Ionic Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

A mixture of acetic acid and sodium acetate acts as an acidic buffer.

Boron is an element of 13 group and contains three electrons in its valence shell. When its compound $${B{H_3}}$$  dimerises, each boron atom carry only 6 electrons that is their octet is incomplete. Hence, $${\left( {B{H_3}} \right)_2}$$   is an electron deficient compound.
In all other given molecules octet of central atom is complete.

$$\eqalign{ & RaS{O_4} \rightleftharpoons R{a^{2 + }} + SO_4^{2 - } \cr & {K_{sp}} = \left[ {R{a^{2 + }}} \right]\left[ {SO_4^{2 - }} \right] \cr} $$
Concentration of $$SO_4^{2 - }$$  from $$N{a_2}S{O_4} = 0.10M$$
$$\eqalign{ & R{a^{2 + }} = \frac{{4 \times {{10}^{ - 11}}}}{{0.10}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4 \times {10^{ - 10}}M \cr} $$

NOTE : Electron acceptors or elements having incomplete octet are Lewis acids.
(i) $$B{F_3}$$  ( $$B$$ has $$6{e^ - }$$  in valance shell ), $$AlC{l_3}$$  ( $$Al$$ has 6 electrons in valance shell ) $$BeC{l_2}$$  ( $$Be$$ has $$4{e^ - }$$  in valance shell ) are electron defecient compounds and hence Lewis acids.
(ii) $$SnC{l_4}$$  has complete octet so it is Lewis base.

$$\eqalign{ & {A_p}{B_q}\left( s \right) \rightleftharpoons p{A^{ + q}} + q{B^{ - p}} \cr & {L_S} = {\left( {pS} \right)^p}.{\left( {qS} \right)^q} = {p^p}.{q^q}.{S^{\left( {p + q} \right)}} \cr} $$

In acidic medium weak acids are unionized due to common ion effect and they are completely ionised in alkaline medium.
Aspirin ( or acetyl salicylic acid ) is unionised in stomach (where $$pH$$ is 2-3 ) and is completely ionised in small intestine (when $$pH$$ is 8).

All are binary salts, hence their solubility is equal to square root of their solubility products.
So, order of solubility is $$CuS > CdS > HgS$$     or greater the value of solubility product, greater will be the solubility.

TIPS/Formulae :
The equilibrium constant for the nuetralization of a weak acid with a strong base is given by
$$K = \frac{{{K_a}}}{{{K_w}}} = \frac{{1.0 \times {{10}^{ - 4}}}}{{1.0 \times {{10}^{ - 14}}}} = 1.0 \times {10^{10}}$$

\[\begin{align} & \text{Given}\,{{K}_{b}}=1.0\times {{10}^{-12}} \\ & \left[ BOH \right]=0.01\,M\,\,\,\,{{\left[ OH \right]}^{-}}=? \\ & \underset{\begin{smallmatrix} t=0 \\ t={{t}_{eq}} \end{smallmatrix}}{\mathop{{}}}\,\,\underset{\begin{smallmatrix} c \\ c\left( 1-\alpha \right) \end{smallmatrix}}{\mathop{BOH}}\,\rightleftharpoons \underset{\begin{smallmatrix} 0 \\ c\alpha \end{smallmatrix}}{\mathop{{{B}^{+}}}}\,+\underset{\begin{smallmatrix} 0 \\ c\alpha \end{smallmatrix}}{\mathop{O{{H}^{-}}}}\, \\ & {{K}_{b}}=\frac{{{c}^{2}}{{\alpha }^{2}}}{c\left( 1-\alpha \right)}=\frac{c{{\alpha }^{2}}}{\left( 1-\alpha \right)} \\ & \Rightarrow 1.0\times {{10}^{-12}}=\frac{0.01{{\alpha }^{2}}}{\left( 1-\alpha \right)} \\ & \text{On calculation, we get,}\,\alpha =1.0\times {{10}^{-5}} \\ & \text{Now,}\left[ O{{H}^{-}} \right]=c\alpha \\ & =0.01\times {{10}^{-5}} \\ & =1\times {{10}^{-7}}mol\,{{L}^{-1}} \\ \end{align}\]

$$\eqalign{ & HA + BOH \rightleftharpoons BA + {H_2}O \cr & \therefore \,\,{K_H} = \frac{{{K_w}}}{{{K_a} \times {K_b}}} \cr & {\text{The inverse of}}\,{K_H}\,{\text{is}}\,{K_c} = \frac{{{K_a} \times {K_b}}}{{{K_w}}} \cr & = \frac{{2 \times {{10}^{ - 5}} \times 5 \times {{10}^{ - 6}}}}{{1 \times {{10}^{ - 14}}}} \cr & = 1.0 \times {10^4} \cr} $$