ch16 - 16 Applications of Aqueous Equilibria 16.1 (a) HNO 2...

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Unformatted text preview: 16 Applications of Aqueous Equilibria 16.1 (a) HNO 2 (aq) + OH S (aq) 6 NO 2 S (aq) + H 2 O(l); NO 2 S (basic anion), pH > 7.00 (b) H 3 O + (aq) + NH 3 (aq) 6 NH 4 + (aq) + H 2 O(l); NH 4 + (acidic cation), pH < 7.00 (c) OH S (aq) + H 3 O + (aq) 6 2 H 2 O(l); pH = 7.00 16.2 (a) HF(aq) + OH S (aq) 6 H 2 O(l) + F S (aq) K n = _ 4 a _14 w 3.5 x 10 K = 1.0 x 10 K = 3.5 x 10 10 (b) H 3 O + (aq) + OH S (aq) 6 2 H 2 O(l) K n = _14 w 1 1 = 1.0 x 10 K = 1.0 x 10 14 (c) HF(aq) + NH 3 (aq) 6 NH 4 + (aq) + F S (aq) K n = _ 4 _5 a b _14 w (3.5 x )(1.8 x ) 10 10 K K = 1.0 x 10 K = 6.3 x 10 5 The tendency to proceed to completion is determined by the magnitude of K n . The larger the value of K n , the further does the reaction proceed to completion. The tendency to proceed to completion is: reaction (c) < reaction (a) < reaction (b) 16.3 HCN(aq) + H 2 O(l) 6 H 3 O + (aq) + CN S (aq) initial (M) 0.025 ~0 0.010 change (M) S x +x +x equil (M) 0.025 S x x 0.010 + x K a = + _ 3 _10 [ ][ ] x(0.010 + x) x(0.010) O CN H = 4.9 x = 10 [HCN] 0.025 _ x 0.025 6 Solve for x. x = 1.23 x 10 S 9 M = 1.2 x 10 S 9 M = [H 3 O + ] pH = S log[H 3 O + ] = S log(1.23 x 10 S 9 ) = 8.91 [OH S ] = _14 w + _9 3 1.0 x 10 K = [ ] 1.23 x O 10 H = 8.2 x 10 S 6 M [Na + ] = [CN S ] = 0.010 M; [HCN] = 0.025 M % dissociation = _9 diss initial [HCN] 1.23 x M 10 x 100% = x 100% [HCN 0.025 M ] = 4.9 x 10 S 6 % 16.4 From NH 4 Cl(s), [NH 4 + ] initial = 0.10 mol 0.500 L = 0.20 M NH 3 (aq) + H 2 O(l) 6 NH 4 + (aq) + OH S (aq) initial (M) 0.40 0.20 ~0 177 change (M) S x +x +x equil (M) 0.40 S x 0.20 + x x K b = + _ 4 _5 3 [ ][ ] (0.20 + x)(x) (0.20)(x) NH OH = 1.8 x = 10 [ ] (0.40 _ x) (0.40) NH 6 Solve for x. x = [OH S ] = 3.6 x 10 S 5 M [H 3 O + ] = _14 w _ _5 1.0 x 10 K = [ ] 3.6 x OH 10 = 2.8 x 10 S 10 M pH = S log[H 3 O + ] = S log(2.8 x 10 S 10 ) = 9.55 16.5 Each solution contains the same number of B molecules. The presence of BH + from BHCl lowers the percent dissociation of B. Solution (2) contains no BH + , therefore it has the largest percent dissociation. BH + is the conjugate acid of B. Solution (1) has the largest amount of BH + and it would be the most acidic solution and have the lowest pH. 16.6 (a) (1) and (3). Both pictures show equal concentrations of HA and A S . (b) (3). It contains a higher concentration of HA and A S . 16.7 HF(aq) + H 2 O(l) 6 H 3 O + (aq) + F S (aq) initial (M) 0.25 ~0 0.50 change (M) S x +x +x equil (M) 0.25 S x x 0.50 + x K a = + _ 3 _ 4 [ ][ ] x(0.50 + x) x(0.50) O H F = 3.5 x = 10 [HF] 0.25 _ x 0.25 6 Solve for x. x = 1.75 x 10 S 4 M = [H 3 O + ] For the buffer, pH = S log[H 3 O + ] = S log(1.75 x 10 S 4 ) = 3.76 (a) mol HF = 0.025 mol; mol F S = 0.050 mol; vol = 0.100 L 100% F S (aq) + H 3 O + (aq) 6 HF(aq) + H 2 O(l) before (mol) 0.050 0.002 0.025 change (mol) S 0.002 S 0.002 +0.002 after (mol) 0.048 0 0.027 [H 3 O + ] = _ 4 a _ [HF] 0.27 = (3.5 x ) 10 K [ ] 0.48 F = 1.97 x 10 S 4 M pH = S log[H 3 O + ] = S log(1.97 x 10 S 4 ) = 3.71 (b) mol HF = 0.025 mol; mol Fmol HF = 0....
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ch16 - 16 Applications of Aqueous Equilibria 16.1 (a) HNO 2...

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