A `0.1molar` solution of weak base `BOH` is `1%` dissociated. If `0.2mol` of `BCl` is added in `1L` solution of `BOH`. The degree of dissociation of `BOH` will become

Options

A. `0.02`
B. `0.005`
C. `5xx10^(-5)`
D. `2xx10^(-3)`

Explanation

To solve the problem, we need to find the new degree of dissociation of the weak base \( BOH \) after adding \( BCl \). Here’s a step-by-step solution:

### Step 1: Calculate the initial concentration of \( BOH \) and its degree of dissociation
Given:
- Initial concentration of \( BOH = 0.1 \, \text{M} \)
- Degree of dissociation \( \alpha = 1\% = 0.01 \)

The amount of \( BOH \) that dissociates can be calculated as:
\[ \text{Dissociated concentration} = 0.1 \times 0.01 = 0.001 \, \text{M} \]

### Step 2: Determine the concentrations of ions at equilibrium
At equilibrium, the concentrations will be:
- Concentration of \( BOH \) remaining:
\[ [BOH] = 0.1 - 0.001 = 0.099 \, \text{M} \]
- Concentration of \( B^+ \) produced from dissociation:
\[ [B^+] = 0.001 \, \text{M} \]
- Concentration of \( OH^- \) produced from dissociation:
\[ [OH^-] = 0.001 \, \text{M} \]

### Step 3: Add \( BCl \) to the solution
When \( 0.2 \, \text{mol} \) of \( BCl \) is added to \( 1 \, \text{L} \) of the solution, the concentration of \( B^+ \) from \( BCl \) will be:
\[ [B^+]_{\text{from } BCl} = \frac{0.2 \, \text{mol}}{1 \, \text{L}} = 0.2 \, \text{M} \]

### Step 4: Calculate the new concentration of \( B^+ \)
The total concentration of \( B^+ \) after adding \( BCl \) will be:
\[ [B^+]_{\text{total}} = [B^+]_{\text{from dissociation}} + [B^+]_{\text{from } BCl} = 0.001 + 0.2 = 0.201 \, \text{M} \]

### Step 5: Use the equilibrium expression to find the new degree of dissociation
The equilibrium expression for the weak base \( BOH \) dissociation is given by:
\[ K_b = \frac{[B^+][OH^-]}{[BOH]} \]
Let \( \alpha' \) be the new degree of dissociation. Then:
\[ [B^+] = 0.201 \, \text{M} \quad \text{and} \quad [OH^-] = \alpha' \, \text{M} \]
The concentration of \( BOH \) will be:
\[ [BOH] = 0.1 - \alpha' \approx 0.1 \, \text{M} \quad (\text{since } \alpha' \text{ will be small}) \]

Substituting into the \( K_b \) expression:
\[ K_b = \frac{(0.201)(\alpha')}{0.1} \]

### Step 6: Calculate \( K_b \)
From the initial dissociation:
\[ K_b = [B^+] \cdot [OH^-] = (0.001)(0.001) = 10^{-6} \]

### Step 7: Solve for \( \alpha' \)
Setting the two expressions for \( K_b \) equal:
\[ 10^{-6} = \frac{(0.201)(\alpha')}{0.1} \]
\[ \alpha' = \frac{10^{-6} \cdot 0.1}{0.201} = \frac{10^{-7}}{0.201} \approx 4.975 \times 10^{-6} \]

### Final Answer
The new degree of dissociation of \( BOH \) after adding \( BCl \) is approximately:
\[ \alpha' \approx 5 \times 10^{-6} \]