Ionic Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

Key concept For a sparingly soluble salt, if $$S$$  is the molar solubility,
$$\eqalign{ & {A_x}{B_y}\left( s \right) + {H_2}O \rightleftharpoons x{A^{y + }} + y{B^{x - }} \cr & {\text{At saturation,}} \cr} $$
$$K\left[ {{A_x}{B_y}} \right] = {\left[ {{A^{y + }}} \right]^x} \times {\left[ {{B^{x - }}} \right]^y}$$     $$ = {\left[ {xS} \right]^x}{\left[ {yS} \right]^y}$$
$${\text{or}}\,{K_{sp}} = {x^y}.\,\,{y^y}{S^{x + y}}$$
Where, the constant $${K_{sp}}$$  is called solubility product.
$$\eqalign{ & A{g_2}{C_2}{O_4}\left( s \right) \rightleftharpoons \mathop {2A{g^ + }}\limits_{2S} + \mathop {{C_2}O_4^{2 - }}\limits_S \cr & {K_{sp}} = {\left[ {A{g^ + }} \right]^2}\left[ {{C_2}O_4^{2 - }} \right] \cr & \,\,\,\,\,\,\,\,\,\, = {\left[ {2S} \right]^2}\left[ S \right] \cr} $$
Given, $$2S = 2.2 \times {10^{ - 4}}\,\,{\text{or}}\,\,S = 1.1 \times {10^{ - 4}}\,M$$
$$\eqalign{ & \therefore \,{K_{sp}} = {\left[ {2.2 \times {{10}^{ - 4}}} \right]^2}\left[ {1.1 \times {{10}^{ - 4}}} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 5.3 \times {10^{ - 12}} \cr} $$

$$NaCN + {H_2}O \rightleftharpoons NaOH + HCN$$
(hydrolysis)
$$NaOH$$   will give $$O{H^ - }\,ions$$   and react with $${H^ + }$$ to form $${H_2}O$$  hence decrease the concentration of $${H^ + }\,ions.$$

The dissociation of a weak electrolyte $$(AgCl)$$  is suppressed on adding the strong electrolyte having a common ion $$\left( {C{l^ - }} \right),$$  hence concentration of $$A{g^ + }\,ion$$   will be less.

Sodium acetate is a salt of strong base and weak acid.
$$\eqalign{ & \therefore \,pH = 7 + \frac{1}{2}p{K_a} + \frac{1}{2}\log c \cr & {\text{where}}\,\,p{k_a} = - \log \,{K_a} = - \log \,{10^{ - 5}} = 5 \cr & \log \,c = \log \,{10^{ - 1}} = - 1 \cr & pH = 7 + \frac{5}{2} - \frac{1}{2} = 9.0 \cr} $$

$$\eqalign{ & \alpha = \sqrt {\frac{{{K_w}}}{{{K_a} \times c}}} \cr & = \sqrt {\frac{{{{10}^{ - 14}}}}{{1.3 \times {{10}^{ - 9}} \times \frac{1}{{80}}}}} \cr & = 2.48\% . \cr} $$

$$\eqalign{ & {\text{As, molarity,}} \cr & = \frac{{Wt.\,\,{\text{of solute per litre of solution}}}}{{Mol.{\text{ }}wt.{\text{ of solute}}}} \cr & {\text{Molarity of}}\,{H_2}O = \frac{{1000}}{{18}}mol/L \cr & {H_2}O\,\,\,\,\,\,\,\,\,\,\, \rightleftharpoons {H^ + } + O{H^ - } \cr & c\left( {1 - \alpha } \right)\,\,\,\,\,\,\,\,\,\,\,c\alpha \,\,\,\,\,\,\,\,\,\,c\alpha \cr & {\text{Thus,}}\,\,{K_a} = \frac{{c{\alpha ^2}}}{{1 - \alpha \,}} = c\alpha {\,^2} \cr & = \frac{{1000}}{{18}} \times {\left( {1.8 \times {{10}^{ - 8}}} \right)^2} \cr & = 1.8 \times {10^{ - 14}} \cr} $$

