Question

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: Degree of dissociation $(\alpha )$ of weak acid and weak base is << 1; degree of hydrolysis of salt $<<1;[H^{+}]$ represents the concentration of H+ ions)
Match each set of hybrid orbitals from LIST-I with complex(es) given in LIST-II.

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 two 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 $\frac{1}{\sqrt{2}}$ times of its initial value on dilution
		5	the value of $[H^{+}]$ changes to $\sqrt{2}$ times of 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

• A. P $\to$ 4; Q $\to$ 2; R $\to$ 3; S $\to$ 1
• B. P $\to$ 4; Q $\to$ 3; R $\to$ 2; S $\to$ 3
• C. P $\to$ 1; Q $\to$ 4; R $\to$ 5; S $\to$ 3
• D. P $\to$ 1; Q → 5; R → 4; S → 1

Answer 1

Answer: (D)

P . $\underset{0.1M,20ml}{C{H}_{3}COOH}+\underset{0.1M,10ml}{\mathrm{NaOH}}⟶C{H}_{3}COONa{H}_{2}O$
$pH=pKa\Rightarrow [H^{+}]$ will not change on dilution correct match : P-1
Q. $\underset{0.1M,20ml}{C{H}_{3}COOH}+\underset{0.1M,20ml}{\mathrm{NaOH}}⟶C{H}_{2}COONa+H_{2}O$
$\left[O{H}^{-}\right]=\sqrt{K_{H}C}=\sqrt{\left(\frac{k_w}{k_a}C\right)}$
${\left[H^{+}\right]}_1=\sqrt{\frac{k_w{k}_a}{C}}$
$\frac{{\left[H^{+}\right]}_2}{{\left[H^{+}\right]}_1}=\sqrt{\frac{C_1}{C_2}}=\sqrt{\frac{0.05}{0.025}}=\sqrt{2}$
correct match : Q-5
R. $\underset{0.1M,20ml}{N{H}_{4}OH}+\underset{0.1M,20ml}{\mathrm{HCl}}⟶\underset{\begin{array}{c}0.05M\end{array}}{N{H}_{4}Cl}$
$\left[H^{+}\right]=\sqrt{K_{H}C}$
$\frac{{\left[H^{+}\right]}_2}{{\left[H^{+}\right]}_1}=\sqrt{\frac{C_2}{C_1}}=\frac{1}{\sqrt{2}}$
S. Because of dilution solubility does not change so $[H^{+}]$ = constant