A solution with a pH=6.00 is 4.0E-3M in each of the metal cations Mn2+, Fe2+, Co2+, Ni2+, and Zn2+ and 0.10M in H2S. Under these conditions, which of the metal cations will be precipitated as a sulfide? Hint: Calculate IP.
General Equilibrium MS(s) + 2H3O+(aq) = M2+(aq) + 2H2O(l) + H2S(aq); Kspa
Metal Sulfide Kspa
MnS 3E10
FeS 6E2
CoS 3
NiS 8E-1
ZnS 3E-2
a.) Mn2+ and Fe2+
b.) Zn2+ only
c.) Zn2+ and Ni2+
d.) none of the above
I know the answer is d, from the back of my book, I just don't know how to solve it. Thanks a lot
To determine which metal cations will be precipitated as a sulfide, we need to compare their ion product (IP) to their respective Ksp values.
The ion product (IP) can be calculated using the concentrations of metal cations and the concentration of H2S. IP for a given metal sulfide is given as follows: IP = [M2+][H2S]^2
In this case, we have concentrations of metal cations Mn2+, Fe2+, Co2+, Ni2+, and Zn2+ as 4.0E-3M, and the H2S concentration is given as 0.10M.
Let's calculate the IP for each metal cation:
For MnS: IP of MnS = [Mn2+][H2S]^2 = (4.0E-3)(0.10)^2 = 4.0E-6
For FeS: IP of FeS = [Fe2+][H2S]^2 = (4.0E-3)(0.10)^2 = 4.0E-6
For CoS: IP of CoS = [Co2+][H2S]^2 = (4.0E-3)(0.10)^2 = 4.0E-6
For NiS: IP of NiS = [Ni2+][H2S]^2 = (4.0E-3)(0.10)^2 = 4.0E-6
For ZnS: IP of ZnS = [Zn2+][H2S]^2 = (4.0E-3)(0.10)^2 = 4.0E-6
Now, let's compare the IP values to the respective Ksp values:
For MnS: IP (4.0E-6) < Ksp (3E10) - No precipitation of MnS
For FeS: IP (4.0E-6) < Ksp (6E2) - No precipitation of FeS
For CoS: IP (4.0E-6) < Ksp (3) - No precipitation of CoS
For NiS: IP (4.0E-6) < Ksp (8E-1) - No precipitation of NiS
For ZnS: IP (4.0E-6) < Ksp (3E-2) - No precipitation of ZnS
Based on the calculations, none of the metal cations will be precipitated as a sulfide. Therefore, the correct answer is d) none of the above.