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: Identify the components of the solution
- HCN is a weak acid.
- KCN is a salt that provides the conjugate base (CN⁻) of the weak acid.

### Step 2: Recognize that this is a buffer solution
Since we have a weak acid (HCN) and its conjugate base (CN⁻ from KCN), the solution acts as a buffer.

### Step 3: Write the dissociation constant expression
The dissociation constant (Ka) for HCN is given as:
\[ K_a = 5 \times 10^{-10} \]

### Step 4: Calculate pKa
To find pKa, we use the formula:
\[ pK_a = -\log(K_a) \]
Calculating this:
\[ pK_a = -\log(5 \times 10^{-10}) \]
Using properties of logarithms:
\[ pK_a = -\log(5) + 10 \]
Using a calculator, we find:
\[ \log(5) \approx 0.699 \]
So:
\[ pK_a \approx 10 - 0.699 = 9.301 \]

### Step 5: Calculate the concentrations of HCN and KCN
- The concentration of HCN:
\[ [HCN] = \frac{1.5 \text{ moles}}{0.5 \text{ dm}^3} = 3 \text{ M} \]
- The concentration of KCN:
\[ [KCN] = \frac{0.15 \text{ moles}}{0.5 \text{ dm}^3} = 0.3 \text{ M} \]

### Step 6: Use the Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation for a buffer solution is:
\[ pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
Where:
- \([A^-]\) is the concentration of the conjugate base (KCN).
- \([HA]\) is the concentration of the weak acid (HCN).

Substituting the values:
\[ pH = 9.301 + \log\left(\frac{0.3}{3}\right) \]
\[ pH = 9.301 + \log(0.1) \]
Since \(\log(0.1) = -1\):
\[ pH = 9.301 - 1 = 8.301 \]

### Step 7: Final pH value
Thus, the pH of the solution is approximately:
\[ \text{pH} \approx 8.301 \]