whether a salt will form an acidic, basic, or pH-neutral solution, the first step is to classify each of its ions as acidic, basic, or pH neutral. For many ions, whether they have acidic, basic or pH-neutral behavior can be predicted with a Table of Acid Strengths and the following Rules for Identifying Ions As pH-Neutral, Acidic, or Basic The ions in bold below are those encountered most often in problems. 1. pH-neutral ions include a. In rows 3-7 of the periodic table, the column 1 alkali metal ions (Na+, K+, Rb+, Cs+, Fr+) plus the column 2 metal ions: Mg2+, Ca2+, Sr2+, Ba2+, Ra2+. b. Conjugates of strong acids, such as Cl─ and NO3─. HCl and HNO3 are strong acids. Conjugates of strong acids or bases are very weak. Very weak acids and bases are considered to be essentially pH neutral in water. 2. If an ion formula is listed in an acid-strength table, its position in the table will identify whether it has acidic, basic, or neutral behavior when dissolved in water. For an acid-strength table (see Lesson 31B), the rules include: • Each line in the table includes an acid-base conjugate pair. • Acids are listed in the left column and bases after the H+ in the right column. • The strongest acids are at the top left, and strongest bases at the bottom right. If one acid or base particle is strong (K>1), its conjugate is very weak (K<10―16). • • If one member of a conjugate pair is moderately weak (with its Ka or Kb between 1 and ~10―16), the other will also be moderately weak. • The stronger is one, the weaker is its conjugate. a. Acidic ions are those that are • stronger acids than H2O in an acid-strength table ≡ ≡ • which are those with a Ka larger than ~10―16. ≡ ≡ H2O ≡ The acidic ions encountered most frequently are NH4+ and R-NHx+ ions, where R is a group containing carbon and hydrogen. © 2011 ChemReview.Net v. 7w Ka = Large Ka = ~10―16 Page 981 Module 32 — pH of Salts b. Another type of acidic ion is the highly charged ion of a small-radius metal atom, such as Fe3+, Al3+, and Sn4+. These ions, when dissolved in water, form hydrated complexes that can bind to the hydroxide ions formed by the auto-ionization of water. This results in a solution that is acidic because it has more free H+ ions than OH─ ions. c. Basic ions are listed in the right-side column. Ions at the top right are conjugates of very strong acids, and those bases are so weak that they are essentially pH neutral. The strongest bases are at the bottom right: the conjugates of the weakest acids. The ions that have basic behavior in water are • ions listed below H2O on the base side. • which are also the conjugate bases of acids ≡ H3O+ H2O that have a Ka below 1.0. • Ka = Large ≡ ≡ ≡ Ka = # x10―x ≡ which are also bases with Kb larger than ~10―16 . Ka = 1.0 ≡ Basic ions include F─, CN─, CH3COO─, and C6H5COO─. Any ion which is the conjugate of a weak acid has basic behavior. Practice B: First learn the rules, then do these problems. 1. Using the acid-strength table in Lesson 31B, but applying the rules in this lesson from memory, label these ions as acidic (A), basic (B), or pH-neutral (N). a. CH3COO─ b. K+ c. NH4+ d. Al3+ e. F─ f. Cl─ g. Ca2+ 2. Which of the ions in Problem 1, if present in a compound, will always result in the compound to be soluble? 3. In this problem, do not consult the acid-strength table. H+ + ________ ( H+ + SO42― H+ + F― C6H5COOH H+ + ________ ( ) b. After the conjugate formula, label it as an acid (A) or base (B). HCN H+ + ________ ( ) c. What is the formula for the table ion (other than H+) that is the strongest acid? H+ + NH3 H2O H+ + ________ ( ) NH3 H+ + ________ ( ) In the partial acid-strength table at the right. a. Fill in the blanks with the conjugate formulas. © 2011 ChemReview.Net v. 7w HCl _________ ( ________ ( ________ ( ) ) ) Page 982 ) Module 32 — pH of Salts d. Which ion in the table is the weakest acid? e. Which ion in the table is the strongest base? f. Which ion in the table is the weakest base? 4. These acids are listed strongest to weakest: HBr > HF > HCO3― > H2O > CH3CH2OH Arrange these bases strongest to weakest: CO 2―, CH CH O―, F―, Br―, OH― 3 3 2 Salt Solutions: Acidic or Basic? Salts must contain both positive and negative ions. To predict the acidity or basicity of salt solutions, both ions must be considered. This results in 4 possibilities, but the rules are logical. In a salt solution, if • Both ions are pH-neutral, the salt solution will be pH-neutral: close to a pH of 7. • One ion is acidic, and the other is pH-neutral, the solution will be acidic (pH < 7). • One is basic, and the other is neutral, the salt solution will be basic (pH > 7). • One is acidic and one is basic: compare Ka for the acidic to the Kb for the basic ion. If Ka is a larger number, the salt solution will be acidic. If Kb is larger, it will be basic. The reaction with the higher K value is the dominant reaction: this K is used to predict the ion concentrations in the solution. Apply those rules to this problem, then check your answer below. Q. Is a solution of ammonium chloride acidic, basic, or neutral? ***** Answer: Ammonium chloride contains NH4+ and Cl― ions. NH4+ is in the left column of the acid-strength table; it is acidic. Cl― ion is at the top right, the conjugate of the strong acid HCl: a very weak base that is essentially pH neutral. An acidic ion combined with a neutral ion form a compound that results in an acidic solution when the compound is dissolved in water. Practice C: Consult the acid-strength table in Lesson 31B as needed to answer these. Check and do half now, and save the rest for your next practice session. 1. Predict whether aqueous solutions of these salts will be acidic (A), basic (B), or neutral (N). a. NaCl b. NH4NO3 e. Barium Chloride h. Ammonium Fluoride © 2011 ChemReview.Net v. 7w c. KCN f. Cesium fluoride d. Sodium acetate g. Ferric Nitrate i. Ammonium Cyanide Page 983 Module 32 — pH of Salts ANSWERS Practice A YES. Compounds with alkali metals dissolve and ionize ~100% in dilute aqueous solutions. 1. a. KCl NO. Two non-metals form a covalent, not ionic, compound. b. CH4 YES. It is ionic (Ra is metal atom) and it is soluble (NO3─ compounds). d. Sodium acetate YES. Is ionic (sodium is metal atom). Is soluble (Na+ = soluble). c. Ra(NO3)2 e. Cl2 NO. May dissolve, but two non-metal atoms form a covalent, not ionic, compound. Calcium phosphate NO. It is ionic, but by solubility rules for phosphate ions, it is insoluble. It will dissolve only slightly in water. g. Ammonium bromide YES. All ammonium (NH4+ ) compounds are soluble. f. h. HCl YES. HCl is a strong acid. Strong acids dissolve and ionize ~100% in water. i. RbOH YES. All compounds with alkali metal atoms dissolve and ionize ~100%. j. Silver chloride NO. AgCl is a well-known precipitate: if it precipitates in water, it is insoluble. k. PbCl2 NO. By solubility rules is insoluble. l. Lead nitrate YES. Nitrates are ionic and soluble. 2. a. A strong base? RbOH Alkali metal compounds ionize ~100%, hydroxide ion is formed. b. An element? Cl2 Only one kind of atom. c. Radioactive? Ra(NO3)2 Ra is radium; all atoms with more than 83 protons are radioactive. 1 NH4+ + 1 Cl─ ; 0.20 M NH4+ and 0.20 M Cl─ b. 0.50 M Ba(NO3)2 1 Ba(NO3)3 1 Ba2+ + 2 NO3─ ; 0.50 M Ba2+ and 1.0 M NO3─ c. 0.25 M potassium cyanide 1 KCN 1 K+ + 1 CN─ ; 0.25 M K+ and 0.25 M CN─ 3. a. 0.20 M NH4Cl 1 NH4Cl d. 0.10 M sodium sulfate 1 Na2SO4 2 Na+ + 1 SO42─ ; 0.20 M Na+ and 0.10 M SO42─ Practice B 1. a. CH3COO─ B Appears in table on right, in the base column. The conjugate of an acid is a base. b. K+ N Ions of alkali metals in rows 3-7 are pH neutral. c. NH4+ A Appears in the acid column as a weak acid. d. Al3+ A Metals ions with a 3+ or 4+ charge form acidic solutions in water. e. F ─ B Appears in table on right side: the base column. f. Cl─ N Appears in table on right side base column, but Cl─ is the conjugate of a strong acid, so is very weak base – essentially pH neutral. g. Ca2+ N Ions of column 2 metals in rows 3-7 are pH neutral. 2. Compounds containing row 3-7 alkali metals (including b. K+ ) and compounds containing ammonium ion (c. NH4+ ) always dissolve more than 0.1 mol/L in water, which is the general definition for “soluble.” Except for silver acetate, acetate compounds ( a. CH COO─ ) are also soluble. 3 © 2011 ChemReview.Net v. 7w Page 984 Module 32 — pH of Salts 3. a, b. H+ + Cl― (B) H+ + SO42― H+ + F― HCl HSO4― (A) HF (A) H+ + C6H5COO― (B) H+ + CN― (B) H+ + NH C6H5COOH HCN NH + (A) 4 3 H+ + OH― (B) H+ + NH2― (B) H2O NH3 HSO4― is the ion closest to top left. Ions are charged particles. NH + Ion closest to the bottom left. c. Strongest acid ion? d. Weakest acid ion? 4 e. Strongest basic ion? f. Weakest base ion? NH2― Strongest bases are at bottom right. Cl― Weakest bases are at top right. Cl― is so weak it is pH neutral. 4. The conjugate of the strongest acid is the weakest base, the conjugate of the weakest acid is the strongest base. From strongest to weakest base: CH CH O― > OH― > CO 2― > F― > Br― 3 2 3 Practice C Both are neutral ions = Neutral b. NH4NO3 NH4+ is acidic in the acid-strength table, NO3─ is pH-neutral, combined: Acidic c. KCN pH-neutral K+, basic CN─ in table = Basic d. Sodium acetate Neutral Na+, basic CH3COO─ = Basic 1. a. NaCl e. Barium Chloride f. Cesium fluoride Both are neutral ions = Neutral Neutral Cs+, basic F─ = Basic Acidic Fe3+ (highly charged metal ion), neutral NO3─ = Acidic h. Ammonium Fluoride Acidic NH4+, basic F─ ; must compare Ka and Kb NH4+ , from the acid-strength table, has Ka = 5.6 x 10―10 Don’t know Kb for F─ , but know Ka for its conjugate HF from acid-strength table. For conjugates, g. Ferric Nitrate Kw = Kb x Ka = 1.0 x 10―14 , Kb of F─ = Kw = 1.0 x 10―14 = 1.5 x 10―11 Ka of HF 6.8 x 10―4 Since the Ka of the acid is larger than the Kb of the base, the solution will be acidic. i. Ammonium Cyanide NH4+ is acidic and CN─ is basic, so we must compare Ka and Kb NH4+ , from the acid-strength table, has a Ka = 5.6 x 10―10 = 0.000 000 000 56 = Ka To find Kb for CN─ , use Ka for its acid conjugate HCN. For conjugates, © 2011 ChemReview.Net v. 7w Page 985 Module 32 — pH of Salts Kw = Kb x Ka ; Kb for CN─ = Kw = 1.0 x 10―14 = 1.6 x 10―5 = 0.000 016 = Kb Ka of HCN 6.2 x 10―10 Since Kb for CN─ is a larger number than Ka for NH4+, the solution is basic. ***** Lesson 32B: Will A Salt Acid-Base React? Applying Brønsted-Lowry Rules With Spectators Present In Lesson 31B, we learned two ways to predict whether an acid or a base, when mixed in roughly equal proportions, will react. 1. Labeling: Using the acid-strength table, label each particle in the reactants and products as stronger acid (sA), sB, wA, or wB. Equilibrium favors the side with the wA and wB: sA and sB react, wA and wB form, and wA and wB will not react. Recall that we use the sA to label the stronger acid when comparing two acids, and use SA to label an acid that is strong in absolute terms (with Ka > 1, such as HCl.) 2. The Diagonal Rule: In the acid-strength table, a particle in the left column (the acids) will react with a particle in the right column (the bases) below it. This means: Particles \ diagonal react to form / diagonals; particles / do not react. These two rules are simply different ways to state that when an acid and base are combined, equilibrium favors the side with the weaker acid and base. So far, all of our acid-base prediction problems have involved particles found in the acid strength table, where many of the particles are ions. We also need to be able to predict whether acid-base reactions go when the equations include salts rather than the ions in the table. To make these predictions, we will use what we will call Brønsted-Lowry Rule 13. If salt formulas are used in acid-base equations, to predict whether a reaction will go and what the products will be, • Re-write ionic solids (salts) as their separated ions; • Label each particle as A, B, or N, using the acid-strength table and N rules. • Apply the Brønsted-Lowry labeling or diagonal rule to A and B particles, ignoring N’s, to decide whether the reaction will go, and if so, which products form. • Rewrite the products as solid formulas, with the N spectators added back in. A short way to state this rule: 13. When predicting whether acid-base reactions with salt formulas will go, take out the pH-neutral spectators, apply the Brønsted-Lowry rules, put the spectators back in. These rules are consistent with our general rule: To predict the behavior of ionic solids, re-write substance formulas as separated ions. © 2011 ChemReview.Net v. 7w Page 986 Module 32 — pH of Salts Try Rule 13 on this example. Q. Will these reactions go? If so, write the products using molecular/ionic solid formulas. Use the acid-strength table in Lesson 31B. a. HCl + NaCN b. KNO3 + HCN ***** Answers a. HCl + NaCN HCl + Na+ + CN― A N (re-write, separating the salt into its ions) B (to label, use the table and rules for pH-neutrals) Use the table to write the conjugates, then label sA, sB, wA, wB, and go or no. HCl + Na+ + CN― Cl― + HCN sA wB N sB wA (will go to the right) Add the products, including spectators, after the initial reactants, then balance. HCl + NaCN NaCl + HCN (Will Go) b. KNO3 + HCN K+ + N NO3― + HCN VwB/N (re-write, separating the salt into its ions) A (use the table and rules for neutrals) Nitrate ion is a very weak base (VwB): essentially pH neutral. Using the table: K+ + NO3― + HCN N VwB wA KNO3 + HCN HNO3 + CN― sA sB (write conjugates; label sA, sB…) (will not go to the right) Will Not Acid-Base React If NO3― is labeled as neutral (very weak bases are essentially neutral), the combination of N, N, and A is also predicted to not react. Practice: Put a check by and do every other letter. Save the rest for your next practice session. Use the table of acid strengths in Lesson 31B as needed. Check answers as you go. 1. Rewrite these reactions with the salts separated into ions, then write a letter below each particle to label it as acidic (A), basic (B), or pH-neutral (N). a. CH3COOH + NaCN b. HF + NaCl © 2011 ChemReview.Net v. 7w CH3COONa + HCN NaF + HCl Page 987 Module 32 — pH of Salts c. NaHSO4 + C6H5COOH H2SO4 + C6H5COONa 2. In the above reactions, label each A or B particle as the stronger acid (sA), the stronger base (sB), the weaker acid (wA), or the weaker base (wB). Then label the reaction as “will go to the right” or “won’t go.” 3. Complete these reactions, using molecular/ionic solid formulas, then label the reaction as will go or won’t go. a. HF + KCN b. KHSO4 + KF c. KCN + K2HPO4 d. NaF + NH4NO3 e. HNO3 + RbH2PO4 f. NaHCO3 + NaHSO4 g. NaOH + NH4NO3 * ** * * © 2011 ChemReview.Net v. 7w Page 988 Module 32 — pH of Salts ANSWERS 1,2 a. CH3COOH + Na+ + CN― CH3COO― + Na+ + HCN sA N sB wB N wA (Will go) b. HF + Na+ + Cl― Na+ + F― + HCl wA N wB N sB sA (Will NOT go) H SO + C H COO― + Na+ c. Na+ + HSO ― + C H COOH 4 N 65 wB 2 wA 4 65 sA sB N (Will NOT go) 3. a. HF + KCN = HF + K+ + CN― K+ + F― + HCN = KF + HCN sA N sB N wB wA (Will go) 2― + 2 K+ + HF = K SO + HF + + HSO ― + K+ + F― SO4 b. = K 4 24 N wA N wB sB N sA (Will NOT go) ― must be a base, so HSO ― , which can be an acid or a base, must be acting as an acid) (F 4 c. = K+ + CN― + 2 K+ + HPO42― PO43― + 3 K+ + HCN = K3PO4 + HCN N wB N wA sB N sA (Will NOT go) (CN― must be a base, so HPO 2― , which can be an acid or a base, must be acting as an acid) 4 d. = Na+ + F― + NH4+ + NO3― N wB wA N e. = HNO3 + Rb+ + H2PO4― sA N HF + NH3 + Na+ + NO3― = HF + NH3 + NaNO3 sA sB N N (Will NOT go) H3PO4 + Rb+ + NO3― = H3PO4 + RbNO3 sB wA N wB (Will go) (HNO3 is a strong acid. Its conjugate NO3― must be a very weak base.) f. Hint: Try the diagonal rule, and look for both non-N reactants on both sides. ***** f. = Na+ + HCO3― + Na+ + HSO4― [ H2CO3 ] + 2 Na+ + SO42― = H2O + CO2↑ + Na2SO4 N sB N sA wA N wB (Will go) (HSO4―as A and HCO3― as B are \ . H2CO3 decomposes to H2O + CO2 (see Lesson 14E).) g. = Na+ + OH― + NH4+ + NO3― N sB sA N H2O + NH3 + Na+ + NO3― = H2O + NH3 + NaNO3 wA wB N N (Will go) NH3 is the conjugate of NH4+. ***** © 2011 ChemReview.Net v. 7w Page 989 Module 32 — pH of Salts Lesson 32C: Calculating the pH of a Salt Solution Timing: Some courses ask you to predict whether a salt solution will be acidic or basic (as in Lesson 32A), but do not assign calculations of the pH of salt solutions. Do this lesson only if you are asked to calculate the [H+] or pH of a salt solution. ***** In solutions of soluble ionic compounds (salts), there are four possible types of mixtures of ions. We can easily calculate the [H+] or pH for three of them: • If all ions in a salt are pH-neutral, assume the pH of the solution is 7. • If a salt consists of a neutral ion plus an acidic ion, calculate [H+] or pH based on the acidic ion data and its Ka. • If a salt consists of a neutral ion plus a basic ion, calculate [H+] or pH using the basic ion reaction with water and its Kb. • If the salt consists of one acidic and one basic ion, the pH can be calculated, but most first-year courses defer those calculations until a more advanced course in chemistry. For the first three types of salts above, when a soluble acidic or basic salt dissolves in water, two reactions take place: • First, the ionic solid separates into ions, • then, one of the ions ionizes or hydrolyzes, behaving as a weak acid or base. In an aqueous solution, to calculate [H+], [OH─], or pH of a soluble salt that includes one acidic or basic ion, we solve in three logical steps. • First determine the [salt ions]: apply the REC steps to the salt. The REC steps calculate the salt ion concentrations after the ions separate, but before they react as an acid or base with water. • Identify and label each ion as acidic (A), basic (B) or pH neutral (N). • Treat the acidic or basic ion as a weak acid or base that reacts with water. Solve by using the WRRECK steps for ionization (Ka) or hydrolysis (Ka or Kb). We will summarize these steps as the “REC, label, WRRECK” rule for salts: For salt solutions, to calculate [H+] or pH, • REC the salt to find the concentration of each ion. • Label each ion as A, B, or N. • To the A or B ion, apply its Ka or Kb WRRECK steps. Using the above rule, try this calculation. Q. What would be the pH of a 0.25 M solution of NH4Cl? (Ka NH4+ = 5.6 x 10―10) * * * ** © 2011 ChemReview.Net v. 7w Page 990 Module 32 — pH of Salts Answer WANT: pH . Find [H+] first. DATA: [NH4Cl]as mixed = 0.25 M pH ≡ ─ log [H+] and [H+] ≡ 10―pH pH prompt: Strategy: Ammonium compounds are salts that are soluble in water: they separate 100% into ions. To calculate [H+] or pH when a salt dissolves, analyze the two reactions that take place: the salt separating into ions, then the non-pH-neutral ion ionizing as an acid or hydrolyzing as a base. For salt pH: REC the salt, label the ions A, B, or N, WRRECK the A or B ion. REC steps: R&E: Conc: Acidic Neutral ^ ^ + + 1 Cl─ (goes 100%) 1 NH4 1 NH4Cl ^ 0.25 M 0M ^ 0.25 M ^ 0.25 M Label the ions as A, B, or N: According to the acid-strength table, the NH4+ ion is acidic, and the Cl─ ion is so weakly basic that it is pH-neutral. WRECK: To find [H+], apply the WRRECK steps to the non-pH-neutral ion. WANT [H+] =x Since the non-pH-neutral ion is acidic, write its Reaction as an acid and its K as a Ka. NH4+ Specific R: General R & E: Conc. at Eq: H+ + NH3 H+ + CB WA ^ [WA]mixed─ x Ka ≡ [H+] eq.[CB]eq. ≡ [WA]at eq. ^ Definition ^ x (goes slightly) (goes slightly) ^ x x2 [WA]mixed ─ x ^ Exact ≈ x2 [WA]mixed ^ Approximate SOLVE the Ka approximation for the WANTED symbol. x2 = (5.6 x 10―10 ) ( 0.25 ) = 1.4 x 10―10 x = estimate 1-2 x 10―5 = © 2011 ChemReview.Net v. 7w 1.2 x 10―5 M = [H+] Page 991 Module 32 — pH of Salts To see if the approximation is acceptable, apply the 5% test. % Dissociation = ● 100% = 1.2 x 10―5 ● 102 % = x [WA or WB]mixed 0.25 = 4.8 x 10─3 % = 0.0048% = less than 5%, so the approximation is OK. Done? After long calculations, look back at the WANTED unit or symbol to make sure you have completed the problem. WANT: pH = ─ log [H+] = ─log(1.2 x 10─5) = 4.? = 4.92 (estimate, then calculate.) Check: this mildly acidic pH is consistent with a salt composed of one acidic and one neutral ion. Practice: Problem 3 is more challenging. 1. Which of these solutions passes the 5% test by the quick rule? a. For a weak base solution, [WB] = 0.020 M and [OH─] = 5.0 x 10─5 M b. For a weak acid solution, [WA] = 0.0010 M and [H+] = 4.0 x 10─5 M 2. Calculate the pH in a 0.10 M C6H5COOK solution (Kb of C6H5COO─ = 1.6 x 10―10). 3. Calculate the [OH─] in a 0.20 M calcium acetate solution. ANSWERS 1a. In scientific notation, [WB] = 0.020 M = 2.0 x 10―2 M and [OH─] = x = 5.0 x 10─5 M . The difference in the exponents is 3 or greater. The ionization is less than 5% and passes the 5% test. 1a. In scientific notation, [WA] = 0.0010 M = 1.0 x 10―3 M and [OH─] = x = 4.0 x 10─5 M . The difference in the exponents is 2 or less. Use the dissociation equation for the 5% test. (4% -- barely passes) 2. WANT: pH DATA: 0.10 M C6H5COOK Kb of C6H5COO─ = 1.6 x 10―10 pH prompt: pH ≡ ─ log [H+] and [H+] = 10―pH Analysis: C6H5COOK (potassium benzoate) is an ionic compound (a salt). In all compounds that contain column one metals, the metal atoms are +1 ions. Compounds with column one metal atoms are soluble in water and separate ~100% into ions. In any [ion] calculation where a compound separates ~100% into ions, begin by writing the REC steps. To find pH in salt solutions, REC the salt, label ions A, B, or N, WRRECK the A or B ion. © 2011 ChemReview.Net v. 7w Page 992 Module 32 — pH of Salts REC: R&E: Neutral Basic + + 1 C H COO─ 1K 65 ^ ^ 0.10 M (N) 0.10 M (B) 1 C6H5COOK ^ 0.10 M 0 M Conc: (goes 100%) Label ions. K+ is neutral and in the acid-strength table, C6H5COO─ is basic. Because C6H5COO─ is basic, the dominant pH-related reaction is C6H5COO─ base hydrolysis governed by Kb . Apply the Kb prompt: write the WRRECK steps. ***** WRRECK: For a basic ion, use Kb steps to find [OH─] first. WANT: pH R and E: specific: C6H5COO─ + H2O OH─ + C6H5COOH (goes slightly) (Kb reactions always have base and water on left and OH─ + acid conjugate on right.) R and E general: ^ [WB]mixed.