The `p^(H)` of the solution of weak base at neutralisation with strong acid is 8. Kb for the base is

Answer

To find the \( K_b \) for the weak base when the pH of the solution at neutralization with a strong acid is given as 8, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the pH**:
The pH of the solution is given as 8.

2. **Calculate pOH**:
We can find the pOH using the relationship:
\[ pOH = 14 - pH \]
Substituting the given pH:
\[ pOH = 14 - 8 = 6 \]

3. **Calculate the concentration of hydroxide ions \([OH^-]\)**:
The concentration of hydroxide ions can be calculated using the formula:
\[ [OH^-] = 10^{-pOH} \]
Substituting the calculated pOH:
\[ [OH^-] = 10^{-6} \, \text{M} \]

4. **Relate \([OH^-]\) to \( K_b \)**:
For a weak base \( B \), the dissociation can be represented as:
\[ B + H_2O \rightleftharpoons BH^+ + OH^- \]
At equilibrium, the concentration of \( OH^- \) is equal to the concentration of \( BH^+ \) produced. Therefore, we can express \( K_b \) as:
\[ K_b = \frac{[BH^+][OH^-]}{[B]} \]
Assuming that the initial concentration of the weak base is much greater than the change due to dissociation, we can approximate:
\[ K_b \approx [OH^-]^2 \]
Thus:
\[ K_b = (10^{-6})^2 = 10^{-12} \]

5. **Final Result**:
Therefore, the \( K_b \) for the weak base is:
\[ K_b = 10^{-12} \]