Question

**Checking each statement:** **Statement A:** For $1.0 \times 10^{-8}$ M HCl, the concentration is comparable to $[\text{H}^+]$ from water, so water's contribution cannot be ignored. Let $x$ = $[\text{H}^+]$ contributed by water. Then: $$[\text{H}^+]_{\text{total}} = 10^{-8} + x$$ Using $K_w = [\text{H}^+][\text{OH}^-] = 10^{-14}$ and charge balance: $$[\text{H}^+]_{\text{total}} = \frac{10^{-8} + \sqrt{(10^{-8})^2 + 4 \times 10^{-14}}}{2} \approx 1.05 \times 10^{-7} \text{ M}$$ $$\text{pH} = -\log(1.05 \times 10^{-7}) \approx 6.98$$ The solution is acidic (pH < 7), not pH = 8. **Statement A is incorrect.** ✗ **Statement B:** The conjugate base is formed by removing one proton: $$\text{H}_2\text{PO}_4^- \rightarrow \text{H}^+ + \text{HPO}_4^{2-}$$ So the conjugate base of $\text{H}_2\text{PO}_4^-$ is $\text{HPO}_4^{2-}$. **Statement B is correct.** ✓ **Statement C:** The autoprotolysis of water: $$2\text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^-$$ This is an endothermic process. Increasing temperature shifts equilibrium forward, increasing $K_w$. (At 25°C, $K_w = 10^{-14}$; at higher temperatures, $K_w > 10^{-14}$.) **Statement C is correct.** ✓ **Statement D:** At the half-neutralisation point, moles of salt = moles of remaining acid: $$[\text{A}^-] = [\text{HA}]$$ Using the Henderson–Hasselbalch equation: $$\text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]} = \text{p}K_a + \log 1 = \text{p}K_a$$ The correct result is $\text{pH} = \text{p}K_a$, not $\text{pH} = \dfrac{1}{2}\text{p}K_a$. **Statement D is incorrect.** ✗ **Answer: (B, C)**