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 acid \( HQ \) given the pH of a 0.1 M solution, we can follow these steps:

### Step 1: Calculate the concentration of hydrogen ions \([H^+]\)
The pH is given as 3. We can use the formula:
\[ \text{pH} = -\log[H^+] \]
To find \([H^+]\), we rearrange this equation:
\[ [H^+] = 10^{-\text{pH}} = 10^{-3} \, \text{M} \]

### Step 2: Set up the expression for the ionization constant \( K_a \)
The ionization of the weak acid \( HQ \) can be represented as:
\[ HQ \rightleftharpoons H^+ + Q^- \]
Let \( x \) be the concentration of \( H^+ \) produced at equilibrium. Since we calculated \([H^+] = 10^{-3} \, \text{M}\), we have:
\[ x = 10^{-3} \, \text{M} \]

### Step 3: Write the expression for \( K_a \)
The ionization constant \( K_a \) is given by:
\[ K_a = \frac{[H^+][Q^-]}{[HQ]} \]
At equilibrium, the concentrations will be:
- \([H^+] = x = 10^{-3} \, \text{M}\)
- \([Q^-] = x = 10^{-3} \, \text{M}\)
- \([HQ] = C - x = 0.1 - 10^{-3} \approx 0.1 \, \text{M}\) (since \( x \) is much smaller than \( C \))

### Step 4: Substitute the values into the \( K_a \) expression
Substituting the values we have:
\[ K_a = \frac{(10^{-3})(10^{-3})}{0.1} = \frac{10^{-6}}{0.1} = 10^{-5} \]

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