JEE MAIN - Chemistry (2023 - 15th April Morning Shift - No. 9)
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:
(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:
$(\mathrm{A}),(\mathrm{B}),(\mathrm{C})$
(B), (C)
(B), (C), (D)
(A), (D)
Explanation
(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.
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.
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