A weak base `(BOH)` with `K_(b) = 10^(-5)` is titrated with a strong acid `(HCl)`, At `3//4` th of the equivalence point, pH of the solution is:

Options

A. `5+log^(30`
B. `5-log^(3)`
C. `9-log^(3)`
D. `8.523`

Answer

To solve the problem of finding the pH of a solution at \( \frac{3}{4} \) of the equivalence point during the titration of a weak base \( BOH \) with a strong acid \( HCl \), we can follow these steps:

### Step 1: Understand the system
We have a weak base \( BOH \) that dissociates as follows:
\[ BOH \rightleftharpoons B^+ + OH^- \]
Given that \( K_b = 10^{-5} \), we can find \( pK_b \):
\[ pK_b = -\log(K_b) = -\log(10^{-5}) = 5 \]

### Step 2: Determine concentrations at \( \frac{3}{4} \) of the equivalence point
At the equivalence point, all the weak base \( BOH \) has reacted with the strong acid \( HCl \). At \( \frac{3}{4} \) of the equivalence point, we have:
- Concentration of \( B^+ \) (from \( BOH \)): \( \frac{3}{4} \)
- Remaining concentration of \( BOH \): \( \frac{1}{4} \)

### Step 3: Use the Henderson-Hasselbalch equation
For the titration of a weak base with a strong acid, the pOH can be calculated using the Henderson-Hasselbalch equation:
\[ pOH = pK_b + \log\left(\frac{[B^+]}{[BOH]}\right) \]
Substituting the known values:
\[ pOH = 5 + \log\left(\frac{3/4}{1/4}\right) = 5 + \log(3) \]

### Step 4: Calculate pH from pOH
We know that:
\[ pH + pOH = 14 \]
Thus, we can find pH:
\[ pH = 14 - pOH = 14 - (5 + \log(3)) = 9 - \log(3) \]

### Step 5: Final calculation
To find the numerical value of \( pH \):
If \( \log(3) \approx 0.477 \):
\[ pH \approx 9 - 0.477 = 8.523 \]

### Conclusion
The pH of the solution at \( \frac{3}{4} \) of the equivalence point is approximately \( 8.523 \).