Question

NH3 is a weak base (Kb = 1.8 × 10–5) and so the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.058 M in NH4Cl at 25 °C?

Answer 1

NH4+ + H2O -----> NH3 + H3O+

0.058                     0         0

(0.058-x)                x         x

Kb = Kw/Ka = (1 * 10^(-14) / 1.8 * 10^(-5)) = 5.55 * 10^(-10)

assuming (0.058-x) is approximately equal to 0.058

pH = -log[H+] = -log(5.673 * 10^(-6)) = 6 - log(5.673) = 5.246