pH of 0.1 M NaA solution is :
Given : `(K_b)_(A^-)=10^(-9)`

Options

A. 5
B. 11
C. 9
D. 8

Answer

To find the pH of a 0.1 M NaA solution where \( K_b(A^-) = 10^{-9} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Dissociation of NaA**:
- Sodium acetate (NaA) dissociates completely in water to give \( Na^+ \) and \( A^- \). The \( Na^+ \) ion is neutral and does not affect the pH.
- Thus, the concentration of \( A^- \) in solution is 0.1 M.

2. **Set Up the Hydrolysis Reaction**:
- The anion \( A^- \) can hydrolyze in water:
\[ A^- + H_2O \rightleftharpoons HA + OH^- \]
- This reaction shows that \( A^- \) acts as a base.

3. **Write the Expression for \( K_b \)**:
- The base dissociation constant \( K_b \) for the reaction is given by:
\[ K_b = \frac{[HA][OH^-]}{[A^-]} \]
- Since we start with 0.1 M \( A^- \) and let \( x \) be the concentration of \( OH^- \) produced, we can express the concentrations at equilibrium:
\[ K_b = \frac{x \cdot x}{0.1 - x} = \frac{x^2}{0.1 - x} \]

4. **Substitute the Known Value of \( K_b \)**:
- Given \( K_b = 10^{-9} \):
\[ 10^{-9} = \frac{x^2}{0.1 - x} \]

5. **Assume \( x \) is Small**:
- Since \( K_b \) is small, we can assume \( x \) is much smaller than 0.1, thus \( 0.1 - x \approx 0.1 \):
\[ 10^{-9} = \frac{x^2}{0.1} \]

6. **Solve for \( x \)**:
- Rearranging gives:
\[ x^2 = 10^{-9} \times 0.1 = 10^{-10} \]
- Taking the square root:
\[ x = 10^{-5} \text{ M} \]
- Here, \( x \) represents the concentration of \( OH^- \).

7. **Calculate \( pOH \)**:
- Using the concentration of \( OH^- \):
\[ pOH = -\log[OH^-] = -\log(10^{-5}) = 5 \]

8. **Calculate \( pH \)**:
- Using the relationship \( pH + pOH = 14 \):
\[ pH = 14 - pOH = 14 - 5 = 9 \]

### Final Answer:
The pH of the 0.1 M NaA solution is **9**.