# Question 0e2fd

Aug 27, 2015

For part (b) $s = 4.0 \cdot {10}^{- 4} \text{M}$

#### Explanation:

I will show you how to solve point (b) and leave point (a) for you to solve as practice.

So, you want to know how what the molar solubility of lead (II) chloride, ${\text{PbCl}}_{2}$, is in a solution that contains calcium chloride, ${\text{CaCl}}_{2}$.

The idea here is that the solubility of lead (II) chloride will actually decrease in this solution because of the common-ion effect.

Calcium chloride is a soluble salt, it will dissociate completely in aqueous solution to form

${\text{CaCl"""_text(2(s]) -> "Ca"_text((aq])^(2+) + color(red)(2)"Cl}}_{\textrm{\left(a q\right]}}^{-}$

Lead (II) chloride is only sparingly soluble in aqueous solution, which means that its dissociation is actually an equilibrium

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

In essence, the molar solubility of lead (II) chloride will decrease because the solution already contains the chloride anions, ${\text{Cl}}^{-}$ that resulted from the dissolution of calcium chloride.

You can use an ICE table to find the molar solubility of lead (II) chloride. Before doing that, calculate what the concentration of the chloride anions will be before adding the lead (II) chloride.

Notice that you have a $1 : \textcolor{red}{2}$ mole ratio between calcium chloride and chloride anions. This means that every mole of calcium chloride will produce 2 moles of chloride anions.

Since you get twice as many moles of chloride anions than you have of calcium chloride in the same volume, the concentration of the chloride anions will be

["Cl"^(-)] = 2 * ["CaCl"""_2] = 2 * "0.10 M" = "0.20 M"

Now write the ICE table - $s$ represents the molar solubility of lead (II) chloride

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

color(purple)("I")" " " "-" " " " " " " " " "0" " " " " " " " " "0.20
color(purple)("C")" " " "-" " " " " " " " (+s)" " " " " " "(+color(blue)(2)s)
color(purple)("E")" " " "-" " " " " " " " " "s" " " " " " "(0.20+color(blue)(2)s)

By definition, the solubility product constant, ${K}_{s p}$, for lead (II) chloride will be

${K}_{s p} = {\left[{\text{Pb"^(2+)] * ["Cl}}^{-}\right]}^{\textcolor{b l u e}{2}}$

K_(sp) = s * (0.20 + 2s)""^2#

Because ${K}_{s p}$ is very small, you can approximate $\left(0.20 + 2 s\right)$ with $0.20$. This will get you

$1.6 \cdot {10}^{- 5} = s \cdot 0.04 \implies s = \frac{1.6 \cdot {10}^{- 5}}{0.04} = \textcolor{g r e e n}{4.0 \cdot {10}^{- 4} \text{M}}$

SIDE NOTE If you choose not to use that approximation, you will end up with a cubic equation that will produce three values for $s$, one positive and two negative.

The only one that has any chemical significant is the positive one, which is equal to

$s = 3.97 \cdot {10}^{- 4} \text{M}$

As you can see, rounding this value to two sig figs, the number of sig figs you gave for the concentration of calcium chloride, the answer will be the same

$s = 4.0 \cdot {10}^{- 4} \text{M}$