─ x Conc. at Eq: Kb: OH─ + acid conjugate weak base + H2O Kb = ^ x [OH─][conjugate] [WB]mixed ─ x ≈ ^ x x2 [WB]mixed (goes slightly) ≈ Kb (rice bottom row) Remember that Ka solves for x = [H+] and Kb solves for x = [OH―]. Make a DATA TABLE to match the symbols in the approximation; solve the approximation for x. DATA: x = [OH─] = ? Kb of C6H5COO─ = 1.6 x 10―10 [WB]mixed = [C6H5COO ─] = 0.10 M based on the REC steps above, SOLVE: WANT: [OH─] = x = ? Solving the Kb approximation: x2 = Kb • [WB] = Kb • [C6H5COO ─] = (1.6 x 10―10 ) ( 0.10 ) = 16 x 10―12 x = 4.0 x 10―6 M = [OH─] But after solving with the approximation, do the Quick 5% test : x = 4.0 x 10―6 M, [WB] = 0.10 M = 1.0 x 10―1 M Since the difference in the exponents is 3 or more, the ionization passes the 5% test, and the approximation may be used. WANT: pH. Since we just found [OH─], one option is to find pOH then use pH + pOH = 14.00 at 25ºC pOH = ─ log [OH─]) = ─log(4.0 x 10―6)) = 5.? = 5.40 (estimate, then calculate.) pH = 14.00 ─ pOH = 14.00 ─ 5.40 = 8.60 , which is a mildly basic pH, as expected. © 2011 ChemReview.Net v. 7w Page 993 Module 32 — pH of Salts [OH─] 3. WANT: DATA: 0.20 M calcium acetate = 0.20 M Ca(CH3COO)2 Analysis: Ca(CH3COO)2 is a salt. Acetates (except silver acetate) are soluble in water. To find pH (or [H+] or [OH─]) in salt solutions, REC salt, label ions, WRRECK A or B ion. In calculations where a compound separates 100% into ions, begin by writing the REC steps. Neutral Basic 1 Ca(CH3COO)2 1 Ca2+ + 2 CH3COO─ ^ ^ ^ 0.20 M 0 M 0.20 M (N) 0.40 M (B) R&E: Conc: (goes 100%) Ca2+ is neutral, CH3COO─ is mildly basic in the acid-strength table. Label: WRRECK: Because CH3COO─ is basic, use Kb and the CH3COO─ base hydrolysis reaction when you write the WRRECK steps. [OH─] WANT: R and E specific: CH3COO─ + H2O CH3COOH + OH─ (goes slightly) (Kb reactions have base on left and OH─ + acid conjugate on right.) R and E general: OH─ + acid conjugate weak base + H2O ^ ^ Conc. at Eq: K: [OH─][conjugate] [WB]mixed ─ x Kb = ≈ ^ x [WB]mixed─ x (goes slightly) x x2 ≈ Kb [WB]mixed To solve, make a DATA TABLE to match the symbols in the approximation: DATA: x = [OH─] = ? = [CH3COOH] Kb = ? We don’t have a K of CH COO─, but can solve from table K of its conjugate. a b 3 Kw = Kb x Ka , Kb = Kw = 1.0 x 10―14 = 5.6 x 10―10 Ka of CH3COOH 1.8 x 10―5 Kb for CH COO ─ = 5.6 x 10―10 3 [WB]mixed = from the REC steps: [CH3COO─] = 0.40 M SOLVE: WANT: [OH─] = x = ? Solving the Kb approximation for the WANTED symbol: x2 = Kb • [WB]mixed = (5.6 x 10―10 ) ( 0.40 ) = 2.24 x 10―10 © 2011 ChemReview.Net v. 7w Page 994 Module 32 — pH of Salts x = 1.5 x 10―5 M = [OH─] Since we used the approximation, do the Quick 5% test : x = 1.5 x 10―5 M, [WB] = 0.40 M = 4.0 x 10―1 M Since the difference in the exponents is 4 = 3 or more, the ionization passes the 5% test, and the approximation may be used. ***** Lesson 32D: Salts That Contain Amphoteric Ions Timing: Do this lesson if you are assigned problems that ask you to predict whether an amphoteric salt will be acidic or basic. Pre-requisite: Complete Lesson 30G on polyprotic acids. ***** Recognizing Amphoteric Ions Particles that are amphoteric can react as acids when mixed with bases, and bases when mixed with acids. Compounds that are amphoteric include water, amino acids, and many metal oxides. Ions, as well as compounds, can be amphoteric. An ion that is amphoteric will be listed in an acid-strength table twice. The ion will be listed in the right column as a base, and then on a lower line in the table on the left side as an acid. For example: find these amphoteric ions in two places in the acid-strength table in Lesson 31B: HPO42― HSO4― H2PO4― Polyprotic acids, when they ionize, always form one or more amphoteric ions. • Diprotic acids (with two acidic hydrogens) form one amphoteric ion when they lose their first proton. Examples: Sulfuric acid forms one amphoteric ion: the hydrogen sulfate ion. H2SO4 H+ + HSO4― The HSO4― ion can react as an acid with a base, or as a base with an acid: H+ from acid + HSO4― H2SO4 OH― from base + HSO4― • SO42― + H2O Triprotic acids such as H3PO4 can successively ionize to form two amphoteric ions. H3PO4 H2PO4― HPO42― PO43― The two middle ions in the series above can react as an acid or a base. When the successive ionizations of a polyprotic acid are written in a series, all of the middle particles in the series (between the first and last particles) will be amphoteric. © 2011 ChemReview.Net v. 7w Page 995 Module 32 — pH of Salts Amphoteric Salt Solutions Whether an amphoteric substance will react as an acid or a base will depend on what it is mixed with. A frequent question is: If a salt that includes a neutral ion and an amphoteric ion is dissolved in water, will it form a solution that is acidic or basic? The rule is An Amphoteric Salt: Acidic or Basic? To determine whether a salt composed of an amphoteric and a neutral ion will form an acidic or basic solution, compare the Ka of the amphoteric ion acting as an acid to its Kb acting as a base. If Ka is a larger number, the solution will be acidic, if Kb is larger, it will be basic. The Kb of an amphoteric ion will often need to be calculated from the Ka of its acid conjugate using Kw = Ka x Kb Use those rules and the acid-strength table in Lesson 31B on the following problem. Q. For a solution of sodium hydrogen carbonate (NaHCO3), a. What is the Ka value for HCO3―? b. What is the Kb value for HCO3―? c. Will the solution be acidic or basic? ***** Answer a. The table lists Ka values for acids in the left column. Find HCO3― in the left column. Ka = 5.6 x 10―11 b. Kb may be found from the Ka of its acid conjugate. The acid conjugate of HCO3― is H2CO3 , with Ka = 4.3 x 10―7. For acid-base conjugates: Kw = Ka x Kb = 1.0 x 10─14 Kb = 1.0 x 10─14 Ka = 1.0 x 10─14 = 0.23 x 10─7 = 4.3 x 10─7 2.3 x 10─8 = Kb c. NaHCO3 ionizes to form Na+ and HCO3― . Na+ is an alkali metal ion in rows 3 -7; those ions are pH neutral. Because HCO3― is amphoteric, to determine if HCO3― is acidic or basic in water, its Ka is compared to its Kb. Since its Kb is higher (above), HCO3― is basic, and the NaHCO3 solution has a basic pH. ***** © 2011 ChemReview.Net v. 7w Page 996 Module 32 — pH of Salts Practice: Learn the rules for amphoteric salts above, then use the acid-strength table in Lesson 31B or a textbook to answer these. 1. Which of these are amphoteric particles? a. HSO3― b. NH4+ c. H2PO4― d. H2SO4 e. PO43― f. H2O 2. The Ka value of H2SO3 is 1.2 x 10―2 . The Kb value of SO32― is 6.0 x 10―8 . What is the Ka value for HSO3― ? 3. Predict whether a solution of a salt composed of H PO ― and a pH neutral atom will 2 4 be acidic (A), basic (B), or neutral (N). 4. Will a solution of K2HPO4 be acidic (A), basic (B), or neutral (N)? 5. For citric acid, H3C6H5O7, Ka = 7.1 x 10―4 , Ka = 1.7 x 10―5, and 1 2 Ka = 4.0 x 10―7. Will a solution of NaH2C6H5O7 be acidic or basic? 3 ANSWERS 1. An amphoteric particle can act as both an acid and a base. If a particle is listed in both columns of the acid-strength table, it is amphoteric. The particle will be listed higher in the base column than in the acid column. Particles in this problem that meet those conditions are a. HSO3―, c. H2PO4― and f. H2O . H2SO4 and PO43― can participate in reactions that form amphoteric ions, but they are not amphoteric, since H SO cannot react as a base, and PO 3― cannot react as an acid. 2 4 4 2. The Ka for HSO3― can be calculated if you know the Kb of its conjugate base. Its conjugate base is SO32―, with Kb = 6.0 x 10―8 . For acid-base conjugates: Kw = Ka x Kb = 1.0 x 10─14 Ka = 1.0 x 10─14 = 1.0 x 10─14 = 0.17 x 10─6 = 1.7 x 10─7 = K a ─8 Kb 6.0 x 10 Quick check: Kb x Ka must estimate to = Kw = 10.0 x 10─15 or 1.0 x 10─14. 3. Since H2PO4― is amphoteric, its Ka as an acid and Kb as a base must be compared. The acid-strength table lists the Ka for H2PO4― as 6.3 x 10―8 To find the Kb of H2PO4―, write its acid conjugate formula: H3PO4 Find the Ka of that acid conjugate in the acid-strength table: 7.2 x 10―3 And apply the rule for conjugate pairs: © 2011 ChemReview.Net v. 7w Kw = Ka x Kb = 1.0 x 10─14 Page 997 Module 32 — pH of Salts SOLVE: Kb = 1.0 x 10─14 = 1.0 x 10─14 = 1.4 x 10─12 = Kb for H2PO4― Ka 7.2 x 10―3 Since the Ka of 6.3 x 10―8 is larger than the Kb of 1.4 x 10─12, H2PO4― is acidic. 4. For ionic compounds dissolving in water, write the reaction for ions separating. K2HPO4 2 K+ + HPO42─ That’s two pH-neutral K+ ions and one amphoteric ion. Since HPO42― is amphoteric, compare its Ka as an acid and Kb as a base. The acid-strength table lists the Ka for HPO42― as 4.2 x 10─13 To find the Kb of HPO42―, write its acid conjugate formula: H2PO4― , find the Ka of that acid conjugate in the acid-strength table: 6.3 x 10―8 And apply the rule for conjugates: Kw = Ka x Kb = 1.0 x 10─14 SOLVE: Kb = 1.0 x 10─14 = 1.0 x 10─14 = 1.6 x 10─7 = Kb for HPO42― Ka 6.3 x 10―8 Since the Ka of 4.2 x 10─13 is smaller than the Kb of 1.6 x 10─7, the solution is basic. 5. For ionic compounds dissolving in water, write the reaction for the ions separating. NaH2C6H5O7 Na+ + H2C6H5O7― Since there are three K values for citric acid, there are 3 H’s that can be lost. This reaction therefore produces one pH-neutral Na+ ion and one amphoteric ion. Because H2C6H5O7― is amphoteric, to determine if the salt solution will be acidic or basic, compare its Ka as an acid and Kb as a base. When successive numeric Ka values are given, it helps to write out the ionization equations. H+ + H2C6H5O7― For: H3C6H5O7 For: H2C6H5O7― H+ + HC6H5O72― For: HC6H5O72― H+ + C6H5O73― Ka = 7.1 x 10―4 1 Ka = 1.7 x 10―5 2 Ka = 4.0 x 10―7 3 For H2C6H5O7― acting as an acid, Ka = 1.7 x 10―5 We need to compare that value to its Kb when acting as a base. To find that value, we need to use the Ka of its acid conjugate and Ka x Kb = 1.0 x 10─14 The acid conjugate of H2C6H5O7― is H3C6H5O7 with a Ka = 7.1 x 10―4 1 SOLVE: Kb = 1.0 x 10─14 = 1.0 x 10─14 = 1.3 x 10─11 = Kb for H C H O ― 2657 Ka 7.1 x 10―4 Since the Ka is larger than the Kb for this amphoteric ion, the solution is acidic. ***** © 2011 ChemReview.Net v. 7w Page 998 Module 32 — pH of Salts SUMMARY – pH of Salts 1. In salt solutions, for ions that are not amphoteric: Neutral ions include • In Rows 3-7, column 1 and 2 ions: Na+, K+, Rb+, Cs+; Ca2+, Sr2+, Ba2+, Ra2+. • Plus conjugates of strong acids and bases, including Cl─, NO3─. Acidic ions are • • Stronger acids than H2O in an acid-strength table OR have a Ka larger than 1.0 x 10―14. Include NH4+, other R-NHx+, Fe3+, Al3+, and Sn4+. ≡ ≡ ≡ ≡ H2O ≡ Basic ions are • Base conjugates of acids below Ka= 1.0; bases below ≡ H2O as base OR with a Kb larger than 1.0 x 10―14 . • include F─, CN─, CH3COO─, C6H5COO─. ≡ H2O ≡ Ka=Very Large Ka = 1.0 ≡ 2. When predicting whether acid-base reactions with salt formulas will go to the right, take out the pH-neutral spectators, apply the Brønsted-Lowry labeling or diagonal rules, then put the spectators back in. 3. For the pH of salt solutions, if • Both ions are neutral ions, pH ≈ 7. • One ion is acidic, and the other is neutral, solution will be acidic (pH <7). • One is basic, and the other is neutral, solution will be basic (pH>7). • One is acidic and one is basic: Compare the Ka and Kb for the two ions. If Ka is larger, solution will be acidic. If Kb is larger, it will be basic. 4. An amphoteric compound reacts as an acid when mixed with a base, and as a base when mixed with an acid. 5. To determine whether a salt with both amphoteric and neutral ions will form an acidic or basic solution, compare the numeric value of the Ka of the amphoteric ion as an acid to its Kb as a base. If Ka is larger, the solution is acidic, if Kb is larger, it is basic. 6. To calculate the pH of salt solutions, if the salt contains: • One neutral ion plus one acidic ion, apply Ka rules to the acidic-ion hydrolysis. • One neutral ion plus one basic ion, apply Kb rules to the basic-ion hydrolysis. ### © 2011 ChemReview.Net v. 7w Page 999 Module 33 — Buffers Module 33 — Buffers Prerequisites: Complete Modules 29, 30, and 32 before starting this module. Lesson 33A: Acid-Base Common-Ions; Buffers Buffer Solutions In chemical and biological systems, it is often important to buffer solutions to resist a change in pH when acids or bases are added. A buffer contains both an acid to react with bases and a base to react with acids. Blood is one example of a buffered solution. As chemicals are added and removed during digestion and metabolism, the buffers in blood assure that its pH stays relatively constant, and the important pH-sensitive reactions that occur in blood can continue to take place. Common-Ion Buffers A common-ion buffer solution is a mixture of two components: a weak acid or base and its conjugate. One component may be ionic and one covalent, or both may be ionic. Both components must be substantially soluble in water. One way to prepare a common-ion buffer is to mix • a weak acid (Ka between 1 and 10―16) or a weak base (Kb between 1 and 10―16) with • a soluble salt composed of a pH-neutral ion and the conjugate ion of the acid or base. Each of the two components of the buffer reacts with water: the weak acid or base hydrolyzes slightly, while the soluble salt separates ~100% into ions. The reactions are: (Weak acid or base) + H2O Conjugate salt in water (H3O+ or OH―) + conjugate ion neutral ion + same conjugate ion (goes slightly) (goes ~ 100%) This mixture is called a common-ion buffer because the two reactions have one product in common: the same conjugate ion is formed in both the reaction of the weak acid or base with water, and the dissolving of the salt in water. In a common-ion buffer, the common ion = the conjugate ion How Buffers Differ From Weak Acid or Weak Base Solutions Weak acid or base solutions contain both the weak acid or base and its conjugate. Buffer solutions also contain both a weak acid or base and its conjugate, but the two types of solutions are different. • In a weak acid or weak base solution, the [weak acid or base] is much higher than the [conjugate], because a weak acid or base reacts only slightly with water to form its conjugate. • In a buffer solution, the [weak acid or base] and [conjugate] are relatively close, because to create the buffer, substantial amounts of the conjugate particles are added to the solution of the weak acid or base. © 2011 ChemReview.Net v. 7w Page 1000 Module 33 — Buffers For example, Acetic acid is a weak acid: it ionizes only slightly and forms only a small concentration of hydrogen and acetate ions. 1 CH3COOH(l) 1 H+(aq) + 1 CH3COO─(aq) (goes slightly) In 0.10 M acetic acid, [CH3COOH] ≈ 0.10 M and [CH3COO―] = 0.0013 M In a buffer solution that contains 0.10 M acetic acid and 0.10 M acetate ions, [CH COOH] ≈ 0.10 M and [CH COO―] ≈ 0.10 M 3 3 The acetic acid concentration is approximately the same in both solutions, but in the buffer, the concentration of the conjugate ion is much higher. In a buffer, the moles and concentrations of the weak acid and the conjugate do not need to be the same, but they need to be relatively high and close, preferably differing by a ratio of less than 3 to 1. Why? For a buffer to be effective at limiting pH changes, it needs a relatively high number of moles of acid to neutralize base added, and a relatively high number moles of base to neutralize acid added. Selecting Buffer Components A buffer is a mixture of the acid and base of a conjugate pair. Since the acid in the conjugate pair must have one more positive charge than the base, if one particle in the pair is electrically neutral, the other must have a charge. Both particles in the conjugate pair can also be charged, so long as their charges differ by the +1 charge on a proton. For example, in the conjugate pair HSO4─ and SO42─, both particles are ions. In short, one or both of the particles in a conjugate pair must be an ion. A buffer may be described as either • the two particles in the conjugate pair, in which at least one is an ion, or • as two electrically neutral compounds that contain the particles in the conjugate pair, in which one or both of the compounds is a soluble salt. For example: The acetic acid/acetate buffer can be represented as or as CH3COOH and CH3COONa CH3COOH and CH3COO─ The first is the weak acid and its conjugate ion, the second is the weak acid and a salt that contains the conjugate ion. To select a conjugate particle that will buffer a weak acid or weak base solution, • find the weak acid or base in a table of acid strength (see Lesson 31B). • The conjugate particle in the opposite column will buffer the solution. • If one of the buffer components is amphoteric, two substances can buffer the solution: the conjugate acid will buffer to a lower (more acidic) pH, and the conjugate base will buffer to a higher pH. If a problem asks for the formula for a neutral substance that will buffer an acid or base, first write the formulas for the conjugate pair. Then, for each particle that is an ion, pair the © 2011 ChemReview.Net v. 7w Page 1001 Module 33 — Buffers ion with a pH-neutral ion that will form a soluble combination, then write formula for that combination in a molecular formula (ionic solid) format. An ionic solid (salt) that will buffer a solution may contain either the acid or the base component of the acid-base conjugate pair. Using the acid-strength table in Lesson 31B if needed, apply those rules to this problem. Q. Write formulas for two ionic solids that would buffer the weak acid HF. ***** The conjugate base is F─. Soluble salts that include F─ include NaF, KF, and RbF. The ion that you pair with F─ must be both pH-neutral and result in a soluble compound. All row 3-7 alkali metal ions are pH neutral, and all compounds that include those ions are water soluble. Answer: To determine whether an ionic compound will buffer a weak acid or base solution, • Write the reaction for the ionic solid separating into its ions. • If one ion is pH neutral, the other ion is a conjugate of the weak acid or base, and the combination is water soluble, then the salt will buffer the solution. Practice. Use the acid-strength table in Lesson 31B as needed. 1. Write the formula for the conjugates of a. NH4+ b. d. HCO3─ acting as a base HNO3 e. c. F─ HCO3─ acting as an acid 2. Write the formula for an ion that would buffer solutions of these particles. a. C6H5COOH b. HCN 3. Write the formula for a compound that would buffer solutions of these particles. b. CH COOH a. C H COO─ 65 3 4. Answer the following questions using these letters. Each question may have multiple letters for its answer. a. KF i. b. NH4NO3 c. NH3 d. NaCN e. NH4Cl Which compound(s) above will serve as a common-ion buffer for an HF solution? ii. Which compound(s) above will serve as a common-ion buffer for NH3 solution? iii. Which compound(s) will form a buffer if mixed into an NH4Cl solution? © 2011 ChemReview.Net v. 7w Page 1002 Module 33 — Buffers ANSWERS 1. In the acid-strength table, the conjugate of a. NH4+ is NH3 b. HNO3 is NO3─ c. F─ is HF d. HCO ─ if acting as a base, is the acid H CO e. HCO ─ as an acid is the base CO 2─. 3 2 3 3 3 2. a. C6H5COO─ is an ion. Since it is also the conjugate base of the acid, it will buffer the solution. b. CN─ is the conjugate base of the acid HCN; it buffers an HCN solution. 3. a. The acid-strength table shows that C6H5COO─ is the conjugate base of C6H5COOH. A buffer is a mixture of an acid-base conjugate pair, so C6H5COOH buffers C6H5COO─. b. The acid-strength table shows that for CH3COOH, CH3COO─ is the conjugate base ion that would buffer the solution. A compound must have a neutral overall charge. Compounds that would buffer the solution include any pH–neutral ion that combines with the acetate ion to form a soluble combination, such as Na+ to form CH3COONa, or Mg2+ to form (CH3COO)2Mg . (Except for silver acetate, metal-acetate combinations are soluble). 4. i. ii. (a). KF is a soluble salt with pH-neutral ion K+ and basic F─. F─ is the conjugate base of HF. (b) and (e). Both are soluble salts that contain the acid-conjugate NH + of the base NH . 