Ph question acid dissociation

Question

The acid dissociation of CH3COOH is 1.74x 10-5mol L-1 use this to find the ph of 0.25mol L-1

Answer 1

CH₃COOH (aq) <==> H⁺ (aq) + CH₃COO^- (aq)

CH₃COOH is acetic acid, which is a weak monoprotic acid. It dissociates only partially in an aqueous solution to form hydrogen ions and acetate ions. As a result, the acid dissociation reaction is written as an equilibrium reaction. First, we need to setup an ICE table to determine the concentration of [H⁺] from the initial concentration of the acid (0.25 mol L^-1), and the K_a value (1.74 x 10^-5).

CH₃COOH (aq) <==> H⁺ (aq) + CH₃COO^- (aq)

I: 0.25 mol L^-1 0 0

C: -x +x +x

E: (0.25 - x) x x

Since K_a is a small # (~10^-5), we will assume that the value of x for the change in the equilibrium concentration of CH₃COOH is negligible.

K_a = [H⁺] [CH₃COO^-] / [CH₃COOH] = 1.74 x 10^-5

(x) (x) / 0.25 = 1.74 x 10^-5

x² = 1.74 x 10^-5 x 0.25

x² = 4.35 x 10^-6

x = √4.35 x 10^-6 = 2.1 x 10^-3 mol L^-1 (rounded to 2 significant figures)

x = [H⁺] = 2.1 x 10^-3 mol L^-1

pH = - log [H⁺] = - log 2.1 x 10^-3 = 2.68