$$\eqalign{ & \left( {\text{i}} \right)\,\,Ag \to A{g^ + } + {e^ - }\,\,\,\,\,\,{E^ \circ } = - 0.800\,V \cr & {\text{(ii)}}Ag + {I^ - } \to AgI + {e^ - }\,\,\,{E^ \circ } = 0.152\,V \cr & {\text{From (i) and (ii) we have,}} \cr & AgI \to A{g^ + } + {I^ - }\,\,\,\,{E^ \circ } = - 0.952\,V \cr & E_{cell}^o = \frac{{0.059}}{n}\log \,K \cr & \therefore \,\, - 0.952 = \frac{{0.059}}{1}\log \,\left[ {A{g^ + }} \right]\left[ {{I^ - }} \right]\,\,\left[ {\because \,\,k = \left[ {A{g^ + }} \right]\left[ {{I^ - }} \right]} \right] \cr & {\text{or}}\,\, - \frac{{0.952}}{{0.059}} = \log \,{K_{sp}}\,\,{\text{or}}\,\, - 16.13 = \log \,{K_{sp}} \cr} $$

(A) It is not correct answer because $$100\,ml\,M/10\,HCl$$     will completely neutralise $$100\,ml\,M/10\,NaOH$$    and the solution will be neutral.
(B) After neutralisation resultant solution will be acidic due to presence of excess of $$HCl.$$
(C) After netralisation resultant solution will be basic due to presence of excess of $$NaOH.$$
(D) $$M.eq.$$  of $$HCl = 75\,N/5 = 15Meq$$
$$\eqalign{ & M.eq.NaOH = 25 \times \frac{1}{5} = 5Meq \cr & \therefore \,M.eq{\text{.}}\,{\text{of}}\,\,HCl\,\,{\text{left}} = 10\,\,\,\therefore \left[ {HCl} \right] = \frac{{10}}{{100}} = M/10 \cr & \therefore \,\,pH = - \log \left[ {{H^ + }} \right] = - \log \left[ {\frac{1}{{10}}} \right] = 1 \cr} $$

$$\eqalign{ & {\text{Solubility of}}\,{M_2}S\,\,{\text{salt is }}3.5 \times {10^{ - 6}}M \cr & \mathop {{M_2}S\,}\limits_{3.5 \times {{10}^{ - 6}}M} \, \rightleftharpoons \mathop {2{M^ + }}\limits_{2 \times 3.5 \times {{10}^{ - 6}}M} + \mathop {{S^{2 - }}}\limits_{3.5 \times {{10}^{ - 6}}M} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{on}}\,\,100\% \,\,{\text{ionisation}}} \right) \cr & \therefore \,\,{K_{sp}}\left( {{\text{solubility product of}}\,{M_2}S} \right) \cr & = {\left[ {{M^ + }} \right]^2}\left[ {{S^{2 - }}} \right] \cr & = {\left( {7.0 \times {{10}^{ - 6}}} \right)^2}\left( {3.5 \times {{10}^{ - 6}}} \right) \cr & = 171.5 \times {10^{ - 18}} \cr & = 1.71 \times {10^{ - 16}}{\left[ M \right]^3} \cr} $$

Milliequivalents of $$NaOH = 10 \times 0.1 = 1$$
Milliequivalents of $${H_2}S{O_4} = 10 \times 0.05 = 0.5$$
$$\mathop {2NaOH}\limits_{2\,equ.} + \mathop {{H_2}S{O_4}}\limits_{1\,equ.} \to N{a_2}S{O_4} + {H_2}O$$
∴ 1 equivalent of $$NaOH$$  reacts with $$0.5\,eq.$$  of $${H_2}S{O_4}$$  to give neutral $$\left( {pH = 7} \right)$$   solution.

$$CaO,\,CaC{O_3}\,{\text{and}}\,Ca{\left( {OH} \right)_2}$$      dissolve in $$C{H_3}COOH$$   due to formation of $${\left( {C{H_3}COO} \right)_2}Ca.$$    But $$Ca{C_2}{O_4}$$  does not dissolve as $$C{H_3}CO{O^ - }$$   is a stronger conjugate base than $${C_2}O_4^{2 - }.$$

