A monoprotic acid in a 0.1 M solution ionizes to `0.001%`. Its ionisation constant is

Options

A. `1.0 xx10^(-3)`
B. `1.0 xx 10^(-6)`
C. `1.0 xx 10^(-8)`
D. `1.0 xx 10^(-11)`

Answer

To find the ionization constant (Ka) of a monoprotic acid that ionizes to 0.001% in a 0.1 M solution, we can follow these steps:

### Step 1: Convert the percentage ionization to a decimal
Given that the acid ionizes to 0.001%, we can convert this percentage to a decimal (α).

\[ \alpha = \frac{0.001}{100} = 0.00001 = 10^{-5} \]

### Step 2: Set up the equilibrium expression
For a monoprotic acid (HA), the dissociation can be represented as:

\[ HA \rightleftharpoons H^+ + A^- \]

Let:
- Initial concentration of HA = C = 0.1 M
- Change in concentration due to ionization = Cα = \(0.1 \times 10^{-5}\)

At equilibrium:
- \([HA] = C - C\alpha = 0.1 - 0.1 \times 10^{-5}\)
- \([H^+] = C\alpha = 0.1 \times 10^{-5}\)
- \([A^-] = C\alpha = 0.1 \times 10^{-5}\)

### Step 3: Calculate the equilibrium concentrations
Since \(C\alpha\) is very small compared to C, we can approximate:

\[ [HA] \approx 0.1 \, \text{M} \]

Thus:
\[ [H^+] \approx 0.1 \times 10^{-5} \, \text{M} \]
\[ [A^-] \approx 0.1 \times 10^{-5} \, \text{M} \]

### Step 4: Write the expression for the ionization constant (Ka)
The ionization constant (Ka) is given by:

\[ K_a = \frac{[H^+][A^-]}{[HA]} \]

Substituting the equilibrium concentrations into the equation:

\[ K_a = \frac{(0.1 \times 10^{-5})(0.1 \times 10^{-5})}{0.1} \]

### Step 5: Simplify the expression
\[ K_a = \frac{(0.1 \times 10^{-5})^2}{0.1} \]
\[ K_a = \frac{0.01 \times 10^{-10}}{0.1} \]
\[ K_a = 0.1 \times 10^{-10} = 1.0 \times 10^{-11} \]

### Final Answer
Thus, the ionization constant (Ka) of the monoprotic acid is:

\[ K_a = 1.0 \times 10^{-11} \]

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