Question

# 10.5mg of solid NaF is added to a 50mL solution that is 0.0050M in Ba(NO3)2 and...

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

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