0.2 molar solution of formic acid is ionised to an extent of 3.2 % its ionisation constant is

Options

A. `9.6 xx 10 ^(-3)`
B. ` 2.1 xx 10 ^(-4)`
C. ` 1.25 xx 10 ^(-6)`
D. ` 4. 8 xx 10 ^(-5)`

Answer

To find the ionization constant (K_a) of formic acid from the given data, we can follow these steps:

### Step 1: Understand the given data
We have:
- Concentration of formic acid (C) = 0.2 M
- Extent of ionization (α) = 3.2%

### Step 2: Convert the extent of ionization to a decimal
The extent of ionization (α) in decimal form is:
\[ \alpha = \frac{3.2}{100} = 0.032 \]

### Step 3: Write the ionization equation
The ionization of formic acid (HCOOH) can be represented as:
\[ \text{HCOOH} \rightleftharpoons \text{H}_3\text{O}^+ + \text{HCOO}^- \]

### Step 4: Set up the equilibrium concentrations
Let:
- Initial concentration of HCOOH = C = 0.2 M
- Change in concentration due to ionization = Cα = 0.2 × 0.032 = 0.0064 M

At equilibrium:
- Concentration of HCOOH = C - Cα = 0.2 - 0.0064 = 0.1936 M
- Concentration of H₃O⁺ = Cα = 0.0064 M
- Concentration of HCOO⁻ = Cα = 0.0064 M

### Step 5: Write the expression for the ionization constant (K_a)
The ionization constant (K_a) is given by:
\[ K_a = \frac{[\text{H}_3\text{O}^+][\text{HCOO}^-]}{[\text{HCOOH}]} \]

Substituting the equilibrium concentrations into the expression:
\[ K_a = \frac{(0.0064)(0.0064)}{0.1936} \]

### Step 6: Calculate K_a
Calculating the numerator:
\[ 0.0064 \times 0.0064 = 0.00004096 \]

Now substituting into the K_a expression:
\[ K_a = \frac{0.00004096}{0.1936} \approx 0.000211 \]

Converting this to scientific notation:
\[ K_a \approx 2.11 \times 10^{-4} \]

### Final Answer
The ionization constant (K_a) of formic acid is approximately:
\[ K_a \approx 2.1 \times 10^{-4} \]