Refer to Appendix \(J\) for solubility products for calcium salts. Determine which of the calcium salts listed is most soluble in moles per liter and which is most soluble in grams per liter.

\(\begin{array}{l}{\rm{Ca}}{({\rm{OH}})_2} \to {\rm{C}}{{\rm{a}}^{2 + }} + 2O{{\rm{H}}^ - }\\{{\rm{K}}_{{\rm{sp}}}} = {\rm{x}}{(2{\rm{x}})^2} = 4{{\rm{x}}^3} = 7.9 \times {10^{ - 6}}\\{\rm{x}} = \sqrt[3]{{\frac{{7.9 \times {{10}^{ - 6}}}}{4}}} = 0.013\;{\rm{M}}\end{array}\)

\(\begin{array}{l}{\rm{CaC}}{{\rm{O}}_3} \to {\rm{C}}{{\rm{a}}^{2 + }} + {\rm{CO}}_3^{2 - }\\{{\rm{K}}_{s{\rm{p}}}} = {{\rm{x}}^2} = 4.8 \times {10^{ - 9}}\\{\rm{x}} = 6.9 \times {10^{ - 5}}\;{\rm{M}}\end{array}\)

\(\begin{array}{l}{\rm{CaS}}{{\rm{O}}_4} \cdot {{\rm{H}}_2}{\rm{O}} \to {\rm{C}}{{\rm{a}}^{2 + }} + {\rm{SO}}_4^{(2 - )}.2{{\rm{H}}_{\rm{2}}}{\rm{O}}\\{{\rm{K}}_{{\rm{sp}}}} = {\rm{xx}}{(2x)^2} = 4{{\rm{x}}^4} = 2.4 \times {10^{ - 5}}\\{\rm{x}} = \sqrt[4]{{\frac{{2.4 \times {{10}^{ - 5}}}}{4}}} = 0.049\;{\rm{M}}\end{array}\)\(\)

\(\begin{array}{l}{\rm{Ca}}{{\rm{C}}_2}{{\rm{O}}_4} \cdot {{\rm{H}}_2}{\rm{O}} \to {\rm{C}}{{\rm{a}}^{2 + }} + {{\rm{C}}_2}{\rm{O}}_4^{2 - }.{{\rm{H}}_2}{\rm{O}}\\{{\rm{K}}_{{\rm{sp}}}} = {\rm{xxx}} = {{\rm{x}}^3} = 2.27 \times {10^{ - 9}}\\{\rm{x}} = \sqrt[3]{{2.27 \times {{10}^{ - 9}}}} = 1.3 \times {10^{ - 3}}\;{{\rm{M}}_{_{}}}\end{array}\)

\(\begin{array}{l}{\rm{C}}{{\rm{a}}_3}{\left( {{\rm{P}}{{\rm{O}}_4}} \right)_2} \to 3{\rm{C}}{{\rm{a}}^{2 + }} + 2{\rm{PO}}_4^{(3 - )}\\{K_{{\rm{sp}}}} = {{\rm{x}}^3}{\left( {\frac{2}{3}{\rm{x}}} \right)^2} = 1 \times {10^{ - 25}}\\{\rm{x}} = \sqrt[5]{{\frac{{1 \times {{10}^{ - 25}} \times 9}}{4}}} = 1 \times {10^{ - 5}}{\rm{M}}\end{array}\)

In this case, solubility equals to one third of\(\left[ {{\rm{C}}{{\rm{a}}^{{\rm{2 + }}}}} \right]\), that is\(4 \times {10^{ - 6}}\).

\(\begin{array}{l}{\rm{CaHP}}{{\rm{O}}_4} \to {\rm{C}}{{\rm{a}}^{2 + }} + {\rm{HPO}}_4^{(2 - )}\\{{\rm{K}}_{{\rm{sp}}}} = {{\rm{x}}^2} = 7 \times {10^{ - 7}}\\{\rm{x}} = 8.4 \times {10^{ - 4}}\;{\rm{M}}\end{array}\)

\(\begin{array}{l}{\rm{Ca}}{{\rm{F}}_2} \to {\rm{C}}{{\rm{a}}^{2 + }} + {{\rm{F}}^{(2 - )}}\\{{\rm{K}}_{{\rm{sp}}}} = {{\rm{x}}^2} = 4 \times {10^{ - 11}}\\{\rm{x}} = 6.3 \times {10^{ - 6}}\;{\rm{M}}\end{array}\)

Finally we get,

a) \(0.013\;\;{\rm{M}}\)

b) \(6.9 \times {10^{ - 5}}\;{\rm{M}}\)

c) \(0.049\;\;{\rm{M}}\)

d) \(1.3 \times {10^{ - 3}}\;{\rm{M}}\)

e) \(4 \times {10^{ - 6}}\;{\rm{M}}\)

f) \(8.4 \times {10^{ - 4}}\;{\rm{M}}\)

g) \(6.3 \times {10^{ - 6}}\;{\rm{M}}\)