Michael S. answered • 08/05/26
B.S. in Chemistry, Indiana University; Organic Chem Teaching Intern
[Ca2+] = 3.00 × 10-3 M, and 94.0% of the calcium can be precipitated.
Table values used: Ksp(CaSO4) = 4.93 × 10-5 and Ksp(Ag2SO4) = 1.20 × 10-5. Check these against your own appendix, since the arithmetic follows the values.
Step 1: which salt precipitates first? You are adding sulfate slowly, so the salt that needs the least sulfate to start forming appears first. Work out the threshold for each.
CaSO4: Ksp = [Ca2+][SO42-], so [SO42-] = 4.93 × 10-5 / 0.0500 = 9.86 × 10-4 M
Ag2SO4: Ksp = [Ag+]2[SO42-], so [SO42-] = 1.20 × 10-5 / (0.0270)2 = 1.646 × 10-2 M
Calcium needs about seventeen times less sulfate, so CaSO4 precipitates first. That is what makes the separation possible at all.
Watch the exponent on silver. Ag2SO4 releases two Ag+ per formula unit, so the silver term is squared. Forgetting that square is the single most common error in this problem, and it is invisible in your final answer, so nothing warns you.
Step 2: how much calcium is left when silver starts? The instant Ag2SO4 begins to form, the sulfate concentration is 1.646 × 10-2 M. Calcium sulfate has been precipitating this whole time and its solution stays saturated, so its Ksp still holds:
[Ca2+] = Ksp(CaSO4) / [SO42-] = 4.93 × 10-5 / 1.646 × 10-2 = 3.00 × 10-3 M
Step 3: percentage removed.
(0.0500 - 0.00300) / 0.0500 × 100 = 94.0%
One judgement worth forming about that number. 94% sounds high, but it means 6% of the calcium is still in solution when silver starts coming down, so anything you filter off after that point is contaminated. Analytical separations are usually considered clean at 99.9% or better. This pair of salts differs in Ksp by only about a factor of four, and that is simply not a big enough gap for a sharp separation. Selective precipitation works well when the two solubility products are separated by several orders of magnitude, and poorly when they are close, which is exactly what this problem is quietly demonstrating.
