Question looking for hydronium ion concentration?

Question

The [H3O+] concentration of a solution of 1.93 mol/L benzoic acid in water is 

a.bcdef? mol/L

Given Ka for Benzoic acid = 6.3 × 10^–5

Answer 1

Benzoic acid is a weak monoprotic acid. Its dissociation reaction in water can be written as:

C₆H₅COOH (aq) + H₂O (l) ⇔ H₃O⁺ (aq) + C₆H₅COO^- (aq)

K_a = [H₃O⁺] [C₆H₅COO^-] / [C₆H₅COOH] = 6.3 x 10^-5

Setup an ICE table for the above reaction.

[C₆H₅COOH]_i = 1.93 mol/L

[C₆H₅COOH]_eq = 1.93 - x ==> Assume that this change in 'x' is negligible!!

[H₃O⁺]_eq = x

[ C₆H₅COO^-]_eq = x

Substitute the above values in the K_a expression, you get

6.3 x 10^-5 = (x) (x) / 1.93

x² = 6.3 x 10^-5 x 1.93

x² = 1.22 x 10^-4

x = √1.22 x 10^-4 = 0.0110

[H₃O⁺] = 0.0110 mol/L