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)#
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
By definition, the solubility product constant for this equilibrium looks like this
#K_(sp) = ["Fe"^(2+)] * ["S"^(2-)]#
Now, let's say that
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,
#"Na"_ 2"S"_ ((aq)) -> 2"Na"_ ((aq))^(+) + "S"_ ((aq))^(2-)#
Since sodium sulfide dissociates in a
#["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
#s = 2.776 * 10^(-17)#
Now, we've said that
In other words, you can only hope to dissolve
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.