Ionic Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & 6 = - \log {10^{ - 5}} + \log \frac{{{\text{salt}}}}{{{\text{acid}}}} = 5 + \log \frac{{{\text{salt}}}}{{{\text{acid}}}} \cr & \log \frac{{{\text{salt}}}}{{{\text{acid}}}}{\text{must be}}\,1. \cr & \therefore \,\,\frac{{{\text{salt}}}}{{{\text{acid}}}} = \frac{{10}}{1}\,{\text{or}}\,10:1. \cr} $$

Key Concept When equal volumes of acid and base are mixed, then resulting solution become alkaline if concentration of base is taken high.
Let normality of the solution after mixing $$0.1\,M\,NaOH$$   and $$0.01\,M\,HCl$$   is $$N.$$
$$\eqalign{ & \therefore \,\,{N_1}{V_1} - {N_2}{V_2} = NV \cr & {\text{or}}\,\,0.1 \times 1 - 0.01 \times 1 = N \times 2 \cr} $$
Since, normality of $$NaOH$$  is more than that of $$HCl.$$
Hence, the resulting solution is alkaline.
$$\eqalign{ & {\text{or}}\,\,\left[ {\bar OH} \right] = N = \frac{{0.09}}{2} = 0.045\,N \cr & {\text{or}}\,\,pOH = - \,{\text{log}}\left( {0.045} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.35 \cr & \therefore \,\,pH = 14 - pOH \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 14 - 1.35 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 12.65 \cr} $$

$$C{H_3}COOH$$   is weak acid while $$NaOH$$  is strong base, so one equivalent of $$NaOH$$  can not be neutralized with one equivalent of $$C{H_3}COOH.$$   Hence the solution of one equivalent of each does not have $$pH$$  value as 7. Its $$pH$$  will be towards basic side as $$NaOH$$  is a strong base hence conc. of $$O{H^ - }$$  will be more than the conc. of $${H^ + }.$$

$$\eqalign{ & AgBr \rightleftharpoons A{g^ + } + B{r^ - } \cr & {K_{sp}} = \left[ {A{g^ + }} \right]\left[ {B{r^ - }} \right] \cr & {\text{For precipitation to occur}} \cr & {\text{Tonic product > Solubility product}} \cr & \left[ {B{r^ - }} \right] = \frac{{{K_{sp}}}}{{\left[ {A{g^ + }} \right]}} = \frac{{5 \times {{10}^{ - 13}}}}{{0.05}} = {10^{-11}} \cr} $$
i.e., precipitation just starts when $${10^{ - 11}}$$  moles of $$KBr$$  is added to $$1\ell \,\,AgN{O_3}$$   solution
∴ Number of moles of $${B{r^ - }}$$  needed from $$KBr = {10^{ - 11}}$$
∴ Mass of $$KBr = {10^{ - 11}} \times 120 = 1.2 \times {10^{ - 9}}\,g$$

$$\eqalign{ & {\text{For a basic buffer, pH}} = 14 - p{K_b} - \log \frac{{\left[ {{\text{salt}}} \right]}}{{\left[ {{\text{base}}} \right]}} \cr & pH = 14 - p{K_b} - \log \frac{{\left[ {{\text{salt}}} \right]}}{{\left[ {{\text{base}}} \right]}} = 14 - \left( { - \log {{10}^{ - 10}}} \right) - \log 1 \cr & \Rightarrow pH = 4 \cr} $$

Metal halide on hydrolysis with water form corresponding hydroxides. The basic strength of hydroxide increases as we move down in a group. This is because of the increase in size which results in decrease of ionization energy which weakens the strength of $$M – O$$  bonds in $$MOH$$  and thus increases the basic strength.
$$\eqalign{ & \mathop {Be{{\left( {OH} \right)}_2}}\limits_{{\text{Amphoteric}}} \,\,\,\,\,\,\,\mathop {Mg{{\left( {OH} \right)}_2}}\limits_{{\text{Weak base}}} \cr & Ca{\left( {OH} \right)_2}\,\,\mathop {Sr{{\left( {OH} \right)}_2}}\limits_{{\text{Strong base}}} \,\,\,Ba{\left( {OH} \right)_2} \cr} $$
Hence, $$Be{\left( {OH} \right)_2}$$   will have lowest $$pH.$$

