Medium MCQ +4 / -1 PYQ · JEE Mains 2023

Which of the following statement(s) is/are correct?

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

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

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

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

Choose the correct answer from the options given below:

  1. A $(\mathrm{A}),(\mathrm{B}),(\mathrm{C})$
  2. B (B), (C) Correct answer
  3. C (B), (C), (D)
  4. D (A), (D)

Solution

(A) The $\mathrm{pH}$ of $1 \times 10^{-8}~ \mathrm{M} ~\mathrm{HCl}$ solution is 8. <br/><br/> 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: <br/><br/> $\mathrm{pH} = -\log [\mathrm{H}^+] = -\log (1 \times 10^{-8}) = 8$ <br/><br/> 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. <br/><br/> (B) The conjugate base of $\mathrm{H}_{2} \mathrm{PO}_{4}^{-}$ is $\mathrm{HPO}_{4}^{2-}$. <br/><br/> This statement is correct. The conjugate base of an acid is formed when it loses one H+ ion: <br/><br/> $\mathrm{H}_{2} \mathrm{PO}_{4}^{-} \rightarrow \mathrm{HPO}_{4}^{2-} + \mathrm{H}^{+}$ <br/><br/> (C) $\mathrm{K}_{\mathrm{w}}$ increases with an increase in temperature. <br/><br/> 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: <br/><br/> $\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}^{+} + \mathrm{OH}^{-}$ <br/><br/> As the temperature increases, the equilibrium shifts towards the formation of more $\mathrm{H}^{+}$ and $\mathrm{OH}^{-}$ ions, leading to an increase in $\mathrm{K}_{\mathrm{w}}$. <br/><br/> (D) When a solution of a weak monoprotic acid is titrated against a strong base at the half-neutralization point, $\mathrm{pH}=\frac{1}{2} \mathrm{pK}_{\mathrm{a}}$ <br/><br/> 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: <br/><br/> $\mathrm{pH} = \mathrm{pK}_{\mathrm{a}} + \log \frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}$ <br/><br/> At the half-neutralization point, the ratio of [A-] to [HA] is 1, so the equation becomes: <br/><br/> $\mathrm{pH} = \mathrm{pK}_{\mathrm{a}} + \log (1) = \mathrm{pK}_{\mathrm{a}}$ <br/><br/> Therefore, the correct answer is: <br/><br/> (B) and (C) are correct.

About this question

Subject: Chemistry · Chapter: Equilibrium · Topic: Chemical Equilibrium and Kc, Kp

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