Question

What will be the value of pH of 0.01 mol dm⁻³
CH₃COOH (Kₐ = 1.74 X10⁻⁵ )?
(a) 3. 4 (b) 3. 6
(c) 3. 9 (d) 3. 0

Answer 1

Explanation:

(a)3.4 is the answer of this value

Answer 2

(a) 3. 4

Explanation:

CH3COOH + H2O ⇌ H3O+ + CH3COO-

Conc. Initially:   0.01       0    0

At equilibrium : 0.01-x    x    x

Ka = [H3O+] [CH3COO-] / [CH3COOH] = x² / 0.01-x

Since x < 0.01, therefore 0.01-x = 0 = 0.01

x²/0.01 = 1.74 * 10^-5

x² = 1.74 * 10^-7

x = 4.2 * 10^-4

pH = -log(4.2 * 10^-4) = 3.4

Option A is the answer.