Ionic Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

Acid indicators are generally weak acid. The dissociation of indicator $$HIn$$  takes place as follows
$$\eqalign{ & HIn \rightleftharpoons {H^ + } + I{n^ - } \cr & \therefore \,\,{K_{In}} = \frac{{\left[ {{H^ + }} \right]\left[ {I{n^ - }} \right]}}{{\left[ {HIn} \right]}} \cr & {\text{or}}\,\,\,\,\left[ {{H^ + }} \right] = {K_{In}}.\frac{{\left[ {HIn} \right]}}{{\left[ {I{n^ - }} \right]}}\,\,\,...{\text{(i)}} \cr & \because \,\,pH = - {\text{log}}\left[ {{H^ + }} \right]\,\,\,...{\text{(ii)}} \cr & {\text{From eq}}{\text{. (i) and (ii) we get,}} \cr & \therefore \,\,pH = - {\text{log}}\left( {{K_{In}}.\frac{{\left[ {HIn} \right]}}{{\left[ {I{n^ - }} \right]}}} \right) \cr & = - {\text{log}}\,{K_{In}} + {\text{log}}\frac{{\left[ {I{n^ - }} \right]}}{{\left[ {HIn} \right]}} \cr & = p{K_{In}} + {\text{log}}\frac{{\left[ {I{n^ - }} \right]}}{{\left[ {HIn} \right]}} \cr & {\text{or}}\,\,\,{\text{log}}\frac{{\left[ {I{n^ - }} \right]}}{{\left[ {HIn} \right]}} = pH - p{K_{In}} \cr} $$

NOTE: Acidic buffer is mixture of weak acid and its salt with common anion.
$$\left( {\text{A}} \right)\,C{H_3}COOH + C{H_3}COON{H_4}$$       is acidic buffer.
$$\left( {\text{B}} \right)\,N{H_4}Cl + N{H_4}OH$$     is basic buffer.
$$\left( {\text{C}} \right)\,{H_2}S{O_4} + N{a_2}S{O_4}$$     is not buffer because both the compounds are strong electrolytes.
$$\left( {\text{D}} \right)\,NaCl + NaOH$$    is not buffer solution because both compounds are strong electrolytes.

The highest $$pH$$ refers to the basic solution containing $$O{H^ - }\,ions.$$   Therefore, the basic salt releasing more $$O{H^ - }\,ions$$   on hydrolysis will give highest $$pH$$ in water.
Only the salt of strong base and weak acid would release more $$O{H^ - }\,ion$$   on hydrolysis. Among the given salts, $$N{a_2}C{O_3}$$  corresponds to the basic salt as it is formed by the neutralisation of $$NaOH$$  [ strong base ] and $${H_2}C{O_3}$$  [ weak acid ].
$$CO_3^{2 - } + {H_2}O \rightleftharpoons HCO_3^ - + O{H^ - }$$

$$\eqalign{ & pH = - \log \left[ {{H^ + }} \right] = \log \frac{1}{{\left[ {{H^ + }} \right]}};\,\,5.4 = \log \frac{1}{{\left[ {{H^ + }} \right]}} \cr & {\text{On solving,}}\,\left[ {{H^ + }} \right] = 3.98 \times {10^{ - 6}} \cr} $$

