Consider the reaction $$\mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q)$$ At \(25^{\circ} \mathrm{C}\), the equilibrium constant is $K_{c}=2.4 \times 10^{-5}\( for this reaction. (a) If excess \)\operatorname{CaSO}_{4}(s)$ is mixed with water at \(25^{\circ} \mathrm{C}\) to produce a saturated solution of \(\mathrm{CaSO}_{4},\) what are the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}^{2-}\) ? (b) If the resulting solution has a volume of \(1.4 \mathrm{~L},\) what is the minimum mass of \(\operatorname{CaSO}_{4}(s)\) needed to achieve equilibrium?

We need to write the expression for the equilibrium constant \(K_c\) for the given reaction: \(\mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q)\) According to the reaction, the formation of 1 mole of \(\mathrm{Ca}^{2+}\) ion is accompanied by the formation of 1 mole of \(\mathrm{SO}_{4}^{2-}\) ion. Therefore, we can simplify the expression for \(K_c\) as follows: \[ K_c = [\mathrm{Ca}^{2+}][\mathrm{SO}_{4}^{2-}] \]

Let the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}^{2-}\) be \(x\). According to the reaction, their concentrations are equal. We then have: \(K_c = x^2\) Given \(K_c = 2.4\times 10^{-5}\), we can now solve for \(x\): \(x^2 = 2.4\times 10^{-5}\) \(x = \sqrt{2.4\times 10^{-5}} = 4.9\times 10^{-3}\, \mathrm{M}\) Thus, the equilibrium concentrations of both \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}^{2-}\) are \(4.9\times 10^{-3}\, \mathrm{M}\).

Given that the solution has a volume of 1.4 L, we can calculate the moles of \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}^{2-}\) at equilibrium: moles of \(\mathrm{Ca}^{2+}\) (and \(\mathrm{SO}_{4}^{2-}\)) = Molar concentration × Volume moles of \(\mathrm{Ca}^{2+}\) (and \(\mathrm{SO}_{4}^{2-}\)) = \(4.9\times 10^{-3}\, \mathrm{M}\times 1.4\, \mathrm{L} = 6.86\times 10^{-3}\, \mathrm{moles}\) Now, we can calculate the minimum mass of \(\mathrm{CaSO}_{4}\) needed using its molar mass. The molar mass of \(\mathrm{CaSO}_{4}\) can be found by adding the individual molar masses of \(\mathrm{Ca}\), \(\mathrm{S}\), and \(\mathrm{O}\): Molar mass of $\mathrm{CaSO}_{4} = 40.08 + 32.06 + 4(16.00) = 136.14\, \mathrm{g/mol}\) Finally, we can calculate the minimum mass by multiplying the moles of \(\mathrm{CaSO}_{4}\) needed by its molar mass: Minimum mass of \(\mathrm{CaSO}_{4}\) = moles × molar mass Minimum mass of \(\mathrm{CaSO}_{4}\) = \(6.86\times 10^{-3}\, \mathrm{moles} \times 136.14\, \mathrm{g/mol} = 0.93\, \mathrm{g}\) Therefore, the minimum mass of \(\mathrm{CaSO}_{4}(s)\) needed to achieve equilibrium in a 1.4 L solution is 0.93 g.