If the equilibrium constant for the reaction of weak acid HA with strong base is `10^(9)`, then pH of `0.1M` Na A is:

Options

A. 5
B. 9
C. 7
D. 8

Explanation

To find the pH of a 0.1 M NaA solution, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The weak acid (HA) reacts with a strong base (NaOH) to form the salt (NaA) and water. The relevant equilibrium constant (K) for the reaction is given as \( K = 10^9 \).

\[ HA + NaOH \rightleftharpoons NaA + H_2O \]

2. **Understanding the Relationship**:
The equilibrium constant for the dissociation of the weak acid HA can be expressed as:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
Since NaA completely dissociates in solution, we can say that:
\[ [A^-] = 0.1 \, M \quad \text{(from NaA)} \]

3. **Calculate \( K_a \)**:
The relationship between \( K \) and \( K_a \) is given by:
\[ K = \frac{K_w}{K_a} \]
where \( K_w \) (the ion product of water) is \( 10^{-14} \) at 25°C. Thus,
\[ K_a = \frac{K_w}{K} = \frac{10^{-14}}{10^9} = 10^{-23} \]

4. **Set Up the Equilibrium Expression**:
For the dissociation of the weak acid HA:
\[ HA \rightleftharpoons H^+ + A^- \]
The equilibrium expression becomes:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]

5. **Assume Initial Concentrations**:
Let \( x \) be the concentration of \( H^+ \) produced from the dissociation of HA. At equilibrium:
- \([H^+] = x\)
- \([A^-] = 0.1 + x \approx 0.1\) (since \( x \) is small)
- \([HA] = C - x \approx C\) (where \( C \) is the initial concentration of HA)

6. **Substituting into the Equilibrium Expression**:
\[ K_a = \frac{x \cdot 0.1}{C} \]
Rearranging gives:
\[ x = \frac{K_a \cdot C}{0.1} \]

7. **Calculate \( x \)**:
Substitute \( K_a = 10^{-23} \) and \( C = 0.1 \):
\[ x = \frac{10^{-23} \cdot 0.1}{0.1} = 10^{-23} \]

8. **Calculate pOH**:
Since \( [OH^-] = x \):
\[ pOH = -\log(10^{-23}) = 23 \]

9. **Calculate pH**:
Using the relationship \( pH + pOH = 14 \):
\[ pH = 14 - pOH = 14 - 23 = -9 \]

### Conclusion:
The calculated pH of the solution is \( -9 \), which indicates a very basic solution due to the presence of the strong base.