AP Ch 15 Acid-Base

AP Ch 15 Acid-Base - Acids and Bases Chapter 15 Acids •...

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Unformatted text preview: Acids and Bases Chapter 15 Acids • Have a sour taste. Vinegar owes its taste to acetic acid. Citrus fruits contain citric acid. • React with certain metals to produce hydrogen gas. • React with carbonates and bicarbonates to produce carbon dioxide gas Bases • Have a bitter taste. • Feel slippery. Many soaps contain bases. 4.3 Some Properties of Acids Produce H+ (as H3O+) ions in water (the hydronium ion is a hydrogen ion attached to a water molecule) r Taste sour r Corrode metals r Electrolytes r React with bases to form a salt and water r pH is less than 7 r Turns blue litmus paper to red “Blue to Red A-CID” Acid Nomenclature Review Anion Ending No Oxygen Acid Name hydro-(stem)-ic acid (stem)-ic acid (stem)-ous acid -ide -ate w/Oxygen w/Oxygen -ite An easy way to remember which goes with which… An “In the cafeteria, you ATE something ICky” ATE something IC Some Properties of Bases Produce OH- ions in water Taste bitter, chalky Are electrolytes Feel soapy, slippery React with acids to form salts and water pH greater than 7 Turns red litmus paper to blue “Basic Blue” asic Acid/Base definitions Definition 1: Arrhenius Arrhenius acid is a substance that produces H+ (H3O+) in water Arrhenius base is a substance that produces OH- in water 4.3 Acid/Base Definitions • Definition #2: Brønsted – Lowry Acids – proton donor Bases – proton acceptor A “proton” is really just a hydrogen atom that has lost it’s electron! A Brønsted-Lowry acid is a proton donor A Brønsted-Lowry base is a proton acceptor base acid conjugate acid conjugate base ACID-BASE THEORIES ACID-BASE The Brønsted definition means NH3 is a BASE BASE in water — and water is itself an ACID in ACID NH3 Base + H2O Acid NH4+ + OHAcid Base Conjugate Pairs Conjugate Learning Check! Label the acid, base, conjugate acid, and Label conjugate base in each reaction: conjugate HCl + OH­ Cl­ + H2O Acid Acid Base Conj. Conj. Base Acid H2O + H2SO4 HSO4­ + H3O+ Base Acid Conj. Base Conj. Acid Acids & Base Definitions Acids Definition #3 – Lewis Lewis acid - a substance that Lewis accepts an electron pair accepts Lewis base - a substance Lewis that donates an electron pair pair Lewis Acids & Bases Formation of hydronium ion is also Formation an excellent example. an H + •• • O—H • H BASE •• H O—H H ACID •Electron pair of the new O-H bond Electron originates on the Lewis base. originates Lewis Acid/Base Reaction Lewis The pH scale is a way of pH expressing the strength of acids and bases. Instead of using very small numbers, we just use the NEGATIVE power of 10 on the Molarity of the H (or OH ) ion. of + - Under 7 = acid 7 = neutral Over 7 = base pH of Common Substances Substances Calculating the pH pH = - log [H+] (Remember that the [ ] mean Molarity) Example: If [H+] = 1 X 10-10 pH = - log 1 X 10-10 pH = - (- 10) pH = 10 Example: If [H+] = 1.8 X 10-5 pH = - log 1.8 X 10-5 pH = - (- 4.74) pH = 4.74 Try These! Try pH = - log [H+] Find the pH of these: pH = - log 0.15 1) A 0.15 M solution of 0.15 pH = - (- 0.82) Hydrochloric acid Hydrochloric pH = 0.82 2) A 3.00 X 10-7 M solution of Nitric acid acid pH = - log 3 X 10-7 pH = - (- 6.52) pH pH = 6.52 pH calculations – Solving for H+ pH If the pH of Coke is 3.12, [H+] = ??? Because pH = - log [H+] then - pH = log [H+] pH Take antilog (10x) of both