4 3 iii. (c). NH4Cl contains the acidic NH4+ ion and the pH-neutral Cl─ ion. The particle needed as a buffer is the base-conjugate of the acid, which is NH3 . ***** Lesson 33B: Buffer Example Simplifying Buffer Calculations Buffers are a mixture of an acid and a base. Buffer calculations can be solved based on either the Ka of the acid or the Kb of the base. One way to simplify buffer calculations is to choose to solve based on Ka of the acid or the Kb and to use this approach consistently. We will use that method in these lessons. Our fundamental rule will be Treat buffer solutions as a weak acid ionization equilibrium to which conjugate base has been added and solve buffer calculations using Ka equations. Reactions in a Buffer An example of a common-ion buffer solution is the combination of aqueous acetic acid (CH3COOH) and the salt sodium acetate. Let’s describe the mixing of an acetic acid/acetate buffer, developing the equations for buffer calculations as we go. 1. First, mix 0.10 moles of the weak acid CH3COOH and about 950 mL of water. Acetic acid is highly soluble and ionizes slightly (Ka = 1.8 x 10―5). © 2011 ChemReview.Net v. 7w Page 1003 Module 33 — Buffers 2. Next, add 0.20 moles of the solid but soluble salt sodium acetate (CH3COONa) to the acetic acid solution. Dissolve the salt, then add a bit more water, with mixing, until the total volume is 1.0 L. Write the concentrations of the two substances, as mixed into this solution, but before any reactions take place. ***** Since the moles of both compounds are dissolved in 1.0 liters of solution, [CH3COOH]as mixed = 0.10 M and [CH3COONa]as mixed = 0.20 mol/L Those are concentrations “as mixed.” However, those values do not represent the actual concentrations in the solution, because both substances react with water. Write the REC steps for the reaction of CH3COOH in water, then CH3COONa in water, and then check your answers below. ***** For the weak acid: 1 CH3COOH ^ Conc@Eq. symbols = [WA]mixed ─ x 1 H+ ^ x Rxn. & Extent: Conc@Eq. 0.10 M ─ x ▐ ▐ x │ + 1 CH3COO─ (goes slightly) ^ │ x x Because the weak acid ionizes slightly, its concentration is written as 0.10 M ─ x. For the salt: Rxn. & Extent: Conc@Eq. values: ^ 0 M 0.20 M 1 Na+ + 1 CH3COO─ 1 CH3COONa ▐ ^ 0.20 M (goes ~ 100%). ^ 0.20 M The actual [CH3COONa] = 0 M, because in dilute aqueous solutions, all Na compounds dissolve and separate into ions essentially 100%. Na+ is a pH-neutral ion. CH COO─ is the conjugate base of the weak acid CH COOH. 3 3 This solution is a buffer because it combines substantial amounts of both a weak acid and its conjugate base. CH3COO─ is a common ion because it is formed in both reactions above that occur in the solution. Ion Concentrations In Buffers Adding the numbers and x’s underlined in the two “concentration at equilibrium” rows above, fill in the chart below with the total values for the concentrations at equilibrium. Totals: Exact Approximate [CH3COOH]at eq. = [WA]eq. = _____________________ ≈ ______________ [H+]at equilibrium = _____________________ ≈ ______________ = _____________________ ≈ ______________ [CH3COO─]at eq. = [CB]eq. ***** © 2011 ChemReview.Net v. 7w Page 1004 Module 33 — Buffers Totals in the buffer solution are Exact Approx. [CH3COOH]at eq. = [WA]eq. = 0.10 M ─ x ≈ 0.10 M [H+]at equilibrium = x ≈ x [CH3COO─]at eq. = [CB]eq. = x + 0.20 M ≈ 0.20 M Approx. Symbol = ______________ = ______________ The [CB] at equilibrium includes the concentration formed when the weak acid ionizes (x), plus the concentration added by the salt used to form the buffer. However, in the approximate concentrations, the small x values that are added or subtracted from larger numbers are ignored. Based on the symbols for WA and CB concentrations used in point 2 above, in the two blanks above, write what you think is an appropriate symbol for each approximate concentration. ***** [CH3COOH]at eq. = [WA]eq. = 0.10 M ─ x [CH3COO─]at eq. = [CB]eq. = 0.20 M + x ≈ 0.10 M = [WA]as mixed ≈ 0.20 M = [CB]as mixed To summarize: In buffer calculations: [WA]eq. ≈ [WA]as mixed and both are large compared to [H+] = x . [CB]eq. ≈ [CB]as mixed and The Buffer Equations In buffer solutions, we are most often interested in the [H+]. For a weak acid alone, the reaction that occurs is 1 WA(aq) 1 H+(aq) + 1 CB(aq) (goes slightly) and the K equation that governs the reaction is: Ka ≡ [H+] eq.[CB]eq. [WA]at eq. ^ Definition ≡ x•x ≈ [WA]mixed ─ x ^ Exact x2 [WA]mixed ≈ Ka ^ Approximation After conjugate base is added to a weak acid solution to create a buffer, the weak-acidionization reaction continues to govern the formation of H+. The Ka equation that allows us to calculate the [H+] has the same definition. However, once additional conjugate is added, the two top terms in the three ratios are no longer equal. In a buffer, the Ka equations become Ka ≡ [H+]eq.[CB]eq. ≡ [WA]at eq. ^ Definition © 2011 ChemReview.Net v. 7w x • ( x + [CB]mixed ) [WA]mixed ─ x ^ Exact ≈ x • [CB]mixed ≈ Ka [WA]mixed ^ Approximate Page 1005 Module 33 — Buffers Calculations for both a weak acid ionization and a buffer solution are based on the same definition, but the exact and approximate equations are different. Note how they are different. Now calculate the [H+] in this 0.10 M acetic acid/0.20 M acetate buffer solution. To do so, solve the boxed buffer approximation equation above. Use the Conc@Eq. approximate data in the totals table above, plus the Ka for CH3COOH = 1.8 x 10―5. ***** Ka ≈ x • [CB]mixed [WA]mixed DATA: = 1.8 x 10―5 Ka x = [H+] =? [CB]mixed [WA]mixed SOLVE: ***** ? = x = [H+] ≈ = 0.20 M = 0.10 M Ka ● [WA]mixed ≈ 1.8 x 10―5 ( 0.10 ) ≈ [CB]mixed 0.20 9.0 x 10―6 M H+ As always, units are omitted during K calculations, but concentrations that are calculated are labeled mol/L (or M). Finally, apply the 5% test to see if the exact buffer equation needs to be solved. ***** Quick 5% test : x = [H+] = 9.0 x 10―6 M, [WA] = 0.10 M = 1.0 x 10―1 M Since the difference in the exponents is 3 or more, the approximation may be used. Done! We won’t need to do all those steps for future buffer calculations. However, exploring in detail the impact of common ions on solution concentrations will be useful in several types of reactions where common ions are added to solutions. Summary 1. A common-ion buffer solution is composed of an acid-base conjugate pair in which • each particle is a weak acid or base; • both particles in the pair have relatively high, but often different, concentrations; • one or both of the particles is an ion. 2. A buffer solution can be viewed as • A weak acid with conjugate base added, with reactions governed by Ka , or • A weak base with acid conjugate added, with reactions governed by Kb . To simplify calculations, these lessons will treat buffers as a weak acid ionization equilibrium to which conjugate base has been added, and solve using Ka equations. © 2011 ChemReview.Net v. 7w Page 1006 Module 33 — Buffers 3. For a buffer, the Ka ratios are Ka ≡ [H+]eq.[CB]eq. ≡ [WA]at eq. x • ( x + [CB]mixed ) ≈ [WA]mixed ─ x ^ Definition x • [CB]mixed ≈ Ka [WA]mixed ^ Exact ^ Approximate In the buffer equations, x = [H+] = small, but [WA] and [CB] are large and often differ. 4. In a buffer solution, [WA]at eq. ≈ [WA]as originally mixed and [CB]at eq. ≈ [CB]as originally mixed into the buffer solution. These “concentration as originally mixed” are easily measured and are usually the data supplied in problems. 5. In buffer equations, the [conjugate base] may be abbreviated as [CB]at eq. ≈ [CB]mixed or [CB]added or [CB] or [base] and the [weak acid] may be abbreviated in equations as [WA]at eq. ≈ [WA]mixed or [WA] or [acid] Practice 1. The general reaction for the ionization of a weak acid is 1 weak acid(aq) 1 H+(aq) + 1 conjugate base(aq) (goes slightly) The K definition and exact equations for that reaction are Ka ≡ [H+]eq. [CB]eq. ≡ x● x [WA]at eq. [WA]mixed ― x ^Definition ^ Exact ≈ ^ Approximation a. Modify the exact equation above to reflect the changes when conjugate base particles are added to make a buffer solution. b. Based on the buffer exact equation, write the approximate buffer equation. 2. Which concentrations are small a. In a weak acid approximation equation? b. In a buffer approximation equation? 3. Which concentrations are equal a. In a weak acid approximation equation? b. In a buffer approximation equation? © 2011 ChemReview.Net v. 7w Page 1007 Module 33 — Buffers 4. For this reaction, write the K definition and exact equations. 2 A+(aq) + 1 B2─(aq) Rxn. & Extent: 1 A2B (aq) Conc@Eq. = ^ [A2B]mixed ─ x ▐ ^ 2x (goes slightly) ^ x │ 5. Write the K definition, exact, and approximation equations for the Problem 2 reaction, modified by adding a substantial amount of B2─ to the solution. ANSWERS 1. Ka ≡ [H+]eq. [CB]eq. [WA]at eq. x ● ( x + [ CB mixed ] ) [WA]mixed ― x ≡ ≈ x ● [CB]mixed [WA]mixed ^Definition ^ Exact ^ Approximation 2. In the weak acid approximation equation, x = [H+] = [CB] = small, In the buffer approximation equation, [H+] is the only concentration that is small, 3 4. In the weak acid approximation equation, [H+] = [CB] In the buffer approximation equation, [H+] is small, and [WA] and [CB] are large and may be equal, but none of the concentrations are required to be equal. K≡ [A+]2eq. [B2─]eq. [A2B]eq. ^Definition 5. K≡ [A+]2eq. [B2─]eq. ≡ [A2B]eq. ^Definition ≡ (2x)2 ● x [A2B]mixed ― x or 4x2 ● x [A2B]mixed ― x ^ Exact 2─ (2x)2 ● ( x + [ B ] as mixed) [A2B]mixed ― x ^ Exact ≈ 2─ 4x2 ● [B mixed] [A2B]mixed ^ Approximate ***** © 2011 ChemReview.Net v. 7w Page 1008 Module 33 — Buffers Lesson 33C: Buffer Components Identifying Buffer Components A key step that we will use to simplify buffer calculations is to fill in the following Buffer Chart: WA formula = _________ CB formula = ___________ mol or [WA]mixed = _______ mol or [CB]mixed = _________ You need to be able to write the four labels for the blanks in the chart from memory. The two formulas, WA and CB, must be a conjugate pair: the acid particle formula must have one more H and one more + charge than the base formula. For some problems, you can fill in the buffer chart by inspection. Let’s try an example. Q. For a buffer that consists of 0.10 M HF and 0.20 M F─, fill in the Buffer Chart: WA formula = ______________ CB formula = ________________ mol or [WA]mixed = _________ mol or [CB]mixed = ___________ WA formula = HF CB formula = F─ mol or [WA]mixed = 0.10 M HF mol or [CB]mixed = 0.20 M F─ ***** Buffer Chart: The buffer is a mixture of the weak acid HF and its conjugate base F─ ions. Salts In Buffers In buffer problems, the ion or the two ions in the conjugate pair may be written as a part of a soluble ionic solid (salt). In problems that supply the concentration of the salt, you will need to write in the buffer chart the concentrations of the ions that form when the salt dissolves. To do so, you must be able to recall from memory the formulas for ions that tend to form. It helps that most salts will include familiar ions that are always soluble. Recall that if a substance formula includes one or more of these groups: • Na K NO3 NH4 then the compound is both soluble and will separate ~100% into ions in dilute aqueous solutions. Other combinations of ions are soluble as well (see solubility rules, Lesson 13A), but the four atoms/groups above are those encountered most frequently. For some calculations in which salt formulas are supplied, you will be able to identify the ions and fill in the buffer chart by inspection. However, if the salts in a buffer solution are complex, it helps to write out the REC steps to list the ions and ion concentrations that form when ionic solids separate, and then fill in the buffer chart. The rule is: If you cannot fill in a buffer chart by inspection, REC the salt(s). Try this problem. © 2011 ChemReview.Net v. 7w Page 1009 Module 33 — Buffers Q. For a buffer consisting of 0.20 M Na2HPO4 and 0.10 M NaH2PO4 , write and fill in the buffer chart. ***** Since both buffer components contain Na, both are ionic compounds (salts) that separate in water ~ 100% into ions. The REC steps for those two reactions are Rxn. & Extent: ^ 0 M 0.20 M Conc@Eq. Rxn. & Extent: 1 HPO42─ ▐ ^ 0.40 M + 1 H2PO4─ ^ 0.10 M (goes ~ 100%) ^ 0.20 M 1 Na+ ▐ 1 NaH2PO4 ^ 0 M 0.10 M Conc@Eq. 2 Na+ + 1 Na2HPO4 ^ 0.10 M (goes ~ 100%) Identify the conjugate pair in the two dissolved salts, then complete the buffer chart. ***** In a buffer chart, the two formulas must be a conjugate pair. One or both formulas may be an ion. In this problem, based on the REC steps above, Buffer Chart: WA formula = H2PO4─ CB formula = HPO42─ mol or [WA]mixed = 0.10 M H2PO4─ mol or [CB]mixed = 0.20 M HPO42─ Practice: Try to solve these by inspection first, but if you are not certain of your answer, REC the salt(s) and/or consult the acid-strength table in Lesson 31B. 1. From the list of ions below, write the formula(s) for the ions that are a. pH-neutral b. Acidic c. Always result in a compound containing the ion to be soluble. (1) Na+ (2) NO3― (3) NH4+ (4) Cl─ (5) K+ 2. The following pairs of compounds can be mixed to make aqueous buffer solutions. Circle the compounds that are soluble salts. a. NaF and HF b. HCN and KCN d. C6H5COOH and C6H5COOK c. NH3 and NH4Br e. NaHCO3 and Na2CO3 3. For each of the following buffer solutions, write and fill in this first line of the buffer chart. Weak acid (WA) formula = _________ Base conjugate (CB) formula = ________ a. 0.20 M CN─ and 0.40 M HCN b. 0.10 M NaHCO3 and 0.20 M Na2CO3 4. For each of the following buffer solutions, write and fill in a complete buffer chart. a. 0.25 M NH3 and 0.30 M NH4Cl © 2011 ChemReview.Net v. 7w Page 1010 Module 33 — Buffers b. 0.10 M CH3COOH and 0.15 M (CH3COO)2Ca c. 0.40 mol Na3PO4 and 0.20 mol Na2HPO4 d. 0.15 M C6H5NH2 and 0.25 M C6H5NH3Cl ANSWERS 1a. pH-neutral: (1) Na+ (2) NO3― (4) Cl─ (5) K+ (For review, see Lesson 31A) 1b. Acidic: (3) NH + 1c. Always soluble: (1) Na+ (2) NO ― (3) NH + (5) K+ (see Lesson 13A) 4 3 4 2. Salts are compounds composed of ions. In buffers, one or both components will be salts that are soluble in water, and one ion in the salt will be acidic or basic. a. NaF -- is a soluble salt. The two ions are Na+ and F─. HF is composed of two non-metal atoms and will have less ionic character. According to the acid-strength table, HF behaves as a weak acid, which means that it ionizes only slightly. b. KCN -- is a salt. Potassium atoms in compounds are always ions, and all potassium compounds are soluble and ionize 100% in dilute aqueous solutions. The two ions are K+ and CN─. c. NH4Br -- is ammonium bromide, a compound composed of ions, which makes it a salt. One ion formed is the ammonium ion NH4+. Ammonium compounds are soluble in water and ionize ~ 100%. NH3 is weak base composed of two non-metals and has primarily covalent character. d. C6H5COOK –- All potassium compounds are ionic, and ionic compounds are salts. K compounds are soluble in water and ionize ~100%. The ions formed are K+ and CH3COO─. e. NaHCO3 and Na2CO3 –- Both contain Na. Both are therefore soluble salts. 3a. Weak acid (WA) formula = HCN Base conjugate (CB) formula = CN─ 3b. Weak acid (WA) formula = HCO3― Base conjugate (CB) formula = CO32― WA formula = NH4+ CB formula = NH3 mol or [WA]mixed = 0.30 M NH4+ 4a. Buffer Chart: mol or [CB]mixed = 0.25 M NH3 If needed, REC the salt(s): R&E: Conc.@Eq. NH4Cl ^ 0.30 M 0 NH4+ + Cl─ ^ ^ 0.30 M 0.30 M (goes ~ 100%) NH3 is a covalent compound and a base, but not a salt. Treat the buffer as WA ionization of NH4+ with conjugate base NH3 added. © 2011 ChemReview.Net v. 7w Page 1011 Module 33 — Buffers WA formula = CH3COOH CB formula = CH3COO─ mol or [WA]mixed = 0.10 M CH3COOH 4b. Buffer Chart: mol or [CB]mixed = 0.30 M CH3COO─ 1 Ca2+ + 2 CH3COO─ (goes ~ 100%) ^ ^ 0.15 M 0.30 M If needed, REC salt(s): R&E: 1 (CH3COO)2Ca ^ Conc.@Eq. 0.15 M 0 (CH3COO)2Ca separates into calcium ion and acetate ions: it is a salt. Acetates (except silver acetate) are soluble in water. CH3COOH is a weak acid and not a salt. WA formula = HPO42─ CB formula = PO43─ mol or [WA]mixed = 0.20 mol HPO42─ 4c. Buffer Chart: mol or [CB]mixed = 0.40 mol PO43─ 1 Na2HPO4 ^ 0 M 0.20 M Conc@Eq. ▐ 2 Na+ + 1 HPO42─ (goes ~ 100%) ^ ^ 0.40 M 0.20 M ▐ 3 Na+ ^ 1.20 M If needed, REC salt(s): R&E: R&E: Conc@Eq. 1 Na3PO4 ^ 0.40 M 0 M + 1 PO43─ ^ 0.40 M (goes ~ 100%) WA formula = C6H5NH3+ CB formula = C6H5NH2 mol or [WA]mixed = 0.25 M C6H5NH3+ 4d. Buffer Chart: mol or [CB]mixed = 0.15 M C6H5NH2 If needed, REC salt(s): R&E: Conc.@Eq. C6H5NH3Cl ^ 0.25 M 0 C6H5NH3+ + Cl─ (goes ~ 100%) ^ ^ 0.25 M 0.25 M ***** © 2011 ChemReview.Net v. 7w Page 1012 Module 33 — Buffers Lesson 33D: Methodical Buffer Calculations Buffer Steps To solve buffer calculations methodically, we use the following steps. 1. Fill in the buffer chart using conjugate-pair formulas. If needed, REC the salt(s). Buffer Chart: WA formula = _____________ CB formula = ________________ mol or [WA]mixed = _______ mol or [CB]mixed = __________ 2. WRECK the WA, adding conjugate base at step C. Treat buffers as a weak acid ionizing, with CB mixed in. For all buffers, the REC steps will be R and E: WA ^ [WA]mixed ─ x Conc@Eq. H+ + ^ x │ ▐ CB (goes slightly) ^ x + [CB]mixed 3. At the K step, write the three buffer Ka equations. Ka ≡ [H+]eq.[CB]eq. ≡ [WA]at eq. x • ( x +[CB]mixed ) ≈ [WA]mixed ─ x ^ Definition x • [CB]mixed ≈ Ka [WA]mixed ^ Exact ^ Approximation In a buffer, the two concentrations in the Ka numerators are not equal. 4. Solve the Ka approximation for the WANTED symbol. To solve the approximation, use the concentrations in the buffer chart. 5. Calculate % dissociation. If >5%, solve the exact Ka buffer equation as a quadratic. In short: To solve buffer calculations, • fill-in the buffer chart, WRECK the WA, solve the buffer approximation. Let’s practice on this example. Q. Find the [H+] in a buffer solution composed of 0.10 M HF and 0.20 M NaF. (Ka of HF = 6.8 x 10―4) Apply the steps, then check your answer below. ***** Answer 1. Buffer Chart: WA formula = HF CB formula = F─ mol or [WA]mixed = 0.10 M HF If needed, REC salt(s): R&E: Conc.@Eq. © 2011 ChemReview.Net v. 7w mol or [CB]mixed = 0.20 M F─ Na+ + NaF ^ 0.20 M 0 ^ 0.20 M F─ (~ 100%) ^ 0.20 M Page 1013 Module 33 — Buffers 3. WRECK the WA, then add conjugate base. WANTED = [H+] = x R and E: Conc@Eq. WA ^ [WA]mixed ─ x H+ + CB (goes slightly) ^ ^ x │ x + [CB added] ▐ 4. SOLVE the buffer approximation. [H+]eq.[CB]eq. ≡ [WA]at eq. Ka ≡ ^ Definition x • ( x + [CB]mixed ) ≈ x • [CB]mixed [WA]mixed ─ x In Buffer: ≈ Ka [WA]mixed ^ Exact ^ Approximate Substitute Ka from the problem and the other values from the buffer chart, then solve for the WANTED symbol. Ka = 6.8 x 10―4 ≈ x • ( 0.20 M) ; x = 3.4 x 10―4 M = [H+] 0.10 M 5. 5% test quick : x = [H+] = 3.4 x 10―4 M, [WA] = 0.10 M = 1.0 x 10―1 M The exponent difference is 3 or greater. The % dissociation is < 5%. The approximation may be used. Practice A: For these rules, you may consult the 6 steps above. Do one now and the other in your next practice session. 1. Calculate the [H+] in a solution containing 0.30 M HCN and 0.45 M KCN. (Ka of HCN = 6.2 x 10―10). 2. In a solution containing 0.10 M NaH2PO4 and 0.20 M Na2HPO4, calculate the [H+]. (Ka of H2PO4─ = 6.7 x 10―8) How the Conjugate Shifts the Equilibrium In solution, a weak acid ionizes slightly, and the [H+] formed = [conjugate base] formed. 1 weak acid(aq) 1 H+(aq) + 1 conjugate base(aq) (goes slightly) When conjugate base is added to a weak acid solution, the [conjugate] becomes relatively large compared to that due to the weak acid ionization alone. The weak-acid-ionization reaction written above continues in both directions, but the concentrations of the reactants and products shift to new values at a new equilibrium. Try these questions. Q1. According to Le Châtelier’s Principle (Lesson 28A), how does increasing the [conjugate] change the [H+] in the reaction above? ***** © 2011 ChemReview.Net v. 7w Page 1014 Module 33 — Buffers Answer Increasing the [conjugate], a particle on the right side of the equation, shifts the equilibrium to the left. This reduces the concentration of the other particle on the right, which is H+. As [conjugate] goes up, [H+] goes down. Q2. According to Le Châtelier’s Principle, how does adding conjugate change the percent dissociation of the weak acid in the solution? ***** Answer As the equilibrium shifts to the left, the [weak acid] goes up. That means less of the weak acid is ionizing (dissociating). The % dissociation becomes lower. This smaller dissociation has an impact on buffer calculations. For dilute solutions of relatively strong weak acids (Ka > 10―6), % dissociation is often >5%, and the exact K quadratic needs to be solved. However, if the same weak acid solution is buffered, the % dissociation is lower. The result is that buffer solutions are less likely to be dissociated > 5%, and you are less likely to need to solve a quadratic to solve a buffer calculation. For buffers, the approximation equation nearly always gives an answer that is within the 5% uncertainty allowed in K calculations. Practice B 1. In a weak acid solution, which will be higher: [H+] or [conjugate]? 2. In a weak acid solution, which will be highest: [weak acid], [H+], or [conjugate]? 3. In a weak acid solution to which a buffer salt has been added, which will be higher: [H+] or [conjugate]? 4. In a weak acid solution, as a salt that buffers the solution is first added, what happens to the solution pH? ANSWERS Practice A 1. Strategy: A weak acid (HCN) and a salt with its conjugate = buffer solution. Use the 5 buffer steps. Buffer Chart: WA formula = HCN CB formula = CN─ mol or [WA]mixed = 0.30 M HCN 1. mol or [CB]mixed = 0.45 M CN ─ If needed, REC the salt: © 2011 ChemReview.Net v. 7w 1 KCN ^ 0 M 0.45 M 1 K+ + ^ 0.45 M 1 CN─ ^ 0.45 M (separates 100%). Page 1015 Module 33 — Buffers 2. WRECK the WA, then add additional conjugate base at step C. WANTED = [H+] = x R and E: WA ^ Conc@Eq. [WA]mixed ─ x ▐ H+ ^ x + CB (goes slightly) ^ x + [CB mixed in] │ 3. At the WA K step, write the three buffer Ka equations. Ka ≡ [H+]eq.