Answer:
The amount of calcium sulfate that precipitates is 6.14 grams.
Explanation:
Step 1: Provided data
We are mixing 500.0 mL of 0.10 M Ca^2+ with 500.0 mL of 0.10 M SO4^2−
The Ksp for CaSO4 is 2.40×10^−5.
Step 2: Determine moles of Ca^2+
Moles of Ca^2+ = Molarity of Ca^2+ * Volume
Moles of Ca^2+ = 0.10 * 0.500 L
Moles of Ca^2+ = 0.05 moles
Step 3: Determine moles of SO4^2-
Moles of SO4^2- = 0.10 * 0.500 L
Moles SO4^2- = 0.05 moles
Step 4: Compute total volume
Total volume = 500.0 mL + 500.0 mL = 1000 mL = 1L
Step 5: Compute Q
Q = [Ca2+] [SO42-]
[Ca2+] = 0.050 M and [SO42-]
Qsp = (0.050)(0.050) = 0.0025 >> Ksp
This indicates that precipitation will take place.
Step 6: Calculate molar solubility
Ksp = 2.40 * 10^-5 = [Ca2+][SO42-] =(x)(x)
2.40 * 10^-5 = x²
x = √(2.40 * 10^-5)
x = 0.0049 M (molar solubility)
Step 7: Determine total dissolved CaSO4
Total CaSO4 dissolved = 0.0049 M * 1 L * 136.14 g/mol = 0.667 g
Step 8: Calculate initial mass of CaSO4
Initial moles of CaSO4 = 0.050
Initial mass of CaSO4 = 0.050 * 136.14 g/mol
Initial mass of CaSO4 = 6.807 grams
Step 9: Calculate precipitate mass
6.807 - 0.667 = 6.14 grams.
The mass of calcium sulfate that will emerge as a precipitate is 6.14 grams.
