10.5mg of solid NaF is added to a 50mL solution that is 0.0050M in Ba(NO3)2 and 0.0050M in Pb(NO3)2. For BaF2(S) Ksp= 1.7x10-6 and for PbF2(S) Ksp= 3.6x10-8.
(1a) Write the equilibrium expression(s) for all possible reactions that could take place in this solution to form insoluble ionic salts:
(1b) Calculate Qsp for each insoluble salt:
(1c) What substance will precipitate from solution if more NaF were added? If so, at what concentration [F-]?
moles of NaF = 10.5 x 10^-3 / 42 = 2.5 x 10^-4
molarity of NaF = 2.5 x 10^-4 / 50 x 10^-3
= 5 x 10^-3 M
[F-] = 5 x 10^-3 M
a )
Ba+2 + 2F <----------------------> BaF2
for this solid equailbrium constant Ksp = [Ba+2] [F-]^2
Pb+2 + 2F- <-------------------> PbF2
for this salt equailbrium constant Ksp = [Pb+2] [F-]^2
b)
for BaF2 : Qsp = [Ba+2] [F-]^2
= 0.005 x (5 x 10^-3)^2
= 1.25 x 10^-7
for PbF2 : Qsp = [Pb+2][F-]^2
= 0.005 x (5 x 10^-3)^2
= 1.25 x 10^-7
c)
if Ksp < Qsp . the precipitate will form
PbF2 Ksp (3.6x10^-8) < PbF2 Qsp (1.25 x 10^-7)
so PbF2 will form precipitate
PbF2 ----------------> Pb+2 + 2 F-
Ksp = [Pb+2] [F-]^2
3.6x10^-8 = 0.005 x [F-]^2
[F-] = 2.68 x 10^-3 M
at [F-] = 2.68 x 10^-3 M PbF2 precipitate will form
Get Answers For Free
Most questions answered within 1 hours.