Ionic Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

$$pH = - \log \left[ {{H^ + }} \right] = $$     $$ - \log \left( {3 \times {{10}^{ - 5}}} \right) = 4.5229$$
$$pOH = 14 - pH,$$   $$14 - 4.5229 = 9.48$$

$$\eqalign{ & {\text{For pure water,}}\,\,\left[ {{H_3}{O^ + }} \right] = \left[ {O{H^ - }} \right] \cr & \Rightarrow {K_w} = {10^{ - 6}} \times {10^{ - 6}} = {10^{ - 12}} \cr} $$

$$\eqalign{ & {\text{Species}}\,\,\,\,{\text{Conjugate acid}}\,\,\,\,{\text{Conjugate base}} \cr & HCO_3^ - \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{H_2}C{O_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,CO_3^{2 - } \cr & HSO_4^ - \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{H_2}S{O_4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,SO_4^{2 - } \cr & N{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,NH_4^ + \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,NH_2^ - \cr & {H_2}O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{H_3}{O^ + }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O{H^ - } \cr} $$

$$\eqalign{ & \frac{{{\alpha _1}}}{{{\alpha _2}}} = \sqrt {\frac{{{K_{{a_1}}}}}{{{K_{{a_2}}}}}} \cr & = \sqrt {\frac{{3.14 \times {{10}^{ - 4}}}}{{1.96 \times {{10}^{ - 5}}}}} \cr & = 4:1 \cr} $$

$$\eqalign{ & BaS{O_4} \rightleftharpoons \mathop {B{a^{2 + }}}\limits_{1 \times {{10}^{ - 4}}} + \mathop {SO_4^{2 - }}\limits_s \cr & {K_{sp}} = 4 \times {10^{ - 10}};4 \times {10^{ - 10}} = 1 \times {10^{ - 4}} \times s \cr & s = \frac{{4 \times {{10}^{ - 10}}}}{{1 \times {{10}^{ - 4}}}} \cr & \,\,\,\, = 4 \times {10^{ - 6}}\,M \cr} $$

$$\eqalign{ & {\text{For the reaction :}} \cr & {H_2}S \rightleftharpoons {H^ + } + H{S^ - };{K_{{a_{_1}}}} \cr & {K_{{a_{_1}}}} = \frac{{\left[ {{H^ + }} \right]\left[ {H{S^ - }} \right]}}{{\left[ {{H_2}S} \right]}}\,\,...\left( {\text{i}} \right) \cr & {\text{For the reaction :}} \cr & H{S^ - } \rightleftharpoons {H^ + } + {S^{2 - }};{K_{{a_{_2}}}} \cr & {K_{{a_{_2}}}} = \frac{{\left[ {{H^ + }} \right]\left[ {{S^{2 - }}} \right]}}{{\left[ {H{S^ - }} \right]}}\,\,...\left( {{\text{ii}}} \right) \cr & {\text{For the reaction :}} \cr & {H_2}S \rightleftharpoons 2{H^ + } + {S^{2 - }};{K_{{a_{_3}}}} \cr & {K_{{a_{_3}}}} = \frac{{{{\left[ {{H^ + }} \right]}^2}\left[ {{S^{2 - }}} \right]}}{{\left[ {{H_2}S} \right]}}\,\,...\left( {{\text{iii}}} \right) \cr} $$
$${\text{By}}\,\,{\text{multiplying eqn}}{\text{.}}\left( {\text{i}} \right){\text{and eqn}}{\text{.}}\left( {{\text{ii}}} \right){\text{,}}$$       $${\text{we get}}$$
$$\eqalign{ & {K_{{a_{_1}}}} \times {K_{{a_{_2}}}} = \frac{{\left[ {{H^ + }} \right]\left[ {H{S^ - }} \right]}}{{\left[ {{H_2}S} \right]}} \times \frac{{\left[ {{H^ + }} \right]\left[ {{S^{2 - }}} \right]}}{{\left[ {H{S^ - }} \right]}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{{\left[ {{H^ + }} \right]}^2}\left[ {{S^{2 - }}} \right]}}{{\left[ {{H_2}S} \right]}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {K_{{a_{_3}}}} \cr} $$

Follow Lowry Bronsted concept for conjugate pair. Acid strength follows the order $${H_2}O > {C_2}{H_2} > N{H_3} > {C_2}{H_6}$$       conjugate base will follow the order
$${C_2}H_5^ - > NH_2^ - > {C_2}{H^ - } > O{H^ - }$$

$$NaCN$$  solution is basic in nature since $$HCN$$  formed is a weak acid and does not hydrolyse.
$$NaCN + {H_2}O \rightleftharpoons NaOH + HCN$$
$$NaOH \rightleftharpoons N{a^ + } + O{H^ - }$$    $${\text{(Basic solution)}}$$

$${K_{sp}}\left( {AgBr} \right) < {K_{sp}}\left( {AgCl} \right)$$
Therefore, $$AgBr$$  will precipitate first and at that time all $$C{l^ - }$$  will be present.

Strong acid with its salt cannot form buffer solution. Hence, $$HCl{O_4}\,$$ and $$NaCl{O_4}$$  is not an acidic buffer.