Precipitation?

Question

A solution is 0.0100 M in each of the metal ions in the table. HCl is added to the solution such the H30^+ concentration is 0.300 M. H2S gas is bubbled through the solution to attain a H2S concentration of 0.100 M. Predict which of the sulfides precipitate under the given conditions.

Metal-------Sulfide Kspa
FeS---------- 6 x 10^2
NiS----------- 8 x 10^-1
PbS---------- 3 x 10^-7
CuS---------- 6 x 10^-16

Which ones will precipitate?

Answer 1

All of theses sulfides are of the type MS with a reaction for dissolving MS <=> M+2 + S-2 and
Ksp = [M+2][S-2]

so the minimum concentration of [S-2] needed for each one to start precipitating is

[S-2]= Ksp/[[M+2]

FeS: [S-2]= 6*10^2 /10^-2 = 6*10^4 M
NiS: [S-2]= 5*10^-1 /10^-2= 50 M
PbS : [S-2] =3*10^-7/ 10^-2= 3*10^-5 M
CuS: [S-2]=6*10^-16/10^-2 = 6*10^-14 M

The values of [S-2] are so big that you know they will not precipitate without calculating how much S-2 you actually have. For the other two we need to know.

H2S is a diprotic acid with Ka1=9.1*10^-8 and Ka2=1.1*10^-12 according to

http://ifs.massey.ac.nz/resources/chemistry/dissociation/inorgacids.htm

You should provide the Ka1, Ka2 to get the answer you are looking for, since on the net you can find a bit different values for them.

If you really want to be accurate it is proven that for a diprotic acid (for derivation of the equation look at http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-acid-base where Ka12= Ka1*Ka2)

[S-2] = Ka1*Ka2*Ca / ([H+]^2+Ka1[H+] + Ka1*Ka2)

we assume that since the dissociation of H2S is small, the [H+] is practically the same as before the addition of H2S (0.3 M) thus

[S-2] = (9.1*10^-8)* (1.1*10^-12)*0.1 /(0.3^2+ 0.3*9.1*(10^-8)+(9.1*10^-8) *(1.1*10^-12) ) =1.11 *10^-19 M

So none will precipitate under these conditions and for these Ka1,Ka2 values..

Answer 2

As per Monica:

"CuS and PbS will precipitate.
FeS and NiS will not precipitate."