In a basic aqueous solution chloromethane undergoes a substitution reaction in which Cl is replaced by `OH^(-)` as.
`CH_3Cl(aq)+OH^(-) hArrCH_3OH(aq)+Cl^(-)(aq)`
The equilibrium constant of above reaction `K_c=1xx10^(16)`.If a solution is prepared by mixing equal volumes of 0.1 M `CH_3Cl` and 0.2M NaOH (100% dissociated) then `[OH^(-)]` concentration at equilibrium in mixture will be :

Options

A. 0.1 M
B. 0.5 M
C. 0.2 M
D. 0.05 M

Explanation

To solve the problem, we need to determine the concentration of hydroxide ions \([OH^-]\) at equilibrium after mixing equal volumes of 0.1 M \(CH_3Cl\) and 0.2 M NaOH.

### Step-by-Step Solution:

1. **Determine the Initial Concentrations:**
- When equal volumes of 0.1 M \(CH_3Cl\) and 0.2 M NaOH are mixed, the concentrations will change due to dilution.
- Let’s assume we mix \(V\) liters of each solution. The total volume after mixing will be \(2V\).

- The concentration of \(CH_3Cl\) after mixing:
\[ [CH_3Cl] = \frac{0.1 \, \text{M} \times V}{2V} = \frac{0.1}{2} = 0.05 \, \text{M} \]

- The concentration of \(NaOH\) after mixing:
\[ [NaOH] = \frac{0.2 \, \text{M} \times V}{2V} = \frac{0.2}{2} = 0.1 \, \text{M} \]

2. **Write the Reaction and Initial Concentrations:**
- The reaction is:
\[ CH_3Cl + OH^- \rightleftharpoons CH_3OH + Cl^- \]
- Initial concentrations before any reaction occurs:
- \([CH_3Cl] = 0.05 \, \text{M}\)
- \([OH^-] = 0.1 \, \text{M}\)
- \([CH_3OH] = 0 \, \text{M}\)
- \([Cl^-] = 0 \, \text{M}\)

3. **Set Up the Change in Concentrations:**
- Let \(x\) be the amount of \(CH_3Cl\) that reacts.
- At equilibrium, the concentrations will be:
- \([CH_3Cl] = 0.05 - x\)
- \([OH^-] = 0.1 - x\)
- \([CH_3OH] = x\)
- \([Cl^-] = x\)

4. **Use the Equilibrium Constant:**
- The equilibrium constant \(K_c\) is given as \(1 \times 10^{16}\):
\[ K_c = \frac{[CH_3OH][Cl^-]}{[CH_3Cl][OH^-]} = 1 \times 10^{16} \]
- Substituting the equilibrium concentrations:
\[ 1 \times 10^{16} = \frac{x \cdot x}{(0.05 - x)(0.1 - x)} \]
\[ 1 \times 10^{16} = \frac{x^2}{(0.05 - x)(0.1 - x)} \]

5. **Assume \(x\) is Small:**
- Since \(K_c\) is very large, we can assume that \(x\) is very close to \(0.05\) (the limiting reagent).
- Therefore, we can approximate:
\[ 0.05 - x \approx 0.05 \quad \text{and} \quad 0.1 - x \approx 0.1 \]
- This simplifies our equation to:
\[ 1 \times 10^{16} = \frac{x^2}{0.05 \cdot 0.1} \]
\[ 1 \times 10^{16} = \frac{x^2}{0.005} \]
\[ x^2 = 1 \times 10^{16} \cdot 0.005 = 5 \times 10^{13} \]
\[ x = \sqrt{5 \times 10^{13}} \approx 7.07 \times 10^{6} \, \text{M} \]

6. **Calculate \([OH^-]\) at Equilibrium:**
- Now, substituting \(x\) back to find \([OH^-]\):
\[ [OH^-] = 0.1 - x \approx 0.1 - 0.05 = 0.05 \, \text{M} \]

### Final Answer:
The concentration of \([OH^-]\) at equilibrium in the mixture will be:
\[ \boxed{0.05 \, \text{M}} \]