Ionic Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & \mathop {{H_2}C{O_3}\left( {aq} \right)}\limits_{0.034 - x} + {H_2}O\left( l \right) \rightleftharpoons \mathop {HCO_3^ - \left( {aq} \right)}\limits_x + \mathop {{H_3}{O^ + }\left( {aq} \right)}\limits_x \cr & {K_1} = \frac{{\left[ {HCO_3^ - } \right]\left[ {{H_3}{O^ + }} \right]}}{{\left[ {{H_2}C{O_3}} \right]}} = \frac{{x \times x}}{{0.034 - x}} \cr & \Rightarrow 4.2 \times {10^{ - 7}} \simeq \frac{{{x^2}}}{{0.034}} \Rightarrow x = 1.195 \times {10^{ - 4}} \cr} $$
As $${H_2}C{O_3}$$  is a weak acid so the concentration of $${H_2}C{O_3}$$  will remain $$0.034$$  as $$0.034 > > x.$$
$$\eqalign{ & x = \left[ {{H^ + }} \right] = \left[ {HCO_3^ - } \right] = 1.195 \times {10^{ - 4}} \cr & {\text{Now,}}\,\,\mathop {HCO_3^ - }\limits_{x - y} \left( {aq} \right) + {H_2}O\left( l \right) \rightleftharpoons \mathop {CO_3^{2 - }}\limits_y \left( {aq} \right) + \mathop {{H_3}{O^ + }}\limits_y \left( {aq} \right) \cr} $$
As $$HCO_3^ - $$ is again a weak acid ( weaker than $${H_2}C{O_3}$$ ) with x >> y.
$${K_2} = \frac{{\left[ {CO_3^{2 - }} \right]\left[ {{H_3}{O^ + }} \right]}}{{\left[ {HCO_3^ - } \right]}} = \frac{{y \times \left( {x + y} \right)}}{{\left( {x - y} \right)}}$$
Note : $$\left[ {{H_3}{O^ + }} \right] = {H^ + }$$   from first step $$(x)$$  and from second step $$\left( y \right) = \left( {x + y} \right)$$
$$\eqalign{ & \left[ {{\text{As}}\,\,x > > y\,\,\,so\,\,\,x + y \simeq x\,\,\,{\text{and}}\,\,\,x - y \simeq x} \right] \cr & {\text{So}},\,{K_2} \simeq \frac{{y \times x}}{x} = y \cr & \Rightarrow {K_2} = 4.8 \times {10^{ - 11}} = y = \left[ {CO_3^{2 - }} \right] \cr & {\text{So the concentration of}}\,\left[ {{H^ + }} \right] \simeq \left[ {HCO_3^ - } \right] = \cr} $$
concentrations obtained from the first step. As the dissociation will be very low in second step so there will be no change in these concentrations. Thus the final concentrations are
$$\left[ {{H^ + }} \right] = \left[ {HCO_3^ - } \right] = 1.195 \times {10^{ - 4}}\& \,\left[ {CO_3^{2 - }} \right] = 4.8 \times {10^{ - 11}}$$

NOTE : For isotonic solutions, osmotic pressure is same.
$$\mathop {N{a_2}3{O_4}}\limits_{\left( {0.004\, - \,x} \right)} \,\,\,\,\,\,\,\,\,\, \rightleftharpoons \mathop {2N{a^ + }}\limits_{2x} + \mathop {SO_4^{2 - }}\limits_x $$
$${\text{Since both solutions are isotonic,}}$$
$${\text{therefore, }}0.004 + 2x\, = 0.01;$$      $${\text{ }}x = 3 \times {10^{ - 3}}$$
$$\therefore \,\,\% \,{\text{Dissociation = }}\frac{{3 \times {{10}^{ - 3}}}}{{0.004}} \times 100$$        $$ = 75\% $$

$$\eqalign{ & {K_{sp}} = \left[ {A{g^ + }} \right]\left[ {C{l^ - }} \right] \cr & \left[ {C{l^ - }} \right] = NaCl = 0.05\,M \cr & A{g^ + } = \frac{{1.5 \times {{10}^{ - 10}}}}{{0.05}} = 3 \times {10^{ - 9}}\,M; \cr & \left[ {A{g^ + }} \right] = {\text{solubility}} = 3 \times {10^{ - 9}}\,M \cr} $$

