Question

When 100 mL each of 2.0 × 10-4 M Ca2+ and 2.0 × 10-2 M F- are mixed, what is the remaining Ca2+ ion concentration ? The solubility product constant of CaF2 is 5.3 × 10-9. (Hint: Determine the limiting reactant, the concentration of excess reactant after precipitation, and how the concentration of excess reactant influences the solubility of CaF2).

Answer 1

Ca²⁺ + 2F^- -------> CaF₂

Given

Molarity of Ca²⁺ = 2.0810-4 M

volume of Ca²⁺ = 100 mL =0.100 L

Molarity of F- = 2.0x10^-2 M

Volume of F- = 0.100 L

Moles of Ca²⁺ =

Moles of Ca²⁺ = 2.0x10^-5 mol

Moles of F^- =

Moles of F- = 2.0*10^-3 mol

Finding the limiting reactant

Moles of CaF₂ from from Ca²⁺ ions =

Moles of CaF₂ formed from Ca²⁺ ions = 2.0x10^-5 mol

Moles of CaF₂ formed from F- ions =

Moles of CaF₂ formed from F- ions = 1.0*10^-3 mol

Moles of CaF₂ formed from Ca²⁺ is less so limiting reactant is Ca²⁺

Moles of excess reactant F^- in the solution = Moles of F^- - moles of CaF₂ formed from Ca²⁺

= 2.0*10^-3 mol -  2.0x10^-5 mol

Moles of excess reactant F^- in the solution = 0.00198 mol F^-

Volume of the solution = 200 mL = 0.2 L

Concetration of F- = 0.00198/0.200

=0.0099 M

CaF₂  <====> Ca²⁺ + 2 F^-

Because of the common ion F- present in the solution the reaction shift forward and forms Ca²⁺ ions

Let us find the concentration of Ca²⁺ ions

Ksp = [Ca²⁺][F^-]²

Let concentration of Ca²⁺ be x

and concentration of F^- is calculated = 0.0099 M

Ksp =5.3 × 10^-9

5.3 × 10^-9 = (0.0099)²(x)

X = 5.41x10^-5 M

Concentration of Ca²⁺ in the solution = 5.41x10^-5 M