Ionic Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

The species formed after adding a proton to the base is known as conjugate acid of the base and the species formed after losing a proton is known as conjugate base of acid. So,
$$\mathop {NH_2^ - }\limits_{{\text{Base}}} + {H^ + } \to \mathop {N{H_3}}\limits_{{\text{Conjugate acid}}} $$

$$\eqalign{ & M{X_4} \rightleftharpoons \mathop {{M^ + }}\limits_s + \mathop {4{X^ - }}\limits_{4s} \cr & {K_{sp}} = s \times {\left( {4s} \right)^4} = 256{s^5} \cr} $$

$$\eqalign{ & pH\,\,{\text{of}}\,\,C{H_3}COOH = 3 \cr & \left[ {{H^ + }} \right] = {10^{ - 3}}\,M\,\,{\text{from}}\,\,C{H_3}COOH \cr & \left[ {N{H_4}OH} \right] = \left[ {C{H_3}COOH} \right] = {10^{ - 3}}\,M \cr & \left[ {O{H^ - }} \right] = {10^{ - 3}}M \cr} $$
$$\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] = {10^{ - 14}};$$     $$\left[ {{H^ + }} \right] = \frac{{{{10}^{ - 14}}}}{{{{10}^{ - 3}}}} = {10^{ - 11}}$$
$$pH\,\,{\text{of}}\,\,N{H_4}OH = - \log \left( {{{10}^{ - 11}}} \right) = 11$$

$$\eqalign{ & Ca{C_2}{O_4} \rightleftharpoons \mathop {Ca}\limits^{2 + } + {C_2}O_4^{2 - };{K_{sp}} = {S^2} \cr & \therefore \,\,S = \sqrt {{K_{sp}}} \cr & = \sqrt {2.5 \times {{10}^{ - 9}}} \cr & = 5 \times {10^{ - 5}}mol/litre \cr & {\text{solubility}} = 5 \times {10^{ - 5}} \times 128 = 0.0064\,g \cr} $$

Acidic buffer has a $$pH$$  of around 4.75 and basic buffer has a $$pH$$  around 9.25.

TIPS/Formulae :
For oxyacids containing similar central atom, the acid strength increases with the increase in the number of oxygen atom attached to the central atom and not attached to any other atom.
TIPS/Formulae:
Higher the oxidation number ofthe central atom, higher is the acidity of the species. Thus acidity follows the order
$${\text{Oxi}}{\text{. No}}{\text{. of }}Cl{\text{ }}\,\,\,\,\,\,\mathop {HCl{O_{}}}\limits_{ + 1} < \,\,\mathop {HCl{O_2}}\limits_{ + 3} < \mathop {HCl{O_3}}\limits_{ + 5} < \mathop {HCl{O_4}}\limits_{ + 7} $$

