Ionic Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & \left[ {{H^ + }} \right]\,{\text{in monobasic acid}} \cr & = {\text{molarity}}\, \times \,{\text{degree of ionisation}} \cr & = 0.1 \times \frac{2}{{100}} \cr & = 2 \times {10^{ - 3}}M \cr & {\text{ionisation constant of water}} \cr & {K_w} = \left( {{H^ + }} \right)\left( {O{H^ - }} \right) \cr & \left[ {O{H^ - }} \right] = \frac{{{K_w}}}{{\left[ {{H^ + }} \right]}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{1 \times {{10}^{ - 14}}}}{{2 \times {{10}^{ - 3}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 5 \times {10^{ - 12}}M \cr} $$

$$BC{l_3}\,{\text{and}}\,AlC{l_3},$$   both have vacant $$p-$$orbital and incomplete octet thus they behave as Lewis acids.
$$SiC{l_4}$$  can accept lone pair of electron in d-orbital of silicon hence it can act as Lewis acid.
Although the most suitable answer is (c). However, both options (c) and (a) can be considered as correct answers.
e.g. hydrolysis of $$SiC{l_4}$$

i.e., option (a) $$AlC{l_3}\,{\text{and}}\,SiC{l_4}$$   is also correct.

\[\begin{align} & \underset{\begin{smallmatrix} 100\times 0.1 \\ 0 \end{smallmatrix}}{\mathop{MgC{{l}_{2}}}}\,+\underset{\begin{smallmatrix} 100\times 0.2 \\ \,0 \end{smallmatrix}}{\mathop{2NaOH}}\,\to \underset{\begin{smallmatrix} 0 \\ 10 \end{smallmatrix}}{\mathop{Mg}}\,\underset{\begin{smallmatrix} 0 \\ 20 \end{smallmatrix}}{\mathop{{{\left( OH \right)}_{2}}}}\,+\underset{\begin{smallmatrix} \text{Initial moles} \\ \text{After mixing} \end{smallmatrix}}{\mathop{2NaCl}}\, \\ & \therefore \,\,{{K}_{sp}}=\left[ \frac{10}{200} \right]{{\left[ \frac{20}{200} \right]}^{2}}=5\times {{10}^{-4}} \\ & \text{But actually}\,{{K}_{sp}}\,\text{of} \\ & Mg{{\left( OH \right)}_{2}}=1.2\times {{10}^{-11}} \\ & \therefore \,\,4{{s}^{3}}=1.2\times {{10}^{-11}} \\ & \text{Find }S\text{ then}\left[ O{{H}^{-}} \right]=2S\,\,\text{which is}\,\,2.8\times {{10}^{-4}} \\ \end{align}\]

$$pH = {\text{log}}\frac{1}{{\left[ {{H^ + }} \right]}}$$
$$pH$$  is inversely proportional to hydrogen ion concentration. As concentration of $${H^ + }$$ decreases $$pH$$  increases and vice-versa.
Ammonium chloride $$\left( {N{H_4}Cl} \right)$$   is a salt of weak base and strong acid. So, its aqueous solution will be acidic as
$$N{H_4}Cl + {H_2}O \to \mathop {N{H_4}OH}\limits_{{\text{Weak base}}} + \mathop {HCl}\limits_{{\text{Strong acid}}} $$
So, $$pH$$ of $${N{H_4}Cl}$$  is less than 7.
Sodium nitrate $$\left( {NaN{O_3}} \right)$$   is the salt of strong acid and strong base. So, its aqueous solution is neutral as
$$NaN{O_3} + {H_2}O \to \mathop {NaOH}\limits_{{\text{Strong base}}} + \mathop {HN{O_3}}\limits_{{\text{Strong acid}}} $$
So, $$pH$$ of $${NaN{O_3}}$$  is 7.
Potassium acetate $$\left( {C{H_3}COOK} \right)$$   is a salt of strong base and weak acid. Its aqueous solution will be basic and $$pH$$  value will be greater than $$7 \approx 8.8$$
$$C{H_3}CO{O^ - }{K^ + } + {H_2}O \to \mathop {C{H_3}COOH}\limits_{{\text{Weak acid}}} + \mathop {KOH}\limits_{{\text{Strong base}}} $$
Sodium carbonate $$\left( {N{a_2}C{O_3}} \right)$$   is a salt of strong base and weak acid. Its aqueous solution is also basic and its $$pH$$  value will be more than 10,
i.e. highest among them.
$$N{a_2}C{O_3} + {H_2}O \to \mathop {2NaOH}\limits_{{\text{Strong base}}} + \mathop {{H_2}C{O_3}}\limits_{{\text{Weak acid}}} $$