Suppose $$S$$  is the molar solubility of silver benzoate in water, then
$${C_6}{H_5}COOA{g_{\left( s \right)}} \rightleftharpoons $$    $${C_6}{H_5}COO_{\left( {aq} \right)}^ - + Ag_{\left( {aq} \right)}^ + $$
$${K_{sp}} = {s^2}\therefore S = \sqrt {2.5 \times {{10}^{ - 13}}} $$       $$ = 5.0 \times {10^{ - 7}}\,M$$
If the solubility of salt of weak acid of ionization constant $${K_a}$$  is $$S',$$  the $${K_{sp}},{K_a}$$   and $$S'$$  are related to each other at $$pH = 3.19.$$
$$\eqalign{ & \therefore \left[ {{H^ + }} \right] = 6.46 \times {10^{ - 4}}M\,\,\,\left( {\because pH = 3.19} \right) \cr & {K_{sp}} = S{'^{\,2}}\left[ {\frac{{{K_a}}}{{{K_a} + \left[ {{H^ + }} \right]}}} \right] \cr & S' = {\left\{ {\frac{{2.5 \times {{10}^{ - 13}}}}{{\left[ {\frac{{6.46 \times {{10}^{ - 5}}}}{{6.46 \times {{10}^{ - 5}} + 6.46 \times {{10}^{ - 4}}}}} \right]}}} \right\}^{\frac{1}{2}}} \cr & S' = {\left\{ {\frac{{2.5 \times {{10}^{ - 13}} \times 7.106 \times {{10}^{ - 4}}}}{{6.46 \times {{10}^{ - 5}}}}} \right\}^{\frac{1}{2}}} \cr & \,\,\,\,\,\,\, = {\left( {2.75 \times {{10}^{ - 12}}} \right)^{\frac{1}{2}}} \cr & \,\,\,\,\,\,\, = 1.658 \times {10^{ - 6}}\,M \cr & \therefore \,{\text{The}}\,\,{\text{ratio}}\,\,{\text{of}}\,\frac{{S'}}{s} = \frac{{1.658 \times {{10}^{ - 6}}}}{{5.0 \times {{10}^{ - 7}}}} = 3.32 \cr} $$
Silver benzoate is 3.32 times more soluble in buffer of $$pH = 3.19$$   than in pure water.

A pair constituent with $$HN{O_2}$$  and $$NaN{O_2}$$  because $$HN{O_2}$$  is weak acid and $$NaN{O_2}$$  is a salt of weak acid $$\left( {HN{O_2}} \right)$$   with strong base $$\left( {NaOH} \right).$$   Hence, it is an example of acidic buffer solution.

The aqueous solution of acetic acid ionise as follows :
$${H_2}O + \mathop {C{H_3}COOH}\limits_{Base} \rightleftharpoons \mathop {C{H_3}CO{O^ - }}\limits_{Acid} + {H_3}{O^ + }$$
So, the aqueous solution of acetic acid contains $$C{H_3}CO{O^ - },{H_3}{O^ + }$$    and $$C{H_3}COOH.$$

$$\eqalign{ & \left[ {{H_3}{O^ + }} \right] = \left[ {{H^ + }} \right] = {10^{ - 10}} \cr & pH + pOH = 14...{\text{(i)}} \cr & {\text{and}}\,\,pH = - {\text{log}}\left[ {{H^ + }} \right] \cr & \,\,\,\,\,\,\,\,\,\,pH = - {\text{log}}\left[ {{{10}^{ - 10}}} \right]...{\text{(ii)}} \cr & \,\,\,\,\,\,\,\,\,\,pH = 10 \cr & {\text{from eq}}{\text{. (i) and (ii), we get}} \cr & pOH + 10 = 14 \cr & pOH = 14 - 10 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 4 \cr} $$