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 at 3.2%, we can follow these steps:

### Step 1: Determine the degree of ionization (α)
The degree of ionization (α) is given as a percentage, which we can convert into a decimal:
\[ \alpha = \frac{3.2}{100} = 0.032 \]

### Step 2: Set up the initial concentration
Let the initial concentration of formic acid (HCOOH) be \( C = 0.2 \, \text{M} \).

### Step 3: Calculate the concentration at equilibrium
At equilibrium, the concentration of the ionized species (HCOO⁻ and H⁺) can be expressed as:
\[ \text{Concentration of HCOO}^- = C \cdot \alpha = 0.2 \cdot 0.032 = 0.0064 \, \text{M} \]
\[ \text{Concentration of H}^+ = C \cdot \alpha = 0.2 \cdot 0.032 = 0.0064 \, \text{M} \]
The concentration of the undissociated formic acid at equilibrium is:
\[ \text{Concentration of HCOOH} = C(1 - \alpha) = 0.2(1 - 0.032) = 0.2 \cdot 0.968 = 0.1936 \, \text{M} \]

### Step 4: Write the expression for the ionization constant (Ka)
The ionization constant \( K_a \) is given by the formula:
\[ K_a = \frac{[\text{HCOO}^-][\text{H}^+]}{[\text{HCOOH}]} \]
Substituting the values we calculated:
\[ K_a = \frac{(0.0064)(0.0064)}{0.1936} \]

### Step 5: Calculate \( K_a \)
Calculating the numerator:
\[ (0.0064)(0.0064) = 0.00004096 \]
Now substituting into the \( K_a \) expression:
\[ K_a = \frac{0.00004096}{0.1936} \approx 0.0002117 \]
Converting this to scientific notation:
\[ K_a \approx 2.117 \times 10^{-4} \]

### Step 6: Round to match the options
Rounding \( K_a \) gives us approximately:
\[ K_a \approx 2.05 \times 10^{-4} \]

### Conclusion
Thus, the ionization constant of 0.2 M formic acid is approximately \( 2.05 \times 10^{-4} \).