sides and get 10--pH = [H+] pH [H [H+] = 10-3.12 = 7.6 x 10-4 M *** to find antilog on your calculator, look for “Shift” or “2nd function” and then the log button or nd function” More About Water H2O can function as both an ACID and a BASE. In pure water there can be AUTOIONIZATION AUTOIONIZATION Equilibrium constant for water = Kw Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC 1.00 More About Water Autoionization OH- H3O+ Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC In a neutral solution [H3O+] = [OH-] and so [H3O+] = [OH-] = 1.00 x 10-7 M pOH • Since acids and bases are Since opposites, pH and pOH are opposites! opposites! • pOH does not really exist, but it is pOH useful for changing bases to pH. useful • pOH looks at the perspective of a pOH base base pOH = - log [OH-] Since pH and pOH are on opposite Since ends, ends, pH + pOH = 14 pH [H ] [OH ] pOH [H3O+], [OH-] and pH [H What is the pH of the What 0.0010 M NaOH solution? [OH-] = 0.0010 (or 1.0 X 10-3 M) [OH-] pOH = - log 0.0010 pOH pOH = 3 pOH pH = 14 – 3 = 11 OR Kw = [H3O+] [OH-] [H3O+] = 1.0 x 10-11 M pH = - log (1.0 x 10-11) = 11.00 The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H ion concentration of the rainwater? = = [H+] = 10-pH 10-4.82 1.5 x 10-5 M The OH ion concentration of a blood sample is + pH = -log [H+] 2.5 x 10 M. What is the pH of the blood? -7 pOH = -log [OH-]= -log (2.5 x 10-7)= 6.60 pH = 14.00 – pOH = 14.00 – 6.60 = 7.40 Calculating [H3O+], pH, [OH-], and pOH Problem 1: A chemist dilutes concentrated hydrochloric acid to make two solutions: (a) 3.0 M and (b) 0.0024 M. Calculate the [H O ], pH, 3 + [OH ], and pOH of the two solutions at 25°C. - Problem 2: What is the [H O ], [OH ], and pOH 3 + - of a solution with pH = 3.67? Is this an acid, Strong and Weak Acids/Bases The strength of an acid (or base) is determined by the amount of IONIZATION. HNO3, HCl, H2SO4 and HClO4 are among the only known strong acids. Strong and Weak Acids/Bases • Generally divide acids and bases into STRONG or Generally WEAK ones. WEAK STRONG ACID: HNO3 (aq) + H2O (l) ---> HNO (aq) H3O+ (aq) + NO3- (aq) (aq) (aq) HNO3 is about 100% dissociated in water. Strong and Weak Acids/Bases • Weak acids are much less than 100% ionized in water. water. *One of the best known is acetic acid = CH3CO2H *One Strong and Weak Acids/Bases • Strong Base: 100% dissociated in water. 100% NaOH (aq) ---> Na+ (aq) + OH- (aq) NaOH (aq) (aq) Other common strong Other bases include KOH and bases CaO Ca(OH) . Ca(OH) CaO (lime) + H2O --> Strong and Weak Acids/Bases • Weak base: less than 100% ionized in water less One of the best known weak bases is ammonia NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) (aq) (l) (aq) (aq) Weak Bases Weak Equilibria Involving Weak Acids and Bases Weak Consider acetic acid, HC2H3O2 (HOAc) HC2H3O2 + H2O H3O+ Acid Acid + C2H3O2 Conj. base [H3O+ ][OAc- ] -5 Ka = = 1.8 x 10 [HOAc] (K is designated Ka for ACID) K gives the ratio of ions (split up) to molecules (don’t split up) Ionization Constants for Acids/Bases Ionization Aci ds Aci Increase strength Conjuga te Ba ses Increase strength E qui l i br i um Constants for Weak Aci ds f or Weak acid has Ka < 1 Leads to small [H3O+] and a pH of 2 - 7 Leads E qui l i br i um Constants for Weak Bases f or Weak base has Kb < 1 Leads to small [OH-] and a pH of 12 - 7 Leads Rel ati on of K a, of K b, [H 3O+] [H and pH and Equilibria Involving A Weak Acid Equilibria