What is the pH of a 0.240 M aqueous solution of potassium cyanide?
K_a (HCN) = 6.2 × 10^-10

Question

What is the pH of a 0.240 M aqueous solution of potassium cyanide? K_a (HCN)= 6.2 × 10^-10 A)11.3 B)2.7 C)4.9 D)9.1 E)9.4

Quizplus · Accepted Answer

Potassium cyanide (KCN) dissociates in water to form K+ and CN-. The CN- ion then reacts with water to form HCN and OH-. Given the Ka of HCN is 6.2 × 10^-10, we can find the Kb of CN- by using the relation Kw = Ka × Kb, where Kw (the ion-product constant for water) is 1.0 × 10^-14 at 25°C. Thus, Kb = Kw / Ka = (1.0 × 10^-14) / (6.2 × 10^-10) = 1.61 × 10^-5. Using the formula for Kb, [OH-] = sqrt(Kb × [CN-]) = sqrt(1.61 × 10^-5 × 0.240), we find [OH-] and then calculate pOH. pH is then found by subtracting pOH from 14. The calculation yields a pH close to 11.3.

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