The `pH` of `0.1 M` solution of cyanic acid `(HCNO)` is `2.34`. Calculate the ionization constant of the acid and its degree of ionisation in the solution.

Explanation

To solve the problem, we need to calculate the ionization constant (Ka) of cyanic acid (HCNO) and its degree of ionization (α) based on the given pH of a 0.1 M solution.

### Step 1: Calculate the concentration of H⁺ ions from pH
The pH of the solution is given as 2.34. We can use the formula:
\[ \text{pH} = -\log[H^+] \]
To find the concentration of H⁺ ions, we can rearrange this formula:
\[ [H^+] = 10^{-\text{pH}} \]

Substituting the given pH:
\[ [H^+] = 10^{-2.34} \]

Calculating this gives:
\[ [H^+] \approx 0.0046 \, \text{M} \]

### Step 2: Set up the ionization equilibrium
Cyanic acid (HCNO) ionizes in water as follows:
\[ \text{HCNO} \rightleftharpoons \text{H}^+ + \text{CNO}^- \]

Let the degree of ionization be α. Initially, we have:
- Concentration of HCNO = 0.1 M
- Change in concentration due to ionization = α × 0.1 M

At equilibrium:
- Concentration of HCNO = \( 0.1 - \alpha \times 0.1 \)
- Concentration of H⁺ = \( \alpha \times 0.1 \)
- Concentration of CNO⁻ = \( \alpha \times 0.1 \)

Since we know the concentration of H⁺ ions at equilibrium is approximately 0.0046 M:
\[ \alpha \times 0.1 = 0.0046 \]
\[ \alpha = \frac{0.0046}{0.1} = 0.046 \]

### Step 3: Calculate the ionization constant (Ka)
The ionization constant (Ka) can be calculated using the formula:
\[ K_a = \frac{[H^+][CNO^-]}{[HCNO]} \]

Substituting the equilibrium concentrations:
\[ K_a = \frac{(0.0046)(0.0046)}{(0.1 - 0.0046)} \]

Calculating the denominator:
\[ 0.1 - 0.0046 = 0.0954 \]

Now substituting the values into the Ka expression:
\[ K_a = \frac{(0.0046)^2}{0.0954} \]

Calculating this gives:
\[ K_a \approx \frac{0.00002116}{0.0954} \approx 0.000221 \]
\[ K_a \approx 2.21 \times 10^{-4} \]

### Summary of Results
- Ionization constant (Ka) = \( 2.21 \times 10^{-4} \)
- Degree of ionization (α) = 0.046 or 4.6%