A 0.2M solution of formic acid is 3.8% ionised. Its ionisation constant is

Options

A. `9.6xx10^(-3)`
B. `2.1xx10^(-4)`
C. `1.25xx10^(-6)`
D. `4.2xx10^(-2)`

Answer

To find the ionization constant (Ka) of formic acid given a 0.2 M solution that is 3.8% ionized, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Concentration (C)**:
The initial concentration of formic acid (HCOOH) is given as 0.2 M.

\[ C = 0.2 \, \text{M} \]

2. **Calculate the Degree of Ionization (α)**:
The degree of ionization (α) is given as 3.8%. To convert this percentage into a decimal, we divide by 100.

\[ \alpha = \frac{3.8}{100} = 0.038 \]

3. **Determine the Concentrations at Equilibrium**:
At equilibrium, the concentrations of the species can be expressed as follows:
- Concentration of undissociated formic acid:
\[ [HCOOH] = C(1 - \alpha) = 0.2(1 - 0.038) = 0.2 \times 0.962 = 0.1924 \, \text{M} \]
- Concentration of ionized formic acid (HCOO⁻ and H₃O⁺):
\[ [HCOO^-] = [H₃O^+] = C \alpha = 0.2 \times 0.038 = 0.0076 \, \text{M} \]

4. **Write the Expression for the Ionization Constant (Ka)**:
The ionization constant (Ka) for the dissociation of formic acid is given by the formula:
\[ K_a = \frac{[HCOO^-][H₃O^+]}{[HCOOH]} \]
Substituting the equilibrium concentrations into the expression:
\[ K_a = \frac{(0.0076)(0.0076)}{0.1924} \]

5. **Calculate Ka**:
Now, calculate the value of Ka:
\[ K_a = \frac{(0.0076)^2}{0.1924} = \frac{0.00005776}{0.1924} \approx 0.000299 \]
Converting this to scientific notation:
\[ K_a \approx 2.99 \times 10^{-4} \]

6. **Final Answer**:
The ionization constant (Ka) of formic acid is approximately:
\[ K_a \approx 2.99 \times 10^{-4} \]