For a weak monobasic acid, if `pK_(a)=4.` then at a concentration of 0.01 M of the acid solution, the van't Hoff factor is

Options

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

Explanation

To solve the problem, we need to find the van't Hoff factor (i) for a weak monobasic acid given its pKa and concentration. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the given data
We are given:
- pKa of the weak monobasic acid = 4
- Concentration of the acid solution (C) = 0.01 M

### Step 2: Calculate Ka from pKa
The relationship between pKa and Ka is given by:
\[ K_a = 10^{-pK_a} \]
Substituting the given pKa value:
\[ K_a = 10^{-4} = 0.0001 \]

### Step 3: Determine the degree of ionization (α)
For a weak acid, the degree of ionization (α) can be calculated using the formula:
\[ \alpha = \sqrt{\frac{K_a}{C}} \]
Substituting the values of Ka and C:
\[ \alpha = \sqrt{\frac{10^{-4}}{0.01}} = \sqrt{10^{-4} \times 10^{2}} = \sqrt{10^{-2}} = 0.1 \]

### Step 4: Determine the number of particles (n)
For a monobasic acid (HA), it ionizes as follows:
\[ HA \rightleftharpoons H^+ + A^- \]
This means that for every mole of acid that ionizes, it produces 1 mole of \(H^+\) and 1 mole of \(A^-\). Therefore, the total number of particles formed (n) is:
\[ n = 2 \]

### Step 5: Calculate the van't Hoff factor (i)
The van't Hoff factor (i) can be calculated using the formula:
\[ i = 1 + (n - 1) \cdot \alpha \]
Substituting the values of n and α:
\[ i = 1 + (2 - 1) \cdot 0.1 = 1 + 0.1 = 1.1 \]

### Conclusion
The van't Hoff factor (i) for the weak monobasic acid at a concentration of 0.01 M is:
\[ \boxed{1.1} \]