What is the value of `pK_(b)(CH_(3)COO^(c-))` if `wedge^(@)._(m)=390 S cm^(-1) mol ^(-1)` and `wedge_(m)=7.8 S cm^(2) mol^(-1)` for `0.04 M` of `CH_(3)COOH` at `25^(@)C`?

Options

A. `9.3`
B. `9.2`
C. `4.7`
D. `4.8`

Explanation

To find the value of \( pK_b(CH_3COO^-) \), we will follow these steps:

### Step 1: Calculate the degree of ionization (\( \alpha \))
The degree of ionization (\( \alpha \)) can be calculated using the formula:
\[ \alpha = \frac{\lambda_m}{\lambda_m^\infty} \]
where:
- \( \lambda_m \) is the molar conductance at the given concentration (7.8 S cm² mol⁻¹),
- \( \lambda_m^\infty \) is the molar conductance at infinite dilution (390 S cm² mol⁻¹).

Substituting the values:
\[ \alpha = \frac{7.8}{390} = 0.02 \]

### Step 2: Calculate the dissociation constant (\( K_a \))
For weak electrolytes like acetic acid, the dissociation constant \( K_a \) can be calculated using the formula:
\[ K_a = c \alpha^2 \]
where \( c \) is the concentration of the acetic acid (0.04 M).

Substituting the values:
\[ K_a = 0.04 \times (0.02)^2 = 0.04 \times 0.0004 = 16 \times 10^{-6} \]

### Step 3: Calculate \( pK_a \)
The \( pK_a \) is calculated using the formula:
\[ pK_a = -\log K_a \]
Substituting the value of \( K_a \):
\[ pK_a = -\log(16 \times 10^{-6}) \]

Calculating \( pK_a \):
\[ pK_a \approx 4.8 \]

### Step 4: Calculate \( pK_b \)
We know that:
\[ pK_a + pK_b = 14 \]
Thus, we can find \( pK_b \) as follows:
\[ pK_b = 14 - pK_a = 14 - 4.8 = 9.2 \]

### Final Answer
The value of \( pK_b(CH_3COO^-) \) is **9.2**.

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