Determine the molar solubility of #PbCl_2# in 0.16 M #HCl(aq)#?
Use #K_(sp)# for #PbCl_2# and #K_f# for #[PbCl_3]^(−)# to determine the molar solubility of #PbCl_2# in 0.16 M #HCl(aq)# .
[Hint: What is the total concentration of lead species in solution?]
Use
[Hint: What is the total concentration of lead species in solution?]
1 Answer
Here's what I got.
Explanation:
The idea here is that the solubility of lead(II) chloride in a hydrochloric acid solution is affected because of the presence of the chloride anions and of the formation of the
When you dissolve lead(II) chloride in water, the following equilibrium is established
#"PbCl"_ (2(s)) rightleftharpoons "Pb" _ ((aq))^(2+) + 2"Cl"_ ((aq))^(-)#
The solubility product constant,
#K_(sp) = 5.89 * 10^(-5)#
https://en.wikipedia.org/wiki/Lead(II)_chloride
Now, hydrochloric acid is a strong acid, which implies that it ionizes completely in aqueous solution to produce hydronium cations and chloride anions in a
#"HCl"_ ((aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#
As you can see, a hydrochloric acid solution contains an excess of chloride anions. This will decrease the solubility of the salt when compared to its solubility in pure water because of the excess chloride anions present in solution
When you dissolve lead(II) chloride in this solution, some the excess chloride anions will combine with the lead(II) cations to form
#"Pb"_ ((aq))^(2+) + 3"Cl"_ ((aq))^(-) rightleftharpoons "PbCl"_ (3(aq))^(-)#
The formation constant for this complex ion,
#K_f = 2.4 * 10^1#
http://occonline.occ.cccd.edu/online/jmlaux/Kf%20Table%20App.%20D.pdf
Notice that
So, you know that two equilibrium reactions are established when lead(II) chloride is dissolved in hydrochloric acid.
Notice that the second equilibrium consumes lead(II) cations and produces
This will cause the first equilibrium to shift to the right in order to produce more lead(II) cations in solution, i.e. more lead(II) chloride will dissolve
This is why the solubility of the salt increases in hydrochloric acid.
You can thus say that you have
#color(white)(aaaaaa.)"PbCl"_ (2(s)) rightleftharpoons color(red)(cancel(color(black)("Pb" _ ((aq))^(2+)))) + 2"Cl"_ ((aq))^(-)" "K_(sp)#
#color(red)(cancel(color(black)("Pb" _ ((aq))^(2+)))) + 3"Cl"_ ((aq))^(-) rightleftharpoons "PbCl"_ (3(aq))^(-)" "K_f#
#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#
#"PbCl"_ (2(s)) + "Cl"_ ((aq))^(-) rightleftharpoons "PbCl"_ (3(aq))^(-)" "K = K_(sp) xx K_f#
Remember, when you add two equilibrium reactions, you have to multiply their equilibrium constants.
#K = K_(sp) * K_f#
#K = 5.89 * 10^(-5) * 2.4 * 10^1 = 1.414 * 10^(-3)#
Notice that
By definition, the equilibrium constant for this combined equilibrium will be
#K = (["PbCl"_3^(-)])/(["Cl"^(-)]) " "color(darkorange)("(*)")#
Keep in mind that we do not include the concentration of solids into the expression of the equilibrium constant!
So, you know that
#["Cl"^(-)]_0 = ["HCl"] = "0.16 M"#
and
#["PbCl"_3^(-)]_0 = "0 M" -># you don't have any complex ions present before the reaction
Now, look at the balanced chemical equation that describes the combined equilibrium
#"PbCl"_ (2(s)) + "Cl"_ ((aq))^(-) rightleftharpoons "PbCl"_ (3(aq))^(-)"#
In order for
If you take
#["Cl"^(-)] = ["Cl"^(-)]_0 - s -># the reaction consumes chloride anions
and
#["PbCl"_3^(-)] = 0 + s -># the reaction produces complex ions
According to equation
#1.414 * 10^(-3) = s/(0.16 - s)#
Rearrange to solve for
#2.26 * 10^(-4) - 1.414 * 10^(-3) * s = s#
#s * (1 + 1.414 * 10^(-3)) = 2.26 * 10^(-4)#
This will get you
#s = (2.26 * 10^(-4))/(1.001414) = 2.256 * 10^(-4)#
Therefore, you can say that the molar solubility of lead(II) chloride in this solution will be equal to
#color(darkgreen)(ul(color(black)("molar solubility" = 2.3 * 10^(-4)"M")))#
The answer is rounded to two sig figs.
SIDE NOTE The molar solubility of lead(II) chloride in pure water is equal to
#"PbCl"_ (2(s)) rightleftharpoons "Pb" _ ((aq))^(2+) + color(red)(2)"Cl"_ ((aq))^(-)#
#K_(sp) = s * (color(red)(2)s)^color(red)(2) = 4s^3#
So
#s = root(3)(K_(sp)/4) = root(3) ((5.89 * 10^(-5))/4) = 2.45 * 10^(-2)"M"#
As you can see, the molar solubility of the salt decreased in the hydrochloric acid solution.
Moreover, the fact that you have