`0.01` M HA (aq.) is `2%` dissociated, `[OH^(-)]` of solution is :

Options

A. `2xx10^(-4)`
B. `10^(-8)`
C. `5xx10^(-11)`
D. `5xx10^(-12)`

Answer

To solve the problem, we need to find the concentration of hydroxide ions \([OH^-]\) in a solution of 0.01 M HA that is 2% dissociated. Here’s a step-by-step solution:

### Step 1: Determine the concentration of \([H^+]\)
Given that the acid HA is 2% dissociated, we can calculate the concentration of \([H^+]\) ions produced from the dissociation.

1. **Initial concentration of HA**:
\[ C = 0.01 \, \text{M} \]

2. **Percentage dissociation**:
\[ \text{Dissociation} = 2\% = \frac{2}{100} = 0.02 \]

3. **Concentration of \([H^+]\)**:
\[ [H^+] = C \times \text{Dissociation} = 0.01 \times 0.02 = 0.0002 \, \text{M} = 2 \times 10^{-4} \, \text{M} \]

### Step 2: Use the ion product of water to find \([OH^-]\)
The product of the concentrations of hydrogen ions and hydroxide ions in water at 25°C is given by:
\[ K_w = [H^+][OH^-] = 10^{-14} \]

4. **Substituting the value of \([H^+]\)**:
\[ [OH^-] = \frac{K_w}{[H^+]} = \frac{10^{-14}}{2 \times 10^{-4}} \]

5. **Calculating \([OH^-]\)**:
\[ [OH^-] = \frac{10^{-14}}{2 \times 10^{-4}} = \frac{10^{-14}}{2} \times 10^{4} = 0.5 \times 10^{-10} \, \text{M} \]
\[ [OH^-] = 5 \times 10^{-11} \, \text{M} \]

### Final Answer
The concentration of hydroxide ions \([OH^-]\) in the solution is:
\[ [OH^-] = 5 \times 10^{-11} \, \text{M} \]

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