The ionization constant of 0.2 M formic acid (ionized 3.2%) is

Options

A. `2.06 xx 10^(-8)`
B. `4.8 xx 10^(-5)`
C. `2.05 xx 10^(-4)`
D. `9.6 xx 10^(-5)`

Answer

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

### Step 1: Determine the concentration of ionized species
Given that the initial concentration of formic acid (HCOOH) is 0.2 M and it ionizes 3.2%, we can calculate the concentration of the ionized species (HCOO⁻ and H⁺).

\[ \text{Ionization} = \text{Initial Concentration} \times \text{Percentage Ionization} \]
\[ \text{Ionization} = 0.2 \, \text{M} \times \frac{3.2}{100} = 0.0064 \, \text{M} \]

### Step 2: Set up the equilibrium concentrations
At equilibrium, the concentrations of the species will be:
- [HCOOH] = Initial concentration - Ionization
- [HCOO⁻] = Ionization
- [H⁺] = Ionization

Calculating these:
\[ [\text{HCOOH}] = 0.2 \, \text{M} - 0.0064 \, \text{M} = 0.1936 \, \text{M} \]
\[ [\text{HCOO}^-] = 0.0064 \, \text{M} \]
\[ [\text{H}^+] = 0.0064 \, \text{M} \]

### Step 3: Write the expression for the ionization constant (Ka)
The ionization constant (Ka) for formic acid can be expressed as:
\[ K_a = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]} \]

### Step 4: Substitute the equilibrium concentrations into the Ka expression
Substituting the values we calculated:
\[ K_a = \frac{(0.0064)(0.0064)}{0.1936} \]

### Step 5: Calculate Ka
Calculating the numerator:
\[ (0.0064)(0.0064) = 0.00004096 \]

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

### Step 6: Final result
Thus, the ionization constant (Ka) of the formic acid is approximately:
\[ K_a \approx 2.11 \times 10^{-4} \]