[CB]eq. ≡ [WA]eq. ^ Definition 4. x • ( x + [CB]mixed ) [WA]mixed ─ x x • [CB]mixed ≈ Ka [WA]mixed ≈ ^ Exact ^ Approximation Solve the buffer approximation. Use values in the buffer chart. WANT: [H+] = x SOLVE: 6.2 x 10―10 ≈ x • 0.45 M ; x ≈ 6.2 x 10―10 • 0.30 ≈ 4.2 x 10―10 M = [H+] 0.30 M 0.45 Check: the exponent of the Ka and the [H+] are within + 1 . In buffers, you mix similar moles of acid and a base. Because the Ka of HCN is smaller than the Kb of its conjugate base (CN―), the solution should be basic, and, since [H+] < 10―7 , it is. 5. % Dissociation quick check: x = [H+] = 4.2 x 10―10 M, [WA] = 0.30 M = 3.0 x 10―1 M = the dissociation is much less than 5%, so the approximation may be used. 2. Strategy: 1. If the two substances include a conjugate pair, the solution is a buffer Buffer Chart: WA formula = H2PO4─ CB formula = HPO42─ mol or [WA]mixed = 0.10 M H2PO4─ mol or [CB]mixed = 0.20 M HPO42─ If needed, REC the salt(s): Rxn. & Extent: Conc@Eq. Rxn. & Extent: Conc@Eq. 1 NaH2PO4 ^ 0 M 0.10 M 1 Na2HPO4 ^ 0 M 0.20 M ▐ 1 Na+ + 1 H2PO4─ ^ ^ 0.10 M 0.10 M ▐ 2 Na+ + 1 HPO42─ (goes ~ 100%). ^ ^ 0.40 M 0.20 M (goes ~ 100%). 2. WRECK the weak acid ionization, then add additional conjugate base. WANTED = [H+] = x H2PO4─ is amphoteric, listed in the acid-strength table twice: on the right as a base, and on the left below as an acid. In this buffer, it is the weak acid in the conjugate pair. In water it will ionize slightly. R& E Conc@Eq. © 2011 ChemReview.Net v. 7w WA ^ 0.10 M ─ x ▐ H+ ^ x + │ CB (goes slightly) ^ x + [0.20 M CB] Page 1016 Module 33 — Buffers 3. At the WA K step, write the three buffer Ka equations. Ka ≡ [H+]eq.[CB]eq. ≡ [WA]at eq. ^ Definition x • ( x + [CB]mixed ) [WA]mixed ─ x ≈ x • [CB]mixed ≈ Ka [WA]mixed ^ Exact ^ Approximation 4. Solve the buffer approximation. Use values in the buffer chart. WANT: [H+] = x SOLVE: 6.7 x 10―8 ≈ x • 0.20 M ; x ≈ 6.7 x 10―8 • 0.10 ≈ 3.4 x 10―8 M = [H+] 0.10 M 0.20 5. % Dissociation quick check: x = [H+] = 3.4 x 10―8 M M, [WA] = 0.10 M = 1.0 x 10―1 M The exponent difference is 3 or greater; so dissociation is < 5% and the approximation may be used. Practice B 1. In a weak acid solution, H+ and conjugate are formed in a 1 to 1 ratio, so [H+] = [conjugate]. 2. In a weak acid solution, the weak acid has a substantially higher concentration than H+ or conjugate. 3. A buffer adds conjugate to the weak acid solution. [Conjugate] becomes higher than [H+] . 4. Adding conjugate ion shifts the weak acid ionization reaction to the left, using up H+. As [H+] goes down, pH goes UP. ***** Lesson 33E: Quick Buffer Calculations Buffer Calculations – the Quick Steps In most cases, two factors allow us to solve buffer calculations in fewer steps than required in the methodical method above. 1. In weak acid solutions, except for a few moderately strong weak acids, ionization is usually less than 5%, and the approximation equation can be used in calculations. In buffers, as noted in the prior lesson, the weak acid ionization is even lower than for the weak acid alone. For this reason, calculations based on the buffer approximation equation pass the 5% test even more often than for weak acid solutions. 2. In most courses, instructors will require that you write out the methodical buffer steps at least once during an assignment to demonstrate that you understand how the buffer approximation is derived. However, for buffer calculations, the generic reaction and Ka equations are the same or very similar, and after those steps are written once, in the calculations that follow you are often permitted to omit those steps. © 2011 ChemReview.Net v. 7w Page 1017 Module 33 — Buffers For these reasons, for most buffer calculations it is acceptable to solve by the following Buffer Quick Steps 1. Write “WANTED = “ and the symbol wanted. 2. As DATA, fill in the buffer chart. If needed, REC the salts. 3. SOLVE the buffer approximation Ka ≈ [H+] [CB]mixed [WA]mixed using the chart data. Try the quick steps on the following problem, then check your answer below. Q. In a buffer solution that contains 0.25 M CN― and 0.50 M HCN, find [H+]. (Ka of HCN = 6.2 x 10―10). ***** Answer 1. WANT: [H+] CB formula = CN― 2. Buffer Chart: WA formula = HCN mol or [WA]mixed = 0.50 M HCN 3. SOLVE: Ka ≈ [H+] [CB]mixed mol or [CB]mixed = 0.25 M CN― ≈ 6.2 x 10―10 ≈ [H+] ( 0.25 M) [WA]mixed [H+] = 6.2 x 10―10 ( 0.50) ; 0.50 M = 1.2 x 10―9 M = [H+] 0.25 Practice A. Use the buffer quick steps. 1. Calculate the [H+] in a solution of 0.60 M C6H5COOH and 0.40 M C6H5COO―. (Ka of C6H5COOH = 6.3 x 10―5). 2. Find [H+] in a mixture of 0.10 M NaOCN and 0.20 M HOCN (Ka HOCN = 3.5 x 10―4) Buffer Calculations Based On Moles The buffer approximation equation can be written as Ka ≈ [H+] • [CB]as mixed [WA]mixed In a buffer, because the acid and base are both in the same liters of solution, liters can be cancelled in the fraction of the equation above. [CB] = [WA] mol CB/L of soln. = mol CB mol WA mol WA/L of soln. © 2011 ChemReview.Net v. 7w Page 1018 Module 33 — Buffers This converts the buffer approximation to the form Ka ≈ [H+] [CB]mixed [WA]mixed OR Ka ≈ [H+] (base moles as mixed) (acid moles as mixed) For most buffer calculations, the data for WA and CB is supplied in moles/liter, but the buffer approximation can be solved if the WA and CB as mixed are measured in mol/L or moles, as long as both terms have the same unit. A Check On Buffer Calculations In most buffer solutions, comparing [WA mixed] and [CB mixed]: one is no more than 10 times the other. In the buffer approximation equation, this means that the fraction [CB ]/[WA] has a value of between 1/10 and 10. Ka ≈ [H+] • [CB]mixed ≈ [WA]mixed [H+] • (10―1 to 10 ) This result means that Using in scientific notation, for solutions mixed to be buffers, the exponent of the [H+] is nearly always within + 1 of the exponent of the Ka . For example, in the previous calculation, Ka = 6.2 x 10―10 ; [H+] = 1.2 x 10―9 M Use this rule as a quick check on your [H+] calculations in buffers. This rule also results in an important General Rule for Buffers: In common-ion buffer solutions, the [H+] will be close in value (within a factor of 10) to the Ka of weak acid component of the buffer. A Special Case: Equal Moles or Equal Concentrations For the special case of a buffer in which [WA as mixed] is the same as [CB as mixed], in the buffer approximation equation, those two terms cancel. This means that if [WA] = [CB], then [H+] can be written by inspection: it is the same as the Ka . Ka ≈ [H+] • [CB]mixed so Ka ≈ x = [H+] if [WA] = [CB] [WA]mixed Similarly, if the moles of WA and CB are in the same, then [H+] ≈ Ka . In buffer solutions, if either [WA] = [CB] or moles WA = moles CB, then [H+] ≈ Ka . Limitations The buffer approximation equations are reliable within 5% as long as the mol/L WA that ionizes = x = [H+] is small compared to [WA] and [CB]. In most buffers, this will be the © 2011 ChemReview.Net v. 7w Page 1019 Module 33 — Buffers case, but if the WA ionization is high, or the solution is very dilute, you may need to solve the buffer quadratic to obtain more accurate results. Summary: Buffer Quick Steps 1. Write “WANTED = “ and the symbol wanted 2. As DATA, fill in the buffer chart. If needed, REC the salts. 3. SOLVE the buffer approximation: using the chart data. Ka ≈ [H+] [CB]mixed [WA]mixed OR Ka ≈ [H+] moles CB mixed moles WA mixed 4. Check: If [WA] and [CB] differ by <10x, [H+] and Ka exponents are within + 1. 5. If [WA] = [CB] or moles WA = moles CB, [H+] ≈ Ka . 6. In a buffer, the [H+] will be close to the Ka of the weak acid component. Practice B. Do the odd numbers first, and the evens if you need more practice. Problem 6 is more challenging. 1. Estimate the [H+] for a. Problem 2 b. Problem 3 Use the Quick Steps for Problems 2-5. 2. Find [H+] for a buffer solution in which KF= 0.20 mol and HF = 0.20 mol. Ka of HF = 6.8 x 10―4 3. Calculate [H+] in a mixture of 0.25 M NaOBr and 0.50 M HOBr (Ka HOBr =2.3 x 10―9) 4. In a solution of 0.25 M NH3 and 0.75 M NH4Cl, find the [OH─]. (Kb NH3 = 1.8 x 10―5). 5. 8.10 g of HCN and 4.90 g of NaCN are mixed into about half a liter of water. What is the pH in this solution? (Ka of HCN = 6.2 x 10―10). 6. To have a pH of 5.00 in a buffer solution of 0.10 M CH3COOH and ? M CH3COONa, what must be the [CH3COONa]? (Ka for CH3COOH = 1.8 x 10―5) © 2011 ChemReview.Net v. 7w Page 1020 Module 33 — Buffers ANSWERS Practice A 1. 1. WANT: [H+] 2. Buffer Chart: WA formula = C6H5COOH mol or [WA]mixed = 0.60 M C6H5COOH CB formula = C6H5COO― mol or [CB]mixed = 0.40 M C6H5COO― Ka ≈ [H+] [CB]mixed [WA]mixed 6.3 x 10―5 ≈ [H+] 0.40 M ; [H+] = 6.3 x 10―5 • 0.60 = 9.4 x 10―5 M = [H+] 0.60 M 0.40 4. Check: the exponent of the Ka and the [H+] are within + 1 . 2. 1. WANT: [H+] 2. Buffer Chart: WA formula = HOCN CB formula = OCN― 3. Solve the buffer approximation: mol or [WA]mixed = 0.20 M HOCN 3. Solve the buffer approximation: mol or [CB]mixed = 0.10 M OCN― Ka ≈ [H+] [CB]mixed [WA]mixed [H+] = 3.5 x 10―4 • 0.20 = 7.0 x 10―4 M = [H+] 0.10 4. Check: the exponent of the Ka and the [H+] are within + 1 . 3.5 x 10―4 ≈ [H+] 0.10 M ; 0.20 M Practice B 1a. In problem 2, Ka = 6.8 x 10―4 , so [H+] in scientific notation must end in ” x 10―3,4,or 5 ” by the check rule. The [H+] must be close to the Ka . 1b. In problem 3, Ka = 2.3 x 10―9 , so [H+] in scientific notation must end in ” x 10―8,9,or 10 ”.. 2. WANT: [H+] Quick method: since 0.20 mol HF = 0.20 mol F―, mol WA mixed cancels mol CB mixed in the approximation, and Ka ≈ [H+] ≈ 6.8 x 10―4 M 3. 1. WANT: [H+] WA formula = HOBr CB formula = OBr― mol or [WA]mixed = 0.50 M HOBr 2. Buffer Chart: mol or [CB]mixed = 0.25 M OBr― 3. Solve the buffer approximation: © 2011 ChemReview.Net v. 7w Ka ≈ [H+] [CB]mixed [WA]mixed Page 1021 Module 33 — Buffers [H+] = 2.3 x 10―9 • 0.50 = 4.6 x 10―9 M = [H+] 0.25 4. Check: the exponent of the Ka and the [H+] are within + 1 . 4. A weak acid (NH4+) and its conjugate (NH3) = a buffer solution. 2.3 x 10―9 ≈ [H+] 0.25 M ; 0.50 M 1. WANT: [OH─] In buffer calculations, find [H+] first. WA formula = NH4+ CB formula = NH3 mol or [WA]mixed = 0.75 M NH4+ 2. Buffer Chart: mol or [CB]mixed = 0.25 M NH3 If needed, REC the salt: 1 NH4+ + 1 Cl─ ^ ^ 0.75 M 0.75 M 1 NH4Cl ^ 0 M 0.75 M 3. Solve the buffer approximation: (separates ~100%). Ka ≈ [H+] [CB]mixed [WA]mixed Since you are given Kb and need Ka , first use Ka x Kb = 1.0 x 10─14 ? = Ka NH4+ = 1.0 x 10─14 = 1.0 x 10─14 Kb 1.8 x 10―5 5.6 x 10─10 = = Ka of NH4+ Ka = 5.6 x 10─10 ≈ x • 0.25 M ; x ≈ 5.6 x 10─10 • 0.75 ≈ 1.7 x 10―9 M = [H+] 0.75 M 0.25 4. Check: the exponent of the Ka and the [H+] are within + 1 . But WANTED = [OH─] . Use [H+][OH─] = 1.0 x 10─14 Kw = [OH─] = 1.0 x 10─14 = 1.0 x 10─14 [H+] 1.7 x 10─9 = 0.59 x 10─5 = 5.9 x 10─6 M OH─ (Kw check: [H+] x [OH─] (circled) must estimate to 10.0 x 10─15 or 1.0 x 10─14). 5. In chemistry, given grams, we convert to moles. Once we know moles, we can fill in the buffer chart. For grams and moles, need molar masses: HCN = 27.0 g/mol ; NaCN = 49.0 g/mol ? mol HCN = 8.10 g HCN ● 1 mol HCN 27.0 g HCN = 0.300 mol HCN ? mol NaCN = 4.90 g NaCN ● 1 mol NaCN = 0.100 mol NaCN 49.0 g NaCN 1. WANT: [H+] WA formula = HCN CB formula = CN― mol or [WA]mixed = 0.300 mol HCN 2. Buffer Chart: mol or [CB]mixed = 0.100 mol CN― 3. Solve the buffer approximation that uses moles: © 2011 ChemReview.Net v. 7w Ka ≈ [H+] (moles of base) moles of acid Page 1022 Module 33 — Buffers [H+] = 6.2 x 10―10 • 0.300 = 1.9 x 10―9 M = [H+] 0.100 4. Check: the exponent of the Ka and the [H+] are within + 1 . Note that if the buffer approximation fraction has moles as units, we do not need to know the solution volume (unless the solution becomes very diluted, in which case the approximation assumption that x is much smaller than [WA] and [CB] may no longer be true. 6.2 x 10―10 ≈ [H+] 0.100 mol ; 0.300 mol 6. 1. WANTED: [CH3COONa] = [CB]mixed (write specific and general symbols WANTED) DATA: pH = 5.00 pH prompt: pH ≡ ─ log [H+] and [H+] = 10―pH So [H+] = 1.0 x 10─5 [CH3COOH]mixed = 0.10 M = [WA mixed] Strategy: A weak acid (CH3COOH) and a salt with its conjugate ion = buffer solution. To find the original salt concentration, find the [CB] as mixed into the solution. To find [CB mixed], use the buffer steps . CB formula = CH3COO― 2. Buffer Chart: WA formula = CH3COOH [WA]mixed = 0.10 M CH3COOH REC the salt: 1 CH3COONa ^ 0 M ? ? ? [CB]mixed [CB]mixed = ?? M C6H5COO― 1 Na+ + 1 CH3COO─ (separates ~100%). ^ ^ ???M ? ? ? [CB]mixed ***** Ka ≈ [H+] [CB]mixed for the WANTED symbol. [WA]mixed ? = [ CB mixed ] = Ka ● [WA]mixed = 1.8 x 10―5 (0.10) = 0.18 M [CH3COO─] [H+] 1.0 x 10―5 3. Solve the buffer approximation: To find the WANTED symbol, substitute back into the salt REC equation. 1 Na+ + 1 CH COO─ REC the salt: 1 CH COONa 3 ^ 0 M 0.18 M ^ 0.18 M 3 (separates ~100%). ^ 0.18 M [CH3COONa]as mixed = 0.18 M ***** © 2011 ChemReview.Net v. 7w Page 1023 Module 33 — Buffers Lesson 33F: The Henderson-Hasselbalch Equation Solving For pH in Buffers In buffer solutions, you are often asked to find the pH rather than [H+]. The pH can be calculated after an [H+] is solved by the buffer approximation, but it is convenient to have a formula that calculates pH directly. We can derive such a formula by the following steps. The buffer approximation equation can be written as Ka ≈ [H+] • [CB]as mixed [WA]mixed Taking the log of both sides, applying the rule that “the log of a product is the sum of the logs,” then multiplying both sides by ─1 results in ─log ( Ka ) ≈ ─ log [H+] ─ log ( [CB]mixed [WA]mixed ) In science, the symbol for ─log is a lower case p , as in ─log [H+] = pH. equation can therefore be written as p Ka ≈ pH ─ log [CB]mixed ( Solving for pH, the equation becomes The above ) [WA]mixed pH ≈ pKa + log [CB]mixed ( [WA]mixed ) This form of the equation that solves for pH is known as the Henderson-Hasselbalch equation. It is usually written as pH ≈ pKa + log [base] () [acid] since [CB] = [base] and [WA] = [acid] in a buffer. The Henderson-Hasselbalch equation is simply the buffer approximation equation modified mathematically to solve for pH. If 1. WANTED = buffer pH 2. Fill in the buffer chart. 3. Write and solve the Henderson-Hasselbalch equation using buffer chart values. Try this example. Q. In a buffer that contains 0.30 M acetic acid and 0.20 M acetate ion, find the pH. (Ka CH3COOH = 1.8 x 10―5) ***** © 2011 ChemReview.Net v. 7w Page 1024 Module 33 — Buffers Answer 1. WANTED = pH of a buffer. WA formula = CH3COOH CB formula = CH3COO ─ [WA]mixed = 0.30 M CH3COOH 2. Buffer Chart: [CB]mixed = 0.20 M CH3COO ─ 3. pH ≈ pKa + log [base] () = ─log (1.8 x 10―5) + log (0.20/0.30) [acid] = ─ ( ─ 4.74 ) + log (0.67) = 4.74 ─ 0.18 = 4.56 = pH (To review taking logs with a calculator, see Lesson 27D). Practice A: Commit the Henderson-Hasselbalch equation and rule above to memory, then do these. 1. If X = 2.0 x 10―12 , pX = ? 2. If Ka = 4.5 x 10―7 , pKa = ? 3. If pKa = 5.50 , Ka = ? 4. Find the pH in a solution that contains 0.15 M HCN and 0.25 M NaCN. (Ka of HCN = 6.2 x 10―10). The Henderson-Hasselbalch Fraction: Moles Or Mol/L The acid and base in a buffer solution are usually measured in terms of their concentrations, and the Henderson-Hasselbalch equation is solved using the form pH ≈ pKa + log ( [base] /[acid] ) However, as with the buffer approximation, since the acid and base are dissolved in the same solution, the liters in which both components are dissolved is the same, and liters can be cancelled in the fraction of the equation. [base] = mol base/L of soln. = mol base [acid] mol acid/L of soln. mol acid This converts the Henderson-Hasselbalch equation to the form pH ≈ pKa + log (mol base / mol acid) In upcoming problems when we solve for pH during titration, measuring the acid and base in moles will speed our calculations. To reflect this option, let us modify our buffer pH rule as follows. © 2011 ChemReview.Net v. 7w Page 1025 Module 33 — Buffers If 1. WANTED = buffer pH, 2. Fill in the buffer chart for moles or mol/L . 3. Solve the Henderson-Hasselbalch equation using the buffer chart values and either pH ≈ pKa + log ([base] /[acid]) or pH ≈ pKa + log (mol base /mol acid) Pick the form of the equation that best matches the data supplied in the problem. Henderson-Hasselbalch Relationships From the Henderson-Hasselbalch equation, we can derive additional relationships that simplify buffer calculations. In a buffer solution, 1. If [acid] = [base], or moles acid = moles base, then pH ≈ pKa . Why? If base/acid in moles or mol/L = 1 , then log(1) = log (100) = 0 , and the Henderson-Hasselbalch equation becomes pH ≈ pKa + 0 ≈ pKa 2. For different solutions of the same conjugate pair, if the mole or mol/L ratio base/acid is the same, the pH is the same. For example, in a buffer that contains 0.10 M NaF and 0.20 M HF, the pH will be the same as in a buffer containing 0.30 mol NaF and 0.60 mol HF. Buffer pH is based on the particle ratio base/acid , and in both of these solutions, that ratio is the same. 3. In buffer solutions, the pH will be close to the pKa of the weak acid component. Why? Most buffers are mixed so that the moles of the acid and base are close, in order for the buffer to be able to neutralize substantial amounts of both acids and bases that may be added to the buffer. For the base and acid in most buffers, the mol/L of the one will not be more than 10 times the other. • If a base/acid ratio is between 1/10 and 10, the log of those numbers is small: Log 1/10 = ─ 1 , log 10 = + 1 . • When one component is double the other, Log 2 = 0.3 , log 0.5 = ─ 0.3 , and the difference between pKa and pH is smaller: + 0.3 . Adding those numbers to the pKa in the Henderson-Hasselbalch equation results in a small change in pH. The math above gives us the Buffer pH check rule: when the acid and base are mixed in close to equal moles or concentrations (as in most buffers), check that the pH is within + 2 of the number after the negative sign of the Ka of the weak acid component of the buffer. For example, in the problem above Practice A, Ka = 1.8 x 10―5 ; pH = 4.56 © 2011 ChemReview.Net v. 7w Page 1026 Module 33 — Buffers The check rule also leads to a Summary of fundamentals for buffers. • A buffer contains a weak acid and its conjugate base, both in substantial concentrations. A buffer resists a change in pH if acid or base is added. • In a buffer, the [H+] is close to the Ka of the weak acid component, and the pH is close to the pKa of the weak acid component. Summary: To Calculate pH In A Buffer 1. Write WANTED = symbol wanted. 2. Fill in the buffer chart using either moles or mol/L. 3. Substitute into the Henderson-Hasselbalch equation in the form pH ≈ pKa + log ([base] /[acid]) or pH ≈ pKa + log (mol base /mol acid) 4. In a buffer, if either [WA] = [CB] [H+] ≈ Ka and pH ≈ pKa . or moles WA = moles CB, then 5. For a given conjugate pair in different buffer solutions, if the ratio [base]/[acid] or mol base/mol acid is the same, the pH is the same, and [H+] is the same, in the two solutions. 6. Buffer pH check rule: when the moles or concentrations of acid and base close to equal (as in most buffers), check that the pH is within + 2 of the number after the negative sign of the Ka . Practice B. Put a check by problem numbers as you do them. Save one or two for your next practice session. 1. In which solution in Questions 2-5 below will pH = pKa ? 2. Using the Check Rule, do a quick whole number estimate of the top and bottom of the range of possible pH values in these buffer solutions. a. 0.45 M HF and 0.15 M NaF (Ka of HF = 6.8 x 10―4) b. 0.15 mol HOCl and 0.30 mol KOCl (Ka of HOCl = 3.5 x 10―8) 3. Find the pH for the buffer solutions in Problem 2. 4. In a buffer consisting of 0.20 mol NH3 and 0.20 mol NH4Cl, what is the pH? (Kb of NH3 = 1.8 x 10―5) 5. A solution consists of 0.30 M K2CO3 and 0.40 M KHCO3. Calculate the pH. For H2CO3 , Ka1 = 4.3 x 10―7 and Ka2 = 5.6 x 10―11 . © 2011 ChemReview.Net v. 7w Page 1027 Module 33 — Buffers 6. If you needed a buffer solution with a pH of 6, which would you choose as the acid conjugate of the buffer: HOBr, NH4+, HCN, or CH3COOH ? ( Ka values: HOBr = 2.3 x 10―9, NH4+ = 5.6 x 10─10 , HCN = 6.2 x 10―10 , and CH3COOH = 1.8 x 10―5 ). ANSWERS Practice A 1. If X = 2.0 x 10―12 , pX = ? = ─log (2.0 x 10―12) = ─ ( ─ 11.70 ) = 11.70 The prefix p is an abbreviation for the function ─log . 2. If Ka = 4.5 x 10―7 , pKa = ? = ─log (4.5 x 10―7) = ─ ( ─ 6.35 ) = 6.35 3. If pKa = 5.50 , Ka = ? Since then pH ≡ ─ log [H+] and [H+] ≡ 10─pH Ka ≡ 10─pKa pKa ≡ ─ log Ka and = ? = 10―5.50 = 3.2 x 10―6 = Ka 4. WANT: pH of a buffer. To find buffer pH directly, fill in the buffer chart, then solve the H-H equation. WA formula = HCN CB formula = CN─ mol or [WA]mixed = 0.15 M HCN Buffer Chart: mol or [CB]mixed = 0.25 M CN ─ pH ≈ pKa + log () [base] = ─log (6.2 x 10―10) + log (0.25 / 0.15) [acid] = 9.21 + log (1.67) = 9.21 + 0.22 = 9.43 = pH Practice B 1. Question 4. If [WA] = [CB] , then pH = pKa . 2. The check rule: the calculated pH and the number after the minus sign in the Ka should be within + 2 . a. Ka of HF = 6.8 x 10―4 , pH should be 4 + 2: between 2 and 6. b. Ka of HOCl = 3.5 x 10―8 , pH should be 8 + 2: between 6 and 10. 