# Question a8b65

May 4, 2017

Here's what I got.

#### Explanation:

To make the calculations easier, let's assume that you're working with a $\text{1.0-L}$ sample of this solution.

By definition, a solution's mass by volume percent concentration, $\text{m/v %}$, tells you the number of grams of solute present for every $\text{100 mL}$ of solution.

In your case, a $\text{1.5% m/v}$ lead(II) chloride solution will contain $\text{1.5 g}$ of lead(II) chloride, the solute, for every $\text{100 mL}$ of solution.

This means that your sample will contain

1.0 color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1.5 g PbCl"_2/(100color(red)(cancel(color(black)("mL solution")))) = "15 g PbCl"_2

Convert the number of grams of lead(II) chloride to moles by using the compound's molar mass

15 color(red)(cancel(color