The freezing point depression of a `0.109M` aq. Solution of formic acid is `-0.21^(@)C`. Calculate the equilibrium constant for the reaction,
`HCOOH (aq) hArr H^(+)(aq) +HCOO^(Theta)(aq)`
`K_(f)` for water `= 1.86 kg mol^(-1)K`

Explanation

To solve the problem of calculating the equilibrium constant for the dissociation of formic acid (HCOOH) given the freezing point depression, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Freezing point depression (ΔTf) = -0.21 °C (which we will take as 0.21 °C for calculations).
- Molarity of formic acid solution (C) = 0.109 M.
- Freezing point depression constant for water (Kf) = 1.86 kg mol⁻¹ K.

2. **Calculate Molality (m):**
- Since the solution is aqueous, we can approximate the molality (m) as equal to the molarity (C) for dilute solutions. Therefore, m ≈ 0.109 mol/kg.

3. **Use the Freezing Point Depression Formula:**
- The formula for freezing point depression is given by:
\[ \Delta T_f = i \cdot K_f \cdot m \]
- Where:
- ΔTf = freezing point depression
- i = van 't Hoff factor (number of particles the solute dissociates into)
- Kf = freezing point depression constant
- m = molality of the solution

4. **Rearranging the Formula:**
- Rearranging the formula to find i:
\[ i = \frac{\Delta T_f}{K_f \cdot m} \]

5. **Substituting Known Values:**
- Substitute ΔTf, Kf, and m into the equation:
\[ i = \frac{0.21}{1.86 \cdot 0.109} \]

6. **Calculating i:**
- Calculate the value of i:
\[ i = \frac{0.21}{0.20274} \approx 1.0358 \]

7. **Determine α (degree of ionization):**
- The van 't Hoff factor (i) can be expressed in terms of α (degree of ionization):
\[ i = 1 + \alpha \]
- Therefore:
\[ \alpha = i - 1 = 1.0358 - 1 \approx 0.0358 \]

8. **Calculate the Equilibrium Constant (Ka):**
- The equilibrium constant for the dissociation of formic acid is given by:
\[ K_a = \frac{C \cdot \alpha^2}{1 - \alpha} \]
- Substitute C and α into the equation:
\[ K_a = \frac{0.109 \cdot (0.0358)^2}{1 - 0.0358} \]

9. **Calculating Ka:**
- Calculate the value:
\[ K_a = \frac{0.109 \cdot 0.00128064}{0.9642} \approx \frac{0.0001395}{0.9642} \approx 1.44 \times 10^{-4} \]

### Final Result:
The equilibrium constant \( K_a \) for the dissociation of formic acid is approximately \( 1.44 \times 10^{-4} \).