The dilution processes of different aqueous solutions with water are given in LIST-I. The effects of dilution of the solutions on [ H + ] are given in LIST-II. (Note: The degree of dissociation α of a

Question

The dilution processes of different aqueous solutions with water are given in LIST-I. The effects of dilution of the solutions on $[H^{+}]$ are given in LIST-II.
(Note: The degree of dissociation $(α)$ of a weak acid and a weak base is $< < 1;$ the degree of hydrolysis of salt is $< < 1; [H^{+}]$ represents the concentration of $H^{+}$ ions)

	LIST –I		LIST -II
A)	$10 mL of 0.1 M NaOH + 20 mL of 0 .1 M$ acetic acid diluted to $60 mL$ .	P)	The value of $[H^{+}]$ does not change on dilution.
B)	$20 mL of 0.1 M NaOH + 20 mL of 0.1 M$ acetic acid diluted to $80 mL$ .	Q)	The value of ${[H}^{+}]$ changes to half of its initial value on dilution.
C)	$20 mL of 0 .1 M HCl + 20 mL of 0 .1 M$ ammonia solution diluted to $80 mL$ .	R)	The value of $[H^{+}]$ changes to two times its initial value on dilution.
D)	$10 mL$ saturated solution of $Ni {(OH)}_{2}$ in equilibrium with excess of solid $Ni {(OH)}_{2}$ is diluted to $20 mL$ (solid $Ni {(OH)}_{2}$ is still present after dilution).	S)	The value of $[H^{+}]$ changes to $\sqrt{}$ times its initial value on dilution.
		T)	The value of $[H^{+}]$ changes to $\sqrt{2}$ times its initial value on dilution.

Match each process given in LIST-I with one or more effect(s) in LIST-II. The correct option is

Options:

Answer:

B

Solution:

$(A)$

$pH = pKa \Rightarrow [H^{+}]$ will not change on dilution.
$(B)$

$[{OH}^{-}] = \sqrt{K_{H} C} = \sqrt{(\frac{k_{w}}{k_{a}} C)}$
${[H^{+}]}_{1} = \sqrt{\frac{k_{w} k_{a}}{C}}$
$\frac{{[H^{+}]}_{2}}{{[H^{+}]}_{1}} = \sqrt{\frac{C_{1}}{C_{2}}} = \sqrt{\frac{0 .05}{0 .025}} = \sqrt{2}$
$(C)$

$[H^{+}] = \sqrt{K_{H} C}$
$\frac{{[H^{+}]}_{2}}{{[H^{+}]}_{1}} = \sqrt{\frac{C_{2}}{C_{1}}} = \frac{1}{\sqrt{2}}$
$(D)$ Because of dilution, the solubility does not change. So, $[H^{+}] = constant$ .

TestHub · Accepted Answer

( A ) pH = pKa ⇒ H + will not change on dilution. ( B ) OH - = K H C = k w k a C H + 1 = k w k a C H + 2 H + 1 = C 1 C 2 = 0 .05 0 .025 = 2 ( C ) H + = K H C H + 2 H + 1 = C 2 C 1 = 1 2 ( D ) Because of dilution, the solubility does not change. So, H + = constant .

Chemistry - Ionic Equilibrium Question with Solution | TestHub

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