Determine the molar solubility of CaSO₄ in a solution containing 0.12 M K₂SO₄. K_sp (CaSO₄)= 2.4 × 10^-5. A) 2.9 × 10^-6 M B) 2.0 × 10^-4 M C) 5.8 × 10^-10M D) 4.9 × 10^-3 M E) 0.20 M

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The solubility product expression for CaSO₄ is: Ksp = [CaSO₄] = [Ca2+][S2-] Let x be the molar solubility of CaSO₄ in the presence of K₂SO₄. Then: [Ca2+] = x [S2-] = x The concentration of K₂SO₄ in the solution is 0.12 M, which means: [K+] = 0.12 M Since CaSO₄ is a sparingly soluble salt, we can assume that the contribution of x to the concentration of K+ is negligible, so: [K+] ≈ [K+](original) The original concentration of K+ is zero, since we are assuming that the salt does not affect the pH of the solution. Therefore: [K+] = 0 M Now we can write the mass balance equation for S2-: [S2-] = x + [K₂SO₄] = x + 0.12 Substituting this expression for [S2-] into the solubility product expression, we get: Ksp = x(x + 0.12) = x^2 + 0.12x The problem gives us the value of Ksp, so we can solve for x: x^2 + 0.12x - Ksp = 0 Using the quadratic formula, we get: x = (-0.12 ± √(0.12^2 + 4Ksp))/2 x = (-0.12 ± √(0.0144 + 9.6×10^-5))/2 x = (-0.12 ± 0.00985)/2 x = 0.05 or x = -0.17 We take the positive solution for x, since the molar solubility can't be negative: x = 0.05 M Therefore, the best choice is B) 2.0 × 10^-4 M.

Determine the Molar Solubility of CaSO4 in a Solution Containing

Determine the Molar Solubility of CaSO₄ in a Solution Containing