The ionisation constant `(K_(a))` of `0.1 M` weak monobasic acid
`(HA)` will be whose degree of ionisation is `50%` ?

Options

A. `25xx10^(-3)`
B. `25xx10^(-4)`
C. `5xx10^(-2)`
D. `5xx10^(-5)`

Explanation

To find the ionization constant \( K_a \) of a weak monobasic acid \( HA \) with a concentration of \( 0.1 \, M \) and a degree of ionization of \( 50\% \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Concentration**:
- The initial concentration of the weak acid \( HA \) is given as \( 0.1 \, M \).

2. **Determine the Degree of Ionization**:
- The degree of ionization is given as \( 50\% \). This means that half of the acid will dissociate into ions.

3. **Calculate the Concentration at Equilibrium**:
- Since \( 50\% \) of the acid dissociates, the concentration of \( HA \) that remains at equilibrium is:
\[ [HA]_{eq} = [HA]_{initial} - [HA]_{dissociated} = 0.1 \, M - 0.05 \, M = 0.05 \, M \]
- The concentration of \( H^+ \) and \( A^- \) ions at equilibrium will both be \( 0.05 \, M \) because they are produced in equal amounts from the dissociation of \( HA \).

4. **Write the Expression for the Ionization Constant \( K_a \)**:
- The expression for the ionization constant \( K_a \) is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]

5. **Substitute the Equilibrium Concentrations into the \( K_a \) Expression**:
- Substitute the values into the equation:
\[ K_a = \frac{(0.05)(0.05)}{0.05} \]

6. **Simplify the Expression**:
- Simplifying the equation gives:
\[ K_a = \frac{0.0025}{0.05} = 0.05 \]

7. **Convert to Scientific Notation**:
- The value \( 0.05 \) can be expressed in scientific notation as:
\[ K_a = 5 \times 10^{-2} \]

### Final Answer:
The ionization constant \( K_a \) of the weak monobasic acid \( HA \) is \( 5 \times 10^{-2} \).

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