(1)

Let the solubility of AgCl in water be $s$ and in KCl be $S$.

$\begin{array}{cccccc}& \mathrm{AgCl}& \to & {\mathrm{Ag}}^{+}& +& {\mathrm{Cl}}^{-}\\ \mathrm{Solubility} \mathrm{in} \mathrm{water}:& & & \mathrm{s}& & \mathrm{s}\\ \mathrm{Solubility} \mathrm{in} \mathrm{KCl}:& & & \mathrm{S}& & \mathrm{S}+0.1\end{array}$

$K_{sp}=\left[\text{Ag}\right]\left[\text{Cl}\right]={10}^{-10}$

For water, $K_{sp}=s^2\Rightarrow s={10}^{-5}$

For KCl, $K_{sp}=S(S+0.1)$

$S(0.1)={10}^{-10}$

$\Rightarrow S={10}^{-9} \text{(as, }S\mathit{<}\mathit{<}\mathit{<}\mathit{<}0.1\text{)}$

$\frac{\text{Solubility of AgCl in KCl}}{\text{Solubility of AgCl in water}}= \frac{{10}^{-9}}{{10}^{-5}}={10}^{-4}$

(2)

pH of the saturated solution of $\mathrm{Ca}{\left(\mathrm{OH}\right)}_2$ = 9

$\therefore$  ${\text{pOH of the saturated solution of Ca(OH)}}_2=14-9=5$

$\Rightarrow \left[{\text{OH}}^{-}\right]={10}^{-5}$    $(\because \text{pH}+\text{pOH}=14)$

$\begin{array}{ccccc}\mathrm{Ca}{\left(\mathrm{OH}\right)}_2& \rightleftharpoons & {\mathrm{Ca}}^{2+}& +& 2{\mathrm{OH}}^{-}\\ & & \mathrm{s}& & 2\mathrm{s}\\ & & \frac{1}{2}\times {10}^{-5}& & {10}^{-5}\end{array}$

$K_{sp}=\left[{\text{Ca}}^{2+}\right]{\left[{\text{OH}}^{-}\right]}^2=[\frac{{\displaystyle 1}}{{\displaystyle 2}}\times {10}^{-5}]{\left[{10}^{-5}\right]}^2$

$=0.5\times {10}^{-15}$

(3)

${\text{CaF}}_2\to \underset{\mathrm{s}}{{\text{Ca}}^{2+}}+\underset{2\mathrm{s}}{2{\text{F}}^{-}}$

$\text{NaF→}\underset{\text{0.1 M}}{{\text{Na}}^{+}}+\underset{\text{0.1 M}}{{\text{F}}^{-}}$

$\left[{\text{Ca}}^{2+}\right]=s, \left[{\text{F}}^{-}\right]=(2s+0.1)\approx 0.1$ $\text{M}$

$K_{sp}=\left[{\text{Ca}}^{2+}\right]{[{\text{F}}^{-}\text{]}}^2$

$5.3\times {10}^{-11}=\left(s\right){(0.1)}^2$

$s=\frac{5.3\times {10}^{-11}}{{(0.1)}^2}=5.3\times {10}^{-9}$ $\text{mol}$ ${\mathrm{L}}^{-1}$

$\therefore$   $\text{Molar solubility is }5.3\times {10}^{-9}$ $\text{mol}$ ${\mathrm{L}}^{-1}$

(1)

Solubility of ${\mathrm{BaSO}}_4$

$s=\frac{2.42\times {10}^{-3}}{233}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}=1.04\times {10}^{-5}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$

${\mathrm{BaSO}}_4$ ionizes completely in the solution as: ${\text{BaSO}}_{4\left(s\right)}\rightleftharpoons {\text{Ba}}_{\left(aq\right)}^{2+}+{{\text{SO}}_4}_{\left(aq\right)}^{2-}$

$K_{sp}=\left[{\text{Ba}}^{2+}\right]\left[{\text{SO}}_4^{2-}\right]=s^2$

$={(1.04\times {10}^{-5})}^2=1.08\times {10}^{-10}$ ${\mathrm{mol}}^2$ ${\mathrm{L}}^{-2}$

(3)

Let solubility of ${\mathrm{Ag}}_2{\mathrm{C}}_2{\mathrm{O}}_4$ be $s$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$ 

$\underset{\mathrm{s}}{{\mathrm{Ag}}_2{\mathrm{C}}_2{\mathrm{O}}_{4\left(\mathrm{s}\right)}}\rightleftharpoons \underset{2\mathrm{s}}{2{\mathrm{Ag}}_{\left(\mathrm{aq}\right)}^{+}}+\underset{\mathrm{s}}{{\mathrm{C}}_2{\mathrm{O}}_{4\left(\mathrm{aq}\right)}^{2-}}$

$K_{\mathrm{sp}}={\left[{\mathrm{Ag}}^{+}\right]}^2\left[{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right]$

$K_{\mathrm{sp}}={\left(2s\right)}^2\left(s\right)=4s^3$

$K_{\mathrm{sp}}=4\times {(1.1\times {10}^{-4})}^3 (\because [{\mathrm{Ag}}^{+}]=2s=2.2\times {10}^{-4})$

$K_{\mathrm{sp}}=5.3\times {10}^{-12}$

(2)

Let $\mathrm{s}$ be the solubility of AgCl in moles per litre.

$\underset{\mathrm{s}}{{\mathrm{AgCl}}_{\left(\mathrm{aq}\right)}}\rightleftharpoons \underset{\mathrm{s}}{{\mathrm{Ag}}_{\left(\mathrm{aq}\right)}^{+}}+\underset{(\mathrm{s}+0.1)}{{\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-}}$

($\because$ 0.1 M NaCl solution also provides 0.1 M ${\mathrm{Cl}}^{-}$ ion)

$K_{\mathrm{sp}}=\left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{Cl}}^{-}\right] ; 1.6\times {10}^{-10}=s(s+0.1)$

$1.6\times {10}^{-10}=s(0.1)$                                    $(\because s\mathit{<}\mathit{<}\mathit{<}\mathit{<}0.1)$

$s=\frac{1.6\times {10}^{-10}}{0.1}=1.6\times {10}^{-9}$ $\mathrm{M}$

(4)

For $MY: K_{\mathrm{sp}}=s_1^2$

$\Rightarrow s_1=\sqrt{K_{\mathrm{sp}}}=\sqrt{6.2\times {10}^{-13}}=7.87\times {10}^{-7}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$

For $N{Y}_3: K_{\mathrm{sp}}=\left(s_2\right){\left(3s_2\right)}^3=27s_2^4$

$\Rightarrow s_2=\sqrt[4]{\frac{6.2\times {10}^{-13}}{27}}=3.89\times {10}^{-4}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$

Hence, molar solubility of $\mathrm{MY}$ in water is less than that of ${\mathrm{NY}}_3$.

(2)
 

Solubility of ${\mathrm{Ag}}_2{\mathrm{CrO}}_4$ is highest, thus it will be precipitated at last.

(2)

$\Delta G°=-2.303RT\log K_{sp}$

$63.3\times {10}^3\text{J}=-2.303\times 8.314\times 298 \log K_{sp}$

$63.3\times {10}^3\mathrm{J}=-5705.84 \log K_{sp}$

$\log K_{sp}=-\frac{63.3\times {10}^3}{5705.84}=-11.09$

$K_{sp}=\text{antilog}(-11.09)=8.128\times {10}^{-12}$