3a. WANT: pH of a buffer. Fill in the buffer chart, then solve the H-H equation. pH ≈ pKa + log WA formula = HF CB formula = F─ mol or [WA]mixed = 0.45 M HF Buffer Chart: mol or [CB]mixed = 0.15 M F ─ () [base] = ─log (6.8 x 10―4) + log ( 0.15 / 0.45 ) [acid] = 3.17 + log (0.33) = 3.17 ─ 0.48 = 3b. WANT: 2.69 = pH Between 2-6? Check. pH of a buffer. Fill in the buffer chart, then solve the H-H equation. Since the buffer data is in moles, complete the buffer chart second line based on moles instead of mol/L. Buffer Chart: WA formula = HOCl © 2011 ChemReview.Net v. 7w CB formula = ClO─ Page 1028 Module 33 — Buffers mol or [CB]mixed = 0.30 mol ClO─ mol or [WA]mixed = 0.15 mol HOCl pH ≈ pKa + log ( mol base / mol acid ) = ─ log (3.5 x 10―8) + log ( 0.30 / 0.15 ) = 7.46 + log (2.00) = 7.46 + 0.30 = 4 7.76 = pH Between 6-10? Check. WANT: pH of a buffer. Solve using the shortcut: Since [WA] = [CB] , pH ≈ pKa . But in this problem, we are given the Kb of the base. Ka x Kb = 1.0 x 10─14 To find Ka of the acid, use the rule for conjugate pairs: ? = Ka NH4+ = 1.0 x 10─14 = 1.0 x 10─14 Kb 1.8 x 10―5 = 5.6 x 10─10 = Ka NH4+ pH ≈ pKa = ─ log (5.6 x 10─10 ) = ─ ( ─ 9.25 ) = 9.25 = pH Check: Ka of NH4+ = 5.6 x 10─10 , pH estimate is 8 -12. Check. 5. Two mixed substances with similar formulas often indicate a buffer. Try to complete the buffer chart by inspection. If needed, REC the salt(s). 1. Buffer Chart: WA formula = HCO3─ mol or [WA]mixed = 0.40 M HCO3─ 2. CB formula = CO32─ mol or [CB]mixed = 0.30 M CO32─ In the Henderson-Hasselbalch equation, Ka is needed. In this problem, the two conjugate ions are formed by the successive ionization of the polyprotic acid H2CO3. The Ka value for the acid in this buffer, HCO3─ , is the K of the second H2CO3 ionization: Ka2 = 5.6 x 10―11 (see Lesson 28G on polyprotic acids). pH ≈ pKa + log ( [base]/[acid] ) = ─ log (5.6 x 10─11 ) + log (0.30 / 0.40) = ─ ( ─ 10.25 ) + log (0.75) = 10.25 + ( ─ 0.12 ) = 10.13 = pH Check: Ka of HCO3─ = 5.6 x 10─11 , pH should be between 9 and 13. 6. The pH of the buffer will be within +2 of the after the negative sign in the Ka of the acid component. Only CH3COOH has a Ka number that is within 2 of 6. ***** SUMMARY – Buffers 1. A buffer solution resists a change in pH when an acid or base is added. A buffer can be made from a weak acid or base and the conjugate of the acid or base. 2. To solve buffer calculations for [H+], the methodical steps are: 1. Fill in the buffer chart. If needed, REC the salt(s). 2. WRECK the WA ionization, then add [CB added]. 3. At the WA K step, write the three buffer Ka equations. © 2011 ChemReview.Net v. 7w Page 1029 Module 33 — Buffers 4. Solve the buffer approximation. 5. Calculate the % dissociation. If >5%, solve the exact buffer quadratic. 3. To solve buffer calculations for [H+], the quick steps are 1. Write the symbol WANTED. 2. Fill in the buffer chart. If needed, REC the salt(s). Buffer Chart: WA formula = _____________ CB formula = ________________ mol or [WA]mixed = _______ mol or [CB]mixed = __________ 3. Write and solve the buffer approximation: Ka ≈ [H+] [CB]mixed [WA]mixed OR Ka ≈ [H+] (base moles as mixed) (acid moles as mixed) 4. To solve buffer calculations directly for pH, • Complete the buffer chart in moles or mol/L, then • Substitute the buffer chart values into the Henderson-Hasselbalch equation: pH ≈ pKa + log [base] () or pH ≈ pKa + log (mol base / mol acid) [acid] 5. In a buffer, if [WA] = [CB] or mol WA = mol CB, then [H+] ≈ Ka and pH ≈ pKa. 6. In general in buffers, the [H+] is close to the Ka of the weak acid component, and the pH is close to the pKa of the weak acid component. 7. Summary of acid-base check rules a. In aqueous solutions, [H+] x [OH─] must = Kw = 10.0 x 10─15 or 1.0 x 10─14. b. In conjugate pairs, Kb x Ka must estimate to = Kw = 10.0 x 10─15 or 1.0 x 10─14. c. If the [H+] is written in scientific notation, the number after the negative sign in the exponent of the [H+] must be within + 1 of the pH. d. The Quick 5% Test to see if an approximation equation can be used • Using scientific notation, compare [WA] to [H+] = x or [WB] to [OH─] = x . • If the exponents differ by 3 or more, the ionization is less than 5%. • If the exponents differ by 2 or less, use the % dissociation equation for the 5% test. e. Buffer pH check rule: when the acid and base are in close to equal moles or concentrations (as in most buffers), check that • the exponent of the Ka and the [H+] are within + 1 . • the pH is within + 2 of the number after the negative sign in the Ka of the acid. ##### © 2011 ChemReview.Net v. 7w Page 1030 Module 34 — pH During Titration Module 34 — pH During Titration Prerequisites: Complete Module 33 before starting this module. ***** Lesson 34A: pH In Mixtures Solving for pH in Acids and Bases, Salts, and Buffers So far, our study of acids and bases has included finding [H+] and pH for solutions of • One acid or base that is strong or weak; • Polyprotic acids; • Salts with one ion that is acidic, basic, or amphoteric; and • A mixture of acid-base conjugates (a buffer). These types of calculations have components in common, but their steps are not the same. In calcuations involving on acids and bases, it will be necessary to identify what type a problem is, then to apply the steps needed to solve that problem type. A summary can helping in learning the similarities and distinctions in the problem types. Your list might include the following. Summary: In aqueous solutions, to calculate [H+], [OH─], or pH, for 1. A strong acid or hydroxide base, use the REC steps based on ~100% ionization or quick steps: [HCl or HNO3]mixed = [H+] and [NaOH or KOH] = [OH─] 2. Weak acids: use the WRRECK steps and and/or Ka approximation. 3. Weak bases: use the WRRECK steps and solve Kb approximation. 4. Conjugate pairs: use Ka x Kb = 1.0 x 10─14 5. Polyprotic acids: use Ka1 and the WRRECK steps if Ka1 is >100 times Ka2. If not, add the contributions of the first two ionizations. 6. Salts with one ion that is a weak acid or base: • REC the salt, write and label the ions as A, B, or N, WRRECK the A or B ion. 7. Buffer solutions: • Treat as a weak acid with conjugate base added. • Complete the buffer chart. • Solve the buffer approximation or the H-H equation using mol or mol/L WA and CB as mixed. © 2011 ChemReview.Net v. 7w Page 1031 Module 34 — pH During Titration Practice A For aqueous solutions of the following, 1. NaF and HF 2. NaCN 7. H2CO3 6. KCl 4. CH3COOH 5. NaHCO3 3. NH3 8. KOH 9. NH4Cl 10. HNO3 Pick one term from the following list that best characterizes the solution. a. Strong acid b. Strong base e. Acidic salt f. pH neutral salt i. Polyprotic acid c. Weak acid g. Basic salt d. Weak base h. Amphoteric salt j. Buffer Types of Acid-Base Solutions So far, all of our [H+], [OH─], and pH calculations have been for solutions at equilibrium: if an acid-base reaction did occur, it is now over, and no further changes in the solution are taking place. In addition, all of the systems we have studied have had either • one pH-dominant component (one strong or weak acid or base), or • have been a buffer: a mixture of one weak acid or base and its conjugate. We now turn our attention to other mixtures of acids and bases. The rules for finding [H+] or pH for mixtures acids or bases include the following. 1. For mixtures of acids and bases, we will adopt the following vocabulary: a. The opposite of an acid will mean a base, and the opposite of a base will mean an acid. b. Strong acids and bases are those that ionize ~100% to form H+ or OH─ ions. c. Weak acids and bases will be those with Ka or Kb values between one and 10―16 according to acid-strength tables (see Lesson 31B). 2. Any strong acid or base will react with its opposite if opposite is present. If a mixture contains a strong acid and any base, or a strong base and any acid, a reaction must take place. The limiting reactant will be used up quickly, and the mixture will then be at equilibrium. We will discuss these reaction cases in a later lesson. However, a strong acid can be at equilibrium in a mixture with other acids or pH-neutral particles. A strong base can be at equilibrium in a mixture with other bases or pHneutral particles. In those mixtures, no further net reaction will occur. 3. In a mixture at equilibrium, the pH is largely determined by the types of particles present in significant concentrations (greater than ~0.001 M). To calculate the pH in a mixture of acidic, basic, and neutral particles, use these general steps. • Label each particle in the mixture as neutral or as a strong or weak acid or base, • Solve for the pH as determined by the pH-dominant particle. © 2011 ChemReview.Net v. 7w Page 1032 Module 34 — pH During Titration 3. Specifically, apply these steps in order to determine the [H+] or pH for a solution at equilibrium. a. Re-write formulas for soluble ionic compounds (salts) in their separated-ions format. b. Label each particle in the mixture as a strong acid (SA), strong base (SB), weak acid (WA), weak base (WB), or pH neutral (N), based on the definitions in point 1 above. These definitions and labels are based on absolute rather than comparative strength: a solution may have more than one SA, SB, WA, WB, or N particle. c. If all particles in a mixture are pH neutral, solution pH = 7 . d. Ignore pH-neutral (N) particles if mixed with acidic or basic particles. Particles that are pH-neutral, including Na+, K+, Cl─, and NO ─, do not change the 3 solution pH. e. If the mixture contains a strong acid or base, its concentration determines the pH. The logic is: if a strong acid or base is present, it is the dominant factor in deciding pH. Contributions by any weak components in the mixture are relatively small and with rare exception can be ignored. To find [H+] in solutions containing strong acids or bases, either write the REC steps for 100% ionization or use the quick steps: [HCl or HNO3]mixed = [H+]in soln. and [NaOH or KOH] = [OH─]in soln. f. If a weak acid or base and its conjugate are present in substantial amounts, the solution is a buffer. Use the buffer chart and Ka or Henderson-Hasselbalch equations to find the pH. g. If the only non-pH-neutral particle present is a weak acid or base, its Ka or Kb determines the pH. Solve the Ka or Kb equation, then determine pH. Summary: Steps For Calculating pH in a Mixture At Equilibrium Apply in this order. a. Re-write soluble salts as separated ions. b. Label each particle in the mixture as SA, SB, WA, WB, or N. c. Ignore pH-neutral (N) particles if mixed with acidic or basic particles. d. If all particles are N, pH = pH of water = 7. e. If SA or SB is present, ignore other particles. Find pH based on quick steps: [HCl or HNO ] = [H+] and [NaOH or KOH] = [OH─] 3 f. If a WA or WB and its conjugate is present, solve a buffer chart and the Henderson-Hasselbalch equation in moles or mol/L. g. If only one WA or WB is present, solve the Ka or Kb equation, then find pH. The logic is: • pH-neutral particles do not change the [H+] and pH. • If a strong acid or base is present, it determines the [H+] and pH. © 2011 ChemReview.Net v. 7w Page 1033 Module 34 — pH During Titration Practice B: Study the steps in the summary until you can write them from memory, then apply the steps to these problems. Save a few for your next practice session. 1. For each mixture a-d below, • List the particles present in the solution in significant concentration. • Label each particle as SA, SB, WA, WB, or N. • Write which step (c, d, e, or f) in the summary for pH in mixtures above should be used to solve for the pH of the solution. a. A mixture of 0.40 M NH4Cl, 0.10 M NaCl, and 0.20 M HCl. b. 0.50 L of solution containing 0.20 mol KCN, 0.20 mol HCN, and 0.10 mol KCl. c. A solution containing 0.45 M HF and 0.50 M KCl. d. 0.20 mol KCl and 0.35 mol KNO3 dissolved in 75 mL of solution. Solve these. 2. A solution contains 0.350 mol KCl in 120.0 mL of solution. Calculate the solution pH. 3. A solution contains 0.010 M KOH, 0.20 M KF , and 0.10 M KCl. Find the pH. 4. A solution contains 0.0050 moles of KCN and 0.010 mol KCl in 25.0 mL of solution. Calculate the pH. (Kb CN─ = 1.6 x 10―5). 5. After a reaction, at equilibrium, a solution contains 0.020 mol CH3COOH and 0.020 mol CH3COO─. Find the pH. (Ka CH3COOH = 1.8 x 10―5) ANSWERS Practice A (j) buffer: a weak acid and its conjugate. 2.. NaCN (g) basic salt: Na+ is pH neutral, CN─ is the CB of the weak acid HCN. 1. NaF and HF 3. NH3 (d) weak base: see the acid strength table. 4. CH3COOH (c) weak acid 5. NaHCO (h) amphoteric salt: HCO ─ is an amphoteric ion. 3 3 (f) pH neutral salt: K+ is pH neutral, Cl─ is the pH-neutral conjugate base of strong acid HCl. 7. H2CO3 (i) polyprotic acid: Forms HCO3─ at 1st ionization and CO32─ 2nd. See strength table. 6. KCl 8. KOH (b) strong base (e) acidic salt: NH4+ is a weak acid (see table); Cl─ is pH neutral. 10. HNO3 (a) strong acid 9. NH4Cl Practice B 1a. The particles present are © 2011 ChemReview.Net v. 7w Na+ N NH4+ WA HCl SA Cl─ N If an SA or SB is present, it decides the pH. Use step e . Page 1034 Module 34 — pH During Titration 1b. The particles present are K+ N 1c. The particles present are K+ N 1d. The particles present are K+ N CN─ WB HF WA Cl─ N HCN WA Cl─ N Cl─ N Contains a weak acid and its base conjugate. Use step f . Ignore pH-neutrals. Weak acid determines the pH. Use step g . NO3─ N All are pH-neutral. Use step c . 2. WANTED = pH. This solution contains only KCl, a pH-neutral salt. The pH = 7 . K+ OH─ F─ and Cl─ . N SB WB N If one strong particle is present, it determines the pH. Use the quick steps for strong bases. For this solution: [SB] = [OH─ ] = 0.010 M = 1.0 x 10―2 M , [H+] = 1.0 x 10―12 M , pH = 12.00 3. WANTED = pH The particles present are 4. WANTED = pH KCN is composed of K+ and CN─. The Kb reflects that CN─ is a weak base. KCl consists of K+ and Cl─. Both are neutral ions. The only ion that determines pH is CN─. Use Kb . See pH? Write pH ≡ ─ log [H+] and [H+] ≡ 10─pH at least once in each problem set or quiz. If a final mixture is a buffer, you may also use the Henderson-Hasselbalch pH approximation. In Kb calculations, begin by solving the Kb approximation. Kb ≈ x2 where x = [OH─] [WB]mixed To solve the approximation for [OH─], you need [WB]mixed = [KCN] = [CN─]mixed in this problem. ? = [WB]mixed = mol CN― = 0.0050 mol CN― • 1 mL = 0.200 M CN─ L soln. 25.0 total mL soln. 10―3 L Kb ≈ x2 Substituting: 1.6 x 10─5 ≈ [WB]mixed x2 0.200 M CN─ x2 = ( 1.6 x 10─5 ) ( 0.200 ) = 0.32 x 10―5 = 3.2 x 10―6 x ≈ (estimate 1-2 x 10―3) ≈ 1.79 x 10―3 M = [OH─] ( In Kb, x = [OH―] ) Quick 5% test : x = 1.79 x 10―3 M, [WB] = 0.20 M = 2.0 x 10―1 M Since the difference in the exponents is 2 or less, do the 5% test: 5% test = ● 100% = 1.79 x 10―3 ● 102 % x [WA or WB]mixed 2.0 x x 10―1 = 0.90 %, which is less than 5%, so approximation is OK -- but pH was wanted. ***** © 2011 ChemReview.Net v. 7w Page 1035 Module 34 — pH During Titration pOH = ─ log [OH─] = ─ log(1.79 x 10─3) = 2.75 = pOH pH = 14.00 ─ pOH = 11.25 = pH It’s a good idea to carry an extra sf until the last step, but the last digit in calculations is doubtful and will often vary +1 due to rounding. The solution of an ion that is a weak base should have a basic pH, and this solution does. 5. WANTED = pH. This solution contains a weak acid and its conjugate: it is a buffer. Buffer pH problems can be solved using the buffer chart and the Henderson-Hasselbalch equation. Since the acid and base data is in moles, solve the buffer chart and H-H fraction in moles (see Lesson 32H). WA formula = CH3COOH CB formula = CH3COO ─ mol WA = 0.020 mol Buffer Chart: mol CB = 0.020 mol Since mol WA = mol CB , the H-H equation: pH ≈ pKa + log (mol base/mol acid) simplifies to pH ≈ pKa + log (1) ≈ pKa + 0 ≈ pKa ≈ ─log (1.8 x 10―5) ≈ ─ (─ 4.74) = 4.74 = pH ***** Lesson 34B: pH After Neutralization Finding pH After A Reaction From Amounts Before the Reaction Calculations to find pH can be divided into two types: • Those in which the moles or mol/L of acidic or basic particles are known for a stable solution: one that is at equilibrium and therefore not changing (our pH calculations so far have been this type), and • Those in which you are given amounts of acid and base in the initial reactants, and you must find the pH of the mixture after the acid-base reaction stops. Let’s turn our attention to the second type: calculating the pH of a solution after a reaction from known amounts of acid and base before the reaction. In neutralization reactions, acid and base reactants are used up (react) if the reactants are a stronger acid and base than the products. To find the pH after the reaction, we need to know the amounts of all of the particles present in the mixture after the reaction: which is when the reaction stops, which is when the reaction reaches equilibrium. The simplest way to calculate all of the amounts present after a reaction is to use our “chemistry accounting system:” a rice moles table. Once we know the composition of a mixture after neutralization, we can solve using the rules for mixture pH. In first-year chemistry courses, our interest is limited to finding the pH of a mixture after neutralization when at least one of the components, either the acid or base, is strong. An acid or base that is strong is highly reactive: if opposite is present, it will react with its opposite until the limiting reactant is ~100% used up. © 2011 ChemReview.Net v. 7w Page 1036 Module 34 — pH During Titration This allows us to use a key rule that simplifies neutralization reactions and rice tables: When an acid and base are reacted, IF one of the components is strong, the reaction will go until the limiting component, either the acid or the base, is 100% used up. Another way to state this rule is At the end of a neutralization, if one of the reactants is strong, the moles of one of the reactants must be zero. Using this rule, we can complete a rice moles table for neutralization, using the same steps that were used in Lessons 10H to find the mixture present at the end of a reaction that goes to completion. Then, by applying the rules for mixtures to the particles in the End/Equilibrium row, pH can be calculated. Steps For Calculating pH After Reaction From Amounts Before Reaction 1. Convert the initial amounts of acid and base to moles (or prefix-moles). 2. Enter those moles in the Initial row of a rice moles table. 3. In the Change row, use the rule: when acid and base are mixed, if one or both are strong, one reactant is 100% used up. 4. Calculate the pH of the mixture at the End of the reaction (in the bottom rice row). The steps can be summarized as To find pH after reaction from acid and base amounts before reaction: Reactants > reactant moles > rice moles > equilibrium mixture pH . Lets apply these steps to an example. Q. A solution containing 0.100 moles of HCl is combined 0.080 moles of NaOH. After mixing, the total volume is 120. mL. What is the pH of the solution after mixing? Solve using the steps above and table below. Reaction Initial Change (use + , ―) End/Equilibrium ***** a. The initial moles of acid and base are known. b. From Lesson 14B: acids reacted with hydroxide bases form water as one product. The balanced equation is: 1 HCl + 1 NaOH 1 H2O + 1 NaCl (goes ~100%) ***** © 2011 ChemReview.Net v. 7w Page 1037 Module 34 — pH During Titration • If a reaction starts with all reactants, the sign in the Change row for the reactants must be negative, since reactant must be used up if there is a reaction. • In the Change row, the signs of the reactants and products must be opposites. • The ratios in the Change row and the Row 1 coefficients must be the same. (For a rice review for reactions that go to completion, see Lesson 10H.) ***** Reaction 1 HCl Initial 1 H2O 1 NaCl solvent 1 NaOH 0 mol 0.100 mol 0.080 mol Change ― 0.080 mol ― 0.080 mol + 0.080 mol + 0.080 mol At End/Equilibrium + 0.020 mol 0 mol solvent + 0.080 mol In this mixture, NaOH moles are limiting: when the NaOH is used up, the reaction stops. The limiting reactant determines how much of the products form. The amount of water initially present can be said to be zero from the reaction, or large because water is the solvent for the reaction. In either case, the amount of water will not affect this type of calculation. Calculate the pH of the solution mixture above present at the End of the reaction. ***** To begin, label the particles present in the bottom row as SA, SB, WA, WB, or N. ***** Water and NaCl ions are pH neutral (N). HCl is a strong acid (SA). If a strong acid or base is in a mixture, it determines the pH. Use the quick SA rule to find [H+], then pH. ***** [H+] in solution = [HCl or HNO3]mixed Calculate the [HCl] in the solution at the end of the reaction, then check your answer below. ***** ? = [HCl] = mol H+ = 0.020 mol HCl • 1 mL = 0.167 M HCl (carrying an extra sf) L soln 120. mL soln. 10―3 L To finish, find the pH. ***** [H+]in solution = [HCl or HNO3]as mixed = 0.167 M ? = pH = ─ log [H+] = ─ log(0.167) = ─ (─ 0.78) = © 2011 ChemReview.Net v. 7w 0.78 = pH Page 1038 Module 34 — pH During Titration Practice A: Learn the rules and steps above, then try these problems. 