Answer:
[Ca2+] = 3.36 * 10^-8 M
Explanation:
Step 1: Data given
A 122.0−mL sample of blood contains 9.70 * 10^−5 g Ca2+/mL.
A technologist treats the sample with 100.0 mL of 0.1550 M Na2C2O4.
Ksp = 2.30 * 10^−9
Step 2: Calculate mass of Ca2+
Mass Ca2+ = 9.70 * 10^−5 g Ca2+/mL * 122 mL
Mass Ca2+ = 0.011834 grams
Step 3: Calculate moles of Ca2+
Moles Ca2+ = mass Ca2+ / molar mass Ca2+
Moles Ca2+ = 0.011834 grams / 40.078 g/mol
Moles Ca2+ = 2.95 *10^-4 moles = 0.000295
Step 4: Calculate moles of C2O4^2-
Moles C2O4^2- = Molarity * volume
Moles C2O4^2- = 0.1550 M * 0.100 L
Moles C2O4^2- = 0.0155 moles
Step 5: Calculate limiting reactant
Ca2+ is the limiting reactant. It will completely be consumed ( 0.000295 moles).
C2O4^2- is in excess. There will 0.000295 moles be consumed. There will remain 0.0155 - 0.000295 = 0.015205 moles of C2O4^2-
Step 6: Calculate total volume
Total volume = 122.0 mL + 100.0 mL = 222.0 mL = 0.222 L
Step 7: Calculate concentration CaC2O4
[CaC2O4] = 0.000295 mol / 0.222 L
[CaC2O4] = 0.00133 M
Step 8: Calculate concentration of C2O4^2-
[C2O4^2-] = 0.015205 mol / 0.222L
[C2O4^2-] = 0.0685 M
Step 9: Calculate [Ca2+]
Ksp = 2.3 * 10^-9 = [Ca2+] [C2O42-]
2.3 * 10^-9 = X * (X+0.0685)
X = [Ca2+] = 3.36 * 10^-8 M
