[diablo]

Hi,

got a prob to solve this:

A solution (25°C) contains Zn(II) and Pb(II) ions ( c_zn= 1,0 and c_pb = 0,1 mol/l)
Now H₂s is added to precipitate the Pb(II) as Pbs
The solubility products (in terms of activities) of ZnS and PbS = 2x10^-24 and1x10^-28 and equilibrium konstant for reaction H₂S = 2H +S
= 1x10^-21

 --> calculate the c(H₂s) to precipitate the maximum amount of PbS whilst at the same time not precipitating any ZnS - also calculate % of Pb(II) precipitated as Pbs at this H₂S concentration...
Further there is a hint: caculate using the solubility product of ZnS then use this to calculate the [H₂S using the equil.konst. Second calculate free[Pb] based on the previous calc. and the solub. product of PbS

Actually i dont know how to solve this prob (even with this hints):

Calculate due to hint would be: 2x10^-24 = c[Zn] x c so 2x10^-24 / c[Zn] = c
Then i calculate [H₂S] = 1x10^-21 / c (is that right?)
and then i don't know how to calculate the free [Pb] and the rest respectively...

Hope anyone can help me

thx
Diabloo

[AWK]

c(Zn²⁺)=c(S^2-)=SQRT(2x10^-24)= 1.41x10^-12

[diablo]

Thx for your answer !

are you sure that this is true ? ( cause c_zn = 1,0 mol/L ?)

greets diabloooo

[AWK]

c(Zn²⁺)=c(S^2-)=SQRT(2x10^-24)= 1.41x10^-12

My post with corrections was lost during forum upgrade. Above answer is competely wrong.
From Ksp of ZnS calculate concentration of S^2-, at which precipitation of ZnS can start.
Then from K1xK2=[H⁺]²[S^2-/[H₂S] you can calculate an equlibrium concentration of H2S
From the same concentration of S^2- you can calculate athe equilibrium concetration of Pb²⁺ then percentage of precipitation of Pb