1. To 0.0250 mol HCl is added 0.0300 mol NaOH. The final volume of the mixture after the reaction is 40.0 mL. What is the pH in the solution after the reaction? Strong-Weak Neutralization Let’s try a second calculation of pH after the reaction, starting from amounts measured before the reaction. This time we will react a strong and a weak component. For any acidbase reaction with one or more strong components, we use the same steps as above. When dealing with moles/liter and milliliters, it is often convenient (but not required) to solve the rice moles table in millimoles. As long as all of the units in the table are the same, our rice moles “accounting system” works in • Moles or prefix-moles, or • Mol/L (M) if all of the particles are contained in the same volume. Let’s add that step to this problem. Q. A solution of 50.0 mL of 0.100 M HCl is mixed with 20.0 mL of 0.150 M KF. a. Calculate the millimoles of each reactant. b. Write the balanced equation for the reaction. c. Complete the rice moles table in millimoles. d. Find the pH in the mixture after the reaction. Start by completing steps a and b. ***** a. For the HCl calculation, use HCl data. ? mmol HCl = 50.0 mL HCl • 10─3 L • 0.100 mol HCl • 1 mmol = 5.00 mmol HCl 1 L HCl 10─3 mol 1 mL A way to calculate millimoles more quickly is to use the rule If you WANT a prefix-unit, and are given the same-prefix unit, cancel the given unit, but don’t cancel the given prefix. For example, to find the mmol KF in this problem: ? mmol KF = ↑ 20.0 mL KF • 0.150 ↑ mol = L 3.00 mmol KF ↑ This form of unit cancellation works because milli- is simply an abbreviation for “ times 10─3 .” When doing the math, an exponential term can be separated from the unit after it, so a prefix may also be treated as separate from the unit after it. © 2011 ChemReview.Net v. 7w Page 1039 Module 34 — pH During Titration An alternative rule for solution millimoles is to memorize mmol = mL x (mol/L) . The logic is the same: a prefix can be separated from its unit: mL • mol = mmol L b. The reactants are HCl + KF Write the products using molecular formulas, then balance the equation. ***** You may be able to solve by inspection. If not, to determine the products, first separate the reactant salt into ions: HCl + KF = HCl + K+ + F─ The products of the reaction are the conjugates of the acid and base. HCl + KF = HCl + K+ + F─ SA N B Cl─ + K+ + H F B N A The acid-strength table supplies the formulas for the conjugates if needed. Will this reaction go to the right? Why or why not? ***** The reaction will go to the right. One reason is • Any time a strong acid (HCl) is mixed with a base of any kind (F─), the reaction will go until one of the reactants is completely used up. Using molecular formulas, the balanced Reaction equation is: 1 HCl + 1 KF 1 HF + 1 KCl c. If you have not already done so, complete the rice moles table using millimoles. ***** 1 HCl 1 KF 1 HF 1 KCl 5.00 mmol 3.00 mmol 0 mol formed 0 mol Change (use + , ―) ― 3.00 mmol ― 3.00 mmol + 3.00 mmol +3.00 mmol At End/Equilibrium + 2.00 mmol 0 mmol + 3.00 mmol + 3.00 mmol Reaction Initial d. At the end of the reaction, the bottom row indicates the mixture present. Analyze the mixture and find the pH. ***** The mixture contains the strong acid HCl, the weak acid HF, and the salt KCl that contains two pH-neutral ions. What determines the pH? ***** The [HCl] determines the pH. If a strong component is present, it’s contribution to the [H+] nearly always overwhelms the others. Find pH based on [HCl]. ***** © 2011 ChemReview.Net v. 7w Page 1040 Module 34 — pH During Titration First find the [HCl]. How many moles of HCl are in the solution? ***** ? mol HCl = 2.00 mmol HCl = 2.00 x 10─3 mol HCl ( milli- means “ x 10─3 “) Those moles of HCl are in how many liters of solution? ***** Hint: How much is the total volume of solution at this point? ***** Though the KF has been entirely used up, and part of the HCl was used up, the water in which the KF and HCl particles were originally dissolved has not been used up. For reactions that take place in a solution, the number of reacting particles is always very small compared to the number of solvent particles. A relatively large amount of solvent is used so that the reacting particles can dissolve completely, move about, collide, and react. The large amount of solvent present determines almost entirely the volume of a solution. When solutions are combined, reacting substances are used up or formed, but the volume of the combined solution is determined by the volumes of solvent combined, and those simply add. Even for reactions run in aqueous solutions in which water is a reactant or product, the volume of water used up or formed is nearly always small compared to the volume of the water present as a solvent, and any change in the volume of water present due to the reaction can nearly always be ignored in calculations. The rule is: When solutions with liquid solvents are combined, their volumes simply add. If needed, complete the calculation of [HCl]. ***** We combined the original 50.0 mL of HCl solution with 20.0 mL of KCl solution. All reactants and products present after the reaction are therefore dissolved in a total solution volume of 70.0 mL. [HCl] = ? ***** Moles HCl were found above. ? = [HCl] = mol H+ = 2.00 x 10─3 mol HCl = 2.00 x 10─3 mol HCl = 0.02857 M HCl L soln 70.0 mL soln. 70.0 x 10─3 L soln. For part d, find the WANTED pH. ***** Using the quick rule: [H+]in solution = [HCl or HNO3]mixed = 0.02857 M ? = pH = ─ log [H+] = ─ log(0.02857) = ─ (─ l.54) = 1.54 = pH ***** © 2011 ChemReview.Net v. 7w Page 1041 Module 34 — pH During Titration Let’s summarize. Rules in this lesson include 1. In a reaction between an acid and a base, IF one of the components is strong, the reaction will go until one component, either the acid or the base, is 100% used up. 2. When solutions are combined, their volumes add. 3. The millimoles shortcut for rice moles tables: millimoles = mmol = mL x (mol/L) . 4. To find the [H+] or pH in a solution, a. Ask: Is the problem about a stable solution at equilibrium, or does it supply amounts of reactants before reaction and ask for [H+] or pH after reaction? b. If about a stable solution, apply the steps for finding mixture pH. c. IF about “reactants before and pH after” reaction, the steps are Reactants > reactant mol > rice moles > then pH of mixture at equilibrium. 5. Once the amounts in a mixture at equilibrium are solved or known, use the Steps For Finding Mixture pH Apply in this order. a. Re-write soluble salts as separated ions. b. Label the particles in the mixture as SA, SB, WA, WB, or N. c. In pH calculations, ignore pH-neutral (N) particles. d. If all particles are N, pH = 7. e. If SA or SB is present, ignore other particles. Find pH with the quick steps: [HCl or HNO3] = [H+] and [NaOH or KOH] = [OH─] f. If a WA or WB and its conjugate is present, solve a buffer chart and Henderson-Hasselbalch equation in moles or mol/L. g. If only one WA or WB is present, solve the Ka or Kb equation. Practice B: Commit the rules summary above to memory, then try these problems. 1. Complete the following rice moles table. Reaction Initial 1 HCl 1 NaCN 4.50 mmol 3.00 mmol 1 HCN 0 mol 1 NaCl 0 mol Change At End/Equilibrium 2. Find the pH for the reaction in Problem 1 if the mixture at equilibrium is in 200. mL of solution. © 2011 ChemReview.Net v. 7w Page 1042 Module 34 — pH During Titration 3. If 50.0 mL of 0.100 M KF is combined with 20.0 mL of 0.150 M HCl, find the pH after the reaction. Solve in millimoles. (Ka HF = 6.8 x 10―4) 4. To a solution containing 200. mmol HOBr (hypobromous acid) and 50. mmol NaOBr is added 150. mmol NaOH. What is the pH of the solution after the reaction? (Ka of HOBr = 2.8 x 10─9 ). ANSWERS Practice A 1. WANTED: pH Begin by asking: what type of problem is this? A stable solution or reactants before and pH after? This is about pH after reaction from amounts before. The steps are Reactants > reactant mol > rice moles > mixture pH Reaction 1 HCl 1 NaOH 1 H2O 1 NaCl Initial 0.0250 mol 0.0300 mol solvent 0 mol ― 0.0250 mol ― 0.0250 mol + 0.0250 mol + 0.0250 mol 0 mol 0.0050 mol Change (use + , ―) At End/Equilibrium 0.0250 mol lots In acid-base reactions, if one component is strong, one reactant must be totally used up. At equilibrium, the mixture contains the strong base NaOH and a pH-neutral salt. Find pH. ***** For strong bases, the quick rule is: [NaOH or KOH] = [OH─] . Find [NaOH] first. ***** DATA: 0.0050 mol NaOH in 40.0 mL total volume ? = [NaOH] = mol NaOH = L soln pH ≡ ─ log [H+] and 0.0050 mol HCl · 1 mL = 0.125 M NaOH = [OH─] 40.0 mL soln. 10―3 L [H+] ≡ 10─pH and pH + pOH = 14.00 pOH = ─ log [OH─] = ─ log (0.125) = 0.90 ; pH = 14.00 ─ 0.90 = 13.10 = pH Practice B 1. Reaction Initial Change (use + , ―) At End/Equilibrium 1 NaCl 1 HCl 1 NaCN 1 HCN 4.50 mmol 3.00 mmol 0 mol 0 mol ― 3.00 mmol ― 3.00 mmol + 3.00 mmol + 3.00 mmol 1.50 mmol 0 mmol 3.00 mmol 3.00 mmol In acid-base reactions, if one component is strong, one reactant must be totally used up. © 2011 ChemReview.Net v. 7w Page 1043 Module 34 — pH During Titration 2. WANTED = pH Begin by asking: what type of problem is this: a stable solution or a reaction? In this part, the mixture is at equilibrium. No net change is occurring. Any reaction is over. Apply the rules for acid-base mixture pH. What type of solution is this? ***** At equilibrium, the mixture contains a strong acid, a weak acid, and a salt. What determines the pH? ***** The ionization of the strong acid is the largest contributor to the [H+], so [SA] determines pH. Find pH. ***** For strong acids, the quick rule is: [HCl or HNO3] = [H+] . Find [HCl] first. ***** DATA: 1.50 mmol HCl in 200. mL soln. ? = [HCl] = mol HCl = L soln 1.50 mmol HCl 200. m L soln. = 0.00750 M HCl A prefix can be treated as independent of the number before and unit after it. Like-prefixes cancel. pH ≡ ─ log [H+] 3. WANTED: and [H+] ≡ 10─pH pH = ─ log [H+] = ─ log (0.00750) = 2.12 = pH pH after reaction Begin by asking: what type of problem is this? Stable solution or reaction? This is about a reaction. For pH after reaction from amounts before, the steps are Reactants > reactant mol or prefix-moles > rice moles > mixture pH ? mmol KF = 50.0 mL KF x 0.100 mol/L KF = 5.00 mmol KF ? mmol HCl = 20.0 mL HCl x 0.150 mol/L HCl = 3.00 mmol HCl Reaction 1 HCl 1 KF 1 HF 1 KCl 3.00 mmol 5.00 mmol 0 mol 0 mol Change (use + , ―) ― 3.00 mmol ― 3.00 mmol + 3.00 mmol +3.00 mmol At End/Equilibrium +0 mmol 2.00 mmol + 3.00 mmol + 3.00 mmol Initial At equilibrium, what type of solution is this? What steps do you follow to find the pH? ***** This mixture contains substantial amounts of both the weak acid HF and its conjugate base F─, which makes the solution a buffer. The solution also contains the pH-neutral ions K+ and Cl─, but pH-neutral ions do not affect pH. Solve for pH using the buffer steps. ***** To find buffer pH, • for the two “at end” substances that are not pH-neutral, fill in a buffer chart, and • find the pH using the Ka or Henderson-Hasselbalch equations. © 2011 ChemReview.Net v. 7w Page 1044 Module 34 — pH During Titration WA formula = HF CB formula = F ─ moles WA = 3.00 mmol HF Buffer Chart: moles CB = 2.00 mmol F ─ Henderson-Hasselbalch: pH ≈ pKa + log (mol base/mol acid) ≈ ─log (6.8 x 10―4) + log (2.00 mmol /3.00 mmol) ≈ ─(─3.17) + log (0.667) ≈ 3.17 ─ 0.17 = 3.00 = pH The units in the base/acid ratio of the Henderson-Hasselbalch equation can be M, mol, or mmol, as long as the top and bottom units are consistent. 4 WANTED: pH Begin by asking: what type of problem is this: a stable solution or reactants before and pH after? This is about reactants before and pH after reaction. The base NaOH will react with the acid HOBr. For pH after reaction from amounts before, the steps are Reactants > reactant mol > rice moles > mixture pH When an acid reacts with a hydroxide base, one product is always water. Reaction 1 HOBr Initial 1 NaOH 1 H2O 1 NaOBr solvent 50 mmol 200. mmol Change (use + , ―) 150. mmol ― 150. mmol ― 150. mmol +150. mmol + 150. mmol 0 mmol lots 200. mmol At End/Equilibrium 50. mmol In acid-base reactions, if one component is strong, one reactant must be totally used up. Apply the rules for acid-base mixture pH. What type of solution is this? ***** This mixture contains substantial amounts of the weak acid HOBr (weak based on its Ka value) and its conjugate base OBr─ , which makes the solution a buffer. The solution also contains the pH-neutral ions Na+ ions that do not affect pH. Solve for pH using the buffer steps. ***** To find buffer pH, • for the two “at end” substances that are not pH-neutral, fill in a buffer chart, and • find the pH using the Henderson-Hasselbalch equation. Buffer Chart: WA formula = HOBr CB formula = OBr─ moles WA = 50. mmol HOBr Henderson-Hasselbalch: moles CB = 200. mmol OBr─ pH ≈ pKa + log (mol base/mol acid) ≈ ─log (2.8 x 10─9) + log (200. mmol / 50. mmol ) ≈ ─ ( ─ 8.552) + log (4.0) ≈ 8.552 + 0.602 ≈ 9.15 = pH ***** © 2011 ChemReview.Net v. 7w Page 1045 Module 34 — pH During Titration Lesson 34C: pH Behavior During Titration Timing: Calculations of how much acid and base are required to reach the endpoint of a titration are neutralization stoichiometry, and they were covered in Module 14. Begin this lesson when you are asked to calculate the pH at points during an acid-base titration. ***** Terminology of Acid-Base Titration Redox, precipitation, and other types of reactions can be studied by titration, but in this module the term titration will refer to acid-base neutralization titration. For purposes of discussing acid-base titration, as with other acid-base mixtures, • The opposite of an acid will mean a base, and the opposite of a base will mean an acid. • Strong acids and bases will be defined as those that ionize ~100% to form H+ or OH─ ions. Weak acids and bases will be those with Ka or Kb values between one and 10―16 according to acid-strength tables (see Lesson 31B). The substance gradually added to a sample during a titration is termed the titrant. The Equivalence Point Acid-base titration was discussed in Lesson 14C. To briefly review: • An acid-base titration is a gradual neutralization. Using burets, volumes of acid and base that have been combined in a reaction vessel are carefully measured when the endpoint of the titration is reached. To detect the endpoint, a small amount of pHsensitive dye called an indicator is added to the reaction mixture. At the endpoint, the indicator changes color. • The endpoint signaled by an indicator is ideally a sharp color change at the precise equivalence point at which the moles of acid and base that have been combined are equal (or, for compounds with more than one acidic or basic group, the moles are in a simple whole-number ratio). Types of Titration Calculations In acid-base titration, we are most often concerned with two questions: 1. How much of one reactant (acid or base) is needed to exactly use up the other? 2. What happens to the pH during the titration? The first question we have answered previously. In an acid-base reaction, if one component is strong, the reaction will go to completion. The point at which the H+ ions of the acid reactant are equal to the moles of the basic reactant is the equivalence point and are solved by conversion stoichiometry. If the amount of one component at the equivalence point is known, the amount of the other can be calculated by the stoichiometry steps: WDBB, units © 2011 ChemReview.Net v. 7w moles moles units Page 1046 Module 34 — pH During Titration Practice A: If additional review of equivalence point calculations is needed, see Lessons 12C and 14C. 1. To 50.0 mL of 0.0500 M KOH is added 0.100 M HNO3. a. Will this reaction go to completion? b. How many mL of acid must be added to neutralize all of the KOH? Indicating the Endpoint Let’s turn our attention to the second question above: What happens to the pH during an acid-base titration? This question is important because during neutralization, different acids and bases have different pH behaviors. In addition, different substances can be used as acid-base indicators, and each indicator changes color in a limited pH range. In order to choose the correct indicator to accurately signal an equivalence point, the pH of the acid-base mixture during a titration must be determined. If a sharp change in the pH of the reaction mixture occurs at the equivalence point, and if the indicator is chosen carefully, allowing for small experimental error, the endpoint shown by the indicator and the equivalence point for the neutralization will be the same. If the equivalence point can be determined, many unknown characteristics of an acid or base can then be determined. pH Behavior During Titration During titration, the following rules apply. 1. The indicator in a titration will accurately identify an equivalence point only if there is a sharp pH change (large and occurring with just a drop or two of titrant) at the equivalence point. 2. For a sharp pH change to occur, at least one reactant must be strong. In most titrations, one reactant is usually an aqueous solution of HCl, HNO3, NaOH, or KOH. 3. Acid-base indicators work best when strong or moderately weak acids and bases (Ka or Kb > 10─7) are combined with strong opposites. The weaker the acid or base being neutralized, the less sharp the pH change at the endpoint will be. If the K for the weaker component is below 10─7, the pH change may not sharp enough for an indicator to accurately signal an equivalence point (though an unwieldy pH meter may do so). Similarly, if a weak acid and a weak base are combined by titration, the pH change at the endpoint is often too small for an indicator dye to accurately detect. 4. Substances with two or more acidic or basic groups will titrate to an equivalence point for each group. The calculation of pH is treated as successive titrations: the conjugate product at the first endpoint is the starting acid or base for the titration to the second endpoint, etc. Some of the endpoints may be sharp while others are not. © 2011 ChemReview.Net v. 7w Page 1047 Module 34 — pH During Titration 5. A titration of substances with single acidic or basic groups can be separated into four stages. a. Before the titration begins, the sample to be titrated is a solution of a strong or moderately weak acid or base. b. When the first drop of strong opposite is added, it reacts with the original acid or base. Since one of the particles is strong, the reaction will go until the reactant with the lowest moles (normally the titrant being added) is completely used up, and products, including conjugates of the acid and base, form. Between the beginning and endpoint, as opposite continues to be added, all of the titrant is used up, the number of particles of the original acid or base goes down, and the particles of products go up. The reaction continues as long as particles of the original acid or base remain to react with the titrant. But between the beginning of the titration and the equivalence point, in the reaction flask is a mixture of the original acid or base, products that include both conjugates, and no titrant. c. At the equivalence point, • there is no original acid or base and no titrant. • The last added titrant has reacted with the last particles of original acid or base, and both reactants have been exactly and completely used up. • In the solution are only the products of the reaction. d. If titrant continues to be added after the endpoint, there is no reaction. There is no opposite left for the titrant to react with. Titrant added after the endpoint simply mixes with the particles of products that were present at the endpoint. Summary: The four stages of an acid-base titration are Stage The solution contains Before titrant is added original acid or base Between the beginning and endpoint original acid or base plus products At the endpoint Products only After the endpoint products plus titrant added after endpoint Practice B. Check your answers after each part. 1. Write the products using molecular (solid) formulas, then balance these equations for reactions in aqueous solution. a. KOH + HNO3 b. HCN + RbOH © 2011 ChemReview.Net v. 7w Page 1048 Module 34 — pH During Titration c. NH3 + HCl d. NaHCO3 + NaOH 2. Which compounds among the 7 products above are not pH-neutral? 3. Assuming for the reactions in problem 1 that the 2nd reactant listed is gradually titrated into the first, write molecular formulas for the substances that will be present in the reaction mixture a. In problem 1a before the titration begins. b. In problem 1b just before the endpoint of a titration. c. In problem 1c at the endpoint. d. In problem 1d three drops after the endpoint. 4. In which answer to question 3 is the solution a buffer? 5. Label each substance formula in the answers to question 3 as a strong acid (SA), strong base (SB), weak acid (WA), weak base (WB), or pH-neutral (N). ANSWERS Practice A 1a. If an acid and base are mixed, and one or both components is strong, the reaction will go to completion. 1b. WANTED: ? mL HNO3 solution DATA: 50.0 mL KOH 0.0500 mol KOH = 1 L KOH soln. 0.100 mol HNO3 = 1 L HNO3 soln. Strategy: This is stoichiometry: how much of one substance is needed to exactly use up another. If you need a review of solution stoichiometry, see Lessons 12C and 14C. When solving for a single unit, the stoichiometry steps are WDBB, units moles moles units 1 H2O + 1 KNO3 Balance: 1 HNO3 + 1 KOH Bridge: 1 mol HNO3 = 1 mol KOH SOLVE: 1 L HNO3 = ? mL HNO3 = 50.0 mL KOH • 0.0500 mol KOH • 1 mol HNO3 • ↑ ↑ 1 L KOH 1 mol KOH 0.100 mol HNO3 = 25.0 mL HNO3 The conversions above use the optional rule: If you WANT a prefix-unit and are given the same-prefix unit, cancel the given unit, but don’t cancel the given prefix. © 2011 ChemReview.Net v. 7w Page 1049 Module 34 — pH During Titration Practice B 1. All coefficients are one. a. KOH + HNO3 c. NH3 + HCl H-OH + KNO3 NH4Cl b. HCN + RbOH d. NaHCO3 + NaOH H-OH + RbCN H2O + Na2CO3 2. RbCN is a salt that is a weak base: it contains the weak conjugate base CN─ and pH-neutral Rb+. NH Cl is a weakly acidic salt containing the weak acid ion NH + and pH-neutral Cl─. 