The `pK_(b)` of `CN^(Theta)` is `4.7`. The `pH` is solution prepared by mixing `2.5 mol` of `2.5 mol` of `KCN` of `2.5 mol` of `HCN` in water and making the total volume upto `500mL` is

Options

A. `10.3`
B. `9.3`
C. `8.3`
D. `4.7`

Explanation

To solve the problem, we need to find the pH of a solution prepared by mixing KCN and HCN. We are given the pKb of CN⁻ and the amounts of KCN and HCN used. Let's go through the solution step by step.

### Step 1: Understand the components of the solution
We have:
- KCN (potassium cyanide), which dissociates in water to give CN⁻ ions.
- HCN (hydrocyanic acid), which is a weak acid.

### Step 2: Identify the relationship between pKa and pKb
We know that:
\[ pK_a + pK_b = 14 \]
Given that \( pK_b = 4.7 \), we can find \( pK_a \):
\[ pK_a = 14 - pK_b = 14 - 4.7 = 9.3 \]

### Step 3: Calculate the concentrations of KCN and HCN
We have 2.5 moles of KCN and 2.5 moles of HCN in a total volume of 500 mL. To find the concentrations, we convert the volume to liters:
\[ \text{Volume in liters} = \frac{500 \text{ mL}}{1000} = 0.5 \text{ L} \]

Now, we can calculate the concentrations:
- Concentration of KCN:
\[ [KCN] = \frac{2.5 \text{ moles}}{0.5 \text{ L}} = 5 \text{ M} \]
- Concentration of HCN:
\[ [HCN] = \frac{2.5 \text{ moles}}{0.5 \text{ L}} = 5 \text{ M} \]

### Step 4: Use the Henderson-Hasselbalch equation
The pH of the solution can be calculated using the Henderson-Hasselbalch equation:
\[ pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
Where:
- \( [A^-] \) is the concentration of the base (CN⁻ from KCN)
- \( [HA] \) is the concentration of the acid (HCN)

Since both concentrations are equal (5 M), we can substitute:
\[ pH = pK_a + \log\left(\frac{5}{5}\right) \]
\[ pH = pK_a + \log(1) \]
Since \( \log(1) = 0 \):
\[ pH = pK_a \]
\[ pH = 9.3 \]

### Final Answer
The pH of the solution is **9.3**.