Ionic Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & {H_2}X \rightleftharpoons {H^ + } + H{X^ - }\left( {{K_{{a_1}}}} \right) \cr & H{X^ - } \rightleftharpoons {H^ + } + {X^{2 - }}\left( {{K_{{a_2}}}} \right) \cr} $$
In I^st equation hydrogen ion is formed from neutral molecule whereas in II^nd equation it comes from negatively charged species. Due to negative charge removal of proton is difficult.
So,  $${K_{{a_1}}} > {K_{{a_2}}}$$

$$\eqalign{ & {\text{Given}}\,\left[ {O{H^ - }} \right] = 5 \times {10^{ - 2}} \cr & \therefore \,\,pOH = - \log \,5 \times {10^{ - 2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - \log \,5 + 2\,\log \,10 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.30 \cr & \because \,\,pH + pOH = 14 \cr & \because \,\,pH = 14 - pOH \cr & = 14 - 1.30 \cr & = 12.70 \cr} $$

$$\eqalign{ & C{O_2} + 2{H_2}O \rightleftharpoons {H_3}{O^ + } + HCO_3^ - ; \cr & {K_c} = \frac{{\left[ {{H_3}{O^ + }} \right]\left[ {HCO_3^ - } \right]}}{{\left[ {C{O_2}} \right]}}; \cr & \frac{{\left[ {HCO_3^ - } \right]}}{{\left[ {C{O_2}} \right]}} \cr & = \frac{{3.8 \times {{10}^{ - 7}}}}{{{{10}^{ - 6}}}} \cr & = 3.8 \times {10^{ - 1}} \cr} $$

$$\eqalign{ & \left[ {B{a^{2 + }}} \right]\left[ {SO_4^{2 - }} \right] \cr & {K_{sp}} = \left( {1.0 \times {{10}^{ - 4}}} \right)\left[ {SO_4^{2 - }} \right] = 4 \times {10^{ - 10}} \cr & \therefore \,\,\left[ {SO_4^{2 - }} \right] = 4 \times {10^{ - 6}}M \cr} $$

$$BaC{l_2}$$  is a salt of strong acid $$HCl$$  and strong base $$Ba{\left( {OH} \right)_2}.$$   So, its aqueous solution is neutral with $$pH$$ $$7.$$  All other salts give acidic solution due to cationic hydrolysis, so their $$pH$$ is less than $$7.$$
Thus, $$pH$$ value is highest for the solution of $$BaC{l_2}.$$

$$p{K_a}\,\,{\text{of}}\,\,NH_4^ + = 9.26$$
$${\text{Hence,}}\,p{K_b}\,\,{\text{of}}\,\,N{H_4}OH$$     $$ = 14 - 9.26 = 4.74$$
$$pOH = 14 - 9.35 = 4.65$$
Each $$1\,mole$$  of $${\left( {N{H_4}} \right)_2}S{O_4}$$   furnishes $$2\,moles$$  of $$NH_4^ + $$  in solution.
$${\text{Let}}\,\left[ {{{\left( {N{H_4}} \right)}_2}S{O_4}} \right] = x\,mol\,{L^{ - 1}}$$
$$\left[ {NH_4^ + } \right] = 2\,x\,mol\,{L^{ - 1}};$$     $$\left[ {N{H_4}OH} \right] = 0.2\,mol\,{L^{ - 1}}$$
using Henderson - Hasselbalch equation,
$$pOH = p{K_b} + \log \,\frac{{\left[ {NH_4^ + } \right]}}{{\left[ {N{H_4}OH} \right]}};$$       $$4.65 = 4.74 + \log \left( {\frac{{2x}}{{0.2}}} \right)$$
This gives $$x = 0.081\,mol\,{L^{ - 1}}$$
$$ = 0.0405\,mol$$    in $$500\,mL$$  ( as required ) $$ = 0.0405 \times 132 = 5.32\,g$$

