⇒Ionic Equilibrium

$\begin{array}{rl}& {\mathrm{AgCl}}_{(\mathrm{s})}\rightleftharpoons {\mathrm{Ag}}_{(\mathrm{sq})}^{+}+{\mathrm{Cl}}_{(\mathrm{sq})}^{-}\\ & x=1,\mathrm{y}=1\\ & {\mathrm{K}}_{\mathrm{sp}}=x^x{\mathrm{y}}^{\mathrm{y}}{\mathrm{S}}^{x+\mathrm{y}}=(1{)}^1(1{)}^1{\mathrm{S}}^{1+1}={\mathrm{S}}^2\\ & \mathrm{S}=\sqrt{{\mathrm{K}}_{\mathrm{sp}}}=\sqrt{1.6\times {10}^{-10}}\\ & =1.26\times {10}^{-5}\mathrm{M}\end{array}$

$\begin{array}{ll}& {\mathrm{Ag}}_2{\mathrm{CrO}}_{4(\mathrm{s})}\rightleftharpoons 2{\mathrm{Ag}}_{(\mathrm{aq})}^{+}+{\mathrm{CrO}}_{4(\mathrm{aq})}^{2-}\\ & \text{ Here, }x=2,\mathrm{y}=1\\ \therefore & {\mathrm{K}}_{\mathrm{sp}}=x^x{\mathrm{y}}^{\mathrm{y}}{\mathrm{S}}^{x+\mathrm{y}}=(2{)}^2(1{)}^1{\mathrm{S}}^{2+1}=4{\mathrm{S}}^3\end{array}$

$\begin{array}{rl}& {\mathrm{AgBr}}_{(\mathrm{s})}\rightleftharpoons {\mathrm{Ag}}_{(\mathrm{aq})}^{+}+{\mathrm{Br}}_{(\mathrm{aq})}^{-}\\ & \therefore x=1,y=1\\ & {\mathrm{K}}_{\mathrm{sp}}=x^x{y}^y{\mathrm{S}}^{x+y}=(1{)}^1(1{)}^1{\mathrm{S}}^{1+1}={\mathrm{S}}^2\\ & \therefore \mathrm{S}=\sqrt{{\mathrm{K}}_{\mathrm{sp}}}=\sqrt{6.4\times {10}^{-13}}=\sqrt{64\times {10}^{-14}}\\ & =8\times {10}^{-7}\mathrm{mol}{\mathrm{L}}^{-1}\end{array}$

$\begin{array}{ll}& \text{ For }{\mathrm{PbI}}_2,\\ & {\mathrm{PbI}}_{2(\mathrm{s})}\rightleftharpoons {\mathrm{Pb}}_{\text{(aq) }}^{2+}+2{\mathrm{I}}_{\text{(aq) }}^{-}\\ & x=1,\mathrm{y}=2\\ \therefore & {\mathrm{K}}_{\text{sp }}=x^x{\mathrm{y}}^{\mathrm{y}}{\mathrm{S}}^{x+y}=(1{)}^1(2{)}^2{\mathrm{S}}^{1+2}=4{\mathrm{S}}^3\\ \therefore & \mathrm{S}=\sqrt[3]{\frac{\mathrm{Ksp}}{4}}=\sqrt[3]{\frac{7.0\times {10}^{-9}}{4}}=1.21\times {10}^{-3}\mathrm{mol}{\mathrm{L}}^{-1}\end{array}$

A salt of strong acid and weak base forms aqueous solution having $\mathrm{pH}$ less than 7 (that is, it forms an acidic solution). ${\mathrm{CuSO}}_4$ is salt of strong acid ${\mathrm{H}}_2{\mathrm{SO}}_4$ and weak base $\mathrm{Cu}(\mathrm{OH}{)}_2$.