Ionic Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

$$C{H_3}COOH + {H_2}O \rightleftharpoons $$     $$C{H_3}CO{O^ - } + {H_3}{O^ + }$$
$${K_a} = \frac{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H_3}{O^ + }} \right]}}{{\left[ {C{H_3}COOH} \right]}}$$
$${K_a} = \frac{{{{\left[ {{H_3}{O^ + }} \right]}^2}}}{{\left[ {C{H_3}COOH} \right]}}$$     $$\left( {\because \left[ {C{H_3}CO{O^ - }} \right] = \left[ {{H_3}{O^ + }} \right]} \right)$$
$$\eqalign{ & {\left[ {{H_3}{O^ + }} \right]^2} = {K_a}\left[ {C{H_3}COOH} \right] \cr & \left[ {{H_3}{O^ + }} \right] = \sqrt {{K_a}\left[ {C{H_3}COOH} \right]} \cr} $$
$${\text{Given:}}\,{K_a} = 1.74 \times {10^{ - 5}},$$     $$\left[ {C{H_3}COOH} \right] = 0.01\,mol\,d{m^{ - 3}}$$
$$\eqalign{ & \left[ {{H_3}{O^ + }} \right] = \sqrt {1.74 \times {{10}^{ - 5}} \times 0.01} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {1.74 \times {{10}^{ - 7}}} \cr & \left[ {{H_3}{O^ + }} \right] = 4.17 \times {10^{ - 4}} \cr & pH = - \log \left[ {{H_3}{O^ + }} \right] \cr & \,\,\,\,\,\,\,\,\, = - \log \left( {4.17 \times {{10}^{ - 4}}} \right) \cr & \,\,\,\,\,\,\,\,\, = 3.379 \approx 3.4 \cr} $$

$$HCl\left( {aq} \right) \to {H^ + }\left( {aq} \right) + C{l^ - }\left( {aq} \right)$$       $$\left[ {S = \sqrt {{K_{sp}}} } \right]$$
$$\eqalign{ & \,\,\,\,\,\,\,\left[ {HCl} \right] = 10M \cr & \Rightarrow \left[ {{H^ + }} \right] = 10\,mol/L \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,pH = - {\text{log}}\left[ {{H^ + }} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - {\text{log}}\,10 \cr} $$
                 $$ = - 1,\,{\text{so}}\,{\text{the}}\,pH\,{\text{is less than zero}}{\text{.}}$$

Among oxyacids of the same type formed by different elements, acidic nature increases with increasing electronegativity. In general, the strength of oxyacids decreases as we go down the family in the periodic table.
$$HOCl\left( {\text{I}} \right) > HOBr\left( {{\text{II}}} \right) > HOI\left( {{\text{III}}} \right)$$
[ In halogen groups $$Cl$$ is above $$Br$$ and $$I$$ ]

$$\eqalign{ & {H_2}A \rightleftharpoons {H^ + } + H{A^ - } \cr & \therefore \,\,{K_1} = 1.0 \times {10^{ - 5}} = \frac{{\left[ {{H^ + }} \right]\left[ {H{A^ - }} \right]}}{{\left[ {{H_2}A} \right]}}\,\,\,\,\,\left( {{\text{Given}}} \right) \cr & H{A^ - } \to {H^ + } + {A^ - } \cr & \therefore \,\,{K_2} = 5.0 \times {10^{ - 10}} = \frac{{\left[ {{H^ + }} \right]\left[ {{A^{ - - }}} \right]}}{{\left[ {H{A^ - }} \right]}}\,\,\,\,\left( {{\text{Given}}} \right) \cr & K = \frac{{{{\left[ {{H^ + }} \right]}^2}\left[ {{A^{2 - }}} \right]}}{{\left[ {{H_2}A} \right]}} = {K_1} \times {K_2} \cr & = \left( {1.0 \times {{10}^{ - 5}}} \right) \times \left( {5 \times {{10}^{ - 10}}} \right) \cr & = 5 \times {10^{ - 15}} \cr} $$

$$\eqalign{ & \left[ {Z{r_3}{{\left( {P{O_4}} \right)}_4}} \right] \rightleftharpoons 3\mathop {Zr_{}^{4 + }}\limits_{3S} + 4P\mathop O\limits_{4S} \,_4^{3 - } \cr & {K_{sp}} = {\left( {3S} \right)^3}{\left( {4S} \right)^4} \cr & = 27{S^3} \times 256\,{S^4} \cr & = 6912\,{S^7}. \cr & \therefore \,\,S = {\left( {\frac{{{K_{sp}}}}{{6912}}} \right)^{\frac{1}{7}}} \cr} $$

The salt $$(AB)$$  given is a salt is of weak acid and weak base. Hence its pH can be calculated by the formula
$$\eqalign{ & \therefore pH = 7 + \frac{1}{2}p{K_a} - \frac{1}{2}p{K_b} \cr & = 7 + \frac{1}{2}\left( {3.2} \right) - \frac{1}{2}\left( {3.4} \right) \cr & = 6.9 \cr} $$

Key Idea $$C{H_3}COOH$$   ( weak acid ) and $$C{H_3}COONa$$   ( conjugated salt ) form acidic buffer and for acidic buffer,
$$\eqalign{ & pH = p{K_a} + \log \frac{{\left[ {{\text{salt}}} \right]}}{{\left[ {{\text{acid}}} \right]}} \cr & {\text{and}}\,\,\left[ {{H^ + }} \right] = - {\text{antilog}}\,\,pH \cr} $$
$$pH = - {\text{log}}\,{K_a} + {\text{log}}\frac{{\left[ {{\text{salt}}} \right]}}{{\left[ {{\text{acid}}} \right]}}$$       $$\left[ {\because \,p{K_a} = - {\text{log}}\,{K_a}} \right]$$
$$\eqalign{ & = - {\text{log}}\left( {1.8 \times {{10}^{ - 5}}} \right) + \log \frac{{\left( {0.20} \right)}}{{\left( {0.10} \right)}} \cr & = 4.74 + {\text{log}}2 \cr & = 4.74 + 0.3010 \cr & = 5.041 \cr & {\text{Now,}}\,\,\left[ {{H^ + }} \right] = {\text{antilog}}\left( { - 5.045} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 9.0 \times {10^{ - 6}}\,mol/L \cr} $$

In qualitative analysis of cations of second group $${H_2}S$$  gas is passed in presence of $$HCl,$$  therefore due to common ion effect, lower concentration of sulphide ions is obtained which is sufficient for the precipitation of second group cations in the form of their sulphides due to lower value of their solubility product $$\left( {{K_{sp}}} \right).$$   Here, fourth group cations are not precipitated because it require more sulphide ions for exceeding their ionic product to their solubility products which is not obtained here due to common ion effect.

$$A{g_{_2}}Cr{O_4} \rightleftharpoons 2A{g^ + } + CrO_4^{2 - }$$
$$A{g^ + } = 1.5 \times {10^{ - 4}}\,M;$$     $$CrO_4^{2 - } = 0.75 \times {10^{ - 4}}\,M$$    $$\left( {\frac{1}{2}\,{\text{of}}\,\,A{g^ + }} \right)$$
$$\eqalign{ & {K_{sp}} = {\left( {1.5 \times {{10}^{ - 4}}} \right)^2} \times \left( {0.75 \times {{10}^{ - 4}}} \right) \cr & \,\,\,\,\,\,\,\,\,\, = 1.687 \times {10^{ - 12}} \cr} $$