(Solved): In a saturated solution of an ionic compound in water, there is an equilibrium between the solid co ...

In a saturated solution of an ionic compound in water, there is an equilibrium between the solid compound and the dissolved ions. The equilibrium expression for this type of reaction is called the solubility product constant, \( K_{\mathrm{sp}} \), which can be calculated from the concentrations, in moles per liter, of each ion in solution. For example, the solubility reaction and solubility product of barium carbonate, \( \mathrm{BaCO}_{3} \), are \[ \mathrm{BaCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{Ba}^{2+}(\mathrm{aq})+\mathrm{CO}_{3}{ }^{2-}(\mathrm{aq}) \] \[ K_{\text {mp }}=\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{CO}_{3}{ }^{2-}\right] \] Similarly, the solubility reaction and solubility product of calcium sulfate, \( \mathrm{CaSO}_{4} \), are \[ \begin{array}{c} \mathrm{CaSO}_{4}(\mathrm{~s}) \rightleftharpoons \mathrm{Ca}^{2+}(\mathrm{aq})+\mathrm{SO}_{4}{ }^{2-}(\mathrm{aq}) \\ K_{\text {sp }}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{SO}_{4}{ }^{2-}\right] \end{array} \] The solubility of barium carbonate, \( \mathrm{BaCO}_{3} \), is \( 0.0100 \mathrm{~g} / \mathrm{L} \). Its molar mass is \( 197.3 \mathrm{~g} / \mathrm{mol} \). What is the \( K_{\text {sp }} \) of barium carbonate? Express your answer numerically. The solubility product constant of calclum sulfate, CaSO \( { }_{4} \), is \( 7.10 \times 10^{-5} \). Its molar mass is \( 136.1 \mathrm{~g} / \mathrm{mol} \). How many grams of calclum sulfate can dissolve in \( 72.0 \mathrm{~L} \) of pure water? Express your answento three significant figures and include the appropriate units.