`K_(b)" for "CH_(2)ClCOO^(-)` is `6.4xx10^(-12).` The pH of `"0.1 M "CH_(2)ClCOO^(-)` in water is :

Options

A. 7.9
B. 6.9
C. 1.9
D. 12.1

Answer

To find the pH of a 0.1 M solution of CH₂ClCOO⁻ in water, we will follow these steps:

### Step 1: Identify the relevant equilibrium reaction
The base CH₂ClCOO⁻ (the conjugate base of the acid CH₂ClCOOH) will react with water to produce CH₂ClCOOH and OH⁻ ions:
\[ \text{CH}_2\text{ClCOO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_2\text{ClCOOH} + \text{OH}^- \]

### Step 2: Write the expression for Kb
The base dissociation constant (Kb) for this reaction is given by:
\[ K_b = \frac{[\text{CH}_2\text{ClCOOH}][\text{OH}^-]}{[\text{CH}_2\text{ClCOO}^-]} \]

### Step 3: Set up the equilibrium concentrations
Let \( \alpha \) be the degree of dissociation. Initially, we have:
- \([\text{CH}_2\text{ClCOO}^-] = 0.1 \, \text{M}\)
- \([\text{CH}_2\text{ClCOOH}] = 0 \, \text{M}\)
- \([\text{OH}^-] = 0 \, \text{M}\)

At equilibrium, we have:
- \([\text{CH}_2\text{ClCOO}^-] = 0.1 - \alpha\)
- \([\text{CH}_2\text{ClCOOH}] = \alpha\)
- \([\text{OH}^-] = \alpha\)

### Step 4: Substitute into the Kb expression
Substituting these values into the Kb expression gives:
\[ K_b = \frac{\alpha \cdot \alpha}{0.1 - \alpha} \]
\[ K_b = \frac{\alpha^2}{0.1 - \alpha} \]

Given that \( K_b = 6.4 \times 10^{-12} \), we can simplify the equation assuming \( \alpha \) is small compared to 0.1:
\[ K_b \approx \frac{\alpha^2}{0.1} \]

### Step 5: Solve for α
Now we can solve for \( \alpha \):
\[ 6.4 \times 10^{-12} = \frac{\alpha^2}{0.1} \]
\[ \alpha^2 = 6.4 \times 10^{-12} \times 0.1 \]
\[ \alpha^2 = 6.4 \times 10^{-13} \]
\[ \alpha = \sqrt{6.4 \times 10^{-13}} \]
\[ \alpha \approx 8.0 \times 10^{-7} \]

### Step 6: Calculate [OH⁻]
Since \([\text{OH}^-] = \alpha\):
\[ [\text{OH}^-] = 8.0 \times 10^{-7} \, \text{M} \]

### Step 7: Calculate pOH
Using the formula for pOH:
\[ \text{pOH} = -\log[\text{OH}^-] \]
\[ \text{pOH} = -\log(8.0 \times 10^{-7}) \]
\[ \text{pOH} \approx 6.1 \]

### Step 8: Calculate pH
Using the relationship \( \text{pH} + \text{pOH} = 14 \):
\[ \text{pH} = 14 - \text{pOH} \]
\[ \text{pH} = 14 - 6.1 \]
\[ \text{pH} \approx 7.9 \]

### Final Answer
The pH of a 0.1 M solution of CH₂ClCOO⁻ in water is approximately **7.9**.