Ionic Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & {\text{Base}}\, + {H^ + } \to {\text{( conjugate acid )}} \cr & NH_2^ - {\text{(base)}} + {H^ + } \to N{H_3}{\text{(conjugate acid)}} \cr} $$

$$\eqalign{ & pH = p{K_a} + \log \frac{{\left[ {{\text{salt}}} \right]}}{{\left[ {{\text{acid}}} \right]}} \cr & = 4.76 + \log \frac{{\frac{{7.5}}{{500}}}}{{\frac{5}{{500}}}} \cr & = 4.7 + \log \,1.5 \cr & = 4.87 \cr & {\text{Hence correct answer is}}\,\,4.76 < pH < 5.0 \cr} $$

$$C{H_3}COONa + {H_2}O \rightleftharpoons $$     $$\mathop {C{H_3}COOH}\limits_{{\text{Weak acid}}} + \mathop {NaOH}\limits_{{\text{Strong base}}} $$
$$NaOH \rightleftharpoons N{a^ + } + O{H^ - }$$    $${\text{(Basic solution)}}$$

Higher the value of $${K_a}$$  lower will be the value of $$p{K_a}$$  i.e. higher will be the acidic nature. Further $$C{N^ - },{F^ - }$$  and $$NO_2^ - $$  are conjugate base of the acids $$HCN,HF$$   and $$HN{O_2}$$  respectively hence the correct order of base strength will be $${F^ - } < NO_2^ - < C{N^ - }$$
( $$\because $$  stronger the acid weaker will be its conjugate base )

Key Idea As solubility of $$AgCl\left( s \right)$$  is asked in $$0.1\,M\,NaCl$$   solution, so in the calculation, solubility of $$C{l^ - }$$ ( from $$NaCl$$  ) must be added to the solubility of $$C{l^ - }$$ ( from $$AgCl$$  ).
Let $$s$$ be the solubility of $$A{g^ + }$$  and $$C{l^ - }$$  in $$AgCl$$  before the addition of $$NaCl.$$
\[\underset{\begin{smallmatrix} 0.1\,M \\ \,\,\,\,\,\,0 \end{smallmatrix}}{\mathop{NaCl\left( aq \right)}}\,\to \underset{\begin{smallmatrix} \,\,\,\,\,\,\,\,0 \\ 0.1\,M \end{smallmatrix}}{\mathop{N{{a}^{+}}\left( aq \right)}}\,+\underset{\begin{smallmatrix} \,\,\,\,\,\,\,\,0 \\ 0.1+s \end{smallmatrix}}{\mathop{C{{l}^{-}}\left( aq \right)}}\,\]
$$\eqalign{ & AgCl\left( s \right) \rightleftharpoons \mathop {A{g^ + }\left( {aq} \right)}\limits_s + \mathop {C{l^ - }\left( {aq} \right)}\limits_{s + 0.1} \cr & {\text{Given,}}\,\,{K_{sp}} = 1.6 \times {10^{ - 10}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {A{g^ + }} \right]\left[ {C{l^ - }} \right] \cr & {\text{or}}\,\,1.6 \times {10^{ - 10}} = s\left( {0.1 + s} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.1\,s + {s^2} \cr} $$
$$\because \,\,{K_{sp}}$$  is small, so $$s$$ is very less in comparison with 0.1. Hence, $${s^2}$$  can be neglected.
Thus, $$1.6 \times {10^{ - 10}} = 0.1s$$
                      or, $$s = 1.6 \times {10^{ - 9}}M$$

$$\eqalign{ & {H^ + } = C\alpha ;\,\alpha = \frac{{\left[ {{H^ + }} \right]}}{C}\,\,\,or\,\,\alpha = \frac{{{{10}^{ - 3}}}}{{0.1}} = {10^{ - 2}} \cr & Ka = C{\alpha ^2} = 0.1 \times {10^{ - 2}} \times {10^{ - 2}} = {10^{ - 5}} \cr} $$

Since $${K_b} > {K_a},$$   the solution will be slightly basic.

$$HSO_4^ - $$  can accept or give a proton.

$$HCl$$  is stronger acid than $$C{H_3}COOH$$   and $$C{l^ - }$$ is a stronger base than $$C{H_3}COOH_2^ + $$   and is the conjugate base of $$HCl.$$
$$\mathop {HCl}\limits_{{\text{aci}}{{\text{d}}_{\text{1}}}} + \mathop {C{H_3}COOH}\limits_{{\text{bas}}{{\text{e}}_{\text{2}}}} \rightleftharpoons \mathop {C{l^ - }}\limits_{{\text{bas}}{{\text{e}}_1}} + \mathop {C{H_3}COOH_2^ + }\limits_{{\text{aci}}{{\text{d}}_2}} $$

(A) is a neutral solution due to both cationic and anionic hydrolysis $$\left( {{K_a} = {K_b} = 1.8 \times {{10}^{ - 5}}} \right)$$
(B) is acidic solution due to cationic hydrolysis
(C) is acidic solution due to cationic hydrolysis
(D) is basic solution due to anionic hydrolysis.