Ionic Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & {\text{Given,}}\,pH\,{\text{of}}\,Ba{\left( {OH} \right)_2} = 12 \cr & {\text{so,}}\,\,\,pOH = 2 \cr & \therefore \,\,\left[ {{H^ + }} \right] = \left[ {1 \times {{10}^{ - 12}}} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,{K_W} = \left( {{H^ + }} \right)\left( {O{H^ - }} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,{K_w} = 1 \times {10^{ - 14}} \cr & \,\,\,\,\,\,O{H^ - } = \frac{{{K_w}}}{{{H^ + }}} \cr} $$
$${\text{and}}\,\,\left[ {O{H^ - }} \right] = \frac{{1 \times {{10}^{ - 14}}}}{{1 \times {{10}^{ - 12}}}}$$       $$\left[ {\because \,\,\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] = 1 \times {{10}^{ - 14}}} \right]$$
                     $$ = 1 \times {10^{ - 2}}\,mol/L$$
$$\eqalign{ & Ba{\left( {OH} \right)_2} \to \mathop {B{a^{2 + }}}\limits_s + \mathop {2O{H^ - }}\limits_{2s} \cr & {K_{sp}} = \left[ {B{a^{2 + }}} \right]{\left[ {O{H^ - }} \right]^2} = \left[ S \right]{\left[ {2S} \right]^2} \cr & \,\,\,\,\,\,\,\,\, = \left[ {\frac{{1 \times {{10}^{ - 2}}}}{2}} \right]{\left( {1 \times {{10}^{ - 2}}} \right)^2} \cr & \,\,\,\,\,\,\,\,\, = 0.5 \times {10^{ - 6}} \cr & \,\,\,\,\,\,\,\,\, = 5.0 \times {10^{ - 6}}{M^3} \cr} $$

The hydrolysis of $$NaCl$$  gives neutral solution because it is salt of strong acid and strong base and hence, its $$pH$$ is 7. $$N{H_4}Cl$$   is salt of weak base and strong acid, so its $$pH$$ is less than 7. $$NaHC{O_3}$$   is also acidic whereas $$N{a_2}C{O_3}$$  is salt of strong base and weak acid, so its $$pH$$ is more than 7.

Base, $$BOH$$  is dissociated as follows
$$BOH \rightleftharpoons {B^ + } + O{H^ - }$$
So, the dissociation constant of $$BOH$$  base
$$\eqalign{ & {K_b} = \frac{{\left[ {{B^ + }} \right]\left[ {O{H^ - }} \right]}}{{\left[ {BOH} \right]}}\,\,\,...{\text{(i)}} \cr & {\text{At equilibrium}}\left[ {{B^ + }} \right] = \left[ {O{H^ - }} \right] \cr & \therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{K_b} = \frac{{{{\left[ {O{H^ - }} \right]}^2}}}{{\left[ {BOH} \right]}} \cr & {\text{Given that}}\,{K_b} = 1.0 \times {10^{ - 12}} \cr & {\text{and}}\,\,\,\,\,\,\,\left[ {BOH} \right] = 0.01\,M \cr & {\text{Thus,}}\,1.0 \times {10^{ - 12}} = \frac{{{{\left[ {O{H^ - }} \right]}^2}}}{{0.01}} \cr & {\left[ {O{H^ - }} \right]^2} = 1 \times {10^{ - 14}} \cr & \,\,\left[ {O{H^ - }} \right] = 1.0 \times {10^{ - 7}}mol\,{L^{ - 1}} \cr} $$

$$\eqalign{ & {\text{To precipitate the}}\,AgCl \cr & \left[ {A{g^ + }} \right]\,{\text{required}} \cr & = \frac{{{K_{sp}}\left( {AgCl} \right)}}{{\left[ {C{l^ - }} \right]}} \cr & = \frac{{1.0 \times {{10}^{ - 10}}}}{{0.1}} \cr & = 1.0 \times {10^{ - 9}}M \cr & \left[ {B{r^ - }} \right]\,{\text{left at this stage}} = \frac{{{K_{sp}}\left( {AgBr} \right)}}{{\left[ {A{g^ + }} \right]}} \cr & = \frac{{1.0 \times {{10}^{ - 13}}}}{{1.0 \times {{10}^{ - 9}}}} \cr & = 1.0 \times {10^{ - 4}}M \cr & \% \,\,{\text{of remaining}} \cr & \left[ {B{r^ - }} \right] = \frac{{1.0 \times {{10}^{ - 4}}}}{{0.1}} \times 100 = 0.1 \cr & \% \,{\text{of}}\,B{r^ - }\,{\text{to be precipitated}} \cr & = 100 - 0.1 \cr & = 99.9 \cr} $$

$$\eqalign{ & {\text{In presence of external}}\,{H^ + } \cr & {H_2}S \rightleftharpoons 2{H^ + } + {S^{2 - }},{K_{{a_1}}}.{K_{{a_2}}} = {K_{eq}} \cr & \therefore \,\,\frac{{{{\left[ {{H^ + }} \right]}^2}\left[ {{S^{2 - }}} \right]}}{{\left[ {{H_2}S} \right]}} = 1 \times {10^{ - 7}} \times 1.2 \times {10^{ - 13}} \cr & \frac{{{{\left[ {0.2} \right]}^2}\left[ {{S^{2 - }}} \right]}}{{\left[ {0.1} \right]}} = 1.2 \times {10^{ - 20}} \cr & \left[ {{S^{2 - }}} \right] = 3 \times {10^{ - 20}} \cr} $$

