A 0.20 M solution of methanoic acid has degree of ionization of 0.032. Its dissociation constant would be

Options

A. `2.1xx10-2`
B. `2.1xx10-4`
C. `1.1xx10-6`
D. `1.6xx10-8`

Answer

To find the dissociation constant (Ka) of methanoic acid from the given information, we can follow these steps:

### Step 1: Write the dissociation reaction of methanoic acid.
The dissociation of methanoic acid (HCOOH) in water can be represented as:
\[ \text{HCOOH} \rightleftharpoons \text{H}^+ + \text{HCOO}^- \]

### Step 2: Set up the initial concentrations.
Given that the initial concentration of methanoic acid (C) is 0.20 M, we can set up the initial concentrations as follows:
- \([ \text{HCOOH} ] = C = 0.20 \, \text{M}\)
- \([ \text{H}^+ ] = 0\)
- \([ \text{HCOO}^- ] = 0\)

### Step 3: Define the degree of ionization (α).
The degree of ionization (α) is given as 0.032. This means that 3.2% of the methanoic acid has dissociated.

### Step 4: Calculate the equilibrium concentrations.
At equilibrium, the concentrations will be:
- \([ \text{HCOOH} ] = C(1 - \alpha) = 0.20(1 - 0.032) = 0.20 \times 0.968 = 0.1936 \, \text{M}\)
- \([ \text{H}^+ ] = C\alpha = 0.20 \times 0.032 = 0.0064 \, \text{M}\)
- \([ \text{HCOO}^- ] = C\alpha = 0.20 \times 0.032 = 0.0064 \, \text{M}\)

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

### Step 6: Substitute the equilibrium concentrations into the Ka expression.
Substituting the equilibrium concentrations into the Ka expression:
\[ K_a = \frac{(0.0064)(0.0064)}{0.1936} \]

### Step 7: Calculate Ka.
Calculating the value:
\[ K_a = \frac{0.00004096}{0.1936} \approx 0.000211 \]
\[ K_a \approx 2.11 \times 10^{-4} \]

### Final Answer:
The dissociation constant (Ka) of methanoic acid is approximately \( 2.1 \times 10^{-4} \).

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