Question

Ka for butyric acid (C₃H₇COOH) is 2 × 10^–5. The pH of 0.2 M solution of butyric acid is ______× 10^–1. (Nearest integer) (Given log2 = 0.30)

Answer 1

Correct Answer: 27

To find the pH of a 0.2 M solution of butyric acid, we start by considering the dissociation equilibrium: C₃H₇COOH ⇌ C₃H₇COO^– + H⁺. The acid dissociation constant (K_a) is given as 2 × 10^–5. The expression for K_a is:

K_a = [H⁺][C₃H₇COO^–] / [C₃H₇COOH]

Assume the initial concentration of butyric acid is 0.2 M, and let x be the concentration of ions at equilibrium. Then:

[H⁺] = x, [C₃H₇COO^–] = x, [C₃H₇COOH] ≈ 0.2 – x ≈ 0.2

Substitute into the K_a expression:

2 × 10^–5 = x² / 0.2

x² = 2 × 10^–5 × 0.2

x² = 4 × 10^–6

Take the square root:

x = √(4 × 10^–6)

x = 2 × 10^–3

The concentration of H⁺ ions is 2 × 10^–3 M. Calculate the pH:

pH = –log [H⁺]

pH = –log(2 × 10^–3)

pH = –(log 2 + log 10^–3)

pH = –(0.30 – 3)

pH = 3 – 0.30

pH = 2.70

Given the question asks for the pH expressed as a multiple of 10^–1, the final value is 27 × 10^–1. The solution 27 falls within the given range [27, 27].