The `pH` of a `0.1` molar solution of the acid `HQ` is `3`. The value of the ionisation constant, `K_(a)` of the acid is

Options

A. `3xx10^(-1)`
B. `1xx10^(-3)`
C. `1xx10^(-5)`
D. `1xx10^(-7)`

Explanation

To find the ionization constant \( K_a \) of the weak acid \( HQ \) given its concentration and pH, we can follow these steps:

### Step 1: Calculate the concentration of \( H^+ \) ions
Given that the pH of the solution is 3, we can calculate the concentration of \( H^+ \) ions using the formula:
\[ \text{pH} = -\log[H^+] \]
From this, we can find:
\[ [H^+] = 10^{-\text{pH}} = 10^{-3} \, \text{M} \]

### Step 2: Set up the equilibrium expression
For the dissociation of the weak acid \( HQ \):
\[ HQ \rightleftharpoons H^+ + Q^- \]
Let \( \alpha \) be the degree of dissociation. Initially, the concentration of \( HQ \) is 0.1 M, and at equilibrium:
- Concentration of \( HQ \) = \( 0.1 - 0.1\alpha \)
- Concentration of \( H^+ \) = \( 0.1\alpha \)
- Concentration of \( Q^- \) = \( 0.1\alpha \)

### Step 3: Relate \( H^+ \) concentration to \( \alpha \)
Since we found that \( [H^+] = 10^{-3} \, \text{M} \), we can set up the equation:
\[ 0.1\alpha = 10^{-3} \]
Solving for \( \alpha \):
\[ \alpha = \frac{10^{-3}}{0.1} = 10^{-2} \]

### Step 4: Write the expression for \( K_a \)
The ionization constant \( K_a \) can be expressed as:
\[ K_a = \frac{[H^+][Q^-]}{[HQ]} \]
Substituting the equilibrium concentrations:
\[ K_a = \frac{(0.1\alpha)(0.1\alpha)}{0.1 - 0.1\alpha} \]
Since \( \alpha \) is small, we can approximate \( 0.1 - 0.1\alpha \approx 0.1 \):
\[ K_a \approx \frac{(0.1\alpha)(0.1\alpha)}{0.1} \]
This simplifies to:
\[ K_a \approx 0.1\alpha^2 \]

### Step 5: Substitute \( \alpha \) into the \( K_a \) expression
Now substituting \( \alpha = 10^{-2} \):
\[ K_a \approx 0.1 \times (10^{-2})^2 = 0.1 \times 10^{-4} = 1 \times 10^{-5} \]

### Final Answer
Thus, the value of the ionization constant \( K_a \) of the acid \( HQ \) is:
\[ K_a = 1 \times 10^{-5} \]

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