If the ionisation constant of acetic acid is `1.8 xx 10^(-5)`, at what concentration will it be dissociated to `2%`?

Options

A. `1M`
B. `0.018M`
C. `0.18M`
D. `0.045M`

Explanation

To find the concentration of acetic acid at which it will dissociate to 2%, we can follow these steps:

### Step 1: Define the dissociation constant and the dissociation fraction
The ionization constant (K) of acetic acid is given as:
\[ K_a = 1.8 \times 10^{-5} \]

The dissociation fraction (α) for 2% dissociation is:
\[ \alpha = \frac{2}{100} = 0.02 \]

### Step 2: Set up the dissociation equation
The dissociation of acetic acid (CH₃COOH) can be represented as:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

Let the initial concentration of acetic acid be \( C \). At equilibrium, the concentrations will be:
- Concentration of CH₃COOH = \( C(1 - \alpha) \)
- Concentration of CH₃COO⁻ = \( C\alpha \)
- Concentration of H⁺ = \( C\alpha \)

### Step 3: Write the expression for the ionization constant
The expression for the ionization constant \( K_a \) is given by:
\[ K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \]

Substituting the equilibrium concentrations into the equation:
\[ K_a = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} \]

### Step 4: Simplify the equation
This simplifies to:
\[ K_a = \frac{C\alpha^2}{1 - \alpha} \]

Since \( \alpha \) is small (0.02), we can approximate \( 1 - \alpha \approx 1 \):
\[ K_a \approx C\alpha^2 \]

### Step 5: Solve for concentration \( C \)
Rearranging the equation gives:
\[ C = \frac{K_a}{\alpha^2} \]

Substituting the known values:
\[ C = \frac{1.8 \times 10^{-5}}{(0.02)^2} \]

Calculating \( (0.02)^2 \):
\[ (0.02)^2 = 0.0004 \]

Now substituting this back into the equation:
\[ C = \frac{1.8 \times 10^{-5}}{0.0004} \]

Calculating \( C \):
\[ C = \frac{1.8 \times 10^{-5}}{4 \times 10^{-4}} = 4.5 \times 10^{-2} \]

### Step 6: Final result
Thus, the concentration of acetic acid required for 2% dissociation is:
\[ C = 0.045 \, \text{M} \]