The pH of a 2 M solution of a weak monobasic acid (HA) is 4. What is the value of the Van’t Hoff factor?

Options

A. 0.00005
B. 1.05
C. 1.005
D. 1.00005

Answer

To solve the problem, we need to determine the Van't Hoff factor (i) for a weak monobasic acid (HA) given that its concentration is 2 M and the pH of the solution is 4.

### Step-by-Step Solution:

1. **Calculate the concentration of H⁺ ions:**
- The pH of the solution is given as 4.
- The concentration of H⁺ ions can be calculated using the formula:
\[ [H^+] = 10^{-\text{pH}} = 10^{-4} \, \text{M} \]

2. **Set up the dissociation of the weak acid:**
- The weak acid (HA) dissociates according to the equation:
\[ HA \rightleftharpoons H^+ + A^- \]
- Let the initial concentration of HA be \( C \) (which is 2 M) and \( \alpha \) be the degree of dissociation.

3. **Express the concentrations at equilibrium:**
- Initially, the concentration of HA is \( C = 2 \, \text{M} \).
- At equilibrium, the concentration of H⁺ ions formed is \( C\alpha \).
- Therefore, at equilibrium:
- Concentration of HA = \( C(1 - \alpha) = 2(1 - \alpha) \)
- Concentration of H⁺ = \( C\alpha = 2\alpha \)
- Concentration of A⁻ = \( C\alpha = 2\alpha \)

4. **Set up the equation for H⁺ concentration:**
- From the previous step, we know that:
\[ [H^+] = 2\alpha \]
- We also know from step 1 that:
\[ [H^+] = 10^{-4} \, \text{M} \]
- Therefore, we can set up the equation:
\[ 2\alpha = 10^{-4} \]

5. **Solve for α:**
- Rearranging the equation gives:
\[ \alpha = \frac{10^{-4}}{2} = 5 \times 10^{-5} \]

6. **Calculate the Van't Hoff factor (i):**
- The Van't Hoff factor for a weak acid that dissociates into H⁺ and A⁻ is given by:
\[ i = 1 + \alpha \]
- Substituting the value of α:
\[ i = 1 + 5 \times 10^{-5} = 1.00005 \]

7. **Final result:**
- The value of the Van't Hoff factor (i) is approximately:
\[ i \approx 1.00005 \]

### Conclusion:
The Van't Hoff factor for the weak monobasic acid (HA) in the given conditions is approximately **1.00005**.

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