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 \(\mathrm{CaSO}_{4}(s)\) 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 \(\mathrm{SO}_{4}^{2-} ?(\mathbf{b})\) If the resulting solution has a volume of \(1.4 \mathrm{~L},\) what is the minimum mass of \(\mathrm{CaSO}_{4}(s)\) needed to achieve equilibrium?

We are given the following information: - Reaction: \(\mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \) - Equilibrium constant, \(K_{c} = 2.4 \times 10^{-5}\) at 25°C - Volume of the solution, \(V = 1.4\) L

According to the reaction, the equilibrium constant can be written as: $$K_{c} = \frac{[\mathrm{Ca}^{2+}][\mathrm{SO}_{4}^{2-}]}{[\mathrm{CaSO}_{4}]}$$ Since CaSO₄ is a solid, we can rewrite the equation as: $$K_{c} = [\mathrm{Ca}^{2+}][\mathrm{SO}_{4}^{2-}]$$

Let the equilibrium concentration of Ca²⁺ be x mol/L. Since the stoichiometry of the reaction is 1:1, the equilibrium concentration of SO₄²⁻ will also be x mol/L. Now, we can substitute their concentrations into the equation for the equilibrium constant: $$K_{c} = x^2$$ $$2.4 \times 10^{-5} = x^2$$ Solve for x: $$x = \sqrt{2.4 \times 10^{-5}}$$ $$x = 4.9 \times 10^{-3} \, \text{mol/L}$$ So, the equilibrium concentrations of Ca²⁺ and SO₄²⁻ are both \(4.9 \times 10^{-3} \, \text{mol/L}\).

We know that the equilibrium concentration of Ca²⁺ is \(4.9 \times 10^{-3} \, \text{mol/L}\). As the stoichiometry of the reaction is 1:1, the equilibrium concentration of CaSO₄ is the same. Since we know the volume of the solution (1.4 L), we can find the moles of CaSO₄ required to reach equilibrium: Moles of CaSO₄ = concentration × volume Moles of CaSO₄ = \(4.9 \times 10^{-3} \, \text{mol/L} \times 1.4 \, \text{L}\) Moles of CaSO₄ = \(6.86 \times 10^{-3} \, \text{mol}\) Now, calculate the molar mass of CaSO₄: Molar mass of CaSO₄ = 40.08 (Ca) + 32.07 (S) + 4 × 16.00 (O) = 136.14 g/mol Finally, find the minimum mass of CaSO₄ needed: Mass of CaSO₄ = moles × molar mass Mass of CaSO₄ = \(6.86 \times 10^{-3} \, \text{mol} \times 136.14 \, \text{g/mol}\) Mass of CaSO₄ = 0.934 g The minimum mass of CaSO₄ needed to achieve equilibrium in a 1.4 L solution is approximately 0.934 g.