Weak acid forms strong conjugate base. In $$HN{O_3},HCl,{H_2}S{O_4}$$     and $$C{H_3}COOH,C{H_3}COOH$$     is weakest acid, so its conjugate base is strongest.
$$C{H_3}COOH \rightleftharpoons C{H_3}CO{O^ - } + {H^ + }$$

\[\begin{align} & \underset{\begin{smallmatrix} \text{(Initial)} \\ \text{(At eq}\text{.)} \end{smallmatrix}}{\mathop{{}}}\,\,\,\,\,\underset{\begin{smallmatrix} 0.001M \\ \left( 0.001-R \right) \end{smallmatrix}}{\mathop{{{A}^{-}}}}\,+{{H}_{2}}O\rightleftharpoons \underset{\begin{smallmatrix} 0 \\ \left( 0.001\times h \right) \end{smallmatrix}}{\mathop{HA}}\,+\underset{\begin{smallmatrix} 0 \\ \left( 0.001\times h \right) \end{smallmatrix}}{\mathop{HA}}\, \\ & {{K}_{h}}=\frac{\left( 0.001\times h \right)\left( 0.001\times h \right)}{\left( 0.001-h \right)} \\ & \text{or,}\,\,{{K}_{h}}=0.001\times {{h}^{2}}\,\left( \text{as,}\,0.001-h\approx 0.001 \right) \\ & {{10}^{-9}}=0.001\times {{h}^{2}} \\ & \left( {{K}_{h}}=\frac{{{K}_{w}}}{{{K}_{a}}}=\frac{{{10}^{-14}}}{{{10}^{-5}}}={{10}^{-9}} \right) \\ & {{h}^{2}}={{10}^{-6}};\,\,\,\,\,\,\therefore \,\,h={{10}^{-3}} \\ \end{align}\]

$$\eqalign{ & {\text{(i)}}\,\,\mathop {{H_3}P{O_4}}\limits_{aci{d_1}} + \mathop {{H_2}{O_4}}\limits_{bas{e_2}} \to \mathop {{H_3}{O^ + }}\limits_{aci{d_2}} + \mathop {{H_2}PO_4^ - }\limits_{bas{e_1}} \cr & {\text{(ii)}}\,\,\mathop {{H_2}PO_4^ - }\limits_{aci{d_1}} + \mathop {{H_2}O}\limits_{bas{e_2}} \to \mathop {HPO_4^{ - - }}\limits_{bas{e_1}} + \mathop {{H_3}{O^ + }}\limits_{aci{d_2}} \cr & {\text{(iii)}}\,\mathop {{H_2}PO_4^ - }\limits_{bas{e_1}} + \mathop {O{H^ - }}\limits_{aci{d_2}} \to \mathop {{H_3}P{O_4}}\limits_{aci{d_1}} + \mathop {{O^{ - - }}}\limits_{bas{e_2}} \cr} $$
Hence only in (ii) reaction $${H_2}PO_4^ - $$  is acting as an acid.

The correct order of $$pH$$ of isomolar solution of sodium oxide $$\left( {p{H_1}} \right),$$  sodium sulphide $$\left( {p{H_2}} \right),$$  sodium selenide $$\left( {p{H_3}} \right)$$  and sodium telluride $$\left( {p{H_4}} \right)$$  is $$p{H_1} > p{H_2} > p{H_3} > p{H_4}$$     because in aqueous solution, they are hydrolysed as follows.
$$\eqalign{ & N{a_2}O + 2{H_2}O \to 2\mathop {NaOH}\limits_{{\text{Base}}} + {H_2}O \cr & N{a_2}S + 2{H_2}O \to \mathop {2NaOH}\limits_{{\text{Strong}}\,{\text{base}}} + \mathop {{H_2}S}\limits_{{\text{Weak acid}}} \cr & N{a_2}Se + 2{H_2}O \to \mathop {2NaOH}\limits_{{\text{Strong}}\,{\text{base}}} + \mathop {{H_2}Se}\limits_{{\text{Weak acid}}} \cr & N{a_2}Te + 2{H_2}O \to \mathop {2NaOH}\limits_{{\text{Strong}}\,{\text{base}}} + \mathop {{H_2}Te}\limits_{{\text{Weak acid}}} \cr} $$
On moving down the group acidic character of oxides increases.
Order of acidic strength
$${H_2}Te > {H_2}Se > {H_2}S > {H_2}O$$
Order of neutralisation of $$NaOH$$
$${H_2}Te > {H_2}Se > {H_2}S > {H_2}O$$
Hence, their aqueous solutions have the following order of basic character due to neutralisation of $$NaOH$$  with $${H_2}O,{H_2}S,{H_2}Se\,$$    and $${H_2}Te.$$
$$N{a_2}O > N{a_2}S > N{a_2}Se > N{a_2}Te$$
( $$\because $$   $$pH$$ of basic solution is higher than acidic or least basic solution )

Higher the value of solubility product, higher is its solubility. In all these compounds the $$Mns$$  is most soluble because its solubility product is maximum.