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 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 \(\mathrm{CaSO}_{4}(s)\) needed to achieve equilibrium?

For the given reaction: \[\mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q)\] The equilibrium constant expression is given by: \[K_c = \frac{[\mathrm{Ca}^{2+}][\mathrm{SO}_{4}^{2-}]}{[\mathrm{CaSO}_{4}]}\] Since the concentration of CaSO₄(s) is considered constant, we can write it as: \[K_c = [\mathrm{Ca}^{2+}][\mathrm{SO}_{4}^{2-}]\] Given, \(K_c = 2.4 \times 10^{-5}\)

Let [Ca²⁺]=[SO₄²⁻]=x mol/L, at equilibrium since in the balanced reaction, for every one mole of CaSO₄ that dissolves, one mole of Ca²⁺ and one mole of SO₄²⁻ ions are produced. So we can write: \[K_c = x^2\] Now, we can solve for x: \[2.4 \times 10^{-5} = x^2\] \[x = \sqrt{2.4 \times 10^{-5}}\] \[x = 4.9 \times 10^{-3}\] Hence, the equilibrium concentrations of Ca²⁺ and SO₄²⁻ are: \[[\mathrm{Ca}^{2+}] = [\mathrm{SO}_{4}^{2-}] = 4.9 \times 10^{-3}\,\text{mol/L}\]

Given the volume of the solution is 1.4 L, we can find the moles of Ca²⁺ and SO₄²⁻ ions in the solution: Moles of Ca²⁺ = Moles of SO₄²⁻ = \([\mathrm{Ca}^{2+}] \times \text{Volume}\) \[= (4.9 \times 10^{-3}\,\text{mol/L}) \times 1.4\,\text{L}\] \[= 6.86 \times 10^{-3}\,\text{mol}\] Now, we need to convert the moles of Ca²⁺ into grams of CaSO₄. The molar mass of CaSO₄: Molar mass = 40.08 (Ca) + 32.07 (S) + 4*16 (O) = 136.14 g/mol Minimum mass of CaSO₄ = moles of Ca²⁺ * molar mass of CaSO₄ \[= 6.86 \times 10^{-3}\,\text{mol} \times 136.14\,\text{g/mol}\] \[= 0.933\,\text{g}\] #Answer# (a) The equilibrium concentrations of Ca²⁺ and SO₄²⁻ are 4.9 x 10⁻³ mol/L. (b) The minimum mass of CaSO₄ needed to achieve equilibrium in 1.4 L of solution is 0.933 g.