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(A) $P$: This is an acidic buffer solution $(CH_3COOH + CH_3COONa)$. The $[H^{+}]$ of a buffer solution remains constant upon dilution. Thus,$P \right

Vedclass · Accepted Answer

$P$ $\rightarrow 1, Q$ $\rightarrow 4, R$ $\rightarrow 3, S$ $\rightarrow 1$

Vedclass · Answer

(A) $P$: This is an acidic buffer solution $(CH_3COOH + CH_3COONa)$. The $[H^{+}]$ of a buffer solution remains constant upon dilution. Thus,$P \rightarrow 1$.
$Q$: This is a salt of a weak acid and a strong base $(CH_3COONa)$. The $[H^{+}]$ is given by $[H^{+}] = \sqrt{K_w K_a / C}$. Since $[H^{+}] \propto 1/\sqrt{C}$,and the concentration $C$ decreases by a factor of $2$ (from $40 \ mL$ to $80 \ mL$),$[H^{+}]$ increases by $\sqrt{2}$. Thus,$Q \rightarrow 4$.
$R$: This is a salt of a strong acid and a weak base $(NH_4Cl)$. The $[H^{+}]$ is given by $[H^{+}] = \sqrt{K_h C} = \sqrt{(K_w/K_b) C}$. Since $[H^{+}] \propto \sqrt{C}$,and the concentration $C$ decreases by a factor of $2$ (from $40 \ mL$ to $80 \ mL$),$[H^{+}]$ decreases by $1/\sqrt{2}$. Thus,$R \rightarrow 3$.
$S$: This is a saturated solution of a sparingly soluble base in equilibrium with its solid. The $[OH^{-}]$ remains constant due to the common ion effect of the solid. Since $[H^{+}][OH^{-}] = K_w$,$[H^{+}]$ remains constant. Thus,$S \rightarrow 1$.
Therefore,the correct matching is $P$ $\rightarrow 1, Q$ $\rightarrow 4, R$ $\rightarrow 3, S$ $\rightarrow 1$.

$LIST-I$	$LIST-II$
$P$. ($10 \ mL$ of $0.1 \ M$ $NaOH$ + $20 \ mL$ of $0.1 \ M$ acetic acid) diluted to $60 \ mL$	$1$. The value of $[H^{+}]$ does not change on dilution
$Q$. ($20 \ mL$ of $0.1 \ M$ $NaOH$ + $20 \ mL$ of $0.1 \ M$ acetic acid) diluted to $80 \ mL$	$2$. The value of $[H^{+}]$ changes to half of its initial value on dilution
$R$. ($20 \ mL$ of $0.1 \ M$ $HCl$ + $20 \ mL$ of $0.1 \ M$ ammonia solution) diluted to $80 \ mL$	$3$. The value of $[H^{+}]$ changes to $1/\sqrt{2}$ times of its initial value on dilution
$S$. $10 \ mL$ saturated solution of $Ni(OH)_2$ in equilibrium with excess solid $Ni(OH)_2$ is diluted to $20 \ mL$ (solid $Ni(OH)_2$ is still present after dilution)	$4$. The value of $[H^{+}]$ changes to $\sqrt{2}$ times of its initial value on dilution

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