A solution of sodium acetate (NaCH3COO) has a pH of 9.64. The acid-dissociation constant (Ka) for acetic acid is 1.8 x 10^-5. What is the molarity of the sodium acetate solution?

• A. 0.10 M
• B. 0.25 M
• C. 1.00 M
• D. 0.50 M

Answer: (B)

Verified step by step guidance

1. Start by understanding that sodium acetate (NaCH3COO) is a salt derived from acetic acid (CH3COOH) and sodium hydroxide (NaOH). In solution, it can hydrolyze to form acetic acid and hydroxide ions, affecting the pH.
2. Use the given pH of the solution, 9.64, to find the concentration of hydroxide ions [OH^-]. The relationship between pH and pOH is given by: \( \text{pOH} = 14 - \text{pH} \). Calculate pOH and then find [OH^-] using \( [\text{OH}^-] = 10^{-\text{pOH}} \).
3. Recognize that the hydrolysis of sodium acetate in water can be represented by the equilibrium: \( \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^- \). The equilibrium constant for this reaction, Kb, can be found using the relation \( K_b = \frac{K_w}{K_a} \), where \( K_w = 1.0 \times 10^{-14} \) is the ion-product constant for water.
4. Set up an expression for Kb using the concentrations at equilibrium: \( K_b = \frac{[\text{CH}_3\text{COOH}][\text{OH}^-]}{[\text{CH}_3\text{COO}^-]} \). Assume that the initial concentration of sodium acetate is \( C \) and that the change in concentration due to hydrolysis is \( x \), where \( x = [\text{OH}^-] \).
5. Solve for \( C \) using the expression \( K_b = \frac{x^2}{C - x} \). Substitute the values for \( K_b \) and \( x \) obtained from previous steps, and solve for \( C \), which represents the molarity of the sodium acetate solution.