Chap 16 Answer To Practice Exercises

Chap 16 Answer To Practice Exercises - Chem 210 / WANG /...

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Unformatted text preview: Chem 210 / WANG / Chapter 16B Acid Base Equilibria (2) SAMPLE PROBLEM 2. SAMPLE PROBLEM 2(c). SAMPLE PROBLEM 3. (a) Write an equation to show how SO2 dissolves in water to form an acid. Find the acid dissociation constants in the appendix for this acid. (c) Find [SO3-2] at equilibrium for the solution in (b). Write the dissociation equation and the dissociation constant for each step of the dissociation of the polyprotic acid, phosphoric acid. SO2 + H2O H2SO3 H2SO3 + H2O HSO3-1 + H3O+1 HSO HSO3-1 + H2O SO3-2 + H3O+1 SO x10-2 Ka1 = 1.4 Ka2 = 6.5 x10-8 (b) Predict the pH of the solution prepared by dissolving 0.500 mole of SO2 in 320.0 mL of water. 0.500 mole /0.3200 L = 1.5625 = 1.563 M H2SO3 + H2O HSO HSO3-1 + H3O+1 Ka1 = 1.4 x10-2 1.563 0 0 -x x x Ka1 = x = 0.141 or -0.155 pH = -log (0.141) = 0.851 0.85 0.85 SAMPLE PROBLEM 3. Ka1 > Ka2 > Ka3 Polyprotic acids acids with more than more ionizable proton H3PO4(aq) + H2O(l) H2PO4-(aq) + H3O+(aq) Ka1 = H2PO4-(aq) + H2O(l) [H3O+][H2PO4-] [H3PO4] = 7.2x10-3 HSO HSO3-1 + H3O+1 Ka1 = 1.4 x10-2 0 0 x x Ka1 = x2/(1.563-x) /(1.563x = 0.141 M = [HSO3-1] H2SO3 + H2O 1.563 -x HSO3-1 + H2O 0.141 -y 0.141-y SO3-2 + SO 0 y y 6.5 Ka2 = 6.5 x10-8 = y = 6.5 × 6.5 H3O+1 Ka2 = 6.5 x10-8 0.141 y_______ 0.141+y y <<< 0.141 !! 10-8 M = [SO3-2] [SO SAMPLE PROBLEM 4. SAMPLE PROBLEM 4. Phenylacetic acid (C6H5CH2COOH, simplified here as HPAc) builds up in the blood of persons with phenylketonuria, an inherited disorder that, if untreated, causes mental retardation and death. A study of the acid shows that the pH of 0.13M HPAc is 2.62. What is the Ka of phenylacetic acid? Phenylacetic acid (C6H5CH2COOH, simplified here as HPAc) builds up in the blood of persons with phenylketonuria, an inherited disorder that, if untreated, causes mental retardation and death. A study of the acid shows that the pH of 0.13M HPAc is 2.62. What is the Ka of phenylacetic acid? Assumptions: At pH of 2.62, the [H3O+]HPAc >> [H3O+]water. [PAc [PAc-] ≈ [H3O+]; since HPAc is weak, [HPAc]initial ≈ [HPAc]initial - [HPAc]dissociation HPO HPO42-(aq) + H3O+(aq) + 2Ka2 =[H3O ][HPO4 ] = 6.3x10-8 [H2PO4-] HPO42-(aq) + H2O(l) Ka3 = SAMPLE PROBLEM 4. Initial Change Equilibrium + H3O+(aq) [H3O+][PO43-] [HPO42-] Ka = = 4.2x10-13 SAMPLE PROBLEM 5. continued HPAc( HPAc(aq) + 0.12 -x 0.120.12-x H2O(l) - H3O+(aq) + PAc-(aq) 1x10-7 0 +x +x x +(<1x10-7) +(<1x10 x [H3O+] = 10-pH = 2.4x10-3 M which is >> 10-7 ([H3O+] of water) which [H3O+] ≈ x = 2.4x10-3 M = [PAc-] [PAc [HPAc]equilibrium = 0.12-x ≈ 0.12 M HPAc] 0.12So Ka = (2.4x10-3) (2.4x10-3) 0.12 Be sure to check for % error. HPAc( HPAc(aq) + H2O(l) PO43-(aq) PO = 4.8 x 10-5 -7 [H3O+]from water; 1x10 M3 x100 = 4x10-3 % 2.4x10- M [HPAc]dissn; HPAc] 2.4x10-3M x100 = 2.0 % 0.12M Fall 2009 / UM-SJTU JI Propanoic acid (CH3CH2COOH, which we simplify as HPr) is an organic acid, the salts of