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

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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)#

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.

##### 1 Answer

#### 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 **mole** of iron(II) cations and **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 **that dissociates** in water to produce a **saturated solution** of iron(II) sulfide, i.e. the number of moles per

According to the equilibrium equation, if

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

However, you're not dealing with pure water here. Your solution contains sodium sulfide, **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 **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"^(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 *concentration*, you can discard the negative solution and say that

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

Now, we've said that **that dissociate** in **molar solubility** of the salt at the given temperature.

In other words, you can only hope to dissolve **moles** of iron(II) sulfide in

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**.