Question

17 A) Solid potassium fluoride is slowly added to 150 mL of a 0.0634 M calcium acetate solution. The concentration of fluoride ion required to just initiate precipitation is ____M.

17B)Solid potassium sulfate is slowly added to 175 mL of a 0.0511 M lead nitrate solution. The concentration of sulfate ion required to just initiate precipitation is ___M.

Answer 1

0.0634 M Calcium acetate Ca(CH3COO)2 will provide 0.0634 M Ca+2 then to just start precipitation of CaF2 we must have Ksp = [Ca+2][F-]² = 3.9 × 10⁻¹¹ (wikipedia)

so, putting Ca+2 value

Ksp = (0.0634)*[F-]²= 3.9 × 10⁻¹¹

F- = sqrt( 3.9 × 10⁻¹¹/0.0634) = 2.48*10^-5 M

b)

0.0511 M Lead nitrate Pb(NO3)2 will provide 0.0511 M Pb+2 then to just start precipitation of PbSO4 we must have Ksp = [Pb+2][SO42-] = 2.13 x 10⁻⁸ (wikipedia)

so, putting Pb+2 value

Ksp = (0.0511)*[SO42-]= 2.13 x 10⁻⁸

SO42- = 2.13 x 10⁻⁸/0.0511 = 4.168*10^-7 M