The `pK_(a)` of `HCN` is 9.30. The pH of a solutin prepared by mixing `2.5` moles of `KCN` and `2.5` moles of `HCN` in water and making up the total volume to `500 mL` is

Options

A. `9.30`
B. `7.30`
C. `10.30`
D. `8.30`

Explanation

To find the pH of the solution prepared by mixing 2.5 moles of KCN (the salt of the weak acid HCN) and 2.5 moles of HCN in water, we can use the Henderson-Hasselbalch equation, which is suitable for buffer solutions.

### Step-by-Step Solution:

1. **Identify the Components:**
- Weak acid: HCN
- Its salt (conjugate base): KCN
- Given: pKa of HCN = 9.30
- Moles of HCN = 2.5 moles
- Moles of KCN = 2.5 moles
- Total volume of the solution = 500 mL = 0.5 L

2. **Calculate the Concentration of HCN and KCN:**
- Concentration of HCN = Moles of HCN / Volume of solution
\[ \text{Concentration of HCN} = \frac{2.5 \text{ moles}}{0.5 \text{ L}} = 5 \text{ M} \]
- Concentration of KCN = Moles of KCN / Volume of solution
\[ \text{Concentration of KCN} = \frac{2.5 \text{ moles}}{0.5 \text{ L}} = 5 \text{ M} \]

3. **Apply the Henderson-Hasselbalch Equation:**
The Henderson-Hasselbalch equation is given by:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
Here, [Salt] = concentration of KCN and [Acid] = concentration of HCN.
\[ \text{pH} = 9.30 + \log\left(\frac{5}{5}\right) \]

4. **Calculate the Logarithmic Term:**
Since the concentrations of the salt and acid are equal:
\[ \log\left(\frac{5}{5}\right) = \log(1) = 0 \]

5. **Final Calculation of pH:**
\[ \text{pH} = 9.30 + 0 = 9.30 \]

### Conclusion:
The pH of the solution is **9.30**.