Calculate the pH of a 0.095 M solution of HCN at 25 °C. [Ka for HCN (aq) = 4.9 x 10-10].?

___________HCN(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CN⁻ ___ Ka = 4.9 × 10⁻¹⁰

Initial: _____ 0.095 M __________ 0 M ____ 0 M

Change: _____ -y M __________ +y M ___ +y M

At eqm: __ (0.095 - y) M ________ y M ____ y M

As Ka is very small, assume that 0.095 ≫ y and thus

[HCN] at eqm = (0.095 - y) M ≈ 0.095 M

Ka = [H₃O⁺] [CN⁻] / [HCN]

y² / 0.095 = 4.9 × 10⁻¹⁰

y = √(0.095 × 4.9 × 10⁻¹⁰) = 6.82 × 10⁻⁶ (The assumption that 0.095 ≫ y is correct.)

pH = -log[H₃O]⁺ = -log(6.82 × 10⁻⁶) = 5.2