Question

A solution has a Pb2+ concentration of 1.9 x 10-3 M, and an F-1 concentration of 3.8 x 10-3 M. The value of Ksp for PbF2 at room temperature is 4 x 10-8. Will this solution form a precipitate?
The solubility equilibrium equation is as follows.



.

Yes

No

Answer 1

A solution has a Pb2+ concentration of 1.9 x 10^-3, and an F-1 concentration of 3.8 x 10^-3 M. The value of Mps for Pbf2 at room temperature is 4 x 10^-8. This solution will NOT form a precipitate.
The solubility equilibrium equation is as follows.
PbF2 (s) <--> Pb2 (aq) + 2F- (aq)

Answer 2

We are going to get the value of Qsp and compare it with the Ksp value:

when Qsp is the concentration of the initial products:

Qsp = [Pb2+][F-]^2

when the [Pb2+] = 1.9 x 10^-3

and [F-] = 3.8 x 10^-3

by substitution:

∴Qsp = (1.9 x 10^-3) * (3.8 x 10^-3)^2

           = 2.7 x 10^-8 

when Ksp = 4 x 10^-8

so Qsp < Ksp

∴ precipitate will not be formed