Scruggs Chap 17 (all)

Scruggs Chap 17 (all) - Chapter 17 Additional Aspects of...

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Unformatted text preview: Chapter 17 Additional Aspects of Aqueous Equilibria Water is a common solvent. Dissolved materials can be involved in different i l d i diff t types of chemical equilibria. Acid-Base Acid Solubility 1 17.1 The Common Ion Effect Metal ions or salts containing a conjugate weak acid or base can shift the pH of a acid or base solution We can predict pH changes using Le Chatelier's principle HA + H2O A- + H3O+ B + H2O HB+ + OH2 Common Ion Effect CH3CO2H + H2O CH3CO2- + H3O+ Adding more CH3CO2- to the solution shifts the equilibrium to the left, making the solution less acidic (higher p ) ( g pH). 0.100 M CH3CO2H 0.100 M CH3CO2H, 0.050 M NaCH3CO2 pH = 2.879 pH = 4.456 3 1 1 pH of Weak Acid Calculation What is the pH of a 1.00 M HF solution? Ka = 7.0 x 10- 4 HF H3O+ FInitial 1.00 1 00 0 0 Change -x +x +x Equil. 1.00 - x x x x2/(1.00 - x) = 7.0 x 10- 4 Assume x << 1.00: x2/1.00 = 7.0 x 10- 4 x = 2.65 x 10-2, so the assumption was okay pH = -log(2.65 x 10-2) = 1.576 4 What is the pH of a 1.0M HF solution after 0.5M NaF is added? 1. 2. 2 3. 4. 5. 1.58 2.58 2 58 2.84 3.76 5.48 0% 1. 0% 2. 0% 3. 0% 4. 0% 5. 5 Group Work What is the pH of 1.00 M HF solution to which is added 0.500 M NaF? Ka = 7.0 x 10-4 NaF? HF H3O+ FInitial 1.00 0 0.500 Change -x +x +x Equil. Equil. 1.00 - x x 0.500 + x x(0.500 + x)/(1.00 - x) = 7.0 x 10- 4 Assume x << 0.500 and 1.00: x(0.500)/1.00 = 7.0 x 10- 4 x = 1.40 x 10-3 (>1%), so assumption was okay pH = -log(1.40 x 10-3) = 2.854 6 2 2 17.2 Buffered Solutions Buffer: Solution of a weak acid and its Buffer: conjugate base in approximately equal concentrations Used to control pH Addition of a strong acid or base to a buffer shifts the pH only slightly 7 Buffer Solutions Make buffer: Mix HF and NaF in water. HF(aq) + H2O(l) H3O+(aq) + F-(aq) HF(aq) aq) aq) Add HCl: Equilibrium shifts to left to consume HCl: some of added H3O+ Add NaOH: Some H3O+ is consumed to neutralize NaOH: the added OH-, and the equilibrium shifts to the right to replace some of the consumed H3O+. The pH change is much smaller than if HCl or NaOH was added to water. 8 Buffer Solutions pH of all body fluids is controlled in this way. In our blood, phosphate, carbonate, and hemoglobin all act as buffers. H2CO3(aq) + H2O(l) HCO3-(aq) + H3O+(aq) aq) aq) aq) Secondary Equilibrium: H2CO3(aq) H2O(l) + CO2(g) aq) 9 3 3 Addition of acid or base to a buffer 10 Buffer Solutions Recall that a solution of 1.00 M HF and 0.500 M NaF has a pH of 2.854 Now add 0.100 M HCl (assume there is no volume change) l h ) To find the new pH, first assume that all the added strong acid reacts to change the Fand HF concentrations. [HF] = 1.00 + 0.100 = 1.100 M [F-] = 0.500 - 0.100 = 0.400 M 11 Buffer Solutions Now solve the weak acid/conjugate base system: HF H3O+ FInitial 1.100 0 0.400 Change -x +x +x Equil. Equil. 1.100 - x x 0.400 + x x(0.400 + x)/(1.100 - x) = 7.0 x 10- 4 Assume x << 0.400 and 1.100: x(0.400)/1.100 = 7.0 x 10- 4 x = 1.93 x 10-3, so the assumption was okay pH = -log(1.93 x 10-3) = 2.714 12 4 4 Buffer Solutions Water: pure 0.100 M H3O+, pH 7 pH 1 Buffer: pH 2.854 pH 2.714 What will happen when a strong base (NaOH) is added to: Water? Buffer? 13 Buffer Solutions If base is added, assuming no volume change, reduce the concentration of HF and increase the concentration of F- by the corresponding amount. Then solve the weak acid/conjugate j g base system. 