Expression for solubility of the reaction: is:
\({{\rm{K}}_{{\rm{sp}}}} = {\left[ {{{\rm{A}}^{{\rm{m + }}}}} \right]^{\rm{x}}}{\left[ {\;{{\rm{B}}^{{\rm{n + }}}}} \right]^{\rm{y}}}\)
For \({\rm{BaS}}{{\rm{O}}_4},\;{{\rm{K}}_{{\rm{sp}}}} = 2.3 \cdot {10^{ - 8}}({\rm{a}}\) known constant, Appendix\({\rm{J}}\))
Let's calculate Molar solubility of\({\rm{BaS}}{{\rm{O}}_4}\)and the Mass of barium in\(1\;{\rm{L}}\)of water saturated with\({\rm{BaS}}{{\rm{O}}_4}\).
Dissociation of \({\rm{BaS}}{{\rm{O}}_4}:{\rm{BaS}}{{\rm{O}}_4}({\rm{s}}){\rm{B}}{{\rm{a}}^{2 + }}({\rm{aq}}) + {\rm{SO}}_4^{2 - }({\rm{aq}})\)
\({{\rm{K}}_{{\rm{sp}}}} = {\rm{x}} \cdot {\rm{x}}\)
\(2.3 \cdot {10^{ - 8}} = {{\rm{x}}^2}\)
\({\rm{x}} = 1.52 \cdot {10^{ - 4}}{\rm{M}}\)
Molarity, represented by M or [ ] brackets, expresses a number of moles of a solute per 1 liter of a solution.
Molar solubility of \({\rm{BaS}}{{\rm{O}}_4}\) is \(1.52 \cdot {10^{ - 4}}{\rm{M}}\) for both \(\left[ {{\rm{B}}{{\rm{a}}^{2 + }}} \right]\)and\(\left[ {{\rm{SO}}_4^{2 - }} \right]\).
Step 2: To find the mass of the barium
Mass of barium:
\({\rm{ Molarity }} = \frac{{{\rm{ moles }}}}{{{\rm{ litres }}}} = \frac{{{\rm{ mass }}}}{{{\rm{ molar mass }}x{\rm{ litres }}}}\)
Mass = molarity \( \times \) molar mass \( \times \) liters
The molar mass of Ba is \(137.33\;{\rm{g}}/{\rm{mol}}\)as we know from the periodic table of elements. If we put that into the above formula:
Mass of \({\rm{Ba}} = 1.52 \cdot {10^{ - 4}}\;{\rm{mol}}/{\rm{l}} \cdot 137.33\;{\rm{g}}/{\rm{mol}} \cdot 1{\rm{l}} = {\bf{0}}.{\bf{02}}{\rm{g}}{\bf{Ba}}\)
Finally we get,
Molar solubility of \({\rm{BaS}}{{\rm{O}}_4}\)is \(1.52 \cdot {10^{ - 4}}{\rm{M}}\) for both \(\left[ {{\rm{B}}{{\rm{a}}^{2 + }}} \right]\)and \(\left[ {{\rm{SO}}_4^{2 - }} \right]\)
Mass of barium in 11 is\(0.02\;{\rm{g}}\).
