Ionic Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

$$AgCl \rightleftharpoons \mathop {A{g^ + }}\limits_s + \mathop {C{l^ - }}\limits_s $$
$${s^2} = 1.5625 \times {10^{ - 10}};$$    $$s = 1.25 \times {10^{ - 5}}\,mol\,{L^{ - 1}}$$
$$\eqalign{ & {\text{Solubility in}}\,\,g\,{L^{ - 1}} \cr & = {\text{Molar mass}}\, \times \,s \cr & = 143.5 \times 1.25 \times {10^{ - 5}} \cr & = 1.79 \times {10^{ - 3}}\,g\,{L^{ - 1}} \cr} $$

TIPS/Formulae :
$$\mathop {{H_2}O}\limits_{\left( {1 + \alpha } \right)c} \rightleftharpoons \mathop {{H^ + }}\limits_{\alpha c} + \mathop {O{H^ - }}\limits_{\alpha c} $$
$$\alpha = 1.9 \times {10^{ - 7}};\,{\text{Density}}\,{\text{of water}}$$       $${\text{ = }}\,\frac{{1.0\,gm}}{{c{m^3}}}$$
$$\therefore \,\,c = \frac{1}{{18}} \times 1000 = 55.56\,{\text{moles/l}}$$
$$\therefore \,\,\left[ {{H^ + }} \right] = 55.56 \times 1.9 \times {10^{ - 9}} = $$       $$1.055 \times {10^{ - 7}}$$
$$\left[ {\because \,1.9 \times {{10}^ - }\% = 1.9 \times {{10}^{ - 9}}} \right]$$
$$\therefore \,\,{K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] = $$     $${\left( {1.055 \times {{10}^{ - 7}}} \right)^2} = 1.0 \times {10^{ - 14}}$$

$$\eqalign{ & BaC{l_2} \rightleftharpoons B{a^{2 + }} + 2C{l^ - } \cr & {K_{sp}} = \left[ {B{a^{2 + }}} \right]{\left[ {C{l^ - }} \right]^2} \cr & \,\,\,\,\,\,\,\,\,\, = x \times {\left( {2x} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\, = 4{x^3} \cr} $$
$$4{x^3} = 3.2 \times {10^{ - 9}} \Rightarrow x = 9.28 \times {10^{ - 4}}$$       $$ = 0.928 \times {10^{ - 3}} \approx 1 \times {10^{ - 3}}$$

\[M{{X}_{2}}\rightleftharpoons {{\underset{s}{\mathop{M}}\,}^{++}}+{{\underset{2s}{\mathop{2X}}\,}^{-}}\]
$${\text{Where s is the solubility of }}M{X_2}$$
$${\text{then}}\,\,{K_{sp}} = 4{s^3};\,s \times {\left( {2s} \right)^2} = 4 \times {10^{ - 12}}$$       $$ = 4{s^3};s = 1 \times {10^{ - 4}}$$
$$\therefore \left[ {{M^{ + + }}} \right] = s = 1\left[ {{M^{ + + }}} \right] = 10 \times {10^{ - 4}}$$

$$\eqalign{ & {K_w} = {K_a} \times {K_b} \cr & {K_b} = {10^{ - 10.83}} \cr & = 1.48 \times {10^{ - 11}} \cr & \therefore \,\,{K_a} = \frac{{{K_w}}}{{{K_b}}} \cr & = \frac{{{{10}^{ - 14}}}}{{1.48 \times {{10}^{ - 11}}}} \cr & = 6.75 \times {10^{ - 4}} \cr} $$

$$\eqalign{ & {\text{Given}}\,\,{\text{N}}{{\text{a}}_2}C{O_3} = 1.0 \times {10^{ - 4}}M \cr & \therefore \,\,\left[ {CO_3^{2 - }} \right] = 1.0 \times {10^{ - 4}}M \cr & i.e.\,\,\,s = 1.0 \times {10^{ - 4}}M \cr & {\text{At equilibrium}} \cr & \left[ {B{a^{2 + }}} \right]\left[ {CO_3^{2 - }} \right] = {K_{sp}}\,{\text{of}}\,BaC{O_3} \cr & \left[ {B{a^{2 + }}} \right] = \frac{{{K_{sp}}}}{{\left[ {CO_3^{2 - }} \right]}} \cr & = \frac{{5.1 \times {{10}^{ - 9}}}}{{1.0 \times {{10}^{ - 4}}}} \cr & = 5.1 \times {10^{ - 5}}M \cr} $$

$$\eqalign{ & \alpha = \sqrt {\frac{{{K_w}}}{{{K_a} \times {K_b}}}} \cr & = \sqrt {\frac{{1 \times {{10}^{ - 14}}}}{{1.8 \times {{10}^{ - 5}} \times 1.8 \times {{10}^{ - 5}}}}} \cr & = 0.55\% \cr} $$

Salts of weak base and strong acid get hydrolysed in aqueous solution forming an acidic solution.
$$\mathop {AlC{l_3}}\limits_{\left( {{\text{weak}}} \right)} + 3{H_2}O \to \mathop {Al{{\left( {OH} \right)}_3}}\limits_{\left( {{\text{strong}}} \right)} + \mathop {3HCl}\limits_{\left( {{\text{acidic}}} \right)} $$

$${M^{2 + }} + {s^{2 - }} \to MS$$
Lower the value of $${K_{sp}},$$  lower will be solubility. Hence, $$HgS$$  will precipitate first.

$$\eqalign{ & {\text{Given,}}\,{K_a}\left( {N{H_4}OH} \right) = 1.77 \times {10^{ - 5}} \cr & N{H_4}OH \rightleftharpoons NH_4^ + + O{H^ - } \cr & {K_a} = \frac{{\left[ {NH_4^ + } \right]\left[ {O{H^ - }} \right]}}{{\left[ {N{H_4}OH} \right]}} \cr & \,\,\,\,\,\,\,\, = 1.77 \times {10^{ - 5}}\,\,\,\,...{\text{(i)}} \cr & {\text{Hydrolysis of}}\,N{H_4}Cl\,{\text{takes place as,}} \cr & N{H_4}Cl + {H_2}O \to N{H_4}OH + HCl \cr & {\text{or}}\,\,NH_4^ + + {H_2}O \to N{H_4}OH + {H^ + } \cr & {\text{Hydrolysis constant,}} \cr & \,\,\,\,\,\,{K_h} = \frac{{\left[ {N{H_4}OH} \right]\left[ {{H^ + }} \right]}}{{\left[ {NH_4^ + } \right]}}...{\text{(ii)}} \cr & {\text{or}}\,{{\text{K}}_h} = \frac{{\left[ {N{H_4}OH} \right]\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]}}{{\left[ {NH_4^ + } \right]\left[ {O{H^ - }} \right]}}...{\text{(iii)}} \cr & {\text{From Eqs}}{\text{. (i), (ii) and (iii)}} \cr & {{\text{K}}_h} = \frac{{{K_w}}}{{{K_a}}}\,\,\,\left[ {\because \,\,\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] = {K_w}} \right] \cr & \,\,\,\,\,\,\,\, = \frac{{{{10}^{ - 14}}}}{{1.77 \times {{10}^{ - 5}}}} \cr & \,\,\,\,\,\,\,\, = 5.65 \times {10^{ - 10}} \cr} $$