QJEEM23Chf7b2: JEE Main 2023 Chemistry - Class 11 JEEM23Chf7b2 Solution: Which of the following statement(s) is/are correct? (A) The of…

(A) The $\mathrm{pH}$ of $1 \times 10^{-8}~ \mathrm{M} ~\mathrm{HCl}$ solution is 8.

This statement is incorrect. For a strong acid like HCl, the concentration of H+ ions will be the same as the concentration of the acid, i.e., $1 \times 10^{-8}~\mathrm{M}$ . The pH can be calculated using the formula:

$\mathrm{pH} = -\log [\mathrm{H}^+] = -\log (1 \times 10^{-8}) = 8$

However, because the concentration is so low, it approaches the range where water auto-ionization becomes significant. In this case, the solution pH will be slightly higher than 7, but not exactly 8.

(B) The conjugate base of $\mathrm{H}_{2} \mathrm{PO}_{4}^{-}$ is $\mathrm{HPO}_{4}^{2-}$ .

This statement is correct. The conjugate base of an acid is formed when it loses one H+ ion:

$\mathrm{H}_{2} \mathrm{PO}_{4}^{-} \rightarrow \mathrm{HPO}_{4}^{2-} + \mathrm{H}^{+}$

(C) $\mathrm{K}_{\mathrm{w}}$ increases with an increase in temperature.

This statement is correct. The ion product of water, $\mathrm{K}_{\mathrm{w}}$ , increases with increasing temperature. This is because the auto-ionization of water is an endothermic process, meaning it absorbs heat:

$\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}^{+} + \mathrm{OH}^{-}$

As the temperature increases, the equilibrium shifts towards the formation of more $\mathrm{H}^{+}$ and $\mathrm{OH}^{-}$ ions, leading to an increase in $\mathrm{K}_{\mathrm{w}}$ .

(D) When a solution of a weak monoprotic acid is titrated against a strong base at the half-neutralization point, $\mathrm{pH}=\frac{1}{2} \mathrm{pK}_{\mathrm{a}}$

This statement is incorrect. At the half-neutralization point, the concentration of the weak acid ([HA]) is equal to the concentration of its conjugate base ([A-]). According to the Henderson-Hasselbalch equation:

$\mathrm{pH} = \mathrm{pK}_{\mathrm{a}} + \log \frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}$

At the half-neutralization point, the ratio of [A-] to [HA] is 1, so the equation becomes:

$\mathrm{pH} = \mathrm{pK}_{\mathrm{a}} + \log (1) = \mathrm{pK}_{\mathrm{a}}$

Therefore, the correct answer is:

(B) and (C) are correct.