`0.2` M solution of formic acid is ionized `3.2%`. Its ionization constant is

Options

A. `9.6 xx 10^(-3)`
B. `2.1 xx 10^(-4)`
C. `1.25xx 10^(-6)`
D. `4.8 xx 10^(-5)`

Answer

To find the ionization constant (Ka) of formic acid given a 0.2 M solution that ionizes at 3.2%, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Ionization Percentage**:
The ionization percentage of formic acid is given as 3.2%. We can express this as a fraction:
\[ \alpha = \frac{3.2}{100} = 0.032 \]

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

3. **Calculate the Change in Concentration**:
Since 3.2% of the acid ionizes, we can find the change in concentration due to ionization:
\[ \text{Change in concentration} = C \times \alpha = 0.2 \times 0.032 = 0.0064 \, \text{M} \]

4. **Equilibrium Concentrations**:
At equilibrium, the concentrations of the species will be:
- For HCOOH (formic acid):
\[ [\text{HCOOH}] = C - \text{Change} = 0.2 - 0.0064 = 0.1936 \, \text{M} \]
- For H⁺ (hydrogen ion) and HCOO⁻ (formate ion):
\[ [\text{H}^+] = [\text{HCOO}^-] = 0.0064 \, \text{M} \]

5. **Write the Expression for Ka**:
The ionization constant (Ka) for formic acid can be expressed as:
\[ K_a = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]} \]

6. **Substitute the Equilibrium Concentrations**:
Plugging in the values we calculated:
\[ K_a = \frac{(0.0064)(0.0064)}{0.1936} \]

7. **Calculate Ka**:
First, calculate the numerator:
\[ (0.0064)(0.0064) = 0.00004096 \]
Now, divide by the equilibrium concentration of formic acid:
\[ K_a = \frac{0.00004096}{0.1936} \approx 0.000211 \]

8. **Express in Scientific Notation**:
\[ K_a \approx 2.11 \times 10^{-4} \]

### Final Answer:
The ionization constant (Ka) for formic acid is approximately:
\[ K_a \approx 2.11 \times 10^{-4} \, \text{M} \]