$$\eqalign{ & {\text{Let the weak manoacidic base be }}BOH{\text{, then the + 3}} \cr & {\text{reaction that occurs during titration is}} \cr & BOH + HCl \to BCl + {H_2}O \cr & {\text{Equilibrium :}}\,\,\,\mathop {{B^ + }\,}\limits_{C\left( {1 - h} \right)} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {H_2}O \rightleftharpoons \mathop {BOH}\limits_{C.h} + \mathop {{H^ + }}\limits_{C.h} \cr & {\text{Using the normality equation}},\,\,\mathop {\,{N_1}{V_1}\,}\limits_{\left( {{\text{acid}}} \right)} \, = \mathop {\left( {{N_2}{V_2}} \right)}\limits_{\left( {{\text{base}}} \right)} \cr & {\text{Substituting various given values, we get}} \cr & \frac{2}{{15}} \times {V_1} = 2.5 \times \frac{2}{5} \cr & {\text{or}}\,{V_1} = 2.5 \times \frac{2}{5} \times \frac{{15}}{2} = 2.5 \times 3 = 7.5\,{\text{ml}} \cr & {\text{Then the concentration of }}BCl{\text{ in resulting solution is given by}} \cr & \left[ {BCl} \right] = \frac{{\frac{2}{{15}} \times 2.5}}{{10}} = \frac{2}{{10}}\,{\text{or}}\,0.1\,M \cr & {\text{Since}}\,{K_h} = \frac{{{K_w}}}{{{K_b}}}\,\,\,\,\,\,\therefore \,\,\,{K_h} = \frac{{1 \times {{10}^{ - 14}}}}{{1 \times {{10}^{ - 12}}}} = {10^{ - 2}} \cr & {\text{Thus}}\,{K_h} = \frac{{0.1{h^2}}}{{\left( {1 - h} \right)}}\,\,{\text{or}}\,\,{10^{ - 2}} = \frac{{0.1{h^2}}}{{\left( {1 - h} \right)}} \cr & {\text{or}}\,{10^{ - 2}} - {10^{ - 2}}h = 0.1{h^2}\,\,\,or\,\,0.1\,{h^2} + {10^{ - 2}}h - {10^{ - 2}} = 0 \cr & {\text{(Solving this quadratic equation for }}h{\text{, we get)}} \cr & h = \frac{{ - {{10}^{ - 2}} \pm \sqrt {{{\left( {{{10}^{ - 2}}} \right)}^2} + 4 \times {{10}^{ - 1}} \times {{10}^{ - 2}}} }}{{2 \times 0.1}} \cr & \left[ {x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}} \right] = \frac{{ - {{10}^{ - 2}} \pm \sqrt {{{10}^{ - 4}} + 4 \times {{10}^{ - 2}}} }}{{2 \times 0.1}} \cr & = \frac{{ - 0.01 \pm \sqrt {.0001 + 0.004} }}{{0.2}} = \frac{{ - 0.01 \pm \sqrt {0.0041} }}{{0.2}} \cr & = \frac{{ - 0.01 \pm 0.64}}{{0.2}} = \frac{{0.54}}{{0.2}}\,\,\,{\text{[ Neglecting the negative term ]}} \cr & {\text{ = 0}}{\text{.27}} \cr & \therefore \,\,\left[ {{H^ + }} \right] = C.\,h = 0.1 \times 0.27 = 2.7 \times \times {10^{ - 2}}M \cr & {\text{Thus the correct answer is [D]}}{\text{.}} \cr} $$

$$\eqalign{ & pH = 8,\,pOH = 6;\left[ {O{H^ - }} \right] = {10^{ - 6}}M; \cr & {\text{Ionic product of}}\,Fe{\left( {OH} \right)_2} = 0.2 \times {\left( {1 \times {{10}^{ - 6}}} \right)^2} \cr & = 2 \times {10^{ - 13}} > {K_{sp}}\left( { = 8.1 \times {{10}^{ - 16}}} \right) \cr} $$

$$\eqalign{ & {\text{Let }}s = {\text{solubility}} \cr & AgI{O_3} \rightleftharpoons \mathop {A{g^ + }}\limits_s + \mathop {IO_3^ - }\limits_s \cr & {K_{sp}} = \left[ {A{g^ + }} \right]\left[ {IO_3^ - } \right] \cr & = s \times s \cr & = {s^2} \cr & {\text{Given}}\,\,{K_{sp}} = 1 \times {10^{ - 8}} \cr & \therefore \,\,s = \sqrt {{K_{sp}}} \cr & = \sqrt {1 \times {{10}^{ - 8}}} \cr & = 1.0 \times {10^{ - 4}}mol/lit \cr & = 1.0 \times {10^{ - 4}} \times 283\,g/lit \cr & \left( {\because \,\,{\text{Molecular mass of}}\,AgI{O_3} = 283} \right) \cr & = \frac{{1.0 \times {{10}^{ - 4}} \times 283 \times 100}}{{1000}}g/100mL \cr & = 2.83 \times 10{\,^{ - 3}}g/100\,mL \cr} $$

If small amount of an acid or alkali is added to a buffer solution, it converts them into unionised acid or base. Thus, its $$pH$$ remains unaffected or in other words its acidity/alkalinity remains constant. e.g.
$$\eqalign{ & {H_3}{O^ + } + {A^ - } \rightleftharpoons {H_2}O + HA \cr & ^ - OH + HA \to {H_2}O + {A^ - } \cr} $$
If acid is added, it reacts with $${A^ - }$$  to form undissociated $$H$$ $$A.$$  Similarly, if base/alkali is added, $$O{H^ - }$$  combines with $$H$$ $$A$$ to give $${H_2}O$$  and $${A^ - }$$  and thus, maintains the acidity / alkalinity of buffer solution.