You have 1.00 M HOAc. Calc. the You equilibrium concs. of HOAc, H3O+, OAc-, equilibrium and the pH. and Step 1. Define equilibrium concs. in ICE table. table. [HOAc] [HOAc] initial change equilib 1.00 -x 1.00-x [H3O+] 0 +x x [OAc-] 0 +x x Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH. of Step 2. Write Ka expression Step Write [H3O+ ][OAc- ] x2 Ka = 1.8 x 10-5 = = [HOAc] 1.00 - x This is a quadratic. Solve using quadratic This formula. formula. or you can make an approximation if x is very or small! (Rule of thumb: 10-5 or smaller is ok) small! Equilibria Involving A Weak Acid Equilibria You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH. of Step 3. Solve Ka expression Solve [H3O+ ][OAc- ] x2 Ka = 1.8 x 10-5 = = [HOAc] 1.00 - x First assume x is very small because First Ka is so small. Ka = 1.8 x 10-5 = x2 1.00 Now we can more easily solve this Now approximate expression. approximate Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH. of Step 3. Solve Ka approximate Step Solve approximate expression a = 1.8 x 10-5 = x2 expression K 1.00 x = [H3O+] = [OAc-] = 4.2 x 10-3 M pH = - log [H3O+] = -log (4.2 x 10-3) = 2.37 2.37 Equilibria Involving A Weak Acid Equilibria Calculate the pH of a 0.0010 M solution of formic Calculate acid, HCO2H. acid, HCO2H + H2O HCO2- + H3O+ HCO Ka = 1.8 x 10-4 Approximate solution Approximate [H3O+] = 4.2 x 10-4 M, pH = 3.37 [H pH Exact Solution [H3O+] = [HCO2-] = 3.4 x 10-4 M [H [HCO2H] = 0.0010 - 3.4 x 10-4 = 0.0007 M [HCO pH = 3.47 E qui l i br i a I nvol vi ng A Weak Base B .ase the pH. You have 0.010 M NH3 Calc. You have 0.010 M NH3. Calc. the pH. NH3 + H2O NH4+ + OHNH Kb = 1.8 x 10-5 [NH3] [NH [NH4+] 0 +x x Step 1. Define equilibrium concs. in ICE table Define [OH-] 0 +x x initial change equilib 0.010 -x 0.010 - x Equilibria Involving A Weak Base Equilibria You have 0.010 M NH3. Calc. the pH. NH3 + H2O NH4+ + OHNH Kb = 1.8 x 10-5 Step 1. Define equilibrium concs. in ICE table Define [NH3] [NH [NH4+] 0 +x x [OH-] 0 +x x initial change equilib 0.010 -x 0.010 - x Equilibria Involving A Weak Base Equilibria You have 0.010 M NH3. Calc. the pH. NH3 + H2O NH4+ + OHNH Kb = 1.8 x 10-5 Step 2. Solve the equilibrium expression Solve + x2 -5 = [NH4 ][OH ] = Kb = 1.8 x 10 [NH3 ] 0.010 - x Assume x is small, so x = [OH-] = [NH4+] = 4.2 x 10-4 M [OH and [NH3] = 0.010 - 4.2 x 10-4 ≈ 0.010 M The approximation is valid ! You have 0.010 M NH3. Calc. the pH. NH3 + H2O NH4+ + OHNH Kb = 1.8 x 10-5 Step 3. Calculate pH Calculate [OH-] = 4.2 x 10-4 M so pOH = - log [OH-] = 3.37 Because pH + pOH = 14, E qui l i br i a I nvol vi ng A Weak Base B ase pH = 10.63 Types of Acid/Base Reactions: Summary Summary Acid-Base Properties of Water H2O (l) H+ (aq) + OH- (aq) autoionization of water H O H +H O H base H2O + H2O acid [ ++H HOH O - H conjugate acid H3O+ + OHconjugate base 15.2 The Ion Product of Water H2O (l) H+ (aq) + OH- (aq) [H+][OH-] Kc = [H2O] [H2O] = constant Kc[H2O] = Kw = [H+][OH-] The ion-product constant (Kw) is the product of the molar concentrations of H+ and OH- ions at a particular temperature. Solution Is neutral acidic basic 15.2 [H+] = [OH-] At 250C Kw = [H+][OH-] = 1.0 x 10-14 [H+] > [OH-] [H+] < [OH-] What is the concentration of OH- ions in a HCl solution whose hydrogen ion concentration is 1.3 M? Kw = [H+][OH-] = 1.0 x 10-14 [H+] = 1.3 M Kw 1 x 10-14 = = 7.7 x 10-15 M [OH-] = [H+] 1.3 15.2 pOH = -log [OH-] [H+][OH-] = Kw = 1.0 x 10-14 -log [H+] – log [OH-] = 14.00 pH + pOH = 14.00 15.3 The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H+ ion concentration of the rainwater? pH = -log [H+] [H+] = 10-pH = 10-4.82 = 1.5 x 10-5 M The OH- ion concentration of a blood sample is 2.5 x 10-7 M. What is the pH of the blood? pH + pOH = 14.00 pOH = -log [OH-] = -log (2.5 x 10-7) = 6.60 pH = 14.00 – pOH = 14.00 – 6.60 = 7.40 15.3 Strong Electrolyte – 100% dissociation NaCl (s) H2O Na+ (aq) + Cl- (aq) Weak Electrolyte – not completely dissociated CH3COOH CH3COO- (aq) + H+ (aq) Strong Acids are strong electrolytes HCl (aq) + H2O (l) HNO3 (aq) + H2O (l) HClO4 (aq) + H2O (l) H2SO4 (aq) + H2O (l) H3O+ (aq) + Cl- (aq) H3O+ (aq) + NO3- (aq) H3O+ (aq) + ClO4- (aq) H3O+ (aq) + HSO4- (aq) 15.4 Weak Acids are weak electrolytes HF (aq) + H2O (l) HNO2 (aq) + H2O (l) HSO4- (aq) + H2O (l) H2O (l) + H2O (l) H3O+ (aq) + F- (aq) H3O+ (aq) + NO2- (aq) H3O+ (aq) + SO42- (aq) H3O+ (aq) + OH- (aq) Strong Bases are strong electrolytes NaOH (s) KOH (s) H2O H2O Na+ (aq) + OH- (aq) K+ (aq) + OH- (aq) Ba2+ (aq) + 2OH- (aq) 15.4 Ba(OH)2 (s) H2O Weak Bases are weak electrolytes F- (aq) + H2O (l) NO2- (aq) + H2O (l) OH- (aq) + HF (aq) OH- (aq) + HNO2 (aq) Conjugate acid-base pairs: • • • The conjugate base of a strong acid has no measurable strength. H3O+ is the strongest acid that can exist in aqueous solution. The OH- ion is the strongest base that can exist in aqeous solution. 15.4 15.4 Strong Acid Weak Acid 15.4 What is the pH of a 2 x 10-3 M HNO3 solution? HNO3 is a strong acid – 100% dissociation. Start 0.002 M HNO3 (aq) + H2O (l) End 0.0 M 0.0 M 0.0 M H3O+ (aq) + NO3- (aq) 0.002 M 0.002 M pH = -log [H+] = -log [H3O+] = -log(0.002) = 2.7 What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution? Ba(OH)2 is a strong base – 100% dissociation. Start 0.018 M Ba(OH)2 (s) End 0.0 M 0.0 M 0.0 M Ba2+ (aq) + 2OH- (aq) 0.018 M 0.036 M 15.4 pH = 14.00 – pOH = 14.00 + log(0.036) = 12.56 Weak Acids (HA) and Acid Ionization Constants HA (aq) + H2O (l) HA (aq) H3O+ (aq) + A- (aq) H+ (aq) + A- (aq) [H+][A-] Ka = [HA] Ka is the acid ionization constant Ka weak acid strength 15.5 15.5 What is the pH of a 0.5 M HF solution (at 250C)? HF (aq) H+ (aq) + F- (aq) HF (aq) Initial (M) Change (M) Equilibrium (M) 0.50 -x 0.50 - x Ka = [H+][F-] = 7.1 x 10-4 [HF] H+ (aq) + F- (aq) 0.00 +x x Ka << 1 0.00 +x x 0.50 – x ≈ 0.50 x = 0.019 M x2 = 7.1 x 10-4 Ka = 0.50 - x Ka ≈ x2 = 7.1 x 10-4 0.50 x2 = 3.55 x 10-4 [H+] = [F-] = 0.019 M [HF] = 0.50 – x = 0.48 M pH = -log [H+] = 1.72 15.5 When can I use the approximation? Ka << 1 0.50 – x ≈ 0.50 When x is less than 5% of the value from which it is subtracted. x = 0.019 0.019 M x 100% = 3.8% 0.50 M Less than 5% Approximation ok. What is the pH of a 0.05 M HF solution (at 250C)? x2 Ka ≈ = 7.1 x 10-4 x = 0.006 M 0.05 More than 5% 0.006 M x 100% = 12% 0.05 M Approximation not ok. Must solve for x exactly using quadratic equation or method of successive approximation. 15.5 Solving weak acid ionization problems: 1. Identify the major species that can affect the pH. • • In most cases, you can ignore the autoionization of water. Ignore [OH-] because it is determined by [H+]. 1. Use ICE to express the equilibrium concentrations in terms of single unknown x. 2. Write Ka in terms of equilibrium concentrations. Solve for x by the approximation method. If approximation is not valid, solve for x exactly. 1. Calculate concentrations of all species and/or pH of the solution. 