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

^{2}α

_{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

_{2}O

PbOH

^{+} + H

^{+} lgβ

_{3}=-7.8

CrO

_{4}^{2-} + H

_{2}O

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

_{4}^{2-}]+[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?