If `pK_(a)=-"log"K_(a)=4`, and `K_(a)=Cx^(2-)` then Van't Hoff factor for weak monobasic acid when `C=0.01M` is:

Options

A. `1.01`
B. `1.02`
C. `1.10`
D. `1.20`

Explanation

To solve the problem, we need to calculate the Van't Hoff factor (i) for a weak monobasic acid given the pKa and the concentration of the acid. Here is the step-by-step solution:

### Step 1: Understand the relationship between pKa and Ka
Given that:
\[ pK_a = -\log K_a = 4 \]
We can find \( K_a \) using the formula:
\[ K_a = 10^{-pK_a} = 10^{-4} \]

### Step 2: Define the weak acid dissociation
Let the weak monobasic acid be represented as \( HA \). It dissociates in water as follows:
\[ HA \rightleftharpoons H^+ + A^- \]

### Step 3: Set up the equilibrium expression
Let \( C \) be the initial concentration of the acid (given as 0.01 M). At equilibrium, if \( \alpha \) is the degree of dissociation, the concentrations will be:
- Concentration of \( HA \): \( C(1 - \alpha) \)
- Concentration of \( H^+ \): \( C\alpha \)
- Concentration of \( A^- \): \( C\alpha \)

### Step 4: Write the expression for \( K_a \)
The expression for \( K_a \) at equilibrium is:
\[ K_a = \frac{[H^+][A^-]}{[HA]} = \frac{C\alpha \cdot C\alpha}{C(1 - \alpha)} \]
This simplifies to:
\[ K_a = \frac{C^2 \alpha^2}{C(1 - \alpha)} = \frac{C\alpha^2}{1 - \alpha} \]

### Step 5: Substitute the value of \( K_a \)
From Step 1, we have:
\[ K_a = 10^{-4} \]
Thus, we can set up the equation:
\[ 10^{-4} = \frac{C\alpha^2}{1 - \alpha} \]

### Step 6: Substitute the concentration \( C \)
Given \( C = 0.01 \) M, we substitute this into the equation:
\[ 10^{-4} = \frac{0.01\alpha^2}{1 - \alpha} \]

### Step 7: Solve for \( \alpha \)
Assuming \( \alpha \) is small (which is typical for weak acids), we can simplify \( 1 - \alpha \approx 1 \):
\[ 10^{-4} \approx 0.01\alpha^2 \]
\[ \alpha^2 \approx \frac{10^{-4}}{0.01} = 10^{-2} \]
Taking the square root:
\[ \alpha \approx 10^{-1} = 0.1 \]

### Step 8: Calculate the Van't Hoff factor (i)
The Van't Hoff factor \( i \) for a weak acid is given by:
\[ i = 1 + \alpha \]
Substituting the value of \( \alpha \):
\[ i = 1 + 0.1 = 1.1 \]

### Final Answer
The Van't Hoff factor for the weak monobasic acid when \( C = 0.01 \, M \) is:
\[ i = 1.1 \]