15.5 What is the pH of a 0.122 M monoprotic acid whose Ka is 5.7 x 10-4? HA (aq) Initial (M) Change (M) Equilibrium (M) 0.122 -x 0.122 - x H+ (aq) + A- (aq) 0.00 +x x Ka << 1 0.00 +x x 0.122 – x ≈ 0.122 x = 0.0083 M x2 = 5.7 x 10-4 Ka = 0.122 - x Ka ≈ x2 = 5.7 x 10-4 0.122 x2 = 6.95 x 10-5 0.0083 M x 100% = 6.8% 0.122 M More than 5% Approximation not ok. 15.5 x2 = 5.7 x 10-4 Ka = 0.122 - x ax2 + bx + c =0 x = 0.0081 HA (aq) Initial (M) Change (M) Equilibrium (M) 0.122 -x 0.122 - x x2 + 0.00057x – 6.95 x 10-5 = 0 -b ± √b2 – 4ac x= 2a x = - 0.0081 H+ (aq) + A- (aq) 0.00 +x x 0.00 +x x [H+] = x = 0.0081 M pH = -log[H+] = 2.09 15.5 percent ionization = Ionized acid concentration at equilibrium Initial concentration of acid x 100% For a monoprotic acid HA Percent ionization = [H+] [HA]0 x 100% [HA]0 = initial concentration 15.5 Weak Bases and Base Ionization Constants NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) [NH4+][OH-] Kb = [NH3] Kb is the base ionization constant Kb weak base strength Solve weak base problems like weak acids except solve for [OH-] instead of [H+]. 15.6 15.6 Ionization Constants of Conjugate Acid-Base Pairs HA (aq) A- (aq) + H2O (l) H2O (l) H+ (aq) + A- (aq) OH- (aq) + HA (aq) H+ (aq) + OH- (aq) KaKb = Kw Weak Acid and Its Conjugate Base Kw Ka = Kb Kw Kb = Ka 15.7 Ka Kb Kw 15.8 Molecular Structure and Acid Strength HX The stronger the bond H+ + XThe weaker the acid • Bond strength • Polarity HF << HCl < HBr < HI 15.9 Molecular Structure and Acid Strength Z δO δ+ H Z O- + H+ The O-H bond will be more polar and easier to break if: • • Z is very electronegative or Z is in a high oxidation state 15.9 Molecular Structure and Acid Strength 1. Oxoacids having different central atoms (Z) that are from the same group and that have the same oxidation number. Acid strength increases with increasing electronegativity of Z •• •• O O •• •• •• •• H O Cl O H O Br O •• •• •• •• •• •• •• •• •• •• •• Cl is more electronegative than Br •• HClO3 > HBrO3 15.9 Molecular Structure and Acid Strength 2. Oxoacids having the same central atom (Z) but different numbers of attached groups. Acid strength increases as the oxidation number of Z increases. HClO4 > HClO3 > HClO2 > HClO 15.9 Acid-Base Properties of Salts Neutral Solutions: Salts containing an alkali metal or alkaline earth metal ion (except Be2+) and the conjugate base of a strong acid (e.g. Cl-, Br-, and NO3-). NaCl (s) Basic Solutions: Salts derived from a strong base and a weak acid. NaCH3COOH (s) H2O H2O Na+ (aq) + Cl- (aq) Na+ (aq) + CH3COO- (aq) CH3COOH (aq) + OH- (aq) 15.10 CH3COO- (aq) + H2O (l) Acid-Base Properties of Salts Acid Solutions: Salts derived from a strong acid and a weak base. NH4Cl (s) NH4+ (aq) H2O NH4+ (aq) + Cl- (aq) NH3 (aq) + H+ (aq) Salts with small, highly charged metal cations (e.g. Al3+, Cr3+, and Be2+) and the conjugate base of a strong acid. Al(H2O)3+(aq) 6 Al(OH)(H2O)2+(aq) + H+ (aq) 5 15.10 Acid-Base Properties of Salts Solutions in which both the cation and the anion hydrolyze: • • • Kb for the anion > Ka for the cation, solution will be basic Kb for the anion < Ka for the cation, solution will be acidic Kb for the anion ≈ Ka for the cation, solution will be neutral 15.10 Oxides of the Representative Elements In Their Highest Oxidation States CO2 (g) + H2O (l) N2O5 (g) + H2O (l) H2CO3 (aq) 2HNO3 (aq) 15.11 ...
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