Ionic Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

The species with the greatest proton affinity will be the strongest base, and its conjugate acid will be the weakest acid. The weakest acid will have the smallest value of $${K_a}.$$  Since $$HI$$  is a stronger acid than $$HF$$  which is a stronger acid than $${H_2}S,$$  a partial order of proton affinity is
$${I^ - } < {F^ - } < H{S^ - }$$
Since $$N{H_3}$$  is a very weak acid, $$NH_2^ - $$  must be a very strong base. Therefore the correct order of proton affinity is
$${I^ - } < {F^ - } < H{S^ - } < NH_2^ - $$

$$\eqalign{ & \because \,\,pH = 1\,;\,{H^ + } = {10^{ - 1}} = 0.1\,M \cr & pH = 2\,;\,{H^ + } = {10^{ - 2}} = 0.01\,M \cr & \therefore {M_1} = 0.1\,\,{V_1} = 1;\,\,{M_2} = 0.01\,\,{V_2} = ? \cr & {\text{From}} \cr & {{\text{M}}_1}{V_1} = {M_2}{V_2};\,0.1 \times 1 = 0.01 \times {V_2} \cr & {V_2} = 10\,litre \cr & \therefore \,\,{\text{volume of water added}} = 10 - 1 = 9\,litre \cr} $$

$$\eqalign{ & Pb{I_2} \rightleftharpoons P{b^{ + + }} + 2{I^ - } \cr & {K_{sp}} = s \times {\left( {2s} \right)^2} = 4{s^3} \cr & = 4 \times {\left( {\frac{{0.7}}{{461.2}}} \right)^3} \cr & = 14.0 \times {10^{ - 9}} \cr} $$

$$\eqalign{ & {\text{Given at}}\,330\,K \cr & {K_w} = {10^{ - 13.6}} \cr & {\text{i}}{\text{.e}}{\text{.}}\,p{K_w} = pH + pOH \cr & \because \,\,pOH = - \log \left[ {O{H^ - }} \right] \cr & 13.6 = pH + pOH \cr & pOH = - \log \,{10^{ - 4}} \cr & pOH = 4 \cr & \therefore pH = 13.6 - 4 = 9.6 \cr} $$

$$\eqalign{ & \left[ {{H^ + }} \right] = C\alpha \cr & = 0.2 \times 0.40 \cr & = 0.08\,N. \cr & \left( {\alpha = \frac{{40}}{{100}} = 0.40} \right) \cr} $$

$${K_a} = C{\alpha ^2}$$
$$\alpha = \frac{{14.5}}{{100}} = 0.145;$$    $${K_a} = 0.01 \times {\left( {0.145} \right)^2} = 2.1 \times {10^{ - 4}}$$

In aqueous solution of $${10^{ - 8}}M\,HCl,\left[ {{H^ + }} \right]$$    $$ion$$  concentration is based upon the concentration of $${{H^ + }}$$ $$ion$$  of $${10^{ - 8}}M\,HCl$$    and concentration of $${{H^ + }}$$ $$ion$$  of water.
$${K_w}\,\,{\text{of}}\,\,{H_2}O = {10^{ - 14}} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]$$
or $$\left[ {{H^ + }} \right] = {10^{ - 7}}M$$   ( due to its neutral behaviour )
So, in aqueous solution of $${10^{ - 8}}M\,HCl,$$
$$\left[ {{H^ + }} \right] = \left[ {{H^ + }} \right]\,{\text{of}}\,\,HCl + \left[ {{H^ + }} \right]$$       of water
$$\eqalign{ & {\text{ = 1}}{{\text{0}}^{ - 8}} + {10^{ - 7}} \cr & = 11 \times {10^{ - 8}}M \approx 1.10 \times {10^{ - 7}}M \cr} $$

Millimoles of $${H^ + }$$ from $$HCl = 0.1 \times 10 = 1$$
Millimoles of $${H^ + }$$ from $${H_2}S{O_4} = 0.2 \times 40 \times 2 = 16$$
$$\eqalign{ & {\text{Conc}}{\text{.of}}\,\,{H^ + } = \frac{{{\text{Millimoles}}}}{{{\text{Volume}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{16 + 1}}{{40 + 10}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{17}}{{50}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.34 \cr} $$
$$\eqalign{ & pH = - \log \left[ {{H^ + }} \right] \cr & \,\,\,\,\,\,\,\,\, = - \log \left( {0.34} \right) \cr & \,\,\,\,\,\,\,\,\, = 0.468 \cr} $$

$$\eqalign{ & pH = 9;\left[ {{H^ + }} \right] = {10^{ - 9}};\left[ {O{H^ - }} \right] = {10^{ - 5}}; \cr & \left[ {M{g^{2 + }}} \right] = 1 \times {10^{ - 3}};\left[ {M{g^{2 + }}} \right]{\left[ {O{H^ - }} \right]^2} = 1 \times {10^{ - 13}} \cr & {\text{given}}\,{K_{sp}}\,{\text{of}}\,Mg{\left( {OH} \right)_2} = 8.9 \times {10^{ - 12}} \cr & {\text{which is more than }}1 \times {10^{ - 13}}. \cr & {\text{Hence}}\,Mg{\left( {OH} \right)_2}\,{\text{will not precipitate}} \cr} $$

For ternary electrolyte; solubility
$$\eqalign{ & = {\left( {\frac{{{K_{sp}}}}{4}} \right)^{\frac{1}{3}}} \cr & = {\left( {\frac{{32 \times {{10}^{ - 12}}}}{4}} \right)^{\frac{1}{3}}} \cr & = 2 \times {10^{ - 4}} \cr} $$