# 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. ${K}_{s p} = 4.9 \cdot {10}^{- 18}$

Feb 23, 2018

$2.4 \cdot {10}^{- 15}$ $\text{g}$

#### 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.

${\text{FeS"_ ((s)) rightleftharpoons "Fe"_ ((s))^(2+) + "S}}_{\left(a q\right)}^{2 -}$

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

${K}_{s p} = \left[{\text{Fe"^(2+)] * ["S}}^{2 -}\right]$

Now, let's say that $s$ $\text{M}$ 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 $\text{1 L}$ of the solution.

According to the equilibrium equation, if $s$ $\text{M}$ of the salt dissociates to produce ions, then the resulting solution will contain $s$ $\text{M}$ of iron(II) cations and $s$ $\text{M}$ 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.

${K}_{s p} = s \cdot s = {s}^{2}$

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 $s$ $\text{M}$.

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

${\text{Na"_ 2"S"_ ((aq)) -> 2"Na"_ ((aq))^(+) + "S}}_{\left(a q\right)}^{2 -}$

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 $s$ $\text{M}$ will dissociate to produce $s$ $\text{M}$ of iron(II) cations and $s$ $\text{M}$ of sulfide anions, but since the solution already contains some sufide anions, their equilibrium concentration will be

${\left[{\text{S"^(2-)] = ["S}}^{2 -}\right]}_{0} + s$

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

${K}_{s p} = s \cdot \left(0.20 + s\right)$

which is equivalent to

${s}^{2} + 0.20 \cdot s - 4.9 \cdot {10}^{- 18} = 0$

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

$s = 2.776 \cdot {10}^{- 17}$

Now, we've said that $s$ $\text{M}$ represents the number of moles of iron(II) sulfide that dissociate in $\text{1 L}$ 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 $2.776 \cdot {10}^{- 17}$ moles of iron(II) sulfide in $\text{1 L}$ 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.