If the Ka of a monoprotic weak acid is 7.2 × 10-6, what is the pH of a 0.22 M solution of this acid?

Denote the monoprotic acid as HA.

_________ HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq) …. Ka = 7.2 × 10⁻⁶
Initial ___ 0.22 M __________ 0 M _____ 0 M
Change ___ -y M __________ +y M ____ +y M
At eqm _ (0.22-y) M ________ y M _____ y M

Since Ka is very small, we can assume that y ≪ 0.22, i.e.
[HA] at eqm = (0.22-y) M ≈ 0.22 M

Ka = [H₃O⁺] [A⁻] / [HA]
y² / 0.22 = 7.2 × 10⁻⁶
y = 1.3 × 10⁻³ (This fulfills the assumption that y ≪ 0.22)

pH = -log[H₃O⁺] = -log(1.3 × 10⁻³) = 2.9

pH = 0.5(pKa - log(0.22)) = 2.9