`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^(-), we will follow these steps:

### Step 1: Identify the relevant acid and base
The given ion CH2ClCOO^(-) is the conjugate base of the acid CH2ClCOOH.

### Step 2: Write the dissociation reaction
The dissociation of the acid can be represented as:
\[ \text{CH}_2\text{ClCOOH} \rightleftharpoons \text{CH}_2\text{ClCOO}^- + \text{H}^+ \]

### Step 3: Define the equilibrium constant (Kb)
The base dissociation constant (Kb) for CH2ClCOO^(-) is given as:
\[ K_b = 6.4 \times 10^{-12} \]

### Step 4: Set up the equilibrium expression
For the dissociation of the acid, we can express Kb as:
\[ K_b = \frac{[\text{CH}_2\text{ClCOO}^-][\text{H}^+]}{[\text{CH}_2\text{ClCOOH}]} \]

Assuming the initial concentration of CH2ClCOOH is C (0.1 M), and letting α be the degree of dissociation, we have:
- At equilibrium:
- [CH2ClCOO^-] = Cα
- [H^+] = Cα
- [CH2ClCOOH] = C(1 - α)

Thus, the equilibrium expression becomes:
\[ K_b = \frac{(Cα)(Cα)}{C(1 - α)} \]

### Step 5: Simplify the expression
Since α is very small compared to 1 (α << 1), we can approximate:
\[ K_b \approx \frac{Cα^2}{C} = α^2 \]

### Step 6: Substitute known values
Substituting the known values into the equation:
\[ 6.4 \times 10^{-12} = (0.1)α^2 \]

### Step 7: Solve for α
Rearranging gives:
\[ α^2 = \frac{6.4 \times 10^{-12}}{0.1} = 6.4 \times 10^{-11} \]

Taking the square root:
\[ α = \sqrt{6.4 \times 10^{-11}} \approx 8.0 \times 10^{-6} \]

### Step 8: Calculate [H^+]
The concentration of H^+ ions is given by:
\[ [H^+] = Cα = 0.1 \times 8.0 \times 10^{-6} = 8.0 \times 10^{-7} \, \text{M} \]

### Step 9: Calculate pH
Finally, we can calculate the pH using the formula:
\[ \text{pH} = -\log[H^+] \]
\[ \text{pH} = -\log(8.0 \times 10^{-7}) \approx 6.096 \]

Thus, the pH of the 0.1 M CH2ClCOO^(-) solution is approximately **6.10**.

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