# Determine the molar solubility of PbCl_2 in 0.16 M HCl(aq)?

## Use ${K}_{s p}$ for $P b C {l}_{2}$ and ${K}_{f}$ for [PbCl_3]^(−) to determine the molar solubility of $P b C {l}_{2}$ in 0.16 M $H C l \left(a q\right)$. [Hint: What is the total concentration of lead species in solution?]

Apr 1, 2017

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 ${\text{PbCl}}_{3}^{-}$ complex ions.

When you dissolve lead(II) chloride in water, the following equilibrium is established

${\text{PbCl"_ (2(s)) rightleftharpoons "Pb" _ ((aq))^(2+) + 2"Cl}}_{\left(a q\right)}^{-}$

The solubility product constant, ${K}_{s p}$, for this solubility equilibrium is equal to

${K}_{s p} = 5.89 \cdot {10}^{- 5}$

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 $1 : 1$ mole ratio

${\text{HCl"_ ((aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

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 $\to$ think common-ion effect.

When you dissolve lead(II) chloride in this solution, some the excess chloride anions will combine with the lead(II) cations to form ${\text{PbCl}}_{3}^{-}$ complex ions

${\text{Pb"_ ((aq))^(2+) + 3"Cl"_ ((aq))^(-) rightleftharpoons "PbCl}}_{3 \left(a q\right)}^{-}$

The formation constant for this complex ion, ${K}_{f}$, is equal to

${K}_{f} = 2.4 \cdot {10}^{1}$

http://occonline.occ.cccd.edu/online/jmlaux/Kf%20Table%20App.%20D.pdf

Notice that ${K}_{f} > 1$, but that its value is not that large. This implies that the solubility of the salt will still decrease in this solution compared with its solubility in pure water.

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 ${\text{PbCl}}_{3}^{-}$ ions. Consequently, the concentration of lead(II) cations will decrease in solution.

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 $\to$ think Le Chatelier's Principle here.

This is why the solubility of the salt increases in hydrochloric acid.

You can thus say that you have

$\textcolor{w h i t e}{a a a a a a .} \text{PbCl"_ (2(s)) rightleftharpoons color(red)(cancel(color(black)("Pb" _ ((aq))^(2+)))) + 2"Cl"_ ((aq))^(-)" } {K}_{s p}$

color(red)(cancel(color(black)("Pb" _ ((aq))^(2+)))) + 3"Cl"_ ((aq))^(-) rightleftharpoons "PbCl"_ (3(aq))^(-)" "K_f
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$

$\text{PbCl"_ (2(s)) + "Cl"_ ((aq))^(-) rightleftharpoons "PbCl"_ (3(aq))^(-)" } K = {K}_{s p} \times {K}_{f}$

Remember, when you add two equilibrium reactions, you have to multiply their equilibrium constants.

$K = {K}_{s p} \cdot {K}_{f}$

$K = 5.89 \cdot {10}^{- 5} \cdot 2.4 \cdot {10}^{1} = 1.414 \cdot {10}^{- 3}$

Notice that $K < 1$, which means that lead(II) chloride will have a very low solubility in this solution.

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

$\text{PbCl"_ (2(s)) + "Cl"_ ((aq))^(-) rightleftharpoons "PbCl"_ (3(aq))^(-)}$

In order for $1$ mole of ${\text{PbCl}}_{3}^{-}$ to be produced, the reaction must consume $1$ mole of lead(II) chloride and $1$ mole of chloride anions.

If you take $s$ to be the molar solubility of lead(II) chloride in this solution, i.e. the maximum number of moles that can be dissolved per liter of solution, you will get

${\left[{\text{Cl"^(-)] = ["Cl}}^{-}\right]}_{0} - s \to$ the reaction consumes chloride anions

and

$\left[{\text{PbCl}}_{3}^{-}\right] = 0 + s \to$ the reaction produces complex ions

According to equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$, you will have

$1.414 \cdot {10}^{- 3} = \frac{s}{0.16 - s}$

Rearrange to solve for $s$

$2.26 \cdot {10}^{- 4} - 1.414 \cdot {10}^{- 3} \cdot s = s$

$s \cdot \left(1 + 1.414 \cdot {10}^{- 3}\right) = 2.26 \cdot {10}^{- 4}$

This will get you

$s = \frac{2.26 \cdot {10}^{- 4}}{1.001414} = 2.256 \cdot {10}^{- 4}$

Therefore, you can say that the molar solubility of lead(II) chloride in this solution will be equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{molar solubility" = 2.3 * 10^(-4)"M}}}}$

The answer is rounded to two sig figs.

SIDE NOTE The molar solubility of lead(II) chloride in pure water is equal to

${\text{PbCl"_ (2(s)) rightleftharpoons "Pb" _ ((aq))^(2+) + color(red)(2)"Cl}}_{\left(a q\right)}^{-}$

${K}_{s p} = s \cdot {\left(\textcolor{red}{2} s\right)}^{\textcolor{red}{2}} = 4 {s}^{3}$

So

$s = \sqrt[3]{{K}_{s p} / 4} = \sqrt[3]{\frac{5.89 \cdot {10}^{- 5}}{4}} = 2.45 \cdot {10}^{- 2} \text{M}$

As you can see, the molar solubility of the salt decreased in the hydrochloric acid solution.

Moreover, the fact that you have $K < 1$ for the overall equilibrium tells you that the formation of the complex ions does not improve the solubility of the salt in this solution in any significant manner.