Chapter 17 Answers. Practice Examples [H3O ] 0.018M, 1a. HF = M. 1b. 30 drops. 2a.

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1 Chapter 17 Answers Practice Examples 1a. + [HO ] 0.018M, 1b. 0 drops [HF] = 0.8 M. [H O + ] = 0.10 M, HF = 0.97 M. a. + HO 1.10 M, CHO = M. b. 15g NaCHO a. The hydronium ion and the acetate ion react to form acetic acid. All that is necessary to form a buffer is to have approximately equal amounts of a weak acid and its conjugate base together in solution. This will be achieved if we add an amount of HCl equal to approximately half the original amount of acetate ion. b. HCl dissociates essentially completely in water and serves as a source of hydronium ion. This reacts with ammonia to form ammonium ion. Because a buffer contains approximately equal amounts of a weak base (NH ) and its conjugate acid (NH + ), to prepare a buffer we simply add an amount of HCl equal to approximately half the amount of NH (aq) initially present. a..9 b..51 5a. 1 g 5b. ph = a. (a).5; (b).5; (c).55. 6b. 1. ml 7a. (a) 0.8; (b) 1.8; (c) 7.00; (d) b. (a) 1.1; (b) 11.79; (c) a. (a).0; (b).70; (c).18; (d) b. (a) 11.15; (b) 9.7; (c) 9.6; (d) a b Integrative Examples A. 0.8 C Copyright 011 Pearson Canada Inc. 1

2 B. 80 g/mol, K b = Exercises 1a M 1 1b M 1c M 1d M a b. No change. The ph changes are not the same because there is an equilibrium system to be shifted in the first solution, whereas there is no equilibrium, just a change in total ionic strength, for the second solution. 5a. 0.05M 5b. 5c M M M 9a..6 9b a. Not a buffer. 11b. Not a buffer. 11c. Buffer 11d. Not a buffer. 11e. Buffer 11f. Not a buffer a..8 g NH SO 15b g NH SO Copyright 011 Pearson Canada Inc.

3 19. Use HCHO and NaCHO to prepare a buffer with ph =.50. V acid =100. ml, V salt =58 ml. 1a. ph =.89 to ph = b mol of acid or base per liter of buffer solution. a..8 b..5 c. 1. [NH ] M 5a. ph = pka log 9.6log [NH ] 0.18M 5b. The Henderson-Hasselbalch equation depends on the assumption that: 5 NH M NH. If the solution is diluted to 1.00 L, NH M =7.010, and NH = M. These concentrations are consistent with the assumption. However, 6 5 if the solution is diluted to L, NH =7.10 M, and NH =1.810 M, and these two concentrations are not consistent with the assumption. Thus, in L of solution, the given quantities of NH and NH will not produce a solution with ph = With sufficient dilution, the solution will become indistinguishable from pure water (i.e.; its ph will equal 7.00). 5c d. 1.1mL 1.00M HCl 7a. Bromphenol blue changes color in acidic solution. Bromcresol green changes color in acidic solution. Bromthymol changes color in neutral solution.,-dinitrophenol changes color in acidic solution. Chlorophenol red changes color in acidic solution. Thymolphthalein changes color in basic solution. 7b. If bromcresol green is green, the ph is probably about ph =.7. If chlorophenol red is orange, the ph is probably about ph = a. In an Acid Base titration, the ph of the solution changes sharply at a definite ph that is known prior to titration. Determining the ph of a solution is more difficult because the ph of the solution is not known precisely in advance. Since each indicator only serves to fix the ph over a quite small region, often less than.0 ph units, several indicators carefully chosen to span the entire range of 1 ph units must be employed to narrow the ph to 1 ph unit or possibly lower. 9b. An indicator is, after all, a weak acid. Its addition to a solution will affect the acidity of that solution. Thus, one adds only enough indicator to show a color change and not enough to affect solution acidity. Copyright 011 Pearson Canada Inc.

