AgOH solution was neutralized using `HNO_(3)` solution. If pH of the solution at 50% neutralization point is 8 at `25^(@)C`, then `pK_(b)` of AgOH is:

Options

A. `4.00`
B. `5.52`
C. `6.00`
D. `9.48`

Explanation

To solve the problem, we need to find the \( pK_b \) of \( AgOH \) given that the pH at the 50% neutralization point is 8.

### Step-by-Step Solution:

1. **Understanding the Neutralization Reaction**:
- \( AgOH \) dissociates into \( Ag^+ \) and \( OH^- \).
- \( HNO_3 \) is a strong acid and completely dissociates into \( H^+ \) and \( NO_3^- \).
- At the 50% neutralization point, half of the \( AgOH \) reacts with \( H^+ \) to form \( Ag^+ \) and \( H_2O \).

2. **Calculating \( [H^+] \) from pH**:
- Given that the pH is 8, we can calculate the concentration of \( H^+ \):
\[ [H^+] = 10^{-pH} = 10^{-8} \, \text{M} \]

3. **Calculating \( [OH^-] \)**:
- Using the relationship \( pH + pOH = 14 \):
\[ pOH = 14 - pH = 14 - 8 = 6 \]
- Therefore, the concentration of \( OH^- \) is:
\[ [OH^-] = 10^{-pOH} = 10^{-6} \, \text{M} \]

4. **Finding \( [Ag^+] \)**:
- At the 50% neutralization point, the concentration of \( Ag^+ \) will be equal to the concentration of \( OH^- \):
\[ [Ag^+] = [OH^-] = 10^{-6} \, \text{M} \]

5. **Calculating \( K_b \)**:
- The base dissociation constant \( K_b \) for \( AgOH \) can be expressed as:
\[ K_b = \frac{[Ag^+][OH^-]}{[AgOH]} \]
- Since \( AgOH \) is a solid, its concentration is considered to be 1. Thus:
\[ K_b = [Ag^+][OH^-] = (10^{-6})(10^{-6}) = 10^{-12} \]

6. **Finding \( pK_b \)**:
- Finally, we can calculate \( pK_b \):
\[ pK_b = -\log(K_b) = -\log(10^{-12}) = 12 \]

### Conclusion:
The \( pK_b \) of \( AgOH \) is **12**.