JEE Main 2023 — Ionic Equilibrium Question with Solution

(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:

${\mathrm{H}}_2{\mathrm{PO}}_4^{-}\to {\mathrm{HPO}}_4^{2-}+{\mathrm{H}}^{+}$

(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.