$$\eqalign{ & C{O_2}\,{\text{with }}{H_2}O\,{\text{forms}}\,{H_2}C{O_3} \cr & C{O_2} + {H_2}O \rightleftharpoons {H^ + } + HCO_3^ - \cr & {K_1} = \frac{{\left[ {{H^ + }} \right]\left[ {HCO_3^ - } \right]}}{{\left[ {C{O_2}} \right]}} \cr & = 4.5 \times {10^{ - 7}} \cr & {\text{Again}}\,pH = - \log \left[ {{H^ + }} \right] = 7.4 \cr & \therefore \,\,\left[ {{H^ + }} \right] = 4.0 \times {10^{ - 8}} \cr & \therefore \,\,\frac{{\left[ {HCO_3^ - } \right]}}{{\left[ {C{O_2}} \right]}} \cr & = \frac{{4.5 \times {{10}^{ - 7}}}}{{4 \times {{10}^{ - 8}}}} \cr & = 11 \cr} $$

Sodium hydride dissolved in water as
$$\eqalign{ & NaH + {H_2}O \to NaOH + {H_2} \cr & {\text{or}}\,\,\,{H^ - }\left( {aq} \right) + {H_2}O\left( l \right) \to O{H^ - } + {H_2} \uparrow \cr} $$
In the above reaction hydride ion take proton from water molecule and hydrogen gas is evolved.

$$\eqalign{ & \mathop {N{a_2}S{O_4}}\limits_{1 - \alpha } \rightleftharpoons \mathop {2N{a^ + }}\limits_{2\alpha } + \mathop {SO_4^{ - - }}\limits_\alpha \cr & {\text{Vant}}{\text{. Hoff}}\,{\text{factor}}\,{\text{i}} = \frac{{1 - \alpha + 2\alpha + \alpha }}{1} = 1 + 2\alpha \cr} $$

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,A{B_2}\left( g \right) \rightleftharpoons 2\,AB\left( g \right) + {B_2}\left( g \right) \cr & {\text{Initial}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{0}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{0}} \cr & {\text{moles}} \cr & {\text{At equil}}{\text{.}}\,\,\,{\text{2}}\left( {1 - x} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \cr & {\text{where, }}x = {\text{degree of dissociation}} \cr & {\text{Total moles at equilibrium}} \cr & = 2 - 2x + 2x + x = \left( {2 + x} \right) \cr & {\text{So,}}\,\,{p_{A{B_2}}} = \frac{{2\left( {1 - x} \right)p}}{{\left( {2 + x} \right)}},\,\,{p_{AB}} = \frac{{2x\,p}}{{\left( {2 + x} \right)}} \cr & {p_{{B_2}}} = \frac{{xp}}{{\left( {2 + x} \right)}} \cr & {K_p} = \frac{{{{\left( {{p_{AB}}} \right)}^2}\left( {{P_{{B_2}}}} \right)}}{{{{\left( {{p_{A{B_2}}}} \right)}^2}}} \cr & = \frac{{{{\left( {\frac{{2x\,p}}{{2 + x}}} \right)}^2}\left[ {\left( {\frac{x}{{2 + x}}} \right)\,p} \right]}}{{{{\left[ {\left( {\frac{{2\left( {1 - x} \right)}}{{2 + x}}} \right)\,p} \right]}^2}}} \cr & = \frac{{4{x^3}{p^3}}}{{{{\left( {2 + x} \right)}^3}}} \times \frac{{{{\left( {2 + x} \right)}^2}}}{{{p^2}4{{\left( {1 - x} \right)}^2}}} \cr & = \frac{{{x^3}p}}{{\left( {2 + x} \right){{\left( {1 - x} \right)}^2}}} \cr & = \frac{{{x^3}p}}{2}\,\,\,\,\left[ {\because \,\,x < < < \,1\,{\text{and}}\,2} \right] \cr & x = {\left( {\frac{{2\,{K_p}}}{p}} \right)^{\frac{1}{3}}} \cr & {\text{So,}}\,\,\left( {1 - x} \right) \approx 1 \cr & \left( {2 + x} \right) \approx 2 \cr} $$