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 . $\underset{0.1 M , 20 ml }{ CH _3 COOH }+\underset{0.1 M , 10 ml }{ NaOH } \longrightarrow CH _3 COONaH _2 O$
$pH = pKa \, \Rightarrow \, [H^{+}]$ will not change on dilution correct match : P-1
Q. $\underset{0.1M, 20 ml}{CH _3 COOH} +\underset{0.1 M , 20 ml }{ NaOH } \longrightarrow CH _2 COONa + H _2 O$
$\left[OH^{-}\right] = \sqrt{K_{H}C} = \sqrt{\left(\frac{k_{w}}{k_{a}} C\right)}$
$ \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. $\underset{0.1 M , 20 ml }{ NH _4 OH }+\underset{0.1 M , 20 ml }{ HCl } \longrightarrow \underset{\substack{0.05 M }}{ NH _4 Cl }$
$\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