4 4 Na2CO3 is a basic salt containing the moderately weak base CO32─ and pH-neutral Na+. All of the other products are pH neutral. 3. a. In 1a, before the reaction begins, is KOH. Neutral water is also present in aqueous solutions. b. In 1b, just before the endpoint of a titration is HCN, H-OH, and RbCN. Before the endpoint, as the strong base RbOH is added, it is 100% used up. c. In 1c, at the endpoint are only products: NH4Cl dissolved in water. d. In 1d, three drops after the endpoint are the products H-OH and Na2CO3 , plus the NaOH added after the endpoint. After the endpoint, the original acid has been all used up, and the added base has nothing to react with. 4. In which answer to question 3 is the solution a buffer? 3b is the only part in which a weak acid (HF) or base is mixed with its conjugate. 5. a. KOH (SB) (All aqueous solution also have H-OH (N) ). b. HCN (WA), H-OH (N) , and RbCN (WB). RbCN is a salt composed of the alkali metal ion Rb+ that is pH neutral and CN─ which is the conjugate of the weak acid HCN. Conjugates of mildly weak acids are mildly weak bases. Salts that combine neutral ions and weak basic ions are weak bases. c. NH4Cl (WA) is a salt composed of NH4+ that is a weak acid and Cl─ which is pH neutral. Salts composed of pH neutral ions and ions that are weak acids behave as weak acids. d. Na2CO3 (WB) is a salt containing the moderately weak carbonate ion (Kb = 1.9 x 10―4); NaOH (SB) ***** © 2011 ChemReview.Net v. 7w Page 1050 Module 34 — pH During Titration Lesson 34D: pH During Strong-Strong Titration Calculating pH During a Titration At all points during a titration, in the reaction flask is a mixture of substances. The pH is determined by the types of particles in the mixture. To calculate the solution pH, apply the rules that solve for the pH of a mixture. In the case of titration, we have a special interest in • Tracking what happens to the pH during the four stages of a titration; and • Developing rules that will simplify calculations. We will develop these rules as we solve examples. Let’s begin with Finding pH During A Strong Acid-Strong Base Titration Q1. A solution of 50.0 mL of 0.100 M HCl is titrated by 0.250 M NaOH. a. What is the pH of the solution before any base is added? b. Write the balanced reaction equation. c. What is the pH at the equivalence point of the titration? d. How many mL NaOH must be added to reach the equivalence point? e. What is the pH after 10.0 mL of base has been added? (Solve in mmol.) Try part a, then check your answer below. ***** a. See pH? Write pH ≡ ─ log [H+] and [H+] ≡ 10─pH The initial solution is a strong acid. Use the strong acid rules to find pH. ***** For any ~100% ionization, REC: HCl 0.100 M 0 M H+ 0.100 M + Cl─ (goes 100%) 0.100 M or by the quick rule: [H+]in solution = [HCl or HNO3]mixed = 0.100 M = [H+] ; pH = ─ log [H+] = ─ log (0.100) = ─ log (1.00 x 10─1) = ─ log (10─1) = ─ (─ l) = 1.00 Try part b, then check your answer below. ***** b. When acid reacts with a hydroxide base, one of the products is always….? ***** (goes ~100%) Water. This reaction is: 1 HCl + 1 NaOH 1 H2O + 1 NaCl ***** © 2011 ChemReview.Net v. 7w Page 1051 Module 34 — pH During Titration c. In any neutralization reaction with one strong component, the rules are At an equivalence point in the neutralization, • the moles acid reacted are equivalent to (the same as) the moles base reacted. • The acid and base reactants are exactly neutralized: both are 100% used up. • Since no reactants remain, only products are present. In this reaction flask is salt water: dissolved NaCl. NaCl is a combination of two pHneutral ions; the solution has a neutral pH of 7 . In a titration of a strong acid and strong base, at the equivalence point: pH = 7 d. Part d is what kind of problem? ***** At the endpoint, all of the acid or base in the original sample is used up, so Part d asks what amount of one reactant is needed to use up all of another reactant. What kind of problem is that? ***** Stoichiometry. We apply stoichiometry steps to reactions which • go essentially to completion (rather than to equilibrium), and • ask how much of one substance is needed or formed when one other substance is completely used up (and is therefore a limiting reactant). In acid-base reactions, if one or both reactants is strong, the reaction goes to completion. Apply the stoichiometry steps, then check below. ***** In stoichiometry, the first four steps are WDBB. Then, if a single unit is WANTED, chain conversions to solve. ***** 1. WANTED: ? mL NaOH (a single unit) 2. DATA: 50.0 mL HCl (the single-unit given) 0.100 mol HCl = 1 L HCl (M prompt) 0.250 mol NaOH = 1 L NaOH (M prompt) (If two substances are involved, each unit must include its substance formula.) 3. Balance. 1 NaOH + 1 HCl 1 HOH + 1 NaCl 4. Bridge. 1 mol NaOH = 1 mol HCl 5. SOLVE. © 2011 ChemReview.Net v. 7w Page 1052 Module 34 — pH During Titration 1 L NaOH ? mL NaOH = 50.0 mL HCl • 0.100 mol HCl • 1 mol NaOH • ↑ ↑ 1 L HCl 1 mol HCl 0.250 mol NaOH = = 20.0 mL NaOH (It is optional but saves steps to use the rule: if you WANT a prefix-unit, and are given the same prefix-unit, cancel the given unit, but don’t cancel its prefix.) This titration will reach its endpoint when 20.0 mL of the NaOH solution is added to the original acid solution. e. What kind of problem is part e? What steps will you use to solve? ***** Amounts of acid and base before the reaction are supplied, and pH after the reaction at a point in the titration that is not necessarily the endpoint, is WANTED. The result of the reaction will be a mixture. To find pH after a reaction, given reactant amounts before the reaction, use Reactants > reactant mol or prefix-mol > rice moles > pH of mixture at equilibrium. In most titration calculations, it is easier to solve rice tables in millimoles. ***** Find mmol HCl using HCl DATA: Want a single unit? Start with a single unit. WANTED = ? mmol HCl = 50.0 mL HCl • 0.100 mol HCl = 5.00 mmol HCl 1 L HCl soln. ***** For NaOH calculations, use NaOH DATA: ? mmol NaOH = 10.0 mL NaOH • 0.250 mol NaOH = 2.50 mmol NaOH 1 L NaOH soln. Now complete the rice table below for the reaction. ***** © 2011 ChemReview.Net v. 7w Page 1053 Module 34 — pH During Titration Reaction 1 HCl 1 NaOH 1 H2O 1 NaCl 5.00 mmol 2.50 mmol 0 mol formed 0 mol Change (use + , ―) ― 2.50 mmol ― 2.50 mmol + 2.50 mmol +2.50 mmol At End/Equilibrium + 2.50 mmol 0 mmol + 2.50 mmol + 2.50 mmol Initial At this point in the reaction, the reaction must stop when the NaOH is used up. The limiting NaOH also determines how much of the products form. Calculate the pH of the solution present for the mixture in the End row above. ***** The HCl at the end determines the pH. Use the quick SA rule to find [H+], then pH. ***** [H+]in solution = [HCl or HNO3]mixed Calculate the [HCl] mixed into the solution, then check your answer below. ***** Hint: How much is the total volume of solution at this point? ***** In part e, we mix the original 50.0 mL of acid solution with 10.0 mL of added base solution. All reactants and products present are therefore dissolved in a total solution volume of 60.0 mL. [HCl] = ? ***** ? = [HCl] = mol H+ = 2.50 mmol HCl = 0.04167 M HCl L soln 60.0 mL soln. Since milli is simply an abbreviation for “ x 10─3 “ it can cancel. If needed, finish part e. ***** Using the quick rule: [HCl or HNO3]mixed = [H+]in solution = 0.04167 M ? = pH = ─ log [H+] = ─ log(0.04167) = ─ (─ l.38) = 1.38 = pH In these calculations, we carried an extra significant figure, but because pH is very temperature dependent, in the lab predictions are rarely accurate to more than two places past the decimal. ***** © 2011 ChemReview.Net v. 7w Page 1054 Module 34 — pH During Titration One of our goals in these calculations is to track what happens to pH during a titration. Let’s summarize what we have calculated so far. For the strong acid-strong base titration in this problem, • in the initial acid solution, the pH was 1.00 • After adding 10.0 mL of this particular base solution, pH = 1.38 • After adding 20.0 mL of this particular base solution, we reached the endpoint/equivalence point, where pH = 7.0 Somewhere during the titration, the pH drops substantially, but these numbers indicate that it hasn’t dropped much by half-way to the endpoint. For this same titration, let’s try a pH calculation much closer to the endpoint. Since the endpoint is at 20.0 mL, let’s try Q2. For the same solutions (50.0 mL of 0.100 M HCl is titrated by 0.250 M NaOH), what is the pH after 19.8 mL of base has been added? (Solve in millimoles) ***** The mmol HCl being titrated is the same, but the mmol NaOH added have increased. ? mmol NaOH = 19.8 mL NaOH • 0.250 mol NaOH = 4.95 mmol NaOH 1 L NaOH soln. ***** Reaction Initial 1 HCl 1 NaOH 1 H2O 1 NaCl 0 mmol 0 mmol 5.00 mmol 4.95 mmol Change ― 4.95 mmol ― 4.95 mmol + 4.95 mmol + 4.95 mmol At End/Equilibrium + 0.05 mmol 0 mmol + 4.95 mmol + 4.95 mmol Calculate the pH in the mixture after the reaction, at equilibrium. ***** The only substance present that is not pH neutral is HCl. The strong acid will determine the pH. Find [HCl], then [H+] , then pH. ***** To 50.0 mL original acid solution has been added 19.8 mL base = 69.8 mL total volume. ***** 0.05 mmol HCl = 0.0007 mol/L HCl pH = ? ? [HCl] = mol HCl = L soln 69.8 total mL soln. ***** © 2011 ChemReview.Net v. 7w Page 1055 Module 34 — pH During Titration ? = pH = ─ log [H+] Since HCl is a strong acid, [H+] = [HCl] = 0.0007 M ? = pH = ─ log [H+] = ─ log 0.0007 = ─ (─ 3.1) = 3.1 = pH Between 0 mL and 19.8 mL base added, the pH has fallen from 1.00 to 3.1 . By 20.0 mL base added at the endpoint, the pH must fall to 7 – a comparatively large change for 0.2 mL of additional added NaOH. Titration Past the Equivalence Point In the titration above, the equivalence point is at 20.0 mL NaOH added. What happens to the pH if we go just two drops past the equivalence point? Each drop contains about 0.05 mL, so two drops past would take us to 20.0 + 0.10 = 20.1 mL NaOH added, so the question is Q3. For the same solutions (50.0 mL of 0.100 M HCl titrated by 0.250 M NaOH), what is the pH after 20.1 mL of base has been added? ***** The mmol HCl being titrated is the same, but the mmol NaOH added has increased. ? mmol NaOH = 20.1 mL NaOH • 0.250 mol NaOH = 5.02 mmol NaOH 1 L NaOH soln. Complete the rice table using the increased initial millimoles of NaOH. ***** Reaction Initial Change At End/Equilibrium 1 HCl 1 NaOH 1 H2O 1 NaCl 0 mmol 0 mmol 5.00 mmol 5.02 mmol ― 5.00 mmol ― 5.00 mmol + 5.00 mmol + 5.00 mmol + 0 mmol + 0.02 mmol + 5.00 mmol + 5.00 mmol Calculate the pH at equilibrium in this mixture. ***** The only substance present that is not pH neutral is the strong base. Find [NaOH], then [OH─] , then pH. ***** To 50.0 mL original acid solution has been added 20.1 mL base = 70.1 mL total volume. ***** © 2011 ChemReview.Net v. 7w Page 1056 Module 34 — pH During Titration ? [NaOH] = mol NaOH = L soln ***** 0.02 mmol NaOH 70.1 total mL soln. = 0.0003 mol/L NaOH pH = ? Since NaOH is a strong base, [NaOH] = [OH─] = 0.0003 M pOH = ─ log [OH─] = ─ log(3 x 10─4) = 3.5 pH = 14.00 ─ pOH = Graphing pH During Titration SA Titrated by SB Q4. In pencil, on the grid at the right, plot the data calculated for the titration above at 0, 10.0, 19.8, 20.0, and 20.1 mL NaOH added. Check your answer at the end of the ANSWER section below. 10.5 = pH 0 2 4 6 pH 8 10 ***** 12 14 16 0 5 10 15 20 mL NaOH added The graph reflects an important fact of acid-base neutralization: If a strong or moderately weak acid or base is titrated by a strong opposite, the pH changes sharply during the titration, but not until very close to the endpoint. Practice: Do problem 1. You may want to save problem 2 for later review. 1. For the titration in the lesson above, calculate the pH after 25.0 mL NaOH has been added. Add this value to the graph above. 2. 50.0 mL of 0.0500 M KOH is titrated by 0.150 M HNO3. a. What is the pH before any acid is added? b. What is the pH at the endpoint? c. Calculate the pH after 18.0 mL of HNO3 has been added. © 2011 ChemReview.Net v. 7w Page 1057 25 Module 34 — pH During Titration ANSWERS 1. The mmol HCl being titrated is the same, but the mmol NaOH added have increased. ? mmol NaOH = 25.0 mL NaOH • 0.250 mol NaOH = 6.25 mmol NaOH 1 L NaOH soln. 1 HCl Reaction Initial 1 H2O 1 NaCl 0 mmol 1 NaOH 0 mmol 5.00 mmol ― 5.00 mmol At End/Equilibrium ― 5.00 mmol + 5.00 mmol + 5.00 mmol + 0 mmol Change 6.25 mmol + 1.25 mmol + 5.00 mmol + 5.00 mmol The NaOH determines the pH. Find [NaOH], then [OH─] , then pH. ***** To 50.0 mL original acid solution has been added 25.0 mL base = 75.0 mL total volume. ***** ? [NaOH] = mol NaOH = L soln ***** = 0.0167 mol/L NaOH 1.25 mmol NaOH 75.0 total mL soln. pH = ? Since NaOH is a strong base, [NaOH] = [OH─] = 0.0167 M pOH = ─ log [OH─] = ─ log(1.67 x 10─2) = 1.78 pH = 14.00 ─ pOH = 12.22 = pH Once you are a few drops past the end point, the pH graph again plateaus: the pH again changes slowly. 2a. WANTED: pH The initial solution contains the strong base KOH that ionize 100% in water. To find the ion concentrations in strong acid or base, either use the REC steps, Or recall the quick rule: [NaOH or KOH]mixed = [OH─]in solution (see Lesson 27E) Since [OH─] = 0.0500 M; pOH = ─ log [OH─] = ─ log(0.0500) = 1.30 = pOH pH = 14.00 ─ pOH = 12.70 = pH 2b. In this reaction: KOH + HNO3 H2O + KNO3 (goes ~100%) At the equivalence point of any neutralization titration, only products are in the mixture. In the reaction flask is KNO3, composed of two pH-neutral ions, K+ and NO3─ , so the solution has a neutral pH of 7. © 2011 ChemReview.Net v. 7w Page 1058 Module 34 — pH During Titration 2c. WANTED: pH after reaction Begin by asking: what type of problem is this: a stable solution or “before and after” reaction? This is a reaction. For pH after reaction from amounts before, the steps are Reactants > reactant mol or mmol > rice moles > mixture pH 1) Convert all amounts of reactants to moles, (or, in titration, millimoles). 2) Enter the initial millimoles in a rice initial row. 3) Complete the rice moles table with one reactant used up. 4) Solve for pH based on the substances present after the reaction (in the bottom rice row). The millimoles of base being titrated in the sample are ? mmol KOH = 50.0 mL KOH • 0.0500 mol KOH = 2.50 mmol KOH 1 L KOH The millimoles of acid added so far are ? mmol HNO3 = 18.0 mL HNO3 • 0.150 mol HNO3 = 2.70 mmol HNO3 1 L HCl soln. Reaction 1 HNO3 1 KOH Initial 2.70 mmol 2.50 mmol ― 2.50 mmol 0.20 mmol Change At End/Equilibrium 1 H2O 1 KNO3 0 mmol 0 mmol ― 2.50 mmol + 2.50 mmol + 2.50 mmol 0 mmol 2.50 mmol 2.50 mmol Label the particles present at the end as SA, SB, WA, WB, or N. ***** In the mixture at this point, we have “overshot the endpoint” by adding more moles of titrant acid than base. Only the HNO3 (SA) titrant is not pH neutral. It will therefore determine the pH. To find the pH, [H+] is needed. To find [H+], [SA] is needed: its moles and its liters. The total volume in the flask is 50.0 mL original base + 18.0 mL acid solution added = 68.0 mL total. ? = [HNO3] = mol HNO3 = 0.20 mmol HNO3 = 0.00294 M HNO3 L soln 68.0 total mL soln. Since HNO3 is a strong acid, [H+] = [HNO3] = 0.00294 M H+ pH = ─ log [H+] = ─ log(0.00294) = 2.53 = pH © 2011 ChemReview.Net v. 7w Page 1059 Module 34 — pH During Titration The graph at the right is the answer to Q4 in the lesson. SA Titrated by SB ***** 0 2 4 6 pH 8 10 12 14 16 0 5 10 15 20 25 mL NaOH added Lesson 34E: pH During Strong-Weak Titration Using Mol/L After Reaction As the Rice Unit Our rice table “accounting system” for reactions can be solved in units of • Moles or prefix-moles, or • Mol/L (M) if all of the particles are dissolved in the same volume; as long as all of the units in the rice table are consistent (the same). So far, we have solved all of our rice tables for acid-base reactions in moles or millimoles, and not mol/L. Combining solutions of two substances does not change the moles of each in the new solution before they react, but it does decrease their concentrations because the volume in which each substance is dissolved increases. When solutions of two different substances are combined, but before they react, the moles of each substance does not change, but the mol/L of both are diluted. Rice moles tables can be solved in moles/liter IF you know the concentrations of the acid and base reactants after they are combined, but before they react. Why? After two solutions of reactants are mixed together, all of the moles present, whether before, during, or after the reaction, are in same volume of solution. The volume in dilute aqueous solutions is determined ~100% by how much of the solvent (water) is present. Once the solutions are combined, the volume of water does not substantially change if an acid-base reaction takes place, even if water is a reactant or product. Rice tables are based on mole ratios: the coefficients of the balanced equation. If all of the moles are divided by the same constant liters, the mole ratios of the rice table remain the same, so that “mol/L after combining” may be used as a consistent unit in a rice table. © 2011 ChemReview.Net v. 7w Page 1060 Module 34 — pH During Titration For solution reactions, IF you enter into a rice table the initial [reactants] after they are combined, but before they react, you can find all mol/L after the reaction. For acid-base reactions, if we know the concentrations of all of the substances after the reaction, we can find the [H+] and the pH. Let’s apply that rule on this example. Q. Complete the following table, then find the pH after the reaction. Reaction Initial (after combining, but before reacting) 1 HCl 1 KOH 0.025 M 0.035 M 0 mol 0 mol Change At End/Equilibrium ***** Since the table shows the initial concentrations after the reactants have been combined, the solution volume is the same at all points during the reaction, and the rice moles table can be solved in moles/liter (M). Reaction Initial (after combining, but before reacting) Change At End/Equilibrium 1 HCl 1 KOH 1 H2O 1 KCl 0.025 M 0.035 M 0 mol 0 mol ―0.025 M ―0.025 M + 0.025 M + 0.025 M 0M + 0.010 M + 0.025 M + 0.025 M After the reaction, the [KOH] at equilibrium determines the pH. [KOH] = [OH─] = 0.010 M = 1.0 x 10─2 M ; pOH = 2.00 ; pH = 12.00 Using Mol/L Before Reaction With A Rice Table The question above was solved relatively quickly. Why haven’t we solved all neutralization calculations this way? The reason is that in most problems, we are not given the acid and base concentrations after they are combined. What we usually given the acid and base concentrations before they are combined. Those concentrations before combining cannot be put directly into a rice table. The acid and base are diluted as they are mixed together. How much they are diluted varies with how much of each solution is mixed together. We can solve rice tables in mol/L if we use this rule: For solution reactions: rice calculations can be solved in mol/L IF all supplied mol/L before combining are converted to mol/L after combining but before reacting, using dilution rules. © 2011 ChemReview.Net v. 7w Page 1061 Module 34 — pH During Titration This gives us two ways to find pH after reaction from amounts before reaction. To find pH after reaction, given amounts of reactants before the reaction, the steps are Reactants > reactant mol or prefix-mol > rice moles > mixture pH steps, OR Reactant M before > reactant M combined > rice M > mixture pH steps. In general, for rice tables, use this rule. The values that go into the initial row of a rice table can be moles, or prefix-moles, or mol/L after combining but before reacting (as long as all units are consistent). Before we use mol/L to solve a rice table, let’s refresh our memory on dilution calculations (from Lesson 12A). Dilution Review The dilution equation is written in symbols as VC x MC = VD x MD in which C means concentrated and D diluted. The dilution equation is memorized by recitation: “In dilution, volume times molarity equals volume times molarity.” To find concentrations after two solutions are combined, you are always solving for the diluted molarity. That equation is always MDiluted – after combining = VConc – before combining x MConc – before combining VDiluted-Total after combining In some problems, [reactant after combining] can be solved by inspection using the rule: In dilution, if V or M changes by an easy multiple , multiply the other by 1/multiple. For example, if equal volumes of solutions of two different substances are combined, the volume containing the particles of each substance is doubled, so the concentration of both substances is cut in half. Practice A: For additional review, see Lesson 12A. 1. If 20.0 mL of 0.100 M NaOH is combined with 20.0 mL of 0.200 M HCl, after mixing but before reacting, [HCl] = [NaOH] = ? 