If `K_(a)` of `HCN =4xx10^(-10)` , then the pH of `2.4 xx 10^(-1)` molar HCN(aq) is

Options

A. `2`x`10^(-5)`
B. 4.7
C. `2.0`
D. `5.0`

Explanation

To find the pH of a 0.24 M solution of HCN given that the dissociation constant \( K_a \) of HCN is \( 4 \times 10^{-10} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- \( K_a \) of HCN = \( 4 \times 10^{-10} \)
- Concentration of HCN = \( 0.24 \, \text{M} \)

2. **Write the Dissociation Equation:**
HCN dissociates in water as follows:
\[ \text{HCN} \rightleftharpoons \text{H}^+ + \text{CN}^- \]

3. **Set Up the Expression for \( K_a \):**
The expression for the acid dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]} \]

4. **Assume the Change in Concentration:**
Let \( x \) be the concentration of \( \text{H}^+ \) ions that dissociate from HCN. At equilibrium, we have:
- \( [\text{H}^+] = x \)
- \( [\text{CN}^-] = x \)
- \( [\text{HCN}] = 0.24 - x \)

5. **Substituting into the \( K_a \) Expression:**
Since HCN is a weak acid, we can assume that \( x \) is very small compared to 0.24 M, so:
\[ K_a \approx \frac{x^2}{0.24} \]
Substituting the value of \( K_a \):
\[ 4 \times 10^{-10} = \frac{x^2}{0.24} \]

6. **Solve for \( x \):**
Rearranging gives:
\[ x^2 = 4 \times 10^{-10} \times 0.24 \]
\[ x^2 = 9.6 \times 10^{-11} \]
\[ x = \sqrt{9.6 \times 10^{-11}} \approx 3.1 \times 10^{-6} \, \text{M} \]

7. **Calculate the pH:**
The pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
Substituting the value of \( x \):
\[ \text{pH} = -\log(3.1 \times 10^{-6}) \approx 5.51 \]

### Final Answer:
The pH of the 0.24 M HCN solution is approximately **5.51**.