which are used to retard mold growth in foods. What is the [H3O+] of 0.10M HPr (Given: Ka = 1.3x10-5)? H3O+(aq) + PAc-(aq) O+][PAc-][PAc [H3 [HPAc] HPAc] SAMPLE PROBLEM 5. Propanoic acid (CH3CH2COOH, which we simplify as HPr) is an organic acid, the salts of which are used to retard mold growth in foods. What is the [H3O+] of 0.10M HPr (Given: Ka = 1.3x10-5)? HPr( HPr(aq) + H2O(l) H3O+(aq) + Pr-(aq) Initial 0.10 0 0 +x Change -x +x Equilibrium 0.10-x 0.10x x Ka = [H3O+][Pr-] = 1.3x10-5 = (x)(x) ~ (x)(x) 0.100.10-x 0.10 [HPr] HPr] Since Ka is small, we will assume that x << 0.10 << [H x2 = 1.3x10- 5 x 0.1 x = ( 0 . 10 )( 1 . 3 x 10 − 5 ) = 1.1x10-3 M = [H3O+] 0.1 Check for % error: Can we ignore the x in (0.10-x)? in 0.10x ÷ 0.10 = (1.1x10-3M ÷ 0.10 M) x 100% = 1.1% 1.1% error negligible negligible error! Page 1 Chem 210 / WANG / Chapter 16B Acid Base Equilibria (2) SAMPLE PROBLEM 6. Ascorbic acid (H2C6H6O6; H2Asc for this problem), known as vitamin C, is a diprotic acid (Ka1 = 1.0x10-5 and Ka2 = 5x10-12) found in citrus fruit. Calculate [H2Asc], [HAsc-], [HAsc [Asc2-], and the pH of 0.050M H2Asc. SAMPLE PROBLEM 6. continued Ascorbic acid (H2C6H6O6; H2Asc for this problem), known as vitamin C, is a diprotic acid (Ka1 = 1.0x10-5 and Ka2 = 5x10-12) found in citrus fruit. Calculate [H2Asc], [HAsc-], [HAsc [Asc2-], and the pH of 0.050M H2Asc. HAsc-(aq) + H2O(l) Asc2-(aq) + H3O+(aq) H2Asc(aq) + H2O(l) HAsc-(aq) + H3O+(aq) Ka1 = [HAsc-][H3O+] [H2Asc] x= SAMPLE PROBLEM 7. SAMPLE PROBLEM 7. Sodium acetate (CH3COONa, or NaAc for this problem) has applications in photographic development and textile dyeing. What is the pH of 0.25M NaAc? Ka of acetic acid (HAc) is 1.8x10-5. Sodium acetate (CH3COONa, or NaAc for this problem) has applications in photographic development and textile dyeing. What is the pH of 0.25M NaAc? Ka of acetic acid (HAc) is 1.8x10-5. Equilibrium 0.25M-x 0 +x Kw Ka Fall 2009 / UM-SJTU JI 0 +x x - - x x x = 7.1x10-4 M pH = -log(7.1x10-4) = 3.15 HAsc-(aq) + H2O(l) 7.1x10-4M -x Change -4 - x Equilibrium 7.1x10 −5 − 12 x = (7.4x10 )(5x10 ) Kb = x Asc2-(aq) + H3O+(aq) 0 0 +x +x x x -8 M = 6x10 [Ac-] = 0.25M-x ≈ 0.25M [HAc][OH-] [Ac-] HAc(aq) + OH-(aq) [HAc][OH-] = [Ac-] 1.0x10-14 Kb = = 5.6x10-10M 1.8x10-5 Kb = (0.050)(1.0 x10−5 ) continued Ac-(aq) + H2O(l) - Initial = 5x10-12 Ka1 >> Ka2 so the first dissociation produces virtually all of the H3O+. Initial 0.25M -x Change 0.050 -x Equilibrium 0.050 - x Ka1 = [HAsc-][H3O+]/[H2Asc] = 1.0x10-5 = (x)(x)/0.050 M [Asc2-][H3O+] Ka2 = [HAsc-] = 1.0x10-5 HAsc-(aq) + H3O+(aq) 0 0 +x +x H2Asc(aq) + H2O(l) Initial Change 5.6x10-10 = x2/0.25M x = 1.2x10-5M = [OH-] Check 1.2x10-5M/0.25M x 100 = 4.8x10-3 % [H3O+] = Kw/[OH-] = 1.0x10-14/1.2x10-5 = 8.3x10-10M pH = -log 8.3x10-10M = 9.08 9.08 Page 2 ...
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This note was uploaded on 07/30/2011 for the course CHEM 210 taught by Professor Zhang during the Spring '09 term at Shanghai Jiao Tong University.

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