Question

Sodium fluoride and lead nitrate are both soluble in water. If 0.0946 g of NaF and 0.362 g of Pb(NO₃)₂are dissolved in sufficient water to make 1.50 L of solution, will a precipitate form? How do you know if there should be a precipitate or not? What is the chemical formula for the possible precipitate? The Ksp for lead fluoride is 3.7 x 10^-8.

Answer 1

molarity = (weight / molar mass ) x 1 /Volume

NaF molarity = (0.0946 / 42 ) x 1/ 1.5

                      = 1.5 x 10^-3 M

Pb(NO₃)₂ = (0.362 / 331) x 1/ 1.5

                      = 7.3 x 10^-4 M

from NaF   concentration of   F -   = 1.5 x 10^-3 M

from Pb(NO₃)₂ concentration of   Pb+2 = 7.3 x 10^-4M

ionic product of PbF2 = [Pb+2][F-]^2

                                   = 7.3 x 10^-4 x (1.5 x 10^-3)^2

                                  = 1.64 x 10^-9

Ksp of PbF2 = 3.7 x 10^-8

Ksp > ionic product .

so there is no precipitate is formed.

if Ksp < ionic product precipitate will form

2 NaF + Pb(NO3)2 ----------------------> PbF2 (s) + 2 NaNO3

possible precipitate = PbF2