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: Principles and Practice

3rd Edition

ISBN:9780534420123

Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Chapter18: Electrochemistry

Section: Chapter Questions

Problem 18.103QE

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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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