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If 500.0mL of 0.10 M Ca2+ is mixed with 500.0mL of 0.10 M SO42?, what mass...

If 500.0mL of 0.10 M Ca2+ is mixed with 500.0mL of 0.10 M SO42?, what mass of calcium sulfate will precipitate?

Ksp for CaSO4 is 2.40

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Answer #1

After mixing, the total volume 1000. mL or 1.00 L

mol Ca2+ = M x L = 0.10 mol/L x 0500 L = 0.0500 mol Ca2+
mol SO42- = M x L = 0.10 mol/L x 0.500 L = 0.0500 mol SO42-
0.0500 mol Ca2+ and 0.0500 mol SO42- forms 0.0500 mol CaSO4

CaSO4 -----> Ca2+ + SO42-
Ksp = [Ca2+][SO42-] = 2.40 x 10^-5
Let s = solubility of CaSO4, then
[Ca2+] = s
(SO42-] = s
2.40 x 10^-5 = (s)(s) = s^2
s = 4.90 x 10^-3 M = solubility of CaSO4

moles that dissolve per L:
mol = M x L = 4.90 x 10^-3 mol/L x 1 L = 4.90 x 10^-3 mol CaSO4

mol CaSO4 present - mol CaSO4 that dissolve = mol CaSO4 that precipitates.
0.0500 - 0.00490 = 0.0451 mol CaSO4 remains as a precipitate

The molar mass of CaSO4 is 136 g/mol.
0.0451 mol CaSO4 x (136 g / 1 mol) = 6.13 g CaSO4

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