The solubility product of calcium fluoride ( CaF2 ) is 3.5×10−11. If 2.0 mL of a 0.10 M solution of NaF is added to 128 mL of a 2.0×10−5M solution of Ca(NO3)2 , will CaF2 precipitate?

Question

The solubility product of calcium fluoride ( CaF2 ) is 3.5×10−11. If 2.0 mL of a 0.10 M solution of NaF is added to 128 mL of a 2.0×10−5M solution of Ca(NO3)2 , will CaF2 precipitate?

Unsure how to do this question, any help would be appreciated!

Answer 1

To answer this, we will have to compare the concentration of CaF₂ present to the Ksp for CaF₂ to determine if a precipitate will or will not occur.

2NaF(aq) + Ca(NO₃)₂(aq) ==> 2NaNO₃(aq) + CaF₂(s) ... balanced equation

moles NaF present = 2.0 ml x 1 L/1000 ml x 0.10 mol/L = 0.0002 = 2x10-4 moles NaF

moles Ca(NO₃)₂ present = 128 ml x 1 L/1000 ml x 2x10^-5 mol/L = 2.56x10^-6 moles Ca(NO₃)₂

Based on these values and the mol ratio of the balanced equation, we see that Ca(NO₃)₂ is LIMITING

Moles of CaF₂ formed = 2.56x10^-6 mol Ca(NO₃)₂ x 1 mol CaF₂/mol Ca(NO₃)₂ = 2.56x10_-6 moles CaF₂

Final volume = 130 ml = 0.130 L

Final [CaF₂] = 2.56x10^-6 mol / 0.13 L = 1.9x10^-5 M

The concentration of CaF₂ is greater than Ksp so a ppt will form.