The pH of 1 M solution of a weak monobasic acid (HA) is 2. Then, the van't Hoff factor is

Options

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

Explanation

To find the van't Hoff factor (i) for the weak monobasic acid (HA) given that the pH of a 1 M solution is 2, we can follow these steps:

### Step 1: Determine the concentration of hydrogen ions ([H⁺])
The pH of the solution is given as 2. We can calculate the concentration of hydrogen ions using the formula:

\[ [H^+] = 10^{-\text{pH}} = 10^{-2} \, \text{M} \]

### Step 2: Set up the dissociation equation
For a weak monobasic acid HA, the dissociation can be represented as:

\[ HA \rightleftharpoons H^+ + A^- \]

From this equation, we can see that for every mole of HA that dissociates, one mole of H⁺ and one mole of A⁻ are produced.

### Step 3: Define the initial concentration and change
Let the initial concentration of HA be C = 1 M. If α is the degree of dissociation, then at equilibrium:

- Concentration of HA = C(1 - α) = 1(1 - α) = 1 - α
- Concentration of H⁺ = Cα = 1α = α
- Concentration of A⁻ = Cα = 1α = α

Since we found that [H⁺] = 0.01 M from Step 1, we can equate:

\[ \alpha = 0.01 \]

### Step 4: Calculate the van't Hoff factor (i)
The van't Hoff factor (i) for a weak acid is given by the formula:

\[ i = 1 + n(1 - \alpha) \]

where n is the number of particles the acid dissociates into. For a monobasic acid, n = 2 (1 H⁺ and 1 A⁻). Therefore:

\[ i = 1 + 2(α) = 1 + 2(0.01) = 1 + 0.02 = 1.02 \]

### Conclusion
Thus, the van't Hoff factor (i) for the weak monobasic acid (HA) is approximately **1.02**.

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