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J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry 4th edition

Problem 16.96

The text gives the following equation for the precipitation of a metal sulfide in the qualitative analysis scheme:

M2+(aq) + H2S(aq) + 2 H2O(l) MS(s) + 2 H3O+(aq)

Use the description of the solubility product constant of MS(s) given in a footnote in Table C.2C (Appendix C), together with the relevant data, to show that the equilibrium constant for the precipitation reaction has the form: Kc = KwKa1/Ksp. What is the value of Kc if the metal sulfide is FeS?




Answer:

This problem requires the use of the Law of Multiple Equilibria.
The footnote describes the reaction:
MS(s) + H2O(l)M2+(aq) + HS(aq) + OH(aq) (1) Ksp
Reversing the reaction gives:
M2+(aq) + HS(aq) + OH(aq)MS(s) + H2O(l) (2) Kc = 1/Ksp
Using the hydrolysis of the weak acid H2S:
H2S(aq) + H2O(l)H3O+(aq) + HS(aq) (3) Ka1
The water autoionization equilibrium is:
2 H2O(l)H3O+(aq) + OH(aq) (4) Kw
Adding reactions (2), (3), and (4) gives the desired reaction Multiply equilibrium constants:
M2+(aq) + H2S(aq) + 2 H2O(l)MS(s) + 2 H3O+(aq) Kc = KwKa1/Ksp

For the specific example of FeS:

Ksp = 6×10–19

Ka1 = 1.0×10–7

Kw = 1.0×10–14

so Kc = (1.0×10–14)(1.0×10–7)/(6×10–19) = 2×10–3