Final answer:
When 0.100 mol/L calcium chloride is mixed with 0.0400 mol/L sodium sulfate, a precipitate of calcium sulfate will form because the ion product exceeds the solubility product.
Step-by-step explanation:
When a student mixes 100.0 mL of an aqueous solution of 0.100 mol/L calcium chloride, CaCl2, with 100.0 mL of an aqueous solution of 0.0400 mol/L sodium sulfate, Na2SO4, a precipitation reaction can occur if the product of the ion concentrations exceeds the solubility product (Ksp) of the potential precipitate. In this case, the potential precipitate is calcium sulfate, CaSO4.
To determine whether a precipitate will form, we must calculate the ion product (Q), which is the product of the concentrations of the calcium ions (Ca2+) and sulfate ions (SO42-) after the solutions are mixed. The concentration of calcium ions from calcium chloride will be 0.100 mol/L, and the concentration of sulfate ions from sodium sulfate will be 0.0400 mol/L, each diluted by the combined volume of the solutions to 200.0 mL. If the ion product (Q) exceeds the known solubility product of calcium sulfate (Ksp), a precipitate will form.
Calculation of Ion Product (Q):
Q = [Ca2+][SO42-] after mixing = (0.100 mol/L / 2)(0.0400 mol/L / 2) = (0.0500 mol/L)(0.0200 mol/L) = 0.00100 mol2/L2
Comparing the ion product (Q) to the solubility product (Ksp) of calcium sulfate will determine the formation of a precipitate. The Ksp for calcium sulfate is approximately 2.4×10-5 mol2/L2. Since 0.00100 mol2/L2 is greater than 2.4×10-5 mol2/L2, a precipitate of calcium sulfate will indeed form when the two solutions are mixed.