# Chapter 14 - Acids and Bases - Questions for Review and Thought - Topical Questions - Page 652c: 59a

$pH = 2.611$

#### Work Step by Step

1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [C_6H_5COO^-] = x$ -$[C_6H_5COOH] = [C_6H_5COOH]_{initial} - x = 0.05 - x$ For approximation, we consider: $[C_6H_5COOH] = 0.05M$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][C_6H_5COO^-]}{ [C_6H_5COOH]}$ $Ka = 1.2 \times 10^{- 4}= \frac{x * x}{ 0.05}$ $Ka = 1.2 \times 10^{- 4}= \frac{x^2}{ 0.05}$ $6 \times 10^{- 6} = x^2$ $x = 2.449 \times 10^{- 3}$ Therefore: $[H_3O^+] = [C_6H_5COO^-] = x = 2.449 \times 10^{- 3}M$ 3. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 2.449 \times 10^{- 3})$ $pH = 2.611$

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