# Question 014bf

Jun 6, 2017

${\text{3.3 g L}}^{- 1}$

#### Explanation:

For starters, look up the solubility product constant, ${K}_{s p}$, for lead(II) chloride, ${\text{PbCl}}_{2}$

${K}_{s p} = 1.6 \cdot {10}^{- 5}$

http://bilbo.chm.uri.edu/CHM112/tables/KspTable.htm

Your strategy here will be to calculate the molar solubility of the salt, which is the molar concentration of lead(II) chloride that dissociates in order to produce a saturated solution at room temperature.

You know that lead(II) chloride is considered insoluble in water, so you can say that the solubility equilibrium that is established in aqueous solution will lie to the left.

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

Notice that every $1$ mole of lead(II) chloride that dissociates produces $1$ mole of lead(II) cations and $\textcolor{red}{2}$ moles of chlride anions.

You can thus say that, at equilibrium, you have

$\left[{\text{Cl"^(-)] = color(red)(2) * ["Pb}}^{2 +}\right]$

If you take $s$ to be the equilibrium concentration of the lead(II) cations, i.e. the molar solubility of the salt, you can write the solubility product constant

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

as

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

This is equivalent to

${K}_{s p} = 4 {s}^{3}$

Rearrange to solve for $s$

$s = \sqrt{{K}_{s p} / 4}$

Plug in your value to find

$s = \sqrt{\frac{1.6 8 {10}^{- 5}}{4}} = 0.0159$

You can thus say that a saturated lead(I)) chloride solution will have

["Pb"^(2+)] = "0.0159 M"

As you know, molarity tells you the number of moles of solute present in $\text{1 L}$ of solution. This implies that $\text{1 L}$ of saturated lead(II) chloride solution will contain $0.0159$ moles of lead(II) cations.

To convert this to grams, use the element's molar mass

0.0159 color(red)(cancel(color(black)("moles Pb"^(2+)))) * "207.2 g"/(1color(red)(cancel(color(black)("mole Pb"^(2+))))) = "3.3 g"#

Therefore, the concentration of the lead(II) cations will be equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{concentration Pb"^(2+) = "3.3 g L}}^{- 1}}}}$

The answer is rounded to two sig figs.