Question

Dilution processes of different aqueous solution, with water, are given in LIST-I. The effects of dilution of the solutions on $\left[H^{+}\right]$ are given in LIST -II.
(Note: Degree of dissociation $\left(\alpha \right)$ of weak acid and weak base is << 1; degree of hydrolysis of salt << 1; $\left[H^{+}\right]$ represents the concentration of $H^{+}$ ions)
$\begin{array}{cc}\text{LIST -I}& \text{LIST-II}\\ \text{P. 10 mL of 0.1 M NaOH +20 mL of}& \text{1. The value of}\left[H^{+}\right]\text{does not}\\ \text{0.1 M acetic acid diluted to 60 mL}& \text{change on dilution}\\ \text{Q. 20 mL of 0.1 M NaOH +20 mL of}& \text{2. The value of}\left[H^{+}\right]\text{changes to half}\\ \text{0.1 M acetic acid diluted to 80 mL}& \text{of its initial value on dilution}\\ \text{R. 20 mL of 0.1 M HCl + 20 mL of 0.1}& \text{3. The value of}\left[H^{+}\right]\text{changes to two}\\ \text{M ammonia solution diluted to 80 mL}& \text{times of its initial value on dilution}\\ \text{S. 10 mL saturated solution of}Ni(OH{)}_2& \text{4. The value of}\left[H^{+}\right]\text{changes to}\frac{1}{\sqrt{2}}\\ \text{in equilibrium with excess solid}Ni(OH{)}_2& \text{times of its initial value on dilution}\\ \text{is diluted to 20 mL (solid}Ni(OH{)}_2\text{is still}& \\ \text{present after dilution).}& \\ & \text{5. The value of}\left[H^{+}\right]\text{changes to}\sqrt{2}\\ & \text{times of its initial value on dilution}\end{array}$
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. $P\to 1;Q\to 5;R\to 4;S\to 1$

Answer 1

The correct option is D $P\to 1;Q\to 5;R\to 4;S\to 1$

$\frac{K_w}{K_a}=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[O{H}^{-}\right]}{\left[C{H}_{3}COOH\right]}$
$\left[O{H}^{-}\right]=(\frac{K_w}{K_a}\times C)$
(P) is a buffer, so $\left[H^{+}\right]$ does not change on dilution, as [salt] =[acid].

(Q) contains only $C{H}_{3}COONa$
So, $C{H}_{3}CO{O}^{-}+H_{2}O\rightleftharpoons C{H}_{3}COOH+O{H}^{-}$
$\left[O{H}^{-}\right]=\sqrt{K_h\times C}\Rightarrow \left[H^{+}\right]decreasesby\sqrt{2}times$

(R) is also salt hydrolysis
So, $N{H}_4^{+}+H_{2}O\rightleftharpoons N{H}_{4}OH+H^{+}$
$\left[H^{+}\right]=\sqrt{\frac{K_w}{K_b}\times C}$
So, C is made $\frac{1}{2}so,\left[H^{+}\right]becomes\frac{1}{\sqrt{2}}$

(S) It is the case of solubility equilibria. So, dilution does not affect $\left[H^{+}\right]$ or $\left[O{H}^{-}\right]$