14 Add 0.l00 M NaOH to the 1.00 M HF/0.500 M NaF buffer. What is the new pH? 1. 2. 3. 4. 5. 13.0 7.6 4.3 3.0 2.5 0% 1. 0% 2. 0% 3. 0% 4. 0% 5. 15 5 5 Group Work Add 0.l00 M NaOH to the 1.00 M HF/0.500 M NaF buffer. What is the new pH? First assume that all OH- reacts with HF: [HF] = 1.00 - 0.100 = 0.900 M [F-] = 0.500 + 0.100 = 0.600 M Now solve the new acid-base system: acid- 16 Group Work HF H3O+ FInitial 0.900 0 0.600 Change -x +x +x Equil. q 0.900 - x x 0.600 + x x(0.600 + x)/(0.900 - x) = 7.0 x 10- 4 Assume x << 0.600 and 0.900: x(0.600)/0.900 = 7.0 x 10- 4 x = 1.05 x 10-3, so the assumption was okay pH = -log(1.05 x 10-3) = 2.979 In water, pH would change from 7 to 13. 17 (started at 2.854) Addn of Strong Acids and Bases to Buffers See Sample and Practice Ex 17.5 and end of chapter problems. 18 6 6 Buffer Capacity Buffers only work within a pH range set by the value of pKa: 0.1 < [HA]/[A-] < 10 Outside this range, we see little buffering effect. 19 Henderson-Hasselbalch Equation Assuming that x << [HA] and x << [A-] for good buffer action, the equilibrium constant expression can be rearranged to give simplified calculations: p pH = pKa + log([A-]/[HA]) Derived p. 727 20 Henderson-Hasselbalch Equation pH = pKa + log([A-]/[HA]) Consider 1.00 M HF, 0.500 M NaF Ka = 7 x 10- 4, pKa = 3.155 3 155 log ([F-]/[HF]) = log (0.500/1.00) = -0.301 pH = 3.155 - 0.301 = 2.854 (same result as before) 21 7 7 Henderson-Hasselbalch Equation pH = pKa + log([A-]/[HA]) Determine the Ka for the following system if a solution of equal concentrations of HA and A- give a pH of 8.0? HA + H2O A- + H3O+ 22 Buffer Range We can examine the buffer range graphically by plotting the fraction of each species vs the pH As we move through the buffer range, one g g , species decreases in concentration as the other increases in concentration Buffer range: pH = pKa + 1 Select an appropriate acid-base pair to get acidwithin the range of the desired pH 23 CH3CO2H/CH3CO2Buffer Range 24 8 8 HF/FBuffer Range 25 NH4+/NH3 Buffer Range NH4+ NH3 26 H3PO4/H2PO4-/HPO42-/PO43- 27 9 9 Which buffer system would be the best to use to keep the pH around 4.5? 1. H2CO3/HCO3(Ka=4.3 x 10-7) 2. 2 HCN/CN(Ka=4.9 x 10-10) 3. Ascorbic Acid/Ascorbate Acid/Ascorbate (Ka=8.0 x 10-5) 4. HClO/ClOHClO/ (Ka=1.1 x 10-2) 0% 1 0% 2 0% 3 0% 4 28 17.3 Acid-Base Titrations Refer to Chapter 4 for titration calculations to determine amounts of acids or bases. Widely used for chemical analysis: acids in plating baths ascorbic acid in vitamin C tablets acetylsalicylic acid in aspirin tablets acetic acid in vinegar and wines phosphoric acid in soft drinks 29 Titrations Equivalence point: point in a titration where point: there are exactly equivalent amounts of the reactants End point: point in a titration where the point: indicator changes color; may not be exactly the same as the equivalence point Indicator: an acid or base which has a Indicator: different color than its conjugate; characterized by a pKa, which determines the pH range of the color change (pKa + 1). 30 10 10 Phenolphthalein HIn + H2O In- + H3O+ Can you find the differences? Color change: colorless to pink pH = 8.2 - 10 Phenolphthalein HIn In 31 Indicators 32 Indicators The indicator should be picked so its color change comes close to the equivalence point. 33 11 11 Strong Acid - Strong Base Note the slow change before and after the equivalence q point and the rapid large change at the equivalence point. 