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. `3 xx 10^(-1)`
B. `1 xx 10^(-3)`
C. `1 xx 10^(-5)`
D. `1 xx 10^(-7)`

Answer

To find the ionization constant \( K_a \) of the acid \( HQ \), we can follow these steps:

### Step 1: Understand the given information
We have a \( 0.1 \) M solution of the acid \( HQ \) with a pH of \( 3 \).

### Step 2: Calculate the concentration of \( H^+ \) ions
The pH is related to the concentration of hydrogen ions (\( H^+ \)) by the formula:
\[ \text{pH} = -\log[H^+] \]
Given that \( \text{pH} = 3 \), we can find the concentration of \( H^+ \):
\[ [H^+] = 10^{-\text{pH}} = 10^{-3} \, \text{M} \]

### Step 3: Write the dissociation equation
The dissociation of the acid \( HQ \) can be represented as:
\[ HQ \rightleftharpoons H^+ + Q^- \]
Let \( x \) be the concentration of \( H^+ \) ions produced at equilibrium. From the pH calculation, we have:
\[ x = [H^+] = 10^{-3} \, \text{M} \]

### Step 4: Set up the equilibrium concentrations
Initially, the concentration of \( HQ \) is \( 0.1 \, \text{M} \). At equilibrium:
- The concentration of \( H^+ \) is \( 10^{-3} \, \text{M} \).
- The concentration of \( Q^- \) is also \( 10^{-3} \, \text{M} \).
- The concentration of \( HQ \) at equilibrium will be:
\[ [HQ] = 0.1 - x \approx 0.1 - 10^{-3} \approx 0.1 \, \text{M} \quad (\text{since } 10^{-3} \text{ is negligible compared to } 0.1) \]

### Step 5: Write the expression for \( K_a \)
The ionization constant \( K_a \) is given by:
\[ K_a = \frac{[H^+][Q^-]}{[HQ]} \]
Substituting the equilibrium concentrations:
\[ K_a = \frac{(10^{-3})(10^{-3})}{0.1} \]

### Step 6: Calculate \( K_a \)
\[ K_a = \frac{10^{-6}}{0.1} = 10^{-6} \times 10^{1} = 10^{-5} \]

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