Ionic Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

The more is the value of equilibrium constant, the more is the completion of reaction or more is the concentration of products i.e. the order of relative strength would be $$O{H^ - } > C{N^ - } > {H_2}O$$

In aqueous solution $$BA$$  (salt) hydrolyses to give
$$\eqalign{ & BA + {H_2}O \rightleftharpoons \mathop {BOH}\limits_{Base} + \mathop {HA}\limits_{acid} \cr & {\text{Now pH is given by}} \cr & pH = \frac{1}{2}p{K_w} + \frac{1}{2}pKa - \frac{1}{2}p{K_b} \cr & {\text{Substituting given values, we get}} \cr & pH = \frac{1}{2}\left( {14 + 4.80 - 4.78} \right) = 7.01 \cr} $$

On hydrolysis sodium borate form sodium hydroxide and boric acid, so the solution will show basic character because sodium hydroxide is strong base and boric acid is weak acid. While solution of sodium sulphate is neutral and that of $$N{H_4}Cl$$  and calcium nitrate is acidic.

$$\eqalign{ & \mathop {M{X_2}}\limits_{{\text{Solubility}}\,0.5 \times {{10}^{ - 4}}M} \to \mathop {{M^{2 + }}}\limits_{0.5 \times {{10}^{ - 4}}M} + \mathop {2{X^ - }}\limits_{2 \times 0.5 \times {{10}^{ - 4}}M} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{on}}\,\,100\% \,\,{\text{ionisation}}} \right) \cr & \therefore \,\,{K_{sp}}\,{\text{of}}\,M{X_2} = \left[ {{M^{2 + }}} \right]{\left[ {{X^ - }} \right]^2} \cr & = \left( {0.5 \times {{10}^{ - 4}}} \right){\left( {1.0 \times {{10}^{ - 4}}} \right)^2} \cr & = 0.5 \times {10^{ - 12}} \cr & = 5 \times {10^{ - 13}}{\left[ M \right]^3} \cr} $$

The two $${K_{sp}}$$  values do not differ very much. So it is a case of simultaneous equilibria, where the concentration of any species can not be neglected.
$$\eqalign{ & \frac{{\left[ {S{r^{2 + }}} \right]{{\left[ {{F^ - }} \right]}^2}}}{{\left[ {S{r^{2 + }}} \right]\left[ {CO_3^{2 - }} \right]}} \cr & = \frac{{{K_{s{p_{Sr{F_2}}}}}}}{{{K_{s{p_{SrC{O_2}}}}}}} \cr & = \frac{{7.9 \times {{10}^{ - 10}}}}{{7.0 \times {{10}^{ - 10}}}} \cr & = 1.128 \cr & \therefore \,\,{\left[ {{F^ - }} \right]^2} = 1.128 \times 1.2 \times {10^{ - 3}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 13.5 \times {10^{ - 4}} \cr & \therefore \,\,\left[ {{F^ - }} \right] = {\left( {13.5 \times {{10}^{ - 4}}} \right)^{\frac{1}{2}}} \cr & = 3.674 \times {10^{ - 2}} \approx 3.7 \times {10^{ - 2}}M. \cr} $$

$$C{H_3}COON{H_4}$$    is a salt of weak acid $$\left( {C{H_3}COOH} \right)$$   and weak base $$\left( {N{H_4}OH} \right).$$
$$pH = 7 + \frac{1}{2}\left( {p{K_a} - p{K_b}} \right)$$
$$p{K_a} = - \log \,{K_a} = - \log \left( {1.8 \times {{10}^{ - 5}}} \right)$$       $$ = 4.74$$
$$p{K_b} = - \log \,{K_b} = - \log \left( {1.8 \times {{10}^{ - 5}}} \right)$$       $$ = 4.74$$
$$pH = 7 + \frac{1}{2}\left( {4.74 - 4.74} \right) = 7.0$$

$$pH = - \log \left[ {{H^ + }} \right]$$    $$ = - \log \left( {3.8 \times {{10}^{ - 3}}} \right) = 2.42$$

$$\left( {\text{A}} \right)C{H_3}COONa \rightleftharpoons $$     $$\mathop {C{H_3}CO{O^ - }}\limits_{{\text{Weak acid}}} + \mathop {N{a^ + }}\limits_{{\text{Strong base}}} \left( {{\text{Basic}}} \right)$$
$$\left( {\text{B}} \right)N{H_4}Cl \rightleftharpoons \mathop {NH_4^ + }\limits_{{\text{Weak base}}} + \mathop {C{l^ - }}\limits_{{\text{Strong acid}}} $$       $$\left( {{\text{Acidic}}} \right)$$
$$\left( {\text{C}} \right)NaN{O_3} \rightleftharpoons \mathop {N{a^ + }}\limits_{{\text{Strong base}}} + \mathop {NO_3^ - }\limits_{{\text{Strong acid}}} $$       $$\left( {{\text{Neutral}}} \right)$$
$$\left( {\text{D}} \right)C{H_3}COON{H_4} \rightleftharpoons $$     $$\mathop {C{H_3}CO{O^ - }}\limits_{{\text{Weak acid}}} + \mathop {NH_4^ + }\limits_{{\text{Weak base}}} \left( {{\text{Almost neutral}}} \right)$$
May be slightly acidic or basic depending upon $$p{K_a}$$  and $$p{K_b}$$  values of acid and base.

The values of dissociation constants for successive stages decrease.

$${K_a}$$  is a measure of the strength of the acid i.e., larger the value of $${K_a},$$  the stronger is the acid.
Thus, the correct order of acidic strength is
\[\begin{matrix} {} \\ {{K}_{a}}: \\ \end{matrix}\begin{matrix} HCOOH \\ 1.8\times {{10}^{-4}} \\ \end{matrix}\begin{matrix} > \\ {} \\ \end{matrix}\begin{matrix} C{{H}_{3}}COOH \\ 1.74\times {{10}^{-5}} \\ \end{matrix}\,\,\begin{matrix} > \\ {} \\ \end{matrix}\begin{matrix} HClO \\ 3.0\times {{10}^{-8}} \\ \end{matrix}\]
Stronger the acid, lesser will be the value of $$pH.$$  Hence, the correct order of $$pH$$  is $$HClO > C{H_3}COOH > HCOOH.$$