The dissociation constant of a weak monobasic acid `K_(a)` is `1xx 10^(-5)` . The pH of 0.1 M of that acid would be

Options

A. 5
B. 1
C. 2
D. 3

Explanation

To find the pH of a 0.1 M solution of a weak monobasic acid with a dissociation constant \( K_a = 1 \times 10^{-5} \), we can follow these steps:

### Step-by-Step Solution

1. **Write the dissociation equation**:
For a weak monobasic acid \( HA \), the dissociation can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]

2. **Set up the initial concentrations**:
The initial concentration of the acid \( [HA] \) is 0.1 M. At the start, the concentrations of \( H^+ \) and \( A^- \) are both 0 M.
\[ [HA] = 0.1 \, M, \quad [H^+] = 0 \, M, \quad [A^-] = 0 \, M \]

3. **Define the change in concentration**:
Let \( \alpha \) be the degree of dissociation. At equilibrium, the concentrations will be:
\[ [HA] = 0.1 - \alpha, \quad [H^+] = \alpha, \quad [A^-] = \alpha \]

4. **Write the expression for the dissociation constant \( K_a \)**:
The dissociation constant is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} = \frac{\alpha \cdot \alpha}{0.1 - \alpha} = \frac{\alpha^2}{0.1 - \alpha} \]

5. **Assume \( \alpha \) is small**:
Since \( K_a \) is small, we can assume \( \alpha \) is small compared to 0.1 M, so \( 0.1 - \alpha \approx 0.1 \). This simplifies our equation to:
\[ K_a \approx \frac{\alpha^2}{0.1} \]

6. **Substitute the known values**:
Given \( K_a = 1 \times 10^{-5} \), we can substitute this into the equation:
\[ 1 \times 10^{-5} = \frac{\alpha^2}{0.1} \]

7. **Solve for \( \alpha^2 \)**:
Rearranging gives:
\[ \alpha^2 = 1 \times 10^{-5} \times 0.1 = 1 \times 10^{-6} \]

8. **Calculate \( \alpha \)**:
Taking the square root:
\[ \alpha = \sqrt{1 \times 10^{-6}} = 1 \times 10^{-3} \]

9. **Find the concentration of \( H^+ \)**:
Since \( [H^+] = \alpha \):
\[ [H^+] = 1 \times 10^{-3} \, M \]

10. **Calculate the pH**:
The pH is calculated using the formula:
\[ pH = -\log[H^+] = -\log(1 \times 10^{-3}) = 3 \]

### Final Answer
The pH of the 0.1 M solution of the weak monobasic acid is **3**.