34 Titration Curve Calculations Calculate the equivalence point pH by assuming complete reaction of the acid and base, then solve any weak acid (or base) equilibrium for the pH. pH Calculate the pH on either side of the equivalence point the same way, except in this case, there will be an excess of one substance. 35 Titration Curve Calculations Strong Acid /Strong Base At the equivalence point, all acid and base has been neutralized. The conjugate acid and base do not affect pH so the pH is 7. HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) 36 12 12 Strong Acid/Strong Base Titration 37 Titration Curve Calculations Strong Acid /Strong Base HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) Before and after the equivalence point, there will be excess moles of acid or base. Determine the number of moles xs and divide by the total volume of solution. What is the [HCl] after 25.00 mL of 0.100 M NaOH has been added to 50.00 mL of 0.100 M HCl? 38 Titration Curve Calculations Strong Acid/Strong Base HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) HCl(aq) NaOH(aq) NaCl(aq) [HCl]xs = 0.002500 mol/0.07500 L = 0.0333 M HCl] What is the pH of the solution? [H3O+]xs = 0.0333 M pH = -log(0.0333) = 1.478 Use this method to calculate the entire titration curve (or measure the curve with a pH meter) 39 13 13 Weak Acid - Strong Base Equivalence point pH is higher than for a strong acid strong base titration, and the range for rapid change is narrower. pH = pKa at halfhalfway point pH = pKa 40 Weak Acid - Strong Base CH3CO2H + NaOH CH3CO2- + H2O Must take into account the weak acid / conjugate base equilibrium. j g q At the equivalence point, all the acid is consumed but we have a solution of the conjugate base (which reacts with water to form an equilibrium with HA) 41 Weak Acid Strong Base 42 14 14 Weak Acid-Strong Base What are [CH3COOH] and [CH3COO- ] after 5.0mL of 0.100M NaOH is added to 20.0 mL of 0.100M acetic acid? CH3CO2H + NaOH CH3CO2- + H2O 43 What is the pH after 10.0 mL of 0.100 M NaOH is added to 20.0 mL of 0.100 M acetic acid? Ka of acetic acid is 1.8x10-5 1. 2. 3. 4. 5. 2.35 3.50 4.26 4.74 5.25 0% 1. 0% 2. 0% 3. 0% 4. 0% 5. 44 Weak Acid - Strong Base pH change in titration curve gets narrower as Ka gets smaller. smaller Equivalence point pH gets higher with weaker acids. 45 15 15 Strong Acid - Weak Base Equivalence point pH is lower than for a strong acid strong base titration, and rapid change is narrowed. NaOH 50.0 mL of 0.100 M NH3 titrated with 0.100 M HCl 46 Strong Acid - Weak Base pH at equivalence point is < 7 0.100 M HCl + 0.100 M NH3 gives pH = 5.27 Phenolphthalein doesn't work for this titration. Want an indicator that changes color in the range of pH 3-7 3 Methyl red is a good choice. 47 Weak Acid - Weak Base Titration curve shows major pH change in a very narrow range. Generally doesn't give a good titration. tit ti 48 16 16 17.4 Solubility Equilibria Some combinations of ions in solution form insoluble salts. Recall the solubility rules from Chapter 4. Solubility of insoluble salts is characterized by the solubility product constant, Ksp. aq) aq) AgCl(s) Ag+(aq) + Cl-(aq) AgCl(s) ][Cl Ksp = [Ag+][Cl-] = 1.70 10-10 Use to determine conditions for precipitation 49 Solubility Product Constant Ag2S(s) 2Ag+(aq) + S2-(aq) aq) aq) Ksp = [Ag+]2[S2-] = 1.0 10-51 How can we use Ksp values? Which is more soluble? Ksp(Ag2SO4) = [Ag+]2[SO42-] = 1.5 10-5 Ksp(Ag2S) = [Ag+]2[S2-] = 1.0 10-51 50 Solubility What if ion numbers are not the same? Use molar solubility of salt (x): AgCl: [Ag+] = x [Cl-] = x AgCl: [ g g Ksp = (x)(x) ( )( ) Ag2S: [Ag+] = 2x [S2-] = x Ksp = (2x)2(x) Fe(OH)3: [Fe3+] = x [OH-] = 3x Ksp = (x)(3x)3 We must compare x values in this case. 51 17 17 Solubility If Ksp is known, can solve for x: For Fe(OH)3: (x)(3x)3 = Ksp = 1.1 10-36 27x4 = 1.1 10-36 x4 = 4.07 10-38 4 07 x = 4.49 10-10 M [Fe3+] = 4.49 10-10 M [OH-] = 3 (4.49 10-10) = 1.35 10-9 If volume is 10.0 L, how much Fe(OH)3 dissolved? 