Lawrie Ryan and Roger Norris Solutions for Chapter: Further Aspects of Equilibria, Exercise 15: Question

Thallium$\left(\mathrm{I}\right)$ chloride is a salt that is sparingly soluble in water. When hydrochloric acid is added to a saturated solution of thallium$\left(\mathrm{I}\right)$ chloride, a precipitate is formed. Explain why a precipitate is formed.

Calcium sulfate is a sparingly soluble salt that can be made by mixing solutions containing calcium and sulfate ions. A $0.00100 \mathrm{mol} {\mathrm{dm}}^{-3}$ solution of aqueous calcium chloride, ${\mathrm{CaCl}}_2$ is mixed with an equal volume of $0.00100 \mathrm{mol} {\mathrm{dm}}^{-3}$ solution of aqueous sodium sulfate, ${\mathrm{Na}}_2{\mathrm{SO}}_4$.

Calculate the concentration of calcium and sulfate ions when equal volumes of these solutions of calcium chloride and sodium sulfate are mixed.

Calcium sulfate is a sparingly soluble salt that can be made by mixing solutions containing calcium and sulfate ions. A $0.00100 \mathrm{mol} {\mathrm{dm}}^{-3}$ solution of aqueous calcium chloride, ${\mathrm{CaCl}}_2$ is mixed with an equal volume of $0.00100 \mathrm{mol} {\mathrm{dm}}^{-3}$ solution of aqueous sodium sulfate, ${\mathrm{Na}}_2{\mathrm{SO}}_4$.

Will a precipitate of calcium sulfate form? (${\mathrm{K}}_{\mathrm{sp}}$ of calcium sulfate = $2.0 \mathrm{x} {10}^{-5} {\mathrm{mol}}^2 {\mathrm{dm}}^{-3}$)

Which one of these is the correct numerical value for the solubility product for bismuth sulfide ${\mathrm{Bi}}_2{\mathrm{S}}_3$? Solubility of ${\mathrm{Bi}}_2{\mathrm{S}}_3$ = $3.6 \mathrm{x} {10}^{-8} \mathrm{mol} {\mathrm{dm}}^{-3}$.