The dissociation constant of HCN is `5 xx 10^(-10)`. The pH of the solution prepared by mixing `1.5` mole of HCN and `0.15` moles of KCN in water and making up the total volume to `0.5 dm^(3)` is

Options

A. `7.302`
B. `9.302`
C. `8.302`
D. `10.302`

Answer

To find the pH of the solution prepared by mixing 1.5 moles of HCN and 0.15 moles of KCN in water with a total volume of 0.5 dm³, we can follow these steps:

### Step 1: Calculate the concentrations of HCN and KCN
- **Concentration of HCN (C₁)**:
\[ C_{HCN} = \frac{\text{Number of moles of HCN}}{\text{Volume in liters}} = \frac{1.5 \, \text{moles}}{0.5 \, \text{L}} = 3 \, \text{M} \]

- **Concentration of KCN (C₂)**:
\[ C_{KCN} = \frac{\text{Number of moles of KCN}}{\text{Volume in liters}} = \frac{0.15 \, \text{moles}}{0.5 \, \text{L}} = 0.3 \, \text{M} \]

### Step 2: Calculate pKa from Ka
- Given that the dissociation constant \( K_a \) of HCN is \( 5 \times 10^{-10} \):
\[ pK_a = -\log(K_a) = -\log(5 \times 10^{-10}) \approx 9.3 \]

### Step 3: Use the Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation for a buffer solution is given by:
\[ \text{pH} = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
Where:
- \([A^-]\) is the concentration of the conjugate base (KCN).
- \([HA]\) is the concentration of the weak acid (HCN).

Substituting the values:
\[ \text{pH} = 9.3 + \log\left(\frac{0.3}{3}\right) \]

### Step 4: Calculate the logarithm
\[ \log\left(\frac{0.3}{3}\right) = \log(0.1) = -1 \]

### Step 5: Substitute back into the pH equation
\[ \text{pH} = 9.3 - 1 = 8.3 \]

### Final Answer
The pH of the solution is approximately **8.3**.

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