A monoprotic acid in `1.00 M` solution is `0.01%` ionised. The dissociation constant of this acid is

Options

A. `1xx10^(-8)`
B. `1xx10^(-4)`
C. `1xx10^(-6)`
D. `10^(-5)`

Explanation

To find the dissociation constant (Ka) of a monoprotic acid given that its 1.00 M solution is 0.01% ionized, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Ionization**:
A monoprotic acid (HA) dissociates in water as follows:
\[ HA \rightleftharpoons H^+ + A^- \]
The degree of ionization (α) is given as 0.01%.

2. **Convert Percentage to Decimal**:
To use α in calculations, convert the percentage to a decimal:
\[ \alpha = \frac{0.01}{100} = 0.0001 \]

3. **Determine Initial Concentration (C)**:
The initial concentration (C) of the acid is given as 1.00 M.

4. **Calculate the Concentration of Ions**:
When the acid is 0.01% ionized, the concentration of the ions produced (H+ and A-) can be calculated as:
\[ [H^+] = [A^-] = \alpha \times C = 0.0001 \times 1.00 = 0.0001 \, \text{M} \]

5. **Calculate the Concentration of the Undissociated Acid**:
The concentration of the undissociated acid (HA) at equilibrium is:
\[ [HA] = C - [H^+] = 1.00 - 0.0001 = 0.9999 \, \text{M} \]

6. **Write the Expression for the Acid Dissociation Constant (Ka)**:
The expression for the dissociation constant (Ka) is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
Substituting the values we have:
\[ K_a = \frac{(0.0001)(0.0001)}{0.9999} \]

7. **Calculate Ka**:
\[ K_a = \frac{0.00000001}{0.9999} \approx 0.00000001001 \approx 1.0001 \times 10^{-8} \]

8. **Final Result**:
Thus, the dissociation constant (Ka) of the acid is approximately:
\[ K_a \approx 1.0 \times 10^{-8} \]