Question

A solution is prepared by mixing 100 mL of .01 M Pb(NO3)2 and 100 mL of .001 M NaF. Will PbF2(s) precipitate in this reaction? Calculate the concentration of F- ions present at equilibrium.

PbF2(s) <--> Pb2+(aq) + 2F-(aq) Kc= 3.7E-8

Please show all work! Thank you!

Answer 1

PbF2(s) <--> Pb2+(aq) + 2F-(aq)

Ksp = [Pb+2][*F-]^2 =

Calculation of Concentration of Pb+2

M1V1 (before mixing) = M2V2 (after mixing)

Toatl volume after mixing = 100 ml + 100 ml = 200 ml

0.01*100 = M2*200

M2 = 0.005 M

there fore concentration of Pb(NO3)2 after mixing is 5.0*10^-3 M

Calculation of Concentration of F-

M1V1 (before mixing) = M2V2 (after mixing)

0.001*100 = M2 * 200

M2 = 0.0005 M

there fore concentration of NaF after mixing is 5.0*10^-4 M

Ksp = [Pb+2][2*F-]^2

Ksp =[5.0*10^-3][2*5.0*10^-4]^2

Ksp = 5.0*10^-9

Ionic prduct = [Pb+2][F-]^2

                  = (5.0*10^-3)*(5.0*10^-4)^2

                 = 1.25*10^-9

since solubility product(5.0*10^-9) is greater than ionic product(1.25*10^-9) PbF2 will not precipitate.

....................................................................................................................................................................

given Kc = 3.7*10^-8 for PbF2(s) <--> Pb2+(aq) + 2F-(aq)

so

.........PbF2(s) <--> Pb2+(aq) + 2F-(aq)

initial ...1 ...................0.................0

at eq. (1-x) ................x.................2x

Kc = [Pb^+2][F-]^2/[PbF2]

Kc*[PbF2] = Ksp = 5.0*10^-9

so

5.0*10^-9 = x*(2x)2

4x^3 = 5.0*10^-9

X = 1.07*10^-3 M

[F-] = 2*x = 2*1.07*10^-3 = 2.14*10^-3 M

at equilibrium concentration of F- = 2.14*10^-3 M