$$\eqalign{ & AgI \to A{g^ + } + {I^ - } \cr & {\text{For binary electrolyte}} \cr & {{\text{K}}_{sp}} = {S^2} \cr & {\text{where,}}\,S = {\text{solubility in}}\,\,mol/L \cr & 1.0 \times {10^{ - 16}} = {S^2} \cr & {\text{or}}\,\,\,S = 1 \times {10^{ - 8}}mol/L \cr & {\text{Normality of}}\,KI\,{\text{solutiuon}} = {10^{ - 4}}N \cr & {\text{Here change is one}} \cr & \therefore \,\,M = {10^{ - 4}}M\,\,\left[ {n = 1} \right] \cr & {\text{or }}S{\text{ for }}KI{\text{ solution}} = {10^{ - 4}}M \cr & {\text{Solubility of}}\,AgI{\text{ in }}KI{\text{ solution}} \cr & {\text{ = 1}} \times {10^{ - 8}} \times {10^{ - 4}} \cr & = 1 \times {10^{ - 12}}mol/L \cr} $$

$$\eqalign{ & {\text{Given,}}\,pH\,{\text{of}}\,Ba{\left( {OH} \right)_2} = 12 \cr & \therefore \,\,pOH = 14 - pH \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 14 - 12 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2 \cr & {\text{We know that,}} \cr & \,\,\,\,\,pOH = - {\text{log}}\left[ {O{H^ - }} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,2 = - {\text{log}}\left[ {O{H^ - }} \right] \cr & \left[ {O{H^ - }} \right] = {\text{antilog}}\,\left( { - 2} \right) \cr & \left[ {O{H^ - }} \right] = 1 \times {10^{ - 2}} \cr & Ba{\left( {OH} \right)_2}\,\,{\text{dissolves in water as}} \cr & \mathop {Ba{{\left( {OH} \right)}_2}}\limits_{s\,mol\,{L^{ - 1}}} \left( s \right) \rightleftharpoons \mathop {B{a^{2 + }}}\limits_s + \mathop {2O{H^ - }}\limits_{2s} \cr & \therefore \,\,\left[ {O{H^ - }} \right] = 2s = 1 \times {10^{ - 2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,s = \frac{{\left[ {O{H^ - }} \right]}}{2}\,\,\,\left[ {B{a^{2 + }} = s} \right] \cr & \left[ {B{a^{2 + }}} \right] = \frac{{\left[ {O{H^ - }} \right]}}{2} = \frac{{1 \times {{10}^{ - 2}}}}{2} \cr & {K_{sp}} = \left[ {B{a^{2 + }}} \right]{\left[ {O{H^ - }} \right]^2} \cr & \,\,\,\,\,\,\,\,\, = \left( {\frac{{1 \times {{10}^{ - 2}}}}{2}} \right){\left( {1 \times {{10}^{ - 2}}} \right)^2} \cr & \,\,\,\,\,\,\,\,\, = 0.5 \times {10^{ - 6}} \cr & \,\,\,\,\,\,\,\,\, = 5 \times {10^{ - 7}} \cr} $$

The correct order of acidic strength of the given species in
$$\eqalign{ & {\text{in}}\mathop {\,\,HS{O_3}F}\limits_{\left( {{\text{iv}}} \right)} > \mathop {{H_3}{O^ + }}\limits_{\left( {{\text{ii}}} \right)} > \mathop {HS{O_4}^ - }\limits_{\left( {{\text{iii}}} \right)} > \mathop {HC{O_3}^ - }\limits_{\left( {\text{i}} \right)} \cr & {\text{or (i) < (iii) < (ii) < (iv)}} \cr} $$
It corresponds to choice (C) which is correct answer.