Hi,
I have a problem(4, page 25): Calculate the product of solubilty of PbCrO
4 , if its solubilty in 1M AcOH is 2,9*10
5.
Here's how I dealt with it: Ks=S
2α
CrO42-α
Pb2+, where S is the aparent solubility, and α the coefficients, defined as the raport of the species it denote and the total concentration (protonated, free, complexated etc) of the species.
I had to take into consideration the following equilibriums:
AcOH + Pb
2+ AcOPb
+ + H
+ , lgβ
AcOPb+=2.68;
2 AcOH + Pb
2+ Ac(OPb)
2 + 2H
+ lgβ
(AcO)2Pb=4.08
Pb
2+ + H
2O
PbOH
+ + H
+ lgβ
3=-7.8
CrO
42- + H
2O
HCrO
4- + HO
-.
pKa
HCrO4-=6.5, we are also given pKa
AcOH=4.76
I did the matter balance: S= [Pb
2+]+[PbOH
+] + [PbAcO
+]+ [Pb(AcO)
2] , and also S= [CrO
42-]+[HCrO
4-]. Assuming the ph stays constant(PH=2.38, 1M AcOH solution),
I have obtained α
CrO42-=7.5852*10
-5.
α
Pb2+=1.44519*10
-9.
Ks≈10
-23, when it actually is ≈10
-14. Where am I wrong?