A weak monobasic acid is 1% ionized in 0.1 M solution at `25^@`C. The percentage of ionization in its 0.025 M solution is :

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to determine the percentage of ionization of a weak monobasic acid in a 0.025 M solution, given that it is 1% ionized in a 0.1 M solution.

### Step-by-Step Solution:

1. **Define the Weak Acid and Initial Conditions:**
Let the weak acid be represented as HA. The dissociation of the acid can be written as:
\[ HA \rightleftharpoons H^+ + A^- \]
For the 0.1 M solution, the initial concentration \(C_1 = 0.1 \, \text{M}\) and the percentage ionization is given as 1%.

2. **Calculate the Ionization Constant \(K_a\):**
The percentage ionization is given by:
\[ \text{Percentage Ionization} = \frac{\text{Ionized Concentration}}{\text{Initial Concentration}} \times 100 \]
For the 0.1 M solution:
\[ \alpha_1 = \frac{1}{100} = 0.01 \quad (\text{1% ionization}) \]
The equilibrium concentrations at 0.1 M are:
- Concentration of \(H^+\) and \(A^-\) at equilibrium: \(C_1 \alpha_1 = 0.1 \times 0.01 = 0.001 \, \text{M}\)
- Concentration of HA at equilibrium: \(C_1(1 - \alpha_1) = 0.1(1 - 0.01) = 0.099 \, \text{M}\)

Now, we can calculate \(K_a\):
\[ K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(0.001)(0.001)}{0.099} = \frac{0.000001}{0.099} \approx 1.01 \times 10^{-5} \]

3. **Set Up for the 0.025 M Solution:**
For the 0.025 M solution, let \(C_2 = 0.025 \, \text{M}\) and the percentage ionization be \(\alpha_2\). The equilibrium concentrations will be:
- Concentration of \(H^+\) and \(A^-\) at equilibrium: \(C_2 \alpha_2 = 0.025 \alpha_2\)
- Concentration of HA at equilibrium: \(C_2(1 - \alpha_2) = 0.025(1 - \alpha_2)\)

4. **Calculate \(K_a\) for the 0.025 M Solution:**
Using the same expression for \(K_a\):
\[ K_a = \frac{(0.025 \alpha_2)(0.025 \alpha_2)}{0.025(1 - \alpha_2)} = \frac{0.025^2 \alpha_2^2}{0.025(1 - \alpha_2)} = \frac{0.025 \alpha_2^2}{1 - \alpha_2} \]

5. **Equate the Two Expressions for \(K_a\):**
Since \(K_a\) is constant, we can set the two expressions equal:
\[ 1.01 \times 10^{-5} = \frac{0.025 \alpha_2^2}{1 - \alpha_2} \]

6. **Assume \(\alpha_2\) is Small:**
Since \(\alpha_2\) is expected to be small, we can approximate \(1 - \alpha_2 \approx 1\):
\[ 1.01 \times 10^{-5} \approx 0.025 \alpha_2^2 \]

7. **Solve for \(\alpha_2\):**
Rearranging gives:
\[ \alpha_2^2 = \frac{1.01 \times 10^{-5}}{0.025} = 4.04 \times 10^{-4} \]
Taking the square root:
\[ \alpha_2 \approx \sqrt{4.04 \times 10^{-4}} \approx 0.0201 \]

8. **Calculate the Percentage Ionization:**
The percentage ionization is:
\[ \text{Percentage Ionization} = \alpha_2 \times 100 = 0.0201 \times 100 \approx 2.01\% \]

### Final Answer:
The percentage of ionization in the 0.025 M solution is approximately **2%**.