Question

Find the pH of a 0.130 M solution of a weak monoprotic acid having Ka= 1.1×10⁻³.

Express your answer to two decimal places.

Answer 1

Dissociation equilibrium of monoprotic acid is

HA(aq) + H2O(l ) <------> H3O⁺(aq) + A^-(aq)

Ka = [H3O⁺][A^-]/[HA] = 1.1 ×10^-3

at equilibrium

[HA] = 0.130 - x

[A^-] = x

[H3O⁺] = x

so,

x²/( 0.130 - x) = 1.1 ×10^-3

solving for x

x = 0.01142

[H3O⁺] = 0.01142M

pH = -log[H3O⁺]

pH = -log(0.01142M)

pH = 1.94