To solve the question, we need to evaluate the truthfulness of three statements regarding chemical equilibrium, particularly concerning water's autoprotolysis constant, the titration of weak acids, and the pH of a dilute HCl solution.
### Step-by-Step Solution:
**Statement 1: Autoprotolysis constant of water increases with the increase in temperature.**
1. **Understanding Autoprotolysis of Water:**
- Autoprotolysis of water refers to the self-ionization of water, which can be represented as:
\[
2H_2O \rightleftharpoons H_3O^+ + OH^-
\]
- The equilibrium constant for this reaction is known as the autoprotolysis constant, denoted as \( K_w \).
2. **Effect of Temperature on \( K_w \):**
- The reaction is endothermic, meaning it absorbs heat.
- According to Le Chatelier's principle, increasing the temperature shifts the equilibrium to favor the endothermic reaction, resulting in an increase in the concentrations of \( H_3O^+ \) and \( OH^- \).
3. **Conclusion:**
- Therefore, as temperature increases, \( K_w \) increases.
- **Verdict:** **True**
---
**Statement 2: When a solution of a weak monobasic acid is titrated with a strong base at half neutralization point, \( pH = pK_a + 1 \).**
1. **Understanding Half Neutralization Point:**
- At the half-neutralization point, half of the weak acid is converted to its conjugate base.
- For a weak acid \( HA \) titrated with a strong base \( NaOH \):
\[
HA + NaOH \rightarrow A^- + H_2O
\]
2. **Using Henderson-Hasselbalch Equation:**
- The Henderson-Hasselbalch equation is given by:
\[
pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)
\]
- At half-neutralization, \([A^-] = [HA]\), thus:
\[
\log\left(\frac{[A^-]}{[HA]}\right) = \log(1) = 0
\]
- Therefore, \( pH = pK_a \).
3. **Conclusion:**
- The statement claims \( pH = pK_a + 1 \), which is incorrect.
- **Verdict:** **False**
---
**Statement 3: The pH of \( 10^{-8} \) m HCl is 8.**
1. **Understanding the pH of Strong Acids:**
- HCl is a strong acid and completely dissociates in solution.
- The concentration of \( H^+ \) ions from \( 10^{-8} \) m HCl is \( 10^{-8} \) m.
2. **Consideration of Water's Contribution:**
- Pure water has a \( [H^+] \) concentration of \( 10^{-7} \) m.
- Therefore, the total \( [H^+] \) in the solution becomes:
\[
[H^+] = 10^{-8} + 10^{-7} = 1.1 \times 10^{-7} \text{ m}
\]
3. **Calculating the pH:**
- The pH is calculated as:
\[
pH = -\log(1.1 \times 10^{-7}) \approx 6.96
\]
- Thus, the pH cannot be 8.
4. **Conclusion:**
- The statement that the pH of \( 10^{-8} \) m HCl is 8 is incorrect.
- **Verdict:** **False**
---
### Final Evaluation of Statements:
- Statement 1: **True**
- Statement 2: **False**
- Statement 3: **False**
### Correct Answer:
**Option B: TFF (True, False, False)**
---
