Question

Question 6 (Extra Credit Challenge):

When the sparingly soluble salt lead fluoride, PbF2, is added to an acidic solution with [H+] = 1.6 x 10–4 M, the resulting concentration of [Pb2+] after the salt dissolves is determined to be 7.75 x 10–5 M.

(a) Determine the equilibrium constant for the dissolving of PbF2 in an acidic solution of H3O+ and

(b) given Ka = 6.8 x 10–4, use the value determined in part a to calculate the Ksp for PbF2 through manipulation of the equilibrium equations. Show work.  

Answer 1

When the sparingly soluble salt lead fluoride, PbF2, is added to an acidic solution with [H+] = 1.6 x 10–4 M, the resulting concentration of [Pb2+] after the salt dissolves is determined to be 7.75 x 10–5 M.

(a) Determine the equilibrium constant for the dissolving of PbF2 in an acidic solution of H3O+.

we have,    PbF2 (s) <===> Pb2+ (aq) + 2 F- (aq)

Change        -x                        +x             + 2x

K = [Pb2+ ] [ F- ]2 / [PbF2 (s)]

[PbF2 (s)] ~ 1

S = [Pb2+] = 7.75 x 10–5 M    ===> [ F- ] = 2S = 7.75 x 10–5 M *2 = 1.55 x 10–4 M

K = [Pb2+ ] [ F- ]2 = 1.9*10-12

(b) given Ka = 6.8 x 10–4, use the value determined in part a to calculate the Ksp for PbF2 through manipulation of the equilibrium equations.

F- is a weak base ,

HF (aq) + H2O (l) <===> F- (aq) + H3O+ ; Ka = 6.8 x 10–4

we have mass balance equation , 2* [Pb2+ ] = [HF] + [F-] = 1.55 x 10–4 M ........1

Ka = [H+][F-]/[HF]
6.8 x 10–4 = 1.6 x 10–4 M*[F-]/[HF]
[HF] = 0.2353 [F-]

from 1 : 0.2353 [F-] + [F-] = 1.55 x 10–4 M   ==>   [F-] = 1.25 x 10–4 M  

thus, Ksp = [Pb2+ ] [ F- ]2 = 7.75 x 10–5 * ( 1.25 x 10–4 ) 2   = 1.22* 10-12