Calculate the pH of a 0.20 M solution of KCN at 25.0 °C, given that the Kb for CN⁻ is 2.0 x 10⁻⁵.

• A. 2.70
• B. 12.70
• C. 7.00
• D. 11.30

Answer: (D)

Verified step by step guidance

1. Identify that KCN is a salt formed from the weak acid HCN and the strong base KOH. Therefore, the CN⁻ ion will undergo hydrolysis in water, affecting the pH.
2. Write the hydrolysis reaction for CN⁻: \( \text{CN}^- + \text{H}_2\text{O} \rightleftharpoons \text{HCN} + \text{OH}^- \).
3. Use the expression for the base dissociation constant \( K_b \) of CN⁻: \( K_b = \frac{[\text{HCN}][\text{OH}^-]}{[\text{CN}^-]} \). Given \( K_b = 2.0 \times 10^{-5} \).
4. Set up an ICE table (Initial, Change, Equilibrium) to determine the equilibrium concentrations of the species involved. Initially, \([\text{CN}^-] = 0.20 \text{ M}\), \([\text{HCN}] = 0\), \([\text{OH}^-] = 0\).
5. Solve for \([\text{OH}^-]\) using the equilibrium expression and the initial concentration of CN⁻. Then, calculate the pOH from \([\text{OH}^-]\) and use \( \text{pH} = 14 - \text{pOH} \) to find the pH of the solution.