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
sorry about this second post,
something was wrong with my computer
I don't know the meaning of IP and Kspa but I would do the problem this way.
H2S ==> 2H^+ + S^-2
k1k2 = (H^+)^2(S^-2)/(H2S).
You know pH, convert that to (H^+). You know (H2S). Use the above k1k2 expression to solve for sulfide.
Then, use that sulfide to calculate the (M^+2)(S^-2)= ?? and compare that number with Ksp (which may not be the same as Kspa since I don't know the meaning of Kspa. If the ?? number is higher than Ksp, there will be a ppt. If not, no ppt. Post your work if you get stuck and please let me know the meaning of IP and Kspa.
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