4 1a. Red 1b. Yellow 1c. Yellow 1d. Yellow 1e. Red 1f. Yellow. The approximate pk HIn =.8. This is a relatively good indicator M a b..0 1a..87 1b..51. The volume of acid and its molarity are the same. Thus, the amount of acid is the same in each case. The volume of titrant needed to reach the equivalence point will also be the same in both cases, since the titrant has the same concentration in each case, and it is the same amount of base that reacts with a given amount (in moles) of acid. At the equivalence point in the titration of a strong acid with a strong base, the solution contains ions that do not hydrolyze. But the equivalence point solution of the titration of a weak acid with a strong base contains the anion of a weak acid, which will hydrolyze to produce a basic (alkaline) solution. 5a. Copyright 011 Pearson Canada Inc.

5 5b. 7a. 11.9mL acid 7b. 17.mL acid 7c. 17.5mL acid ph Volume of 0.10 M NaOH added (ml) 9a. bromthymol blue, 9b. See sketch in 9a. Thymol blue. 9c. See sketch in 9a. Alizarin yellow Basic. This alkalinity is created by the hydrolysis of the sulfide ion, the anion of a very weak acid. 55a. H PO aq +CO aq H PO aq +HCO aq H PO aq +CO aq HPO aq +HCO aq HPO aq + OH aq PO aq + H O(l) Copyright 011 Pearson Canada Inc. 5

6 55b. CO is not a strong enough base to remove the third proton from HPO. 57. K a = , K 1 a = a M 59b. 0.9 M 61a. No 61b. No 61c. Yes 61d. No 61e. Yes 61f. No Integrative and Advanced Exercises 6a. HSO (aq) OH (aq) SO (aq) H O(l) 6b..8% 6c. bromthymol blue or phenol red. 6a. 7.8 ml 6b.. ml 6c. 50 ml 66a. No 66b. 18% 69a. ph = pk a + log (f /(1 f )) 69b a b. 6. g KHPO, 6 g Na HPO 1HO. 71. ~ drops NH 7a. Equation (1): K = Equation (): K = Copyright 011 Pearson Canada Inc. 6

7 7b. The extremely large size of each equilibrium constant indicates that each reaction goes essentially to completion. 77a. 77b. 77c. 119 ml 77d. 189 ml 77e. 8.% 81a. A buffer solution is able to react with small amounts of added acid or base. When strong acid is added, it reacts with formate ion. CHO (aq) H O (aq) HCHO (aq) H O Added strong base reacts with acetic acid. HC H O (aq) OH (aq) H O(l) C H O (aq) Therefore neither added strong acid nor added strong base alters the ph of the solution very much. Mixtures of this type are referred to as buffer solutions. Copyright 011 Pearson Canada Inc. 7

8 81b..0 81c..8 8a. Bromthymol blue or phenol red. 8b. The H PO /HPO - system. 8c a. K CO g Ca 1 HO KKK 89b a b c. Copyright 011 Pearson Canada Inc. 8

9 Feature Problems 9a. The two curves cross the point at which half of the total acetate is present as acetic acid and half is present as acetate ion. This is the half equivalence point in a titration, where ph = K =.7. p a 9b. f (H PO ) f (H PO - ) f (HPO - ) f (PO - ) 9c. Copyright 011 Pearson Canada Inc. 9

10 Self-Assessment Exercises 98a. HCHO + OH CHO + H O, CHO + H O + HCHO + H 98b. C 6 H 5 NH + + OH C 6 H 5 NH + H O, C 6 H 5 NH + H O + C 6 H 5 NH + + H O 98c. H PO + OH HPO - + H O, HPO - + H O + H PO + H O 99a. The ph at the equivalence point is 7. Use bromthymol blue. ph Vol of Titrant (ml) 99b. The ph at the equivalence point is ~5. for a 0.1 M solution. Use methyl red. ph Vol of Titrant (ml) Copyright 011 Pearson Canada Inc. 10

11 99c. The ph at the equivalence point is ~8.7 for a 0.1 M solution. Use phenolphthalein. ph Vol of Titrant (ml) 99d. The ph for the first equivalence point (NaH PO to Na HPO ) for a 0.1 M solution is right around ~7. Use bromthymol blue. ph a b c d The answer is (c). 10. The answer is (d). 10. The answer is (b). Vol of Titrant (ml) Copyright 011 Pearson Canada Inc. 11

12 10. The answer is (b) The answer is (a) The answer is (b) The answer is (b). 110a. ph > 7 110b. ph < 7 110c. ph = 7 Copyright 011 Pearson Canada Inc. 1