Calculate the pH of a solution prepared by dissolving 0.482 mol of benzoic acid and 0.104 mol of sodium benzoate in water sufficient to yield 1.00 L of solution. The K, of benzoic acid is 6.30 × 105. CH;COOH(aq) + H2O(1) = H3O*(aq) + CóH5CO2 (aq) K, = 6.30 × %3D 105 [A'] pH = pK, + log- [HA] %3D

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Calculate the pH of a solution prepared by dissolving 0.482 mol of
benzoic acid and 0.104 mol of sodium benzoate in water sufficient to
yield 1.00 L of solution. The K, of benzoic acid is 6.30 x 105.
CHSCOOH(aq) + H2O(1) = H3O*(aq) + C6H5CO2 (aq)
K = 6.30 ×
10-5
A
pH = pK, + log;
[HA]
[* ]
%3D
Transcribed Image Text:Calculate the pH of a solution prepared by dissolving 0.482 mol of benzoic acid and 0.104 mol of sodium benzoate in water sufficient to yield 1.00 L of solution. The K, of benzoic acid is 6.30 x 105. CHSCOOH(aq) + H2O(1) = H3O*(aq) + C6H5CO2 (aq) K = 6.30 × 10-5 A pH = pK, + log; [HA] [* ] %3D
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