The concentration of `[H^(+)]` and concentration of `[OH^(-)]` of a `0.1` aqueous solution of `2%` ionised weak acid is [Ionic product of water `=1xx10^(-14)`]

Options

A. `0.02xx10^(-3)M` and `5xx10^(-11)M`
B. `1xx10^(-3)M` and `3xx10^(-11)M`
C. `2xx10^(-3)M` and `5xx10^(-12)M`
D. `3xx10^(-2)M` and `4xx10^(-13)M`

Explanation

To solve the problem, we need to find the concentrations of \([H^+]\) and \([OH^-]\) in a 0.1 M aqueous solution of a weak acid that is 2% ionized. We also know that the ionic product of water (\(K_w\)) is \(1 \times 10^{-14}\).

### Step-by-Step Solution:

**Step 1: Determine the concentration of the weak acid.**
- The concentration of the weak acid \(HA\) is given as \(C = 0.1 \, \text{M}\).

**Step 2: Calculate the degree of ionization (\(\alpha\)).**
- The degree of ionization is given as 2%, which can be expressed as a fraction:
\[ \alpha = \frac{2}{100} = 0.02 \]

**Step 3: Calculate the concentration of \([H^+]\).**
- The concentration of \([H^+]\) at equilibrium can be calculated using the formula:
\[ [H^+] = C \cdot \alpha \]
Substituting the values:
\[ [H^+] = 0.1 \, \text{M} \cdot 0.02 = 0.002 \, \text{M} = 2 \times 10^{-3} \, \text{M} \]

**Step 4: Use the ionic product of water to find \([OH^-]\).**
- The ionic product of water is given by:
\[ K_w = [H^+] \cdot [OH^-] \]
We can rearrange this to find \([OH^-]\):
\[ [OH^-] = \frac{K_w}{[H^+]} \]
Substituting the known values:
\[ [OH^-] = \frac{1 \times 10^{-14}}{2 \times 10^{-3}} = 5 \times 10^{-12} \, \text{M} \]

### Final Answer:
- The concentration of \([H^+]\) is \(2 \times 10^{-3} \, \text{M}\).
- The concentration of \([OH^-]\) is \(5 \times 10^{-12} \, \text{M}\).