Question

A 12.0 L solution contains 4.15 M Ba(NO3)2. Adding a sufficient amount of Na3PO4 to the solution will precipitate Ba3PO4. Will a precipitate form if 15.00 mg Na3PO4 is added to a given Ba(NO3)2 solution? Determine the minimum concentration of Na3PO4 needed to precipitate Ba(PO4)2 from the given Ba(NO3)2 solution.

Answer 1

Answer – Given, volume = 12.0 L , [Ba(NO₃)₂] = 4.15 M

Mass of Na₃PO₄ = 15.00 mg = 0.015 g

Moles of Na₃PO₄ = 0.015 g / 163.94 g.mol^-1

                             = 9.15*10^-5 moles

[Na₃PO₄] = 9.15*10^-5 moles / 12.0 L

                =7.62*10^-6 M

[PO₄^3-] = 7.62*10^-6 M

[Ba²⁺] = 4.15 M

So, Qsp = [Ba²⁺]³[PO₄^3-]²

               = (4.15 M)³ * (7.62*10^-6 M)²

               = 4.16*10^-9

We know Ksp for the Ba₃(PO₄)₂ = 3.0*10^-23

So, Qsp > Ksp so precipitate will form.

Now we know the Ksp for the Ba₃(PO₄)₂ and [Ba²⁺], so

Ksp = [Ba²⁺]³[PO₄^3-]²

3.0*10^-23 = (4.15 M)³ * (2x)²

                 = 71.47 *4x²

4x² = 3.0*10^-23/71.47

      = 4.20*10^-25

x² = 1.05*10^-25

So, x = 3.24*10^-13 M

So, minimum concentration of Na3PO4 needed to precipitate Ba(PO4)2 from the given Ba(NO3)2 solution is 3.24*10^-13 M