2. If 10.0 mL of 0.45 KOH is combined with 20.0 mL of 0.300 M HCN, after combining, but before reacting, [HCN] = © 2011 ChemReview.Net v. 7w [KOH] = ? Page 1062 Module 34 — pH During Titration 3. If 25.0 mL of 0.150 M HF is combined with 21.5 mL of 0.120 M KOH, a. [HF] after combining, before reacting = ? b. [KOH] after combining, before reacting = ? As the practice above shows, finding mol/L after dilution is relatively easy when dilution can be done by inspection, but it takes time when the numbers are complex. In acid-base rice tables, which units should be used: moles, millimoles, or mol/L? All three methods work, but a rule that may save time is • If [acid] and [base] after combining but before reacting can be solved by inspection, solve in mol/L; • If the dilutions are time-consuming, or if calculations must be done at several points of the same titration, solve in moles or millimoles. If it’s a tossup, solve in moles or millimoles. In chemistry, the unit that best makes sense of most processes is the unit that counts the particles: moles or prefix-moles. pH During Weak-Strong Titration During the titration of a moderately weak acid or base by a strong opposite, most rules are the same as for strong-strong titration. • The moles or mol/L present at any point can be tracked by a rice table. • At the equivalence point, the solution contains only products. The differences include 1. Before the titration begins, the solution contains a weak acid or base. The solution pH is calculated using Ka or Kb . 2. Between the beginning and equivalence point of a weak-strong titration, both the original weak acid and base and its conjugate are present, so the reaction mixture is a buffer. The pH is determined using buffer methods: write the buffer chart and then the Ka or Henderson-Hasselbalch equation. 3. At the equivalence point, only one non-pH-neutral product is present: the conjugate of the original weak acid or base. That conjugate is also a weak acid or base, and the pH is calculated based on its Ka or Kb. 4. Past the endpoint, the solution contains strong titrant and the weak conjugate. The [strong titrant] added after the endpoint will nearly always determine the pH. You should be aware of the points above, but you should not need to memorize them. The bottom row of the rice table identifies the mixture present at each step of a titration. To find pH, simply apply the rules for mixture pH to the end row of the rice table. To find pH after reaction, given amounts of reactants before the reaction, the steps are Reactants > reactant mol or prefix-mol > rice moles > mixture pH steps, OR Reactant M before > reactant M after combining > rice M at end > mixture pH. © 2011 ChemReview.Net v. 7w Page 1063 Module 34 — pH During Titration Let’s apply those steps to a weak-strong example. Q. A 20.0 mL sample of a 0.200 M HF is titrated with 0.120 M KOH. Calculate the pH after 20.0 mL of KOH solution has been added (Ka of HF = 6.8 x 10―4). Solve the rice table in mol/L. ***** When any acid is reacted with any hydroxide (OH─) base, one product is H-OH. ***** In the rice table, [initial] must be after mixing, but before reacting. When equal volumes of solutions of two different substances are combined, the concentration of both substances is cut in half. After mixing, by inspection, [HF] = 0.100 M and [KOH] = 0.0600 M . ***** Reaction 1 HF 1 KOH Initial 0.100 M 0.0600 M 1 H2O 1 KF 0M 0M Change ― 0.0600 M ― 0.0600 M + 0.0600 M + 0.0600 M At End/Equilibrium + 0.040 M 0M + 0.0600 M + 0.0600 M What type of solution is this? ***** A buffer. Solve for buffer pH. ***** Buffer Chart: ***** H-H equation: WA formula = HF CB formula = F ─ moles or [WA] = 0.040 M HF moles or [CB] = 0.0600 M F ─ pH ≈ pKa + log ([base] / [acid]) ≈ ─log (6.8 x 10―4) + log (0.0600 M /0.040 M) ≈ ─ ( ─ 3.17) + log (1.5) ≈ 3.17 + 0.18 = 3.35 = pH Buffer check: Ka = 6.8 x 10―4 , buffer pH estimate is 4+2 = 2 to 6. Check. © 2011 ChemReview.Net v. 7w Page 1064 Module 34 — pH During Titration A Special Case: The Half-Way Point When a weak acid or base is titrated by a strong opposite, between the initial solution and the endpoint, the solution is a buffer. By definition, when the titration is half-way to the endpoint, half of the moles of the initial weak acid or base have been neutralized and converted to its conjugate, and half of the moles of the original substance remain. If the moles of the original weak acid or base and its conjugate are both half the original concentration, the moles of both are equal. We know that In buffer solutions, if either [WA] = [CB] or moles WA = moles CB, then [H+] ≈ Ka and pH ≈ pKa. At the half-way point, this rule can solve some weak acid or base titration calculations by inspection: Half-way to the endpoint in the neutralization of a weak acid or base, a buffer exists in which moles WA = moles CB, and [H+] ≈ Ka and pH ≈ pKa. Limitations The Henderson-Hasselbalch equation is an approximation because it is derived from the buffer approximation. Buffer approximations generally supply answers within the range of experimental error for K and pH except • for relatively strong weak acids and bases, and • very close to the beginning of the titration, and very close to the endpoint, when [WA or WB] is not large compared to ionization (x). In those cases, if more accurate predictions are needed, use the exact buffer quadratic to find [H+], then pH definitions to find pH. Summary: Strong-Strong vs. Weak Strong Titration Stage [H+] and pH Determined By If Strong by Strong If Weak Titrated by Strong Before start [Initial SA or SB] [Weak Acid or Base] & K Between start & endpoint [SA or SB remaining] Buffer Halfway to endpoint 1/2 [Initial SA or SB] [H+] ≈ Ka and pH ≈ pKa At endpoint pH = 7 [Conjugate] and its K After endpoint [titrant] [titrant] Strong-strong and weak-strong titration calculations differ due to the differences between the substances being titrated. Strong acids and bases hydrolyze 100% and form pH-neutral conjugates. Weak acids and bases hydrolyze slightly and form non-pH-neutral conjugates. © 2011 ChemReview.Net v. 7w Page 1065 Module 34 — pH During Titration In some titration calculations, the summary chart above may allow solving by inspection. In all cases, a rice table and mixture rules will solve for pH. Practice B: Apply the rules above from memory. Solve problems 1 and 2. Save Problem 3 for your next practice session. Problem 4 is more challenging. 1. 25.0 mL of 0.0700 M HCl is combined with 25.0 mL of 0.100 M NaCN. Find the pH after the reaction. (Ka of HCN = 6.2 x 10―10). (Solve the rice table using molarity). 2. What is the [H+] half-way to the equivalence point. a. When a strong acid is titrated by a strong base? b. When a weak acid is titrated by a strong base? c. When a weak base is titrated by a strong acid? 3. 20.0 mL of 0.200 M CH3COOH (Ka = 1.8 x 10―5) is titrated by 0.200 M NaOH. a. How many mL of NaOH is required to reach the endpoint? b. When pH = pKa , how many mL of NaOH have been added? c. Calculate the pH in the solution after 19.0 mL NaOH is added (solve in millimoles). d. Calculate the pH at the endpoint of the titration. 4. To determine the Ka of a water soluble weak acid with one acidic hydrogen (call it HBc), 20.0 mL of a 0.150 M HBc solution is combined with 10.0 mL of 0.200 M KOH. The solution pH of this “before the endpoint” mixture is found to be 5.25. What is the new acid’s Ka value? ANSWERS Practice A 1. When equal volumes are combined, the concentrations of both substances are cut in half. [HCl] = 0.100 M ; [NaOH] = 0.0500 M 2. The total volume after mixing is 30.0 mL. The volume of the KOH is tripled, so its concentration is 1/3 the original. [KOH] = 0.45 M x 1/3 = 0.15 M after mixing. If needed, for [HCN], you can use VC x MC = VD x MD ? MD HCN = VC x MC = 20.0 mL x 0.300 M HCN = 0.200 M HCN diluted = [HCN] after mixing. VD 30. mL total Using the dilution equation with M, the units and substances will cancel properly, but the labels C and D often will not. © 2011 ChemReview.Net v. 7w Page 1066 Module 34 — pH During Titration 3a. To use the dilution equation, VC x MC = VD x MD , set up a data table with DATA: VC = 25.0 mL HF MC = 0.150 M (C before mixing) (in wording, goes with 25.0 mL) VD = 25.0 + 21.5 = 46.5 mL soln. = Total, diluted volume MD = ? = WANTED SOLVE: ? MD = VC x MC = 25.0 mL x 0.150 M HF = 0.806 M HF diluted VD 46.5. mL total Using the dilution equation with M, the units will cancel properly, but the C and D labels often will not. Check: from 25 mL to 46.5 mL nearly doubles the volume; [HF] should be about cut in half, and is. 3b. To use the dilution equation, VC x MC = VD x MD , set up a data table with DATA: VC = 21.5 mL KOH MC = 0.120 M KOH (C before mixing) VD = 25.0 + 21.5 = 46.5 mL soln. = Total, diluted volume MD = ? = WANTED SOLVE: ? MD = VC x MC = 21.5 mL x 0.120 M KOH = 0.0555 M KOH diluted VD 46.5. mL Check: 21.5 mL to 46.5 mL about doubles the volume; [KOH] should be about cut in half and is. Practice B 1. The rice table needs initial concentrations after mixing, but before reacting. Since equal volumes of the acid and base are combined, the initial mol/L of each is cut in half as they are mixed together. 1 HCN 1 NaCl 0 mol 0 mol ―0.0350 M + 0.0350 M + 0.0350 M + 0.0150 M + 0.0350 M + 0.0350 M Reaction 1 HCl Initial 0.0350 M 0.0500 M ― 0.0350 M 0M Change At End/Equilibrium 1 NaCN The solution mixes CN─ and HCN. It is a buffer. ***** H-H equation: WA formula = HCN CB formula = CN─ moles or [WA] = 0.0350 M HCN Buffer Chart: moles or [CB] = 0.0150 M CN ─ pH ≈ pKa + log ([base] / [acid]) ≈ ─log (6.2 x 10―10) + log (0.0150 M /0.0350 M) ≈ ─ ( ─ 9.21) + log (0.429) ≈ 9.21 ─ 0.37 = 8.84 = pH Check: Ka = 5.6 x 10─10 , buffer pH estimate 10+2 = 8 -12. Check. © 2011 ChemReview.Net v. 7w Page 1067 Module 34 — pH During Titration 2a. Half way, half of the acid is neutralized, and all products of an SA+SB titration are pH neutral. [H+] = 1/2 [SA initial] 2b. Between the beginning and end of WA or WB titration by strong opposite, the solution is a buffer. Half way to the equivalence point, [H+] ≈ Ka 2c. Between the beginning and end of WA or WB titration by strong opposite, the solution is a buffer. Half way to the equivalence point, you have the same mixture as in the titration of a weak acid by a strong base: What exists is half weak acid and half its conjugate base. As in part 2c, [H+] ≈ Ka of the conjugate of the weak base. All buffers can be treated as a weak acid mixed with its conjugate base. 3. WANTED: mL NaOH to neutralize the acetic acid. In this reaction, one mole of base neutralizes one mole of acid: 1 CH3COOH + 1 NaOH 1 H2O + 1 CH3COONa Since both acid and base have the same concentration, the moles of acid and base particles will be the same when their volumes are the same. mL of 0.200 M NaOH to neutralize this 20.0 mL of 0.200 M acid = 20.0 mL NaOH 3b. pH = pKa half-way to the end point. so pH = pKa at 10.0 mL NaOH added, based on part a. 3c. The Ka conveys that acetic acid is a weak acid. To find pH after reaction, given amounts of reactants before the reaction, the steps are Reactants > reactant mol or prefix-mol > rice moles > mixture pH steps, OR Reactant M before > reactant M after mixing > rice M > mixture pH steps. If we solve in mmol, the moles of acid and base before reacting are ? mmol CH3COOH = 20.0 mL CH3COOH • 0.200 mol CH3COOH = 4.00 mmol CH3COOH 1 L CH3COOH soln. ? mmol NaOH = 19.0 mL NaOH • 0.200 mol NaOH = 3.80 mmol NaOH 1 L NaOH soln. Reaction Initial 1 CH3COOH 4.00 mmol Change ―3.80 mmol At End/Equilibrium + 0.20 mmol 1 NaOH 3.80 mmol ―3.80 mmol 0 mmol 1 H2O 1 CH3COONa - - + 3.80 mmol + 3.80 mmol + 3.80 mmol + 3.80 mmol Label the particles at the end as SA, SB, WA, WB, or N. ***** The pH-not-neutrals are CH3COOH and its conjugate CH3COO─. This is a buffer. ***** © 2011 ChemReview.Net v. 7w Page 1068 Module 34 — pH During Titration WA formula = CH3COOH CB formula = CH3COO ─ moles WA = 0.20 mmol CH3COOH Buffer Chart: moles CB = 3.80 mmol CH3COO ─ Henderson-Hasselbalch: pH ≈ pKa + log (mol base/mol acid) ≈ ─log (1.8 x 10―5) + log (3.80 mmol /0.20 mmol) ≈ ─(─4.74) + log (19) ≈ 4.74 + 1.28 = 6.02 = pH Buffer check: Ka = 1.8 x 10―5 , buffer pH estimate is 5+2 = 3 -7. Check. 3d. At the endpoint, the acid and base have neutralized each other: their moles are zero. The mixture contains water and a dissolved salt (CH3COONa ). Recall the rules for pH of salt solutions? ***** The salt is composed of two ions. If one ion is not pH neutral, find pH based on its reaction as a weak acid or weak base. CH3COO─ is the conjugate of the weak acetic acid, and is therefore a weak base. WRRECK it, and\or solve the Kb approximation. ***** WANT: pH of CH3COO ─ solution R+E: specific CH3COO─ + H2O WB + H2O ^ R+E: general Conc. at Eq: │ Kb: Kb = OH─ + CH3COOH OH─ + acid conjugate ^ [WB]mixed─ x [OH─][conjugate] [WB]mixed ─ x ▐ ≈ x (goes slightly) (goes slightly) ^ │ x2 [WB]mixed x (rice bottom row) ≈ Kb Solve the Kb approximation for x. ***** To solve the approximation, you need Kb and [WB]mixed. Find Kb first. ***** You don’t know Kb, but you know that Ka of acid conjugate = 1.8 x 10―5 For acid-base conjugates: Kw = Ka x Kb = 1.0 x 10─14 Kb = Kw = 1.0 x 10─14 = 1.0 x 10─14 = 0.556 x 10─9 = 5.56 x 1010 = Kb Ka Ka 1.8 x 10―5 Now find [WB]mixed . ***** [WB]mixed = [CH3COO─]at the endpoint . At and after the endpoint, ? = mol salt ion formed = mol of initial weak acid used up = 0.00400 mol CH3COO─ In how much aqueous solution is the salt ion dissolved at the endpoint? ***** © 2011 ChemReview.Net v. 7w Page 1069 Module 34 — pH During Titration 20.0 mL initial acid + 20.0 added of base (see part a) = 40.0 mL total soln. ─ ─ ? [CH3COO─] = mol CH3COO = 0.00400 mol CH3COO • 1 mL = 0.100 M CH3COO─ L soln 40.0 total mL soln. 10―3 L Solve the approximation. ***** Kb x2 [WB]mixed ≈ Substituting: 5.6 x 10─10 ≈ x2 0.100 M x2 = ( 5.6 x 10─10 ) ( 0.100 ) = 5.56 x 10―11 = 55.6 x 10―12 x ≈ (estimate 7-8 x 10―6) ≈ 7.46 x 10―6 M = [OH─] Since the approximation was used, do the: Quick 5% test : x = 7.46 x 10―6 M, [WB] = 0.10 M = 1.0 x 10―1 M Since the exponent difference is 3 or more, the ionization passes the 5% test, and the approximation may be used. But pH was WANTED. ***** pOH = ─ log [OH─] = ─ log(7.46 x 10─6) = 5.13 = pOH pH = 14.00 ─ pOH = 8.87 = pH 4. WANTED = Ka of HBc You know the pH. You want Ka . You are titrating a weak acid with a strong base, before the endpoint. What equation involves these variables and conditions? ***** Before the endpoint, a titrated weak acid or base produces a buffer solution. The buffer equation that includes pH and Ka is the Henderson-Hasselbalch, which can be solved in moles or mol/L. If you choose moles, pH ≈ pKa + log (mol base/mol acid) To solve for pKa then Ka , you need to find the moles of acid and base at the point where the pH = 5.25 You can solve in moles, but if volumes are given in milliliters, you may want to solve in millimoles. To find moles of acid and base in the solution at a stated point in a titration, use a rice moles table. ***** For the balanced equation, you can write either HBc + KOH H2O + KBc or HBc + OH─ H2O + Bc─ Both give the same results. Find the initial mmol, before the reactants react. ***** ? mmol KOH = 10.0 mL KOH • 0.200 mol KOH = 2.00 mmol KOH = 2.00 mmol OH─ 1 L KOH soln. The millimoles of acid mixed is ? mmol HBc = 20.0 mL HBc • 0.150 mol HBc = 3.00 mmol HBc 1 L HCl soln. © 2011 ChemReview.Net v. 7w Page 1070 Module 34 — pH During Titration 1 H2O 1 Bc─ 0 mmol 0 mmol ―2.00 mmol + 2.00 mmol + 2.00 mmol 0 mmol + 2.00 mmol + 2.00 mmol Reaction 1 HBc 1 OH─ Initial 3.00 mmol 2.00 mmol Change ― 2.00 mmol At End/Equilibrium + 1.00 mmol Label the end particles at this point as SA, SB, WA, WB, or N. ***** This is a buffer. The pH-not-neutral components are HBc (WA) and its conjugate Bc─. Solve H-H. H-H: WA formula = HBc CB formula = Bc─ moles or [WA] = 1.00 mmol HBc Buffer Chart: moles or [CB] = 2.00 mmol Bc─ pH ≈ pKa + log (mol base/mol acid) You want Ka. Solve for pKa first. From the above equation, pKa ≈ pH ─ log (mol base/mol acid) ≈ 5.25 ─ log (2.00 mmol /1.00 mmol) ≈ 5.25 ─ log (2.00) ≈ ≈ 5.25 ─ 0.30 ≈ 4.95 = pKa But we WANT Ka . Since pKa ≡ ─ log Ka and pH ≡ ─ log [H+] Ka ≡ 10─pKa and [H+] ≡ 10─pH Finish from here. ***** ? = Ka ≡ 10─pKa = 10─4.95 = ?? x 10─5 = 1.1 x 10─5 = Ka of HBc Check: Buffer pH and ─Ka exponent should be + 2. pH is 5.25. Ka estimate = ? x 10―3 to 7 . Check. ***** © 2011 ChemReview.Net v. 7w Page 1071 Module 34 — pH During Titration SUMMARY – pH During Neutralization and Titration 1. When an acid and base are reacted, IF one of the components is strong, the reaction will go until one component, either the acid or the base, is 100% used up. Another way to express this rule: At the end of a neutralization reaction, if one of the reactants is strong, the moles of one of the reactants must be zero. 2. Steps For Finding Mixture pH Apply in this order. a. Re-write soluble salts as separated ions. b. Label the particles in the mixture as SA, SB, WA, WB, or N, based on Ka or Kb. c. In pH calculations, ignore pH-neutral (N) particles. d. If all particles are N, pH = 7. e. If SA or SB is present, ignore other particles. Find pH based on 100% ionization: [HCl or HNO3] = [H+] f. and [NaOH or KOH] = [OH─] If a WA or WB and its conjugate is present, solve a buffer chart and the Henderson-Hasselbalch equation in moles or mol/L. g. If only one WA or WB is present, use Ka or Kb approximation, then the 5% test. 3. Rice tables must have consistent units. The values that go into the initial row of a rice table can be moles, or prefix-moles, or mol/L after combining but before reacting. 4. To Calculate pH After Reaction From Amounts Before Reaction a. Convert the initial amounts of acid and base to moles, or prefix-moles, or mol/L after combining but before reacting. b. Enter those values in the initial row of a rice moles table. c. In the change row, use the rule: when acid and base are combined, if one or both are strong, the one reactant is 100% used up. d. Calculate pH based on the mixture at the end of the reaction (in bottom rice row). Those four steps can be summarized as: To find pH after reaction, given amounts of reactants before the reaction, Reactants > reactant mol or prefix-mol > rice moles > mixture pH steps, OR Reactant M before > reactant M after mixing > rice M > mixture pH steps. © 2011 ChemReview.Net v. 7w Page 1072 Module 34 — pH During Titration 5. To solve calculations for [H+] or pH, a. Ask: Is the problem about a stable solution at equilibrium, or reactants reacting that then become a stable solution? b. If about a stable solution, apply the steps for finding mixture pH. c. IF about a reaction use the steps for pH after from amounts before combining. 6. The millimoles shortcut for rice moles tables: millimoles = mmol = mL x (mol/L) . 7. When solvent solutions are combined, their volumes add. 8. For reactions, to choose between using stoichiometry steps or the rice moles table: a. For any calculation involving amounts involved in chemical reactions, if you know the amount of one reactant that is 100% used up (the limiting reactant), and you want to know how much of one other reactant will react or one other product will form, use the stoichiometry steps: WDBB, units moles moles units Amounts needed for exact neutralization at the equivalence point can be solved by stoichiometry IF one or both of the acid-base reactants is strong. b. For all other types of reaction calculations, a rice moles table (or, for simple reactions, the abbreviated version of the rice table used in the WRECK steps) is needed to determine amounts used up, formed, and present in the final mixture at equilibrium. These cases include: • IF you know the amounts for two reactants present before the reaction, but you do not know which is limiting, or • the reaction does not go essentially to completion, or • you need the amounts of all of the particles in the mixture after the reaction stops. 9. Special rules for titration calculations a. At the equivalence point in any titration, • the moles of acid added to the reaction mixture are equivalent to (the same as) the moles base added. • The acid and base reactants are exactly neutralized: both are 100% used up. • Since there are no reactants left, there are only products present. b. If a strong or moderately weak acid or base is titrated by a strong opposite, the pH changes sharply during the titration, but not until very close to the endpoint. c. For the titration of any strong acid and strong base, at the endpoint: pH = 7 . d. Half-way to the endpoint in the titration of a weak acid or base, a buffer solution exists in which moles WA = moles CB, and [H+] ≈ Ka and pH ≈ pKa. © 2011 ChemReview.Net v. 7w Page 1073 Module 34 — pH During Titration 10. Summary: Strong-Strong vs. Weak Strong Titration Stage [H+] and pH Determined By If Strong By Strong If Weak By Strong Before start [Strong Acid or Base] [Weak Acid or Base] & K Between start & endpoint [Strong Acid or Base] Buffer Half-way to endpoint 1/2 [Initial SA or SB] [H+] ≈ Ka and pH ≈ pKa At endpoint pH = 7 [Conjugate] and its K After endpoint [strong titrant] [strong titrant] ##### © 2011 ChemReview.Net v. 7w Page 1074 ...
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This note was uploaded on 11/16/2011 for the course CHM 1025 taught by Professor J during the Summer '09 term at Santa Fe College.

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