$$MX \to \mathop {{M^ + }}\limits_s + \mathop {{X^ - }}\limits_s $$     $${\text{(Where sis the solubility)}}$$
$$\eqalign{ & {\text{Then}}\,{K_{sp}} = {S^2}\,\,\,\,{\text{or}}\,s = \sqrt {{K_{sp}}} \cr & {\text{Similarly for}}\,M{X_2} \to {M^{2 + }} + 2{X^ - } \cr & {K_{sp}} = s \times {\left( {2s} \right)^2} = 4{s^3}\,\,\,{\text{or}}\,\,s = {\left[ {\frac{{{K_{sp}}}}{4}} \right]^{\frac{1}{3}}} \cr} $$
\[\text{and}\,\text{for}\,\,{{M}_{3}}X\to 3{{\underset{3s}{\mathop{M}}\,}^{+}}+{{\underset{s}{\mathop{X}}\,}^{-3}}\]
$${K_{sp}} = {\left( {3s} \right)^3} \times s = 27{s^4}\,\,\,s = {\left[ {\frac{{{K_{sp}}}}{{27}}} \right]^{\frac{1}{4}}}$$
From the given values of $${{K_{sp}}}$$  for $$MX,M{X_2}\,{\text{and}}\,{M_3}X$$     we can find the solubilities of those salts at temperature, $$T.$$
$$\eqalign{ & {\text{Solubility}}\,{\text{of}}\,MX{\text{ = }}\sqrt {4 \times {{10}^{ - 8}}} = 2 \times {10^{ - 4}} \cr & {\text{Solubility}}\,{\text{of }}M{X_2} = {\left[ {\frac{{3.2 \times {{10}^{ - 14}}}}{4}} \right]^{\frac{1}{3}}}\,\,\,{\text{or }}{\left[ {\frac{{32}}{4} \times {{10}^{ - 15}}} \right]^{\frac{1}{3}}} \cr & = {\left[ {8 \times {{10}^{ - 15}}} \right]^{\frac{1}{3}\,\,}}\,\,\,\,\,\,{\text{or}}\,\,\,\,\,2 \times {10^{ - 15}} \cr & {\text{Solubility of}}\,{M_3}X = {\left[ {\frac{{2.7 \times {{10}^{ - 15}}}}{{27}}} \right]^{\frac{1}{4}}} \cr & = {\left[ {{{10}^{ - 16}}} \right]^{\frac{1}{4}}}\,\,\,{\text{or}}\,\,\,\,{10^{ - 4}} \cr} $$
Thus the solubilities are in the order $$MX > {M_3}X > M{X_2}$$
i.e the correct t anser is (D) .

TIPS/Formulae :
For precipitation to occur ionic product > solubility products
$$\eqalign{ & {\text{Given,}}\,\,{K_{sp}}Ca{F_2} = 1.7 \times {10^{ - 10}} \cr & Ca{F_2} \rightleftharpoons C{a^{2 + }} + 2{F^ - } \cr & {\text{Ionic product of}}\,\,{\text{Ca}}{{\text{F}}_2} = \left[ {C{a^{2 + }}} \right]{\left[ {{F^ - }} \right]^2} \cr & {\text{Calculate LP}}{\text{. in each case}} \cr & {\text{(a) I}}{\text{.P}}{\text{. of Ca}}{{\text{F}}_2} = \left( {{{10}^{ - 4}}} \right) \times {\left( {{{10}^{ - 4}}} \right)^2} = {10^{ - 12}} \cr & {\text{(b) I}}{\text{.P}}{\text{. of Ca}}{{\text{F}}_2} = \left( {{{10}^{ - 2}}} \right) \times {\left( {{{10}^{ - 3}}} \right)^2} = {10^{ - 8}} \cr & {\text{(c) I}}{\text{.P}}{\text{. of Ca}}{{\text{F}}_2} = \left( {{{10}^{ - 5}}} \right) \times {\left( {{{10}^{ - 3}}} \right)^2} = {10^{ - 11}} \cr & {\text{(d) I}}{\text{.P}}{\text{. of Ca}}{{\text{F}}_2} = \left( {{{10}^{ - 3}}} \right) \times {\left( {{{10}^{ - 5}}} \right)^2} = {10^{ - 13}} \cr & \because \,\,{\text{I}}{\text{.P}} > {\text{solubility in choice (b) only}}{\text{.}} \cr & \therefore \,\,{\text{ppt of}}\,Ca{F_2}\,{\text{is obtained in case of choice (b) only}} \cr} $$

$${\text{Let}}\,s = {\text{solubility}}$$
\[AgI{{O}_{3}}\rightleftharpoons {{\underset{s}{\mathop{Ag}}\,}^{+}}+\underset{s}{\mathop{I{{O}_{3}}^{-}}}\,\]
$$\eqalign{ & {K_{sp}} = \left[ {A{g^ + }} \right]\left[ {I{O_3}^ - } \right] = s \times s = {s^2} \cr & {\text{Given}}\,\,{K_{sp}} = 1 \times {10^{ - 8}} \cr & \therefore \,\,s = \sqrt {{K_{sp}}} = \sqrt {1 \times {{10}^{ - 8}}} \cr & = 1.0 \times {10^{ - 4}}\,mol/lit = 1.0 \times {10^{ - 4}} \times 283\,g/lit \cr & \left( {\because \,\,{\text{Molecular mass of}}\,Ag\,I{O_3} = 283} \right) \cr & = \frac{{1.0 \times {{10}^{ - 4}} \times 283 \times 100}}{{1000}}gm/100\,ml \cr & = 2.83 \times {10^{ - 3}}gm/100\,ml \cr} $$

$${\text{Conc}}{\text{. of}}\,\,NaOH = \frac{{0.05}}{5}$$     $$ = 0.01\,\,{\text{or}}\,\,{10^{ - 2}}M$$
$$\eqalign{ & \left[ {O{H^ - }} \right] = {10^{ - 2}}M \cr & \left[ {{H^ + }} \right] = \frac{{{K_w}}}{{\left[ {O{H^ - }} \right]}} = \frac{{{{10}^{ - 14}}}}{{{{10}^{ - 2}}}} = {10^{ - 12}} \cr} $$
$$pH = - \log \left[ {{H^ + }} \right] = $$     $$ - \log {10^{ - 12}} = 12$$

$${\left[ {Cu{{\left( {{H_2}O} \right)}_4}} \right]^{2 + }} + 4N{H_3} \rightleftharpoons $$       $${\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }} + 4{H_2}O$$      involves lose and gain of electrons. $${H_2}O$$  is coordinated to $$Cu$$  by donating electrons $$(LHS).$$   It is then removed by withdrawing electrons.