Answer:
The mass of calcium sulfate that will precipitate is 6.14 grams
Explanation:
Step 1: Data given
500.0 mL of 0.10 M Ca^2+ is mixed with 500.0 mL of 0.10 M SO4^2−
Ksp for CaSO4 is 2.40*10^−5
Step 2: Calculate moles of Ca^2+
Moles of Ca^2+ = Molarity Ca^2+ * volume
Moles of Ca^2+ = 0.10 * 0.500 L
Moles Ca^2+ = 0.05 moles
Step 3: Calculate moles of SO4^2-
Moles of SO4^2- = 0.10 * 0.500 L
Moles SO4^2- = 0.05 moles
Step 4: Calculate total volume
500.0 mL + 500.0 mL = 1000 mL = 1L
Step 5: Calculate Q
Q = [Ca2+] [SO42-]
[Ca2+]= 0.050 M [O42-]
Qsp = (0.050)(0.050 )=0.0025 >> Ksp
This means precipitation will occur
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: Calculate total CaSO4 dissolved
total CaSO4 dissolved = 0.0049 M * 1 L * 136.14 mol/L = 0.667 g
Step 8: Calculate initial mass of CaSO4
Since initial moles CaSo4 = 0.050
Initial mass of CaSO4 = 0.050 * 136.14 g/mol
Initial mass of CaSO4 = 6.807 grams
Step 9: Calculate mass precipitate
6.807 - 0.667 = 6.14 grams
The mass of calcium sulfate that will precipitate is 6.14 grams