Calculate the degree of ionisation of `0.04 M HOCl` solution having ionisation constant `1.25 xx 10^(-4)` ?

Options

A. `0.025`
B. `0.5`
C. `0.25`
D. `0.055`

Explanation

To calculate the degree of ionization of a 0.04 M HOCl solution with an ionization constant of \( K_a = 1.25 \times 10^{-4} \), we can follow these steps:

### Step 1: Write the ionization equation
The ionization of hypochlorous acid (HOCl) can be represented as:
\[ \text{HOCl} \rightleftharpoons \text{H}^+ + \text{OCl}^- \]

### Step 2: Set up the equilibrium expression
Let the degree of ionization be represented by \( \alpha \). Initially, the concentration of HOCl is \( C = 0.04 \) M. At equilibrium, the concentrations will be:
- \([HOCl] = C(1 - \alpha) = 0.04(1 - \alpha)\)
- \([H^+] = C\alpha = 0.04\alpha\)
- \([OCl^-] = C\alpha = 0.04\alpha\)

### Step 3: Write the expression for the ionization constant
The ionization constant \( K_a \) is given by:
\[ K_a = \frac{[\text{H}^+][\text{OCl}^-]}{[\text{HOCl}]} \]
Substituting the equilibrium concentrations:
\[ K_a = \frac{(0.04\alpha)(0.04\alpha)}{0.04(1 - \alpha)} \]
This simplifies to:
\[ K_a = \frac{0.04^2 \alpha^2}{0.04(1 - \alpha)} = \frac{0.04\alpha^2}{1 - \alpha} \]

### Step 4: Substitute the known values
We know \( K_a = 1.25 \times 10^{-4} \). Thus, we can set up the equation:
\[ 1.25 \times 10^{-4} = \frac{0.04\alpha^2}{1 - \alpha} \]

### Step 5: Assume \( \alpha \) is small
Since \( \alpha \) is expected to be small, we can approximate \( 1 - \alpha \approx 1 \):
\[ 1.25 \times 10^{-4} \approx 0.04\alpha^2 \]

### Step 6: Solve for \( \alpha^2 \)
Rearranging gives:
\[ \alpha^2 = \frac{1.25 \times 10^{-4}}{0.04} \]
Calculating this:
\[ \alpha^2 = \frac{1.25 \times 10^{-4}}{0.04} = 3.125 \times 10^{-3} \]

### Step 7: Calculate \( \alpha \)
Taking the square root to find \( \alpha \):
\[ \alpha = \sqrt{3.125 \times 10^{-3}} \approx 0.0558 \]

### Step 8: Conclusion
Thus, the degree of ionization of the 0.04 M HOCl solution is approximately:
\[ \alpha \approx 0.0558 \]