Hydrocyanic acid, HCN, dissociates in water according to the chemical equation: HCN (aq) ⇆ H+ (aq) + CN- (aq)

Question

Calculate the pH of a 0.345 M solution of hydrocyanic acid. 

Answer 1

To solve this problem, we need to know the Ka for HCN. Looking it up, I find a value of Ka = 6.17x10^-10

HCN(aq) <==> H⁺(aq) + CN^-(aq)

Ka = [H⁺][CN^-] / [HCN]

6.17x10^-10 = (x)(x) / 0.345 - x and assuming x is small relative to 0.345 we can ignore it in the denominator

6.17x10^-10 = x² / 0.345

x² = 2.11x10^-10

x = 1.45x10^-5 M = [H⁺] (note: this is small relative to 0.345 so our above assumption was valid)

pH = -log [H⁺] = -log 1.45x10-5

pH = 4.84