Calculate the ion product for mixing 100.0 mL of 0.0030 M CaCl2 with 100 mL of 0.0020 M Na2CO3. Ksp for CaCO3 is 1.4 10–8. Does a precipitate form? CaCO3 (s) ⬄ Ca2+ (aq) + CO32- (aq)

Question

 Calculate the ion product for mixing 100.0 mL of 0.0030 M CaCl2 with 100 mL of 0.0020 M Na2CO3. Ksp for CaCO3 is 1.4 10–8. Does a precipitate form?

CaCO3 (s) ⬄ Ca2+ (aq) + CO32- (aq)

Answer 1

CaCl₂(aq) + Na₂CO₃(aq) ==> CaCO₃(s) + 2NaCl(aq) ... balanced equation

mols CaCl₂ present = 100.0 ml x 1 L / 1000 ml x 0.0030 mol / L = 0.000300 mols Ca²⁺

mols Na₂CO₃ present = 100 ml x 1 L / 1000 ml x 0.002 mol/L = 0.0002 mols CO₃^2-

Total volume = 100 ml + 100 ml = 200 ml = 0.2 L

Q = ion product = [Ca²⁺][CO₃^2-]

[Ca²⁺] = 0.000300 mols / 0.2 L = 0.0015 M

[CO₃^2-] = 0.0002 mols / 0.2 L = 0.001 M

Q = [Ca²⁺][CO₃^2-] = (0.0015)(0.001)

Q = 1.5x10^-6

Q > Ksp therefore a precipitate will form