$$\eqalign{ & \mathop {N{a_2}C{O_3}}\limits_{1 \times {{10}^{ - 4}}M} \to \mathop {2N{a^ + }}\limits_{1 \times {{10}^{ - 4}}M} + \mathop {C{O_3}^{2 - }}\limits_{1 \times {{10}^{ - 4}}M} \cr & {K_{sp\left( {BaC{O_3}} \right)}} = \left[ {B{a^{2 + }}} \right]\left[ {CO_3^{2 - }} \right] \cr & \left[ {B{a^{2 + }}} \right] = \frac{{5.1 \times {{10}^{ - 9}}}}{{1 \times {{10}^{ - 4}}}} = 5.1 \times {10^{ - 5}}M \cr} $$

$$\begin{array}{l}\text{For a precipitation to occur}\\ \text{Solubility product < Ionic product}\\ \text{Given}{K}_{sp}=1.8\times {10}^{-10}\\ \text{Calculating ionic products in each}\\ \text{Ionic}\text{product}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]\\ =\frac{{10}^{-4}}{2}\times \frac{{10}^{-4}}{2}\\ =2.5\times {10}^{-9}\\ \text{which is greater than}{K}_{sp}\left(1.8\times {10}^{-10}\right).\\ \left(\begin{array}{l}\because \text{equal volumes are mixed,}\\ \text{so}\left[A{g}^{+}\right]=\frac{{10}^{-4}}{2}\text{and}\left[C{l}^{-}\right]=\frac{{10}^{-4}}{2}\end{array}\right)\\ \text{step}-2\end{array}$$

Number of milliequivalents of $$HCl$$
$$ = 20 \times 0.050 \times 1 = 1$$
Number of milliequivalents of $$Ba{\left( {OH} \right)_2}$$
$$ = 2 \times 30 \times 0.10 = 6$$
$$\eqalign{ & \left[ {O{H^ - }} \right]\,{\text{of final solution}} \cr & {\text{Milliequivalents of}}\,Ba{\left( {OH} \right)_2} \cr & = \frac{{ - \,{\text{milliequivalents of}}\,\,HCl}}{{{\text{Total volume}}}} \cr & = \frac{{6 - 1}}{{50}} \cr & = 0.1\,M \cr} $$

$$\eqalign{ & {\alpha _1} = 0.005 = \sqrt {{K_a} \times {C_1}} \cr & {\text{Molarity of diluted solution;}}\,2 \times 1 = 32 \times M, \cr & M = \frac{1}{{16}}\left( {{C_2}} \right) \cr & {\alpha _2} = \sqrt {\frac{{{K_a}}}{{{C_2}}}} \cr & = 0.005\sqrt {16} \cr & = 0.02 \cr & {H_3}\mathop O\limits^ + = {C_2}{\alpha _2} \cr & = \frac{{0.02}}{{16}} \cr & = 1.25 \times {10^{ - 3}}M \cr} $$

In $$0.1\,mol\,d{m^{ - 3}}\,N{H_4}OH$$     and $$0.05\,mol\,d{m^{ - 3}}\,HCl,$$     total amount of $$HCl$$  reacts with $$N{H_4}OH$$   to form $$N{H_4}Cl$$   and some $$N{H_4}OH$$   will be left unreacted. Thus, the resultant solution contains $$N{H_4}Cl$$   and $$N{H_4}OH$$   which will produce a buffer solution.

$$\eqalign{ & NaCl + AgN{O_3} \to AgCl + NaN{O_3} \cr & {K_{sp}} = 1.5 \times {10^{ - 10}} \cr & \left[ {A{g^ + }} \right] = \frac{1}{2} \times {10^{ - 2}}\,M \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.5 \times {10^{ - 2}}\,M \cr & \left[ {C{l^ - }} \right] = \frac{1}{2} \times {10^{ - 2}}\,M \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.5 \times {10^{ - 2}}M \cr & {K_{ip}} = \left[ {A{g^ + }} \right]\left[ {C{l^ - }} \right] \cr & = \left( {0.5 \times {{10}^{ - 2}}} \right) \times \left( {0.5 \times {{10}^{ - 2}}} \right) \cr & = 2.5 \times {\bf{1}}{0^{ - 5}} \cr & {K_{ip}} > {K_{sp}} \cr} $$
When $${K_{ip}} > {K_{sp}},$$   it results in precipitation.

$${K_a} = C{\alpha ^2} = 0.1 \times {\left( {{{10}^{ - 5}}} \right)^2} = {10^{ - 11}}$$