The solubility product of \(CaS{O_4}.2{H_2}O\) is\(2.4 \times 1{0^{ - 5}}\). What mass of this salt will dissolve in \(1.0\;L\) of \(0.010M\) \(SO{4^{2 - }}\)?

\(CaS{O_4}.2{H_2}O\)

Because of the common ion, the concentration of\({\rm{SO}}_4^{2 - }\)present in the solution is a limiting factor to the dissolving of \({\rm{CaS}}{{\rm{O}}_4} \times 2{{\rm{H}}_2}{\rm{O}}\).

\({\rm{CaS}}{{\rm{O}}_4}({\rm{s}}) \to {\rm{C}}{{\rm{a}}^{2 + }}({\rm{aq}}) + {\rm{SO}}_4^{2 - }({\rm{aq}})\)

\(\begin{array}{l}{{\rm{K}}_{{\rm{sp}}}} = \left[ {{\rm{C}}{{\rm{a}}^{{\rm{2 + }}}}} \right]\left[ {{\rm{SO}}_{\rm{4}}^{{\rm{2 - }}}} \right] = {\rm{x}} \times ({\rm{x}} + 0.01) = 2.4 \times {10^{ - 5}}\\\end{array}\)

\({{\rm{x}}^2} + 0.01x - 2.4 \times {10^{ - 5}} = 0\)

\({\rm{x}} = \frac{{ - 0.01 \pm \sqrt {1 \times {{10}^{ - 4}} + 9.6 \times {{10}^{ - 5}}} }}{2} = 2 \times {10^{ - 3}}\)

This value corresponds to the concentration of \({\rm{CaS}}{{\rm{O}}_4} \times 2{{\rm{H}}_2}{\rm{O}}\) that has dissolved.

To find the Mass of the salt in \(1\;{\rm{L}}\)of the solution calculations are done in the following way

\({\rm{m}} = 2 \times {10^{ - 3}}\;{\rm{mol}}/{\rm{L}} \times 172.16\;{\rm{g}}/{\rm{mol}} = 0.34\;{\rm{g}}/{\rm{L}}\)

Finally, we get,

\({\rm{ Mass of dissolved salt }} = 0.34\;{\rm{g}}/{\rm{L}}\)