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

Explanation

To find the concentrations of \([H^+]\) and \([OH^-]\) in a 0.1 M aqueous solution of a weak acid that is 2% ionized, we can follow these steps:

### Step 1: Determine the concentration of the weak acid
Given that the solution is 0.1 M and the weak acid is 2% ionized, we can calculate the concentration of the weak acid.

\[ \text{Concentration of weak acid (C)} = 0.1 \, \text{M} \]

### Step 2: Calculate the degree of ionization
The degree of ionization (\(\alpha\)) is given as 2%. We convert this percentage into a decimal for calculations:

\[ \alpha = \frac{2}{100} = 0.02 \]

### Step 3: Calculate the concentration of \([H^+]\)
The concentration of \([H^+]\) produced by the ionization of the weak acid can be calculated using the formula:

\[ [H^+] = C \cdot \alpha \]

Substituting the values we have:

\[ [H^+] = 0.1 \, \text{M} \cdot 0.02 = 0.002 \, \text{M} = 2 \times 10^{-3} \, \text{M} \]

### Step 4: Calculate the concentration of \([OH^-]\)
Using the ionic product of water at 25°C, which is given as \(K_w = [H^+][OH^-] = 1 \times 10^{-14}\), we can find the concentration of \([OH^-]\):

\[ [OH^-] = \frac{K_w}{[H^+]} \]

Substituting the values:

\[ [OH^-] = \frac{1 \times 10^{-14}}{2 \times 10^{-3}} = 5 \times 10^{-12} \, \text{M} \]

### Final Results
Thus, the concentrations are:
- \([H^+] = 2 \times 10^{-3} \, \text{M}\)
- \([OH^-] = 5 \times 10^{-12} \, \text{M}\)