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

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

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

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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