Ch15 - 15 Aqueous Equilibria Acids and Bases 15.1(a H 2 SO 4(aq H 2 O(l 5 H 3 O(aq HSO 4 S(aq conjugate base(b HSO 4 S(aq H 2 O(l 5 H 3 O(aq SO 4 2

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Unformatted text preview: 15 Aqueous Equilibria: Acids and Bases 15.1 (a) H 2 SO 4 (aq) + H 2 O(l) 5 H 3 O + (aq) + HSO 4 S (aq) conjugate base (b) HSO 4 S (aq) + H 2 O(l) 5 H 3 O + (aq) + SO 4 2 S (aq) conjugate base (c) H 3 O + (aq) + H 2 O(l) 5 H 3 O + (aq) + H 2 O(l) conjugate base (d) NH 4 + (aq) + H 2 O(l) 5 H 3 O + (aq) + NH 3 (aq) conjugate base 15.2 (a) HCO 3 S (aq) + H 2 O(l) 5 H 2 CO 3 (aq) + OH S (aq) conjugate acid (b) CO 3 2 S (aq) + H 2 O(l) 5 HCO 3 S (aq) + OH S (aq) conjugate acid (c) OH S (aq) + H 2 O(l) 5 H 2 O(l)(aq) + OH S (aq) conjugate acid (d) H 2 PO 4 S (aq) + H 2 O(l) 5 H 3 PO 4 (aq) + OH S (aq) conjugate acid 15.3 HCl(aq) + NH 3 (aq) 5 NH 4 + (aq) + Cl S (aq) acid base acid base conjugate acid-base pairs 15.4 (a) HF(aq) + NO 3 S (aq) 5 HNO 3 (aq) + F S (aq) HNO 3 is a stronger acid than HF, and F S is a stronger base than NO 3 S (see Table 15.1). Because proton transfer occurs from the stronger acid to the stronger base, the reaction proceeds from right to left. (b) NH 4 + (aq) + CO 3 2 S (aq) 5 HCO 3 S (aq) + NH 3 (aq) NH 4 + is a stronger acid than HCO 3 S , and CO 3 2 S is a stronger base than NH 3 (see Table 15.1). Because proton transfer occurs from the stronger acid to the stronger base, the reaction proceeds from left to right. 15.5 (a) Both HX and HY have the same initial concentration. HY is more dissociated than HX. Therefore, HY is the stronger acid. (b) The conjugate base (X S ) of the weaker acid (HX) is the stronger base. (c) HX + Y S 5 HY + X S ; Proton transfer occurs from the stronger acid to the stronger base. The reaction proceeds to the left. 15.6 [H 3 O + ] = _14 w _ _6 1.0 x 10 K = [ ] 5.0 x OH 10 = 2.0 x 10 S 9 M Because [OH S ] > [H 3 O + ], the solution is basic. 131 15.7 K w = [H 3 O + ][OH S ]; In a neutral solution, [H 3 O + ] = [OH S ] At 50 o C, [H 3 O + ] = [OH S ] = _14 w = 5.5 x 10 K = 2.3 x 10 S 7 M 15.8 (a) [H 3 O + ] = _14 w _ _6 1.0 x 10 K = [ ] 1.58 x OH 10 = 6.3 x 10 S 9 M pH = S log[H 3 O + ] = S log(6.3 x 10 S 9 ) = 8.20 (b) pH = S log[H 3 O + ] = S log(6.0 x 10 S 5 ) = 4.22 15.9 (a) [H 3 O + ] = 10 S pH = 10 S 7.40 = 4.0 x 10 S 8 M [OH S ] = _14 w + _8 3 1.0 x 10 K = [ ] 4.0 x O 10 H = 2.5 x 10 S 7 M (b) [H 3 O + ] = 10 S pH = 10 S 2.8 = 2 x 10 S 3 M [OH S ] = _14 w + _3 3 1.0 x 10 K = [ ] 2 x O 10 H = 5 x 10 S 12 M 15.10 (a) Because HClO 4 is a strong acid, [H 3 O + ] = 0.050 M. pH = S log[H 3 O + ] = S log(0.050) = 1.30 (b) Because HCl is a strong acid, [H 3 O + ] = 6.0 M. pH = S log[H 3 O + ] = S log(6.0) = S 0.78 (c) Because KOH is a strong base, [OH S ] = 4.0 M. [H 3 O + ] = _14 w _ 1.0 x 10 K = [ ] 4.0 OH = 2.5 x 10 S 15 M pH = S log[H 3 O + ] = S log(2.5 x 10 S 15 ) = 14.60 (d) Because Ba(OH) 2 is a strong base, [OH S ] = 2(0.010 M) = 0.020 M....
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This note was uploaded on 08/27/2008 for the course CHEM 1280 taught by Professor Schmidt during the Fall '07 term at Toledo.

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Ch15 - 15 Aqueous Equilibria Acids and Bases 15.1(a H 2 SO 4(aq H 2 O(l 5 H 3 O(aq HSO 4 S(aq conjugate base(b HSO 4 S(aq H 2 O(l 5 H 3 O(aq SO 4 2

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