TIPS/Formulae :
(i) $$pH$$ of acid cannot be more than 7.
(ii) While calculating $$pH$$ in such case, consider contribution $${\left[ {{H^ + }} \right]}$$  from water also.
Molar conc. of $$HCl = {10^{ - 8}}.$$   (given)
∴ $$pH = 8.$$  But this cannot be possible as pH of an acidic solution can not be more than 7. So we have to consider $${\left[ {{H^ + }} \right]}$$  coming from $${{\text{H}}_2}O.$$
$$\eqalign{ & {\text{Total}}\,\left[ {{H^ + }} \right] = {\left[ {{H^ + }} \right]_{HCl}} + {\left[ {{H^ + }} \right]_{{H_2}O}} \cr & {\text{Ionisation of}}\,{H_2}O:\,{H_2}O \rightleftharpoons {H^ + } + O{H^ - } \cr & {K_w} = {10^{ - 14}} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] \cr & {\text{Let }}x{\text{ be the cone}}{\text{. of}}\,\left[ {{H^ + }} \right]\,{\text{from}}\,{H_2}O \cr & {\text{or}}\,\left[ {{H^ + }} \right] = x = \left[ {O{H^ - }} \right]\,\,\,\,\,\,\,\,\,\left[ {\because \left[ {{H^ + }} \right] = {{\left[ {OH} \right]}^ - }{\text{in water}}} \right] \cr & \therefore \,\,{10^{ - 14}} = \left( {x + {{10}^{ - 8}}} \right)\left( x \right)\,{\text{or}}\,x = 9.5\, \times {10^{ - 8}}M \cr & [{\text{For quadratic equation }}x{\text{ = }}\frac{{ - b \pm \sqrt {4ac} }}{{2a}}] \cr & \therefore \,\,{\text{Total}}\,\left[ {{H^ + }} \right] = {10^{ - 8}} + 9.5 \times {10^{ - 8}} = 10.5 \times {10^{ - 8}}\,{\text{or}}\,{\text{pH}} \cr & = - {\text{log}}\left( {10.5 \times {{10}^{ - 8}}} \right) = 6.98 \cr & {\text{It is between 6 and 7}}{\text{.}} \cr} $$

$$\mathop {{N_3}H}\limits_{{\text{Hydrazoic acid}}} \rightleftharpoons N_3^ - + {H^ + }$$
i.e, conjugate base of hydrazoic acid is $$N_3^ - .$$

Concentration of $$SO_4^{2 - }$$  in $$BaS{O_4}$$  solution
$$\eqalign{ & {M_1}{V_1} = {M_2}{V_2} \cr & 1 \times 50 = {M_2} \times 500 \cr & {M_2} = \frac{1}{{10}} \cr & {\text{For just precipitation}} \cr & {\text{Ionic product = }}{K_{sp}} \cr & \left[ {B{a^{2 + }}} \right]\left[ {SO_4^{2 - }} \right] = {K_{sp}}\left( {BaS{P_4}} \right) \cr & \left[ {B{a^{2 + }}} \right] \times \frac{1}{{10}} = {10^{ - 10}} \cr & \left[ {B{a^{2 + }}} \right] = {10^{ - 9}}M\,{\text{in}}\,{\text{500}}\,ml\,{\text{solution}} \cr & {\text{Thus}}\,\left[ {B{a^{2 + }}} \right]{\text{in}}\,{\text{original}}\,{\text{solution}}\,\left( {450mL} \right) \cr & \Rightarrow {M_1} \times 450 = {10^{ - 9}} \times 500 \cr & \left[ {{\text{where}}\,{M_1} = {\text{molarity of original solution}}} \right] \cr & {M_1} = \frac{{500}}{{450}} \times {10^{ - 9}} = 1.11 \times {10^{ - 9}}M \cr} $$

$$pH$$ of an acidic solution should be less than 7. The reason is that from $${H_2}O.\left[ {{H^ + }} \right] = {10^{ - 7}}M$$   which cannot be neglected in comparison to $${10^{ - 8}}M.$$   The $$pH$$ can be calculated as.
$$\eqalign{ & {\text{from acid,}}\,\left[ {{H^ + }} \right] = {10^{ - 8}}M. \cr & {\text{from}}\,{H_2}O,\left[ {{H^ + }} \right] = {10^{ - 7}}M \cr & \therefore \,\,{\text{Total}}\,\left[ {{H^ + }} \right] = {10^{ - 8}} + {10^{ - 7}} = {10^{ - 8}}\left( {1 + 10} \right) = 11 \times {10^{ - 8}} \cr & \therefore \,\,pH = - \log \left[ {{H^ + }} \right] = - \log 11 \times {10^{ - 8}} = - \left[ {\log 11 + 8\log 10} \right] \cr & = - \left[ {1.0414 - 8} \right] \cr & = 6.9586 \cr} $$

$$\eqalign{ & h = \sqrt {\frac{{{K_w}}}{{{K_a} \times c}}} = \sqrt {\frac{{{{10}^{ - 14}}}}{{{{10}^{ - 5}} \times 0.1}}} = {10^{ - 4}}; \cr & {\text{Hence, }}\% {\text{ hydrolysis = }}\,{10^{ - 4}} \times 100 = 0.01 \cr} $$

The rain water after thunderstorm contains dissolved acid and therefore the $$pH$$ is less than rain water without thunderstorm.