What is the pH of a 0.400 M solution of NaF, given that the Kb of F- is 1.4 x 10^-11?

• A. 6.90
• B. 7.00
• C. 8.10
• D. 5.60

Answer: (C)

Verified step by step guidance

1. Identify that NaF is a salt formed from a strong base (NaOH) and a weak acid (HF). Therefore, the solution will be basic due to the hydrolysis of F⁻ ions.
2. Write the hydrolysis reaction for F⁻: \( \text{F}^- + \text{H}_2\text{O} \rightleftharpoons \text{HF} + \text{OH}^- \).
3. Use the expression for the base dissociation constant \( K_b \) to find the concentration of OH⁻: \( K_b = \frac{[\text{HF}][\text{OH}^-]}{[\text{F}^-]} \).
4. Assume \([\text{OH}^-] = x\) and \([\text{HF}] = x\), and \([\text{F}^-] = 0.400 - x\). Substitute these into the \( K_b \) expression: \( 1.4 \times 10^{-11} = \frac{x^2}{0.400 - x} \).
5. Solve for \( x \) to find \([\text{OH}^-]\), then calculate the pOH using \( \text{pOH} = -\log[\text{OH}^-] \). Finally, find the pH using \( \text{pH} = 14 - \text{pOH} \).