Question

If the minerals bismuthinite (Bi2S3) and stibnite (Sb2S3 are both in equilibrium with the same aqueous solutionn, what will the ratio of the concentrations of Bi3+ to the concentration of Sb3+ ({Bi3+}/{Sb3+}) in the solution? Assume that bismuth and antimony are only present as 3+ cations, that sulfer is only present as S2- and that the solution is ideal. The Solubility product (Ksp) for bismuthinite is 10^-100.02 and that for stibnite is 10^-90.81

Answer 1

We need --> Ratio = [Bi+3] / [Sb+3]

First, write down the equilbiriums of salt

Bi2S3(s) <-> 2Bi+3(aq) + 3S-2(aq)

Sb2S3(s) <-> 2Sb+3(aq) + 3S-2(aq)

Now, write down the Equilibirum epxression constants:

Ksp Bi2S3= [Bi+3]^2 * [S-2]^3

Ksp Sb2S3= [Sb+3]^2 * [S-2]^3

Note that

Ratio -->  [Bi+3] / [Sb+3]

Divide Ksp Bi2S3 by Sb2S3

Ksp Bi2S3 / Ksp Sb2S3 = ([Bi+3]^2 * [S-2]^3 )/([Sb+3]^2 * [S-2]^3)

Ksp Bi2S3 / Ksp Sb2S3 = ([Bi+3]/[Sb+3])^2

get rid of ^2 by sqrt

(Ksp Bi2S3 / Ksp Sb2S3)^0.5 = [Bi+3]/[Sb+3]

finally, substitute Ksp values

(10^-100.02) /(10^-90.81))^0.5 = [Bi+3]/[Sb+3]

(6.1659*10^-10)^0.5= [Bi+3]/[Sb+3]

0.00002483= [Bi+3]/[Sb+3]

[Bi+3]/[Sb+3] = 0.00002483