Ionic Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & pH = - \log \left[ {{H^ + }} \right] \cr & 2 = - \log \left[ {{H^ + }} \right];\left[ {{H^ + }} \right] = 1 \times {10^{ - 2}} \cr} $$

$$\eqalign{ & {\text{Let solubility of}}\,AgCl = x\,mole/L \cr & AgCl \rightleftharpoons A{g^ + } + C{l^ - } \cr & i.e.,\,{K_{sp\left( {AgCl} \right)}} = x \times x \cr & KCl \to {K^ + } + \mathop {C{l^ - }}\limits_{0.1} \cr & \left[ {C{l^ - }} \right]\,{\text{from}}\,KCl = 0.1\,m \cr & {\text{Total}}\,\left[ {C{l^ - }} \right]\,{\text{in solution}} = x + 0.1 \cr & {K_{sp}}\left( {AgCl} \right) = \left[ {A{g^ + }} \right]\left[ {C{l^ - }} \right] = x\left( {x + 0.1} \right) \cr & 1.0 \times {10^{ - 10}} = x\left( {x + 0.1} \right) \cr & 1.0 \times {10^{ - 10}} = {x^2} + 0.1x \cr & 1.0 \times {10^{ - 10}} = 0.1x\left( {{\text{as}}\,\,{x^2} < < 1} \right) \cr & x = 1.0 \times {10^{ - 9}}\,mol/L \cr} $$

Let the volume of each acid $$=V$$
$$pH$$  of first, second and third acids = 3, 4 and 5 respectively
$$\left[ {{H^ + }} \right]$$  of first acid $$\left( {{M_1}} \right) = 1 \times {10^{ - 3}}\left[ {\because \,{H^ + } = 1 \times {{10}^{ - pH}}} \right]$$
$$\left[ {{H^ + }} \right]$$  of second acid $$\left( {{M_2}} \right) = 1 \times {10^{ - 4}}$$
$$\left[ {{H^ + }} \right]$$  of third acid $$\left( {{M_3}} \right) = 1 \times {10^{ - 5}}$$
Total $$\left[ {{H^ + }} \right]$$  concentrated of mixture
$$\eqalign{ & \left( M \right) = \frac{{{M_1}{V_1} + {M_2}{V_2} + {M_3}{V_3}}}{{{V_1} + {V_2} + {V_3}}} \cr & = \frac{{1 \times {{10}^{ - 3}} \times V + 1 \times {{10}^{ - 4}} \times V + 1 \times {{10}^{ - 5}} \times V}}{{V + V + V}} \cr & = \frac{{1 \times {{10}^{ - 3}} \times V\left( {1 + 0.1 + 0.01} \right)}}{{3V}} \cr & = \frac{{1.11 \times {{10}^{ - 3}}}}{3} \cr & = 3.7 \times {10^{ - 4}}M \cr} $$

$${K_{sp}}$$  of $$A{s_2}{S_3}$$  is less than $$ZnS.$$  In acidic medium ionisation of $${H_2}S$$  is suppresed (common ion effect) and $${K_{sp}}$$  of $$ZnS$$  does not exceed.

$$\left( {\text{A}} \right):Fe{\left( {OH} \right)_3} \rightleftharpoons \mathop {F{e^{3 + }}}\limits_s + \mathop {3O{H^ - }}\limits_{{{\left( {3s} \right)}^3}} ;$$       $${K_{sp}} = 27{s^4}$$
$$\left( {\text{B}} \right):A{g_{_2}}Cr{O_4} \rightleftharpoons \mathop {2A{g^ + }}\limits_{{{\left( {2s} \right)}^2}} + \mathop {CrO_4^{2 - }}\limits_s ;$$       $${K_{sp}} = 4{s^3}$$
$$\left( {\text{C}} \right):C{H_3}COOAg \rightleftharpoons $$     $$\mathop {C{H_3}CO{O^ - }}\limits_s + \mathop {A{g^ + }}\limits_s ;{K_{sp}} = {s^2}$$
$$\left( {\text{D}} \right):C{a_3}{\left( {P{O_4}} \right)_2} \rightleftharpoons $$     $$\mathop {3C{a^{2 + }}}\limits_{{{\left( {3s} \right)}^3}} + \mathop {2PO_4^{3 - }}\limits_{{{\left( {2s} \right)}^2}} ;{K_{sp}} = 108{s^5}$$

TIPS/Formulae :
(i) Higher the electronegatively of central atom higher will be the acidic strength.
(ii) In case of same atom higher the value of oxidation state of the metal, higher will be its acidic strength.
The electronegativity of $$Cl > S.$$
Oxidation no. of $$Cl$$ in $$Cl{O_3}\left( {OH} \right) = + 7$$
Oxidation no. of $$Cl$$ in $$Cl{O_2}\left( {OH} \right) = + 5$$
Oxidation no. of $$S$$ in $$SO{\left( {OH} \right)_2} = + 4$$
Oxidation no. of $$S$$ in $$S{O_2}{\left( {OH} \right)_2} = + 6$$
∴ $$Cl{O_3}\left( {OH} \right)$$   is the strongest acid.

In molten state the cations and anions become free and flow of current is due to migration of these ions in opposite directions in the electric field.

$$AgCl + 2N{H_3} \to \mathop {\left[ {Ag{{\left( {N{H_3}} \right)}_2}} \right]Cl}\limits_{{\text{(Soluble)}}} $$
Thus, $$AgCl$$  is most soluble in aqueous ammonia.

$$\eqalign{ & \mathop {Ca{F_2}}\limits_{2 \times {{10}^{ - 4}}M} \rightleftharpoons \mathop {C{a^{2 + }}}\limits_{2 \times {{10}^{ - 4}}M} + \mathop {2{F^ - }}\limits_{2 \times 2 \times {{10}^{ - 4}}M} \cr & {K_{sp}}\,\,{\text{of}}\,\,Ca{F_2} = \left[ {C{a^{2 + }}} \right]{\left[ {{F^ - }} \right]^2} \cr & = \left[ {2 \times {{10}^{ - 4}}} \right]{\left[ {4 \times {{10}^{ - 4}}} \right]^2} \cr & = 32 \times {10^{ - 12}}{\left( {mol/L} \right)^2} \cr} $$

$$\eqalign{ & {K_2}C{O_3} + 2{H_2}O \rightleftharpoons \mathop {{H_2}C{O_3}}\limits_{{\text{Weak acid}}} + \mathop {2KOH}\limits_{{\text{Strong base}}} \cr & KOH \rightleftharpoons {K^ + } + O{H^ - }\left( {{\text{Basic solution}}} \right) \cr} $$