Determine the pH of a 0.55 M NaCN solution. The Ka of hydrocyanic acid, HCN, is 4.9 x 10^-10.

• A. 3.47
• B. 11.53
• C. 4.78
• D. 7.00

Answer: (B)

Verified step by step guidance

1. Identify that NaCN is a salt formed from a weak acid (HCN) and a strong base (NaOH). Therefore, the solution will be basic due to the hydrolysis of CN⁻ ions.
2. Write the hydrolysis equation for CN⁻: \( \text{CN}^- + \text{H}_2\text{O} \rightleftharpoons \text{HCN} + \text{OH}^- \).
3. Use the relationship between Ka and Kb: \( K_w = K_a \times K_b \), where \( K_w = 1.0 \times 10^{-14} \). Calculate Kb for CN⁻ using \( K_b = \frac{K_w}{K_a} \).
4. Set up the expression for the equilibrium constant Kb: \( K_b = \frac{[\text{HCN}][\text{OH}^-]}{[\text{CN}^-]} \). Assume \([\text{OH}^-] = x\) and \([\text{HCN}] = x\), and \([\text{CN}^-] = 0.55 - x\).
5. Solve for \( x \) (\([\text{OH}^-]\)) using the Kb expression, then calculate pOH using \( \text{pOH} = -\log[\text{OH}^-] \). Finally, find pH using \( \text{pH} = 14 - \text{pOH} \).