52 Calculate the Ksp for an unknown compound, given the information below: Formula = XCl2 [Cl-] = 1 x 10-6 1. 1 2. 3. 4. 5. 5 x 10-13 1 x 10-18 1 x 10-6 5 x 10-12 5 x 10-19 1 2 3 4 5 20% 20% 20% 20% 20% 53 Group work 2x = 1.010-6 M 1.0 x = 5.010-7 M 5.0 5.0 Ksp = 4x3 = 5.01019 54 18 18 17.5 Factors that Affect Solubility Modify solubility to dissolve minerals and ores, to precipitate ions from solution, to separate and purify ions. Examples: p remove hardness from water by adding Na2CO3 remove Ag+ from water by adding Cl- to recover Ag metal dissolve Cu(OH)2.CuCO3 to mine Cu separate the halide ions 55 Common Ion Effect AgCl(s) Ag+(aq) + Cl-(aq) AgCl(s) aq) aq) Common ion effect: add more Cl- to precipitate more Ag+ from solution. Saturated solution of AgCl: AgCl: Ksp = 1.70 10-10 = [Ag+][Cl-] = x2 ][Cl x = [Ag+] = [Cl-] = 1.30 10-5 M [Cl If 0.100 M Cl- is added, what is the new [Ag+]? (>> 1.30 x 10-5, so assume [Cl-]= 0.100 M) [Cl Ksp = 1.70 10-10 = [Ag+](0.100) [Ag+] = 1.70 10-9 M 56 Insoluble Basic Salts Metal hydroxide solubility depends on the pH. aq) aq) Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) j p j y Adjust the pH to adjust the solubility. 126 Would you add NaOH or HCl increase the solubility of Mg(OH)2? 57 19 19 Insoluble Basic Salts AgCN(s) Ag+(aq) + CN-(aq) AgCN(s) aq) aq) Ksp = 1.6 x 10-14 Add HNO3 (why not HCl?) to dissolve AgCN. HCl?) AgCN. CN-(aq) + H3O+(aq) HCN(aq) + H2O(l) aq) aq) HCN(aq) K = 2.5 x 109 58 Adding HNO3 to which of the following compounds will NOT increase the solubility? BaF2 1. 1 2. 3. 4. 5. Mg(OH)2 AgCl 20% CaCO3 20% 20% Ag2S 20% 20% BaF2 Mg(OH)2 AgCl CaCO3 Ag2S 1 2 3 4 5 59 How will the solubility of CaCO3 change when the following compounds are added? HNO3 NaOH CaCl2 Na2CO3 20% CaCO3 20% 20% 20% 20% 1. 1 2. 3. 4. 5. +, + + +, -, -, NC +, -, +, -, +, -, NC, -, +, -, -, -, NC -, +, +, -, 1 2 3 4 5 60 20 20 Formation of Complex ions Lewis acid-base reactions can also reach a acidstate of equilibrium Metal ion + ligand complex ion Kf = [complex ion]/[metal ion][ligand] ion][ligand] (formation constant) aq) aq) aq) Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq) Kf = [Ag(NH3)2+ ]/[Ag+][NH3]2 = 1.7107 1.7 Kf values are usually >> 1 (see T 17.1) 61 Formation of Complex ions Other examples: Fe3+ + NCS- FeNCS2+ Cu2+ + 4NH3 Cu(NH3)42+ 62 Cu2+/CuNH32+/Cu(NH3)22+/Cu(NH3)32+/Cu(NH3)42+ 63 21 21 Solubility and Complex Ions Formation of complex ions can also increase solubility of a related salt. Can dissolve AgCl with NH3. AgCl(s) aq) aq) AgCl(s) Ag+(aq) + Cl-(aq) Ksp = [Ag+][Cl-] = 1.70 x 10-10 ][Cl Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq) aq) aq) aq) Kf = [Ag(NH3)2+]/[Ag+][NH3]2 = 1.7x107 Mixture of AgCl and AgBr can be separated: 6 M NH3 dissolves AgCl, but not AgBr. AgCl, AgBr. 64 22 22 ...
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This note was uploaded on 04/06/2011 for the course CHM 116 taught by Professor Unknown during the Spring '08 term at ASU.

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