Question

How many grams of Iron(II) sulfide will dissolve per liter in a `"0.20-M"` sodium sulfide solution?

I don't get how to do this type of problem, I can do other Ksp stuff but don't understand the process for this.

`K_(sp) = 4.9 * 10^(-18)`

Answer 1

`2.4 * 10^(-15)` `"g"`

Explanation:

You know that iron(II) sulfide is considered insoluble in water, which implies that when you dissolve this salt in water, a dissociation equilibrium is established between the undissolved solid and the dissolved ions.

`"FeS"_ ((s)) rightleftharpoons "Fe"_ ((s))^(2+) + "S"_ ((aq))^(2-)`

Notice that every mole of iron(II) sulfide that dissociates produces `1` mole of iron(II) cations and `1` mole of sulfide anions.

By definition, the solubility product constant for this equilibrium looks like this

`K_(sp) = ["Fe"^(2+)] * ["S"^(2-)]`

Now, let's say that `s` `"M"` represents the concentration of iron(II) sulfide that dissociates in water to produce a saturated solution of iron(II) sulfide, i.e. the number of moles per `"1 L"` of the solution.

According to the equilibrium equation, if `s` `"M"` of the salt dissociates to produce ions, then the resulting solution will contain `s` `"M"` of iron(II) cations and `s` `"M"` of sulfide anions.

You can thus say that in pure water, the expression of the solubility product constant will be--I'll write the equation without added units.

`K_(sp) = s * s = s^2`

This is the case because pure water doesn't contain any iron(II) cations or sulfide anions, so the two ions will have an equal concentration of `s` `"M"`.

However, you're not dealing with pure water here. Your solution contains sodium sulfide, `"Na"_2"S"`, a soluble ionic compound that exists as ions in aqueous solution.

`"Na"_ 2"S"_ ((aq)) -> 2"Na"_ ((aq))^(+) + "S"_ ((aq))^(2-)`

Since sodium sulfide dissociates in a `1:1` mole ratio to produce sulfide anions, you can say that your solution has

`["S"^(2-)]_ 0 = ["Na"_ 2"S"] = "0.20 M"`

So now when you add the iron(II) sulfide, you say that `s` `"M"` will dissociate to produce `s` `"M"` of iron(II) cations and `s` `"M"` of sulfide anions, but since the solution already contains some sufide anions, their equilibrium concentration will be

`["S"^(2-)] = ["S"^(2-)]_ 0 + s`

which is

`["S"^(2-)] = overbrace("0.20 M")^(color(blue)("already in solution")) + overbrace(s quad "M")^(color(blue)("what dissociates"))`

`["S"^(2-)] = (0.20 + s) quad "M"`

So in this case, the expression of the solubility equilibrium constant looks like this

`K_(sp) = s * (0.20 + s)`

which is equivalent to

`s^2 + 0.20 * s - 4.9 * 10^(-18) = 0`

This quadratic equation will produce two solutions, one positive and one negative. Since `s` is supposed to represent concentration, you can discard the negative solution and say that

`s = 2.776 * 10^(-17)`

Now, we've said that `s` `"M"` represents the number of moles of iron(II) sulfide that dissociate in `"1 L"` of solution to produce a saturated solution of iron(II) sulfide, i.e. the molar solubility of the salt at the given temperature.

In other words, you can only hope to dissolve `2.776 * 10^(-17)` moles of iron(II) sulfide in `"1 L"` of solution to have a saturated solution of this salt.

To find the solubility in grams per liter, use the molar mass of iron(II) sulfide

`(2.776 * 10^(-17) color(red)(cancel(color(black)("moles FeS"))))/("1 L solution") * "87.91 g"/(1color(red)(cancel(color(black)("mole FeS")))) = color(darkgreen)(ul(color(black)(2.4 * 10^(-15) quad "g L"^(-1))))`

The answer is rounded to two sig figs.