Ionic Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

Since $$HCl$$  is stronger than $$C{H_3}COOH$$   hence acts as acid. On the other hand $$C{l^ - }$$ is a stronger base than $$C{H_3}COOH_2^ + $$  and is the conjugate base of $$HCI.$$
$$\eqalign{ & HCl + C{H_3}COOH \rightleftharpoons C{l^ - } + C{H_3}COOH_2^ + \cr & {\text{aci}}{{\text{d}}_{\text{1}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{bas}}{{\text{e}}_{\text{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{bas}}{{\text{e}}_{\text{1}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{aci}}{{\text{d}}_{\text{2}}} \cr} $$

All other can accept electrons. $$P{H_3}$$  is not electron deficient.

$$\eqalign{ & {K_a} = c{\alpha ^2} \cr & = 0.1 \times {\left( {\frac{{1.34}}{{100}}} \right)^2} \cr & = 1.8 \times {10^{ - 5}} \cr} $$

$$HSO_4^ - $$  accepts a proton to form $${H_2}S{O_4}.$$
Thus $${H_2}S{O_4}$$  is the conjugate acid of $$HSO_4^ - .$$
\[\underset{\text{base}}{\mathop{HSO_{4}^{-}}}\,\xrightarrow{+{{H}^{+}}}\underset{\begin{smallmatrix} \text{conjugate acid} \\ \text{of}\,HSO_{4}^{-} \end{smallmatrix}}{\mathop{{{H}_{2}}S{{O}_{4}}}}\,\]

$$\eqalign{ & \alpha = \sqrt {\frac{{{K_a}}}{C}} \,\,{\text{or}}\,\,\alpha {\text{ = }}\sqrt {\frac{{{{10}^{ - 5}}}}{{0.1}}} = {10^{ - 2}} \cr & \% \,\,{\text{dissociation}} = {10^{ - 2}} \times 100 = 1\% \cr} $$

$${K_{sp}}\,\,{\text{for}}\,\,Mg{\left( {OH} \right)_2} = 1.0 \times {10^{ - 11}}$$
$$\left[ {M{g^{2 + }}} \right]{\left[ {O{H^ - }} \right]^2} = \left( {0.1} \right) \times {\left[ {O{H^ - }} \right]^2}$$       $$ = 1.0 \times {10^{ - 11}}$$
$${\left[ {O{H^ - }} \right]^2} = \frac{{1 \times {{10}^{ - 11}}}}{{0.1}} = {10^{ - 10}}$$
$$\left[ {O{H^ - }} \right] = {10^{ - 5}};$$   $$\left[ {{H^ + }} \right] = \frac{{{{10}^{ - 14}}}}{{{{10}^{ - 5}}}} = {10^{ - 9}};\,pH = 9$$

$$2HN{O_3}\left( {aq} \right) + {\left[ {Ag{{\left( {N{H_3}} \right)}_2}} \right]^ + } + C{l^ - } \to AgCl\left( s \right) \downarrow + 2N{H_4} + 2NO_3^ - $$
When nitric acid is added to amine solution, solution is made acidic and the complex $$ion$$  dissociates and liberate silver ion to recombine with chloride $$ion.$$  This is the conformatory test for silver in group 1.

Aqueous solution of sugar does not conduct electricity.

$$\eqalign{ & C{H_3}COOH \rightleftharpoons C{H_3}CO{O^ - } + {H^ + } \cr & {K_a} = \frac{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H^ + }} \right]}}{{\left[ {C{H_3}COOH} \right]}} \cr & {\text{Given that,}} \cr & \left[ {C{H_3}CO{O^ - }} \right] = \left[ {{H^ + }} \right] = 3.4 \times {10^{ - 4}}M \cr & \,\,\,\,\,\,\,\,{K_a}\,{\text{for}}\,C{H_3}COOH = 1.7 \times {10^{ - 5}} \cr} $$
$$C{H_3}COOH$$   is weak acid, so in it $$\left[ {C{H_3}COOH} \right]$$   is equal to initial concentration. Hence,
$$\eqalign{ & 1.7 \times {10^{ - 5}} = \frac{{\left( {3.4 \times {{10}^{ - 4}}} \right)\left( {3.4 \times {{10}^{ - 4}}} \right)}}{{\left[ {C{H_3}COOH} \right]}} \cr & \left[ {C{H_3}COOH} \right] = \frac{{3.4 \times {{10}^{ - 4}} \times 3.4 \times {{10}^{ - 4}}}}{{1.7 \times {{10}^{ - 5}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6.8 \times {10^{ - 3}}M \cr} $$

When ionic product i.e. the product of the concentration of ions in the solution exceeds the value of solubility product, formation of precitpiate occurs.
$$\eqalign{ & Ca{F_2} \rightleftharpoons C{a^{2 + }} + 2{F^ - } \cr & {\text{Ionic product}} = \left[ {C{a^{2 + }}} \right]{\left[ {{F^ - }} \right]^2} \cr & {\text{when,}}\,\left[ {C{a^{2 + }}} \right] = 1 \times {10^{ - 2}}M \cr & {\left[ {{F^ - }} \right]^2} = {\left( {1 \times {{10}^{ - 3}}} \right)^2}M \cr & = 1 \times {10^{ - 6}}M \cr & \therefore \left[ {C{a^{2 + }}} \right]{\left[ {{F^ - }} \right]^2} \cr & = \left( {1 \times {{10}^{ - 2}}} \right)\left( {1 \times {{10}^{ - 6}}} \right) \cr & = 1 \times {10^{ - 8}} \cr & {\text{In this case,}} \cr & {\text{Ionic product}}\left( {1 \times {{10}^{ - 8}}} \right) > {\text{solubility product}}\left( {1.7 \times {{10}^{ - 10}}} \right) \cr} $$