A weak acid `HA` and a weak base `BOH` are having same value of dissociation constants . If `pH` of `0.01M HA` is 4, then `pH` of `0.01 M BOH` will be

Options

A. 3
B. 4
C. 10
D. 11

Explanation

To solve the problem, we need to find the pH of a weak base `BOH` given that the weak acid `HA` has a pH of 4 and both have the same dissociation constant.

### Step-by-Step Solution:

1. **Understanding the Weak Acid**:
- The weak acid `HA` dissociates in water as follows:
\[ HA \rightleftharpoons H^+ + A^- \]
- The dissociation constant \( K_a \) for the weak acid is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]

2. **Finding the Concentration of \( H^+ \)**:
- Given that the pH of the 0.01 M solution of `HA` is 4, we can find the concentration of \( H^+ \):
\[ pH = -\log[H^+] \]
\[ [H^+] = 10^{-pH} = 10^{-4} \, \text{M} \]

3. **Setting Up the Equilibrium Expression for the Weak Acid**:
- Let \( x \) be the concentration of \( H^+ \) at equilibrium:
\[ K_a = \frac{x^2}{0.01 - x} \]
- Since \( x \) is small compared to 0.01 M, we can approximate:
\[ K_a \approx \frac{x^2}{0.01} \]
- Substituting \( x = 10^{-4} \):
\[ K_a \approx \frac{(10^{-4})^2}{0.01} = \frac{10^{-8}}{0.01} = 10^{-6} \]

4. **Understanding the Weak Base**:
- The weak base `BOH` dissociates in water as follows:
\[ BOH \rightleftharpoons B^+ + OH^- \]
- The dissociation constant \( K_b \) for the weak base is given by:
\[ K_b = \frac{[B^+][OH^-]}{[BOH]} \]
- Since \( K_a = K_b \), we have \( K_b = 10^{-6} \).

5. **Finding the Concentration of \( OH^- \)**:
- For the weak base `BOH`, let \( y \) be the concentration of \( OH^- \) at equilibrium:
\[ K_b = \frac{y^2}{0.01 - y} \]
- Again, we can approximate:
\[ K_b \approx \frac{y^2}{0.01} \]
- Substituting \( K_b = 10^{-6} \):
\[ 10^{-6} \approx \frac{y^2}{0.01} \]
\[ y^2 = 10^{-6} \times 0.01 = 10^{-8} \]
\[ y = 10^{-4} \, \text{M} \]

6. **Calculating the pOH**:
- The concentration of \( OH^- \) is \( 10^{-4} \) M, thus:
\[ pOH = -\log[OH^-] = -\log(10^{-4}) = 4 \]

7. **Finding the pH of the Weak Base**:
- We know that:
\[ pH + pOH = 14 \]
- Therefore:
\[ pH = 14 - pOH = 14 - 4 = 10 \]

### Final Answer:
The pH of the 0.01 M solution of `BOH` is **10**.