Key Idea The molecule with lone pair at centre atom, will behave as Lewis base.
In the given molecules, only $$P{F_3}$$  has lone pair at $$P$$ as shown below :

Thus, $$P{F_3}$$  acts as a Lewis base ( electron-pair donor ) due to presence of lone pair on $$P$$ - atom.

Lewis acid is acceptor of a pair of electrons while Lewis base is donor of a pair of electrons.

The percentage of pyridine can be equal to the percentage of dissociation of pyridinium ion and pyridine solution as shown below :

As pyridinium is a weak base, so degree of dissociation is given as
$$\eqalign{ & \alpha = \sqrt {\frac{{{K_b}}}{C}} \cr & \,\,\,\,\, = \sqrt {\frac{{1.7 \times {{10}^{ - 9}}}}{{0.10}}} \cr & \,\,\,\,\, = \sqrt {1.7 \times {{10}^{ - 8}}} \cr & \,\,\,\,\, = 1.3 \times {10^{ - 4}} \cr & {\text{or, percentage of dissociation}} \cr & = \left( {\alpha \times 100} \right)\% \cr & = \left( {1.3 \times {{10}^{ - 4}}} \right) \times 100 \cr & = 0.013\% \cr} $$

Solution of $$N{H_4}OH$$  and $$N{H_4}Cl$$   acts as a basic buffer solution. For basic buffer solution
$$\eqalign{ & pOH = p{K_b} + {\text{log}}\frac{{\left[ {{\text{salt}}} \right]}}{{\left[ {{\text{base}}} \right]}} \cr & POH = 14 - pH \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 14 - 9.25 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4.75 \cr & 4.75 = p{K_b} + {\text{log}}\frac{{0.1}}{{0.1}} \cr & p{K_b} = 4.75 \cr} $$

\[\begin{align} & M{{X}_{4}}\rightleftharpoons {{\underset{S}{\mathop{M}}\,}^{4+}}+\underset{4S}{\mathop{4{{X}^{-}}}}\, \\ & {{K}_{sp}}=\left[ s \right]{{\left[ 4s \right]}^{4}}=256\,{{s}^{5}}\,\,\,\therefore \,\,s={{\left( \frac{{{K}_{sp}}}{256} \right)}^{\frac{1}{5}}} \\ \end{align}\]

The $$pH$$  of neutral water at $$25{\,^ \circ }C$$  is 7.0.
$$\therefore \left[ {{H^ + }} \right] = \left[ {O{H^ - }} \right] = {10^{ - 7}}$$         $$\left( {\because pH = - \log \left[ {{H^ + }} \right]} \right)$$
$${\text{Now}},\,{K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]$$      $$ = {\left( {1 \times {{10}^{ - 7}}} \right)^2} = 1 \times {10^{ - 14}}$$
As the temperature increases, ionisation of water increases, thus $$\left[ {{H^ + }} \right]$$  and $$\left[ {O{H^ - }} \right]$$  increases equally. Now
$${K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] > 1 \times {10^{ - 14}}$$         $$\left( {\because \left[ {{H^ + }} \right] = \left[ {O{H^ - }} \right]} \right)$$
$$\eqalign{ & {\text{or}}\,\,{\left[ {{H^ + }} \right]^2} > 1 \times {10^{ - 14}} \cr & \therefore \left[ {{H^ + }} \right] > 1 \times {10^{ - 7}}\,\,{\text{and}}\,\,pH < 7 \cr} $$