Ionic Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & m.{\text{ }}mol{\text{ of}}\,\,{H_2}S{O_4} = 20 \times 0.1 = 2 \cr & m.{\text{ }}mol{\text{ of}}\,\,N{H_4}OH = 30 \times 0.2 = 6 \cr} $$

\[\underset{_{\text{final}}^{\text{Initial}}}{\mathop{{}}}\,\,\,\,\underset{_{\begin{smallmatrix} \left( 2-2 \right) \\ =0\,m\,mol \end{smallmatrix}}^{\,\,\,\,2\,m\,mol}}{\mathop{{{H}_{2}}S{{O}_{4}}}}\,+\underset{_{\begin{smallmatrix} \left( 6-2\times 2 \right) \\ =2\,m\,mol \end{smallmatrix}}^{\,\,\,\,6m\,mol}}{\mathop{2N{{H}_{4}}OH}}\,\to \underset{_{2\,m\,\,mol}^{\,\,\,\,\,\,0}}{\mathop{{{\left( N{{H}_{4}} \right)}_{2}}S{{O}_{4}}}}\,+2{{H}_{2}}O\]

$$\eqalign{ & {\left[ {N{H_4}OH} \right]_{left}} = 2\,m\,mol \cr & \left[ {{{\left( {N{H_4}} \right)}_2}S{O_4}} \right] = 2\,m\,mol \cr & \left[ {\mathop N\limits^ + {H_4}} \right] = 2 \times 2 = 4\,m\,mol \cr & {\text{Total Volume = 30 + 20 = 50 mL}} \cr & pOH = p{K_b} + \log \left[ {\frac{{{\text{Salt}}}}{{{\text{Base}}}}} \right] \cr & = 4.7 + \log \frac{{\frac{4}{{50}}}}{{\frac{2}{{50}}}} \cr & = 4.7 + \log 2 \cr & = 5 \cr & pH = 14 - pOH \cr & pH = 14 - 5 = 9 \cr} $$

$$\eqalign{ & {H_2}S{O_4} \rightleftharpoons 2{H^ + } + SO_4^{2 - } \cr & \left[ {{H^ + }} \right] = 2 \times 1 \times {10^{ - 4}}\,M \cr & pH = - \log \left( {2 \times {{10}^{ - 4}}} \right) = 3.70 \cr} $$

$$\eqalign{ & HCO_3^ - \to {H^ + } + CO_3^{2 - } \cr & {K_2} = \frac{{\left[ {{H^ + }} \right]\left[ {CO_3^{2 - }} \right]}}{{\left[ {HCO_3^ - } \right]}} = 4.8 \times {10^{ - 11}} \cr & \left[ {{H^ + }} \right] = \frac{{4.8 \times {{10}^{ - 11}}\left[ {HCO_3^ - } \right]}}{{\left[ {CO_3^{2 - }} \right]}} \cr & = 4.8 \times {10^{ - 11}}\left( {0.1/0.2} \right) \cr & pH = - \log \left[ {{H^ + }} \right] \cr & = - \log \left( {4.8 \times {{10}^{ - 11}} \times 0.5} \right) \cr & = 10.62 \cr} $$

$$\mathop {Ca{F_2}}\limits_{{\text{Solubility}}\,\,{\text{(cone):}}} \rightleftharpoons \mathop {C{a^{2 + }}}\limits_{0.5 \times {{10}^{ - 4}}} + \mathop {2{F^ - }}\limits_{{{\left( {2 \times 0.5 \times {{10}^{ - 4}}} \right)}^2}} $$
$$\eqalign{ & {K_{sp}} = 0.5 \times {10^{ - 4}} \times 1 \times {10^{ - 8}} \cr & \,\,\,\,\,\,\,\,\,\, = 0.5 \times {10^{ - 12}}\,\,{\text{or}}\,\,5 \times {10^{ - 13}} \cr} $$

$$\eqalign{ & {\text{For}}\,\,MY, \cr & \mathop {MY}\limits_0 \rightleftharpoons \mathop {{M^ + }}\limits_s + \mathop {{Y^ - }}\limits_s \cr} $$
where, $$s =$$  solubility and $${K_{sp}} = $$  solubility product.
$$\eqalign{ & \therefore \,\,{K_{sp}} = \left[ {{M^ + }} \right]\left[ {{Y^ - }} \right] \cr & \,\,\,\,\,\,\,{K_{sp}} = {s^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,s = \sqrt {{K_{sp}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {6.2 \times {{10}^{ - 13}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 7.874 \times {10^{ - 7}} \cr & {\text{Similarly, for }}N{Y_3}, \cr & \mathop {N{Y_3}}\limits_0 \rightleftharpoons \mathop {{N^ + }}\limits_s + \mathop {3{Y^ - }}\limits_{3s} \cr & \therefore \,\,{K_{sp}} = \left[ {{N^ + }} \right]{\left[ {{Y^ - }} \right]^3} = s \times {\left( {3s} \right)^3} \cr & \,\,\,\,\,\,\,{K_{sp}} = 27{s^4} \cr & \therefore \,\,\,\,\,\,\,\,s = \root 4 \of {\frac{{{K_{sp}}}}{{27}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \root 4 \of {\frac{{6.2 \times {{10}^{ - 13}}}}{{27}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3.89 \times {10^{ - 4}} \cr} $$

$$\eqalign{ & A{X_2}\,{\text{is ionised as follows}} \cr & A{X_2}\,\,\,\,\,\,\,\,\, \rightleftharpoons {A^{2 + }} + \,\,2{X^ - } \cr & S\,mol\,{L^{ - 1}}\,\,\,\,S\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2S \cr & {\text{Solubility product of}}\,A{X_2}, \cr & {K_{sp}} = \left[ {{A^{2 + }}} \right]{\left[ {{X^ - }} \right]^2} \cr & \,\,\,\,\,\,\,\,\, = S \times {\left( {2S} \right)^2} \cr & \,\,\,\,\,\,\,\,\, = 4{S^3} \cr & \because \,\,{K_{sp}}\,{\text{of}}\,A{X_2} = 3.2 \times {10^{ - 11}} \cr & \therefore \,\,\,\,\,3.2 \times {10^{ - 11}} = 4{S^3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{S^3} = 0.8 \times {10^{ - 11}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 8 \times {10^{ - 12}} \cr & {\text{Solubility}} = 2 \times {10^{ - 4}}\,mol/L \cr} $$

$$\eqalign{ & A{g_2}Cr{O_4} \rightleftharpoons 2A{g^ + } + CrO_4^{2 - } \cr & {\text{Solubility product}} \cr & {K_{sp}} = {\left( {2s} \right)^2} \times S = 4{s^3} \cr & {K_{sp}} = \left( {1.1 \times {{10}^{ - 12}}} \right)\,\,\,\,\,\left( {{\text{given}}} \right) \cr & S = \root 3 \of {\frac{{{K_{sp}}}}{4}} = 0.65 \times {10^{ - 4}} \cr & AgCl \rightleftharpoons A{g^ + } + C{l^ - } \cr & {K_{sp}} = S \times S\,\,\,\,\,\left( {{K_{sp}} = 1.8 \times {{10}^{ - 10}}} \right) \cr & s = \sqrt {{K_{sp}}} = 1.34 \times {10^{ - 5}} \cr & AgBr \rightleftharpoons A{g^ + } + B{r^ - } \cr & {K_{sp}} = S \times S\,\,\,\,\,\,\,\left( {{K_{sp}} = 5 \times {{10}^{ - 13}}} \right) \cr & S = \sqrt {{K_{sp}}} = 0.71 \times {10^{ - 6}} \cr & AgI \rightleftharpoons A{g^ + } + {I^ - } \cr & {K_{sp}} = S \times S\,\,\,\,\left( {{K_{sp}} = 8.3 \times {{10}^{ - 17}}} \right) \cr & \,\,\,\,\,s = \sqrt {{K_{sp}}} = 0.9 \times {10^{ - 8}} \cr & \because \,\,{\text{Solubility of}}\,A{g_2}Cr{O_4}\,{\text{is highest}}{\text{.}} \cr & {\text{So, it will precipitate last}}{\text{.}} \cr} $$

$$\eqalign{ & NaZ\,{\text{is salt of}}\,\,{W_A}/{S_B} \cr & \therefore \,\,pH = \frac{1}{2}\left( {p{K_w} + p{K_a} + \log C} \right) \cr & 8.9 \times 2 = 14 + p{K_a} + \log \,0.1 \cr & 17.8 = 14 + p{K_a} - 1 \cr & p{K_a} = 4.8, \cr & {K_a} = {\text{Antilog}}\left( { - 4.8} \right) \cr & = 1.585 \times {10^{ - 5}} \approx 1.6 \times {10^{ - 5}}. \cr} $$

$$\eqalign{ & {\text{Key Idea (i) For basic buffer,}} \cr & pOH = p{K_b} + {\text{log}}\frac{{\left[ {{\text{salt}}} \right]}}{{\left[ {{\text{base}}} \right]}} \cr & \left( {{\text{ii}}} \right)\,pH + pOH = 14 \cr & {\text{Given,}}\,{K_b} = 1 \times {10^{ - 10}},\left[ {{\text{salt}}} \right] = \left[ {{\text{base}}} \right] \cr & pOH = - {\text{log}}\,{K_b} + {\text{log}}\frac{{\left[ {{\text{salt}}} \right]}}{{\left[ {{\text{base}}} \right]}} \cr & \therefore \,\,pOH = - {\text{log}}\,\left( {1 \times {{10}^{ - 10}}} \right) + {\text{log}}\,1 = 10 \cr & pH + pOH = 14\left[ {\because \,\,{\text{concentration}}\,{\text{of}}\,\left[ {{B^ - }} \right] = \left[ {HB} \right]} \right] \cr & pH = 14 - 10 \cr & \,\,\,\,\,\,\,\,\, = 4 \cr} $$

Hydrolysis of a salt is reverse reaction of acid-base neutralization reaction.
$$\eqalign{ & \therefore \,\,{K_h} = \frac{{{K_w}}}{{{K_a}}} \cr & = \frac{{{{10}^{ - 14}}}}{{{{10}^{ - 7}}}} \cr & = {10^{ - 7}} \cr & \left[ {O{H^ - }} \right] = ch \cr & = c \times \sqrt {\frac{{{K_h}}}{c}} \cr & = \sqrt {c \times {K_h}} \cr & = \sqrt {{{10}^{ - 8}}} \cr & = {10^{ - 4}} \cr & \Rightarrow pO{H^ - } = - \log \left[ {O{H^ - }} \right] \cr & = - \log \left[ {{{10}^{ - 4}}} \right] \cr & = 4 \cr} $$