Determine the pH of a solution made by dissolving 6.1 g of sodium cyanide, NaCN, in enough water to make a 500.0 mL of solution. (MW of NaCN = 49.01 g/mol). The K_a value of HCN is 4.9 × 10⁻¹⁰.

• A. 2.648
• B. 13.389
• C. 5.294
• D. 11.352

Answer: (D)

Verified step by step guidance

1. Calculate the number of moles of NaCN by dividing the mass of NaCN (6.1 g) by its molar mass (49.01 g/mol). Use the formula: \( \text{moles of NaCN} = \frac{6.1 \text{ g}}{49.01 \text{ g/mol}} \).
2. Determine the concentration of NaCN in the solution by dividing the number of moles of NaCN by the volume of the solution in liters (0.500 L). Use the formula: \( \text{[NaCN]} = \frac{\text{moles of NaCN}}{0.500 \text{ L}} \).
3. Recognize that NaCN is a salt that dissociates completely in water to form Na⁺ and CN⁻ ions. The CN⁻ ion is the conjugate base of the weak acid HCN, and it will undergo hydrolysis in water to form OH⁻ ions, increasing the pH.
4. Use the hydrolysis equation for CN⁻: \( \text{CN}^- + \text{H}_2\text{O} \rightleftharpoons \text{HCN} + \text{OH}^- \). The equilibrium expression for this reaction is \( K_b = \frac{[\text{HCN}][\text{OH}^-]}{[\text{CN}^-]} \). Calculate \( K_b \) using the relation \( K_w = K_a \times K_b \), where \( K_w = 1.0 \times 10^{-14} \) and \( K_a = 4.9 \times 10^{-10} \).
5. Set up an ICE table (Initial, Change, Equilibrium) for the hydrolysis reaction to find the concentration of OH⁻ at equilibrium. Use the expression for \( K_b \) to solve for \([\text{OH}^-]\), and then calculate the pOH of the solution. Finally, convert pOH to pH using the relation \( \text{pH} = 14 - \text{pOH} \).