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Given a solution with a $\ce{S^{2-}}$ (sulfide ion) concentration of $2.76 \times 10^{-15}$ mol per litre, and $K_s(\ce{FeS})=4.90 \times 10^{-18}$, calculate the solubility of $\ce{FeS}$ in this solution.

I calculated this by letting the concentration of $\ce{Fe^{2+}} = x$, therefore the concentration of $\ce{S^{2-}} = 2.76 \times 10^{-15} + x$.

Therefore $4.90 \times 10^{-18} = x(2.76 \times 10^{-15} + x)$.

Upon solving this as a quadratic equation, I arrive at the result of $2.14 \times 10^{-9}$ mol per litre. [EDIT: $2.21 \times 10^{-9}$ mol per litre]

The answers to this problem state to ignore the contribution of $\ce{S^{2-}}$ from the $\ce{FeS}$ as it is negligible, and state that the answer is actually $1.78\times10^{-3}$ mol per litre.

Which method is correct?

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The value of the answer you were given is incorrect, both in its value and its explanation of what can be ignored.

As you note:

$K_{sp} = 4.90\times10^{-18}=[\ce{Fe^{2+}}][\ce{S^{2-}}]$

Just eyeballing the equation, we would expect a solution of $\ce{FeS}$ in pure water to contain equal quantities of iron(II) and sulfide ions, at a concentration of $2.21\times10^{-9} M$ (the square root of the solubility product.) This is on the order of a million times more concentrated than the solution described in the problem, so we can ignore the sulfide concentration in the original solution. For the purposes of this problem, the iron sulfide is dissolving in pure water.

Additionally, solving the quadratic equation you provided gives much the same result. The solution rounds to $x = 2.21\times10^{-9} M$; it only differs in the sixth place from the square root of the solubility product. It's easy enough to make a small numerical error with scientific notation and solving a quadratic; I have to assume that's what happened to you here.

The only way that your instructor's answer would be correct would be if the question intended us to understand that there was some chemistry going on to keep the concentration of sulfide ions at the given concentration. Without that, it just doesn't make sense.

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  • $\begingroup$ Yes, solving the quadratic again does result in x=2.21×10^−9. Thank you, a bit shocking considering the question is from an example exam paper. $\endgroup$
    – Nic Steyn
    Nov 10, 2015 at 4:10
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Given a solution with a S2- (sulfide ion) concentration of 2.76x10^-15 mol per litre, and Ks(FeS)=4.90x10^-18, calculate the solubility of FeS in this solution.

(a) $\ce{FeS_{(s)} <-> [Fe^{2+}] [S^{2-}] \quad\quad K_{sp} =} 4.90*10^{-18}$

Given that $\ce{[S^{2-}]}$ is specified as 2.76x10^-15 mol there is no ambiguity in the $\ce{[S^{2-}]}$ concentration. So:

(b) $\ce{[Fe^{2+}]} = \dfrac{4.90*10^{-18}}{\ce{[S^{2-}]}} = \dfrac{4.90*10^{-18}}{2.76*10^{-15}} = 1.78*10^{-3}$

Chemically what is means is that something else must be reacting in the solution to remove almost all of the $\ce{[S^{2-}]}$ that the dissolution of the iron sulfide liberates. The book statement to "ignore" the sulfide that the dissolution creates is very misleading.

The gist I believe is this. Hydrogen sulfide, $\ce{H2S}$ has two protons and would seem to have two pKa values. The first depprotonation is what would be expected with $pK_{a1} \approx 7$ in aqueous solution. The second deprotonation doesn't "really happen" in water. The $pK_{a2} > 14$ which means that the $\ce{[S^{2-}]}$ ion is a very strong Arrhenius base in water.

So there is another reaction happening which removes most of the $\ce{[S^{2-}] species:

(c) $\ce{[S^{2-}] + H2O -> HS^{-} + OH^{-}}$

For calculation purposes the second deprotonation has an effective $pK_{a2} = 17.4$ in a paper by Migdisov et al.

Migdisov, A.A., Williams-Jones, A.E., Lakshtanov, L.Z., Alekhin, Y.V., 2002. Estimates of the second dissociation constant of H2S from the surface sulfidation of crystalline sulfur. Geochimica et Cosmochimica Acta 66 (10), 1713–1725.

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