`K_(a)` for `HCN` is `5 xx 10^(-10)` at `25^(@)C`. For maintaining a constant `pH` of `9.0`, the volume of `5M KCN` solution required to be added to `10mL` of `2M HCN` solution is

Options

A. `9.3 mL`
B. `7.95 mL`
C. `4mL`
D. `2mL`

Explanation

To solve the problem, we need to determine the volume of 5M KCN solution required to maintain a constant pH of 9.0 when mixed with 10 mL of 2M HCN solution. We will use the Henderson-Hasselbalch equation for this purpose.

### Step-by-Step Solution:

1. **Identify Given Values**:
- \( K_a \) for HCN = \( 5 \times 10^{-10} \)
- \( \text{pH} = 9.0 \)
- Volume of HCN solution = 10 mL
- Molarity of HCN = 2 M
- Molarity of KCN = 5 M

2. **Calculate \( pK_a \)**:
\[ pK_a = -\log(K_a) = -\log(5 \times 10^{-10}) \]
Using logarithmic properties:
\[ pK_a = -\log(5) - \log(10^{-10}) = -\log(5) + 10 \]
Approximating \( \log(5) \approx 0.7 \):
\[ pK_a \approx 10 - 0.7 = 9.3 \]

3. **Use the Henderson-Hasselbalch Equation**:
The equation is given by:
\[ \text{pH} = pK_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
Here, the salt is KCN and the acid is HCN. Plugging in the values:
\[ 9.0 = 9.3 + \log\left(\frac{[\text{KCN}]}{[\text{HCN}]}\right) \]

4. **Rearranging the Equation**:
\[ 9.0 - 9.3 = \log\left(\frac{[\text{KCN}]}{[\text{HCN}]}\right) \]
\[ -0.3 = \log\left(\frac{[\text{KCN}]}{[\text{HCN}]}\right) \]
Converting from logarithmic form:
\[ \frac{[\text{KCN}]}{[\text{HCN}]} = 10^{-0.3} \approx 0.5 \]

5. **Calculate Concentrations**:
Let the volume of KCN solution added be \( V \) mL. The moles of KCN and HCN can be expressed as:
- Moles of KCN = \( 5 \times \frac{V}{1000} \) (since \( V \) is in mL)
- Moles of HCN = \( 2 \times 10 = 20 \) mmol (from 10 mL of 2M solution)

The concentration of KCN after dilution will be:
\[ [\text{KCN}] = \frac{5 \times \frac{V}{1000}}{V + 10} \]
The concentration of HCN will be:
\[ [\text{HCN}] = \frac{20}{V + 10} \]

6. **Set Up the Ratio**:
From the ratio we derived:
\[ \frac{5 \times \frac{V}{1000}}{20/(V + 10)} = 0.5 \]
Cross-multiplying gives:
\[ 5V(V + 10) = 10 \times 20 \]
Simplifying:
\[ 5V^2 + 50V = 200 \]
\[ V^2 + 10V - 40 = 0 \]

7. **Solve the Quadratic Equation**:
Using the quadratic formula \( V = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ V = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 1 \cdot (-40)}}{2 \cdot 1} \]
\[ V = \frac{-10 \pm \sqrt{100 + 160}}{2} \]
\[ V = \frac{-10 \pm \sqrt{260}}{2} \]
\[ V = \frac{-10 \pm 16.12}{2} \]
Taking the positive root:
\[ V \approx \frac{6.12}{2} \approx 3.06 \text{ mL} \]

### Final Answer:
The volume of 5M KCN solution required to be added to 10 mL of 2M HCN solution is approximately **3.06 mL**.