Ionic Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

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

NOTE : Conjugate acid-base differ by $${H^ + }$$
\[\mathop {{H_2}PO_4^ - }\limits_{{\text{Acid}}} \xrightarrow{{ - {H^ + }}}\mathop {HPO_4^{ - - }}\limits_{{\text{conjugate base}}} \]

The $$pH$$  of mixture is due to the $$HCl$$  only, because $$C{H_3}COOH$$    is negligibly ionised due to common ion effect. Thus,
$$\eqalign{ & \left[ {{H^ + }} \right] = 0.05\,M = 5 \times {10^{ - 2}}M \cr & pH = - \log \left( {5 \times {{10}^{ - 2}}} \right) \cr & = - \left[ {\log \,5 + \log \,{{10}^{ - 2}}} \right] \cr & pH = 1.3 \cr} $$

TIPS/Formulae :
(i) Lower the oxidation state of central atom, weaker will be oxy acid.
(ii) Weaker the acid, stronger will be its conjugate base.
Oxidation state of $$Cl$$ in $$HClO$$  is + 1, in $$HCl{O_2}$$  is + 3, in $$HCl{O_3}$$ is + 5, and in $$HCl{O_4}$$  is + 7
∴ $$HClO$$  is the weakest acid and so its conjugate base $$Cl{O^ - }$$  is the strongest Bronsted base.

$${\text{For acidic buffer }}pH = $$     $$p{K_a} + \log \left[ {\frac{{{\text{salt}}}}{{{\text{acid}}}}} \right]$$
$${\text{or}}\,\,pH = p{K_a} + \log \frac{{\left[ {{A^ - }} \right]}}{{\left[ {HA} \right]}}$$
Given $$p{K_a} = 4.5$$   and acid is 50% ionised.
$$\left[ {HA} \right] = \left[ {{A^ - }} \right]$$   ( when acid is 50% ionised )
$$\eqalign{ & \therefore \,\,pH = p{K_a} + \log \,{\text{1}}\,\,\,\therefore \,pH = p{K_a} = 4.5 \cr & pOH = 14 - pH = 14 - 4.5 = 9.5 \cr} $$

$$\left[ {{H^ + }} \right] = \sqrt {C \times {K_a}} = \sqrt {0.001 \times 1.8 \times {{10}^{ - 4}}} $$          for formic acid
$$\left[ {{H^ + }} \right] = \sqrt {{C_2} \times 1.8 \times {{10}^{ - 5}}} $$       for acetic acid Equating and solving, $${C_2} = 0.01\,M.$$

$$\eqalign{ & C{r_2}{\left( {S{O_4}} \right)_3} \rightleftharpoons \mathop {2C}\limits_{2s} {r^{3 + }} + \mathop {3S}\limits_{3s} O_4^{2 - } \cr & {K_{sp}} = {\left( {2s} \right)^2}{\left( {3s} \right)^3} = 4{s^2} \times 27{s^3} = 108{s^5} \cr & s = {\left( {\frac{{{K_{sp}}}}{{108}}} \right)^{\frac{1}{5}}} \cr & H{g_2}C{l_2} \rightleftharpoons 2\mathop {Hg_{}^{2 + }}\limits_{2s} + 2\mathop {C{l^ - }}\limits_{2s} \cr & {K_{sp}} = {\left( {2s} \right)^2} \times {\left( {2s} \right)^2} = 16{s^4} \cr & s = {\left( {\frac{{{K_{sp}}}}{{16}}} \right)^{\frac{1}{4}}} \cr & BaS{O_4} \rightleftharpoons \mathop {Ba_{}^{2 + }}\limits_s + \mathop {SO}\limits_s \,_4^{2 - } \cr & {K_{sp}} = {s^2} \cr & s = \sqrt {{K_{sp}}} \cr & CrC{l_3} \rightleftharpoons C\mathop {r_{}^{3 + }}\limits_s + 3\mathop {C{l^ - }}\limits_{3s} \cr & {K_{sp}} = s \times {\left( {3s} \right)^3} = 27{s^4} \cr & s = {\left( {\frac{{{K_{sp}}}}{{27}}} \right)^{\frac{1}{4}}} \cr} $$
Hence the correct order of solubilities of salts is
$$\sqrt {{K_{sp}}} > {\left( {\frac{{{K_{sp}}}}{{16}}} \right)^{\frac{1}{4}}} > {\left( {\frac{{{K_{sp}}}}{{27}}} \right)^{\frac{1}{4}}} > {\left( {\frac{{{K_{sp}}}}{{108}}} \right)^{\frac{1}{5}}}$$

$${\left( {HS{O_4}} \right)^ - }$$  can accept and donate a proton
$$\eqalign{ & {\left( {HS{O_4}} \right)^ - } + {H^ + } \to {H_2}S{O_4}\,\,{\text{(acting as base)}} \cr & {\left( {HS{O_4}} \right)^ - } - {H^ + } \to S{O_4}^{2 - }.\,\,\,\,{\text{(acting as}}\,{\text{acid)}} \cr} $$

$$\eqalign{ & pH = 5,\left[ {{H^ + }} \right] = {10^{ - 5}}\,M \cr & {\text{After dilution}} = \frac{{{{10}^{ - 5}}}}{{1000}} = {10^{ - 8}}\,M \cr & {\text{Total}}\left[ {{H^ + }} \right] = {10^{ - 8}} + {10^{ - 7}} = 1.1 \times {10^{ - 7}} \cr & pH = - \log \left[ {{H^ + }} \right] = - \log \left( {1.1 \times {{10}^{ - 7}}} \right) = 6.96 \cr} $$

$$\eqalign{ & Ba{\left( {OH} \right)_2} \rightleftharpoons B{a^{2 + }} + 2O{H^ - } \cr & \left[ {O{H^ - }} \right] = 2 \times 1 \times {10^{ - 3}}\,M \cr} $$
$$pOH = - \log \left[ {O{H^ - }} \right]$$    $$ = - \log \left( {2 \times {{10}^{ - 3}}} \right) = 2.7$$
$$pOH + pH = 14 \Rightarrow pH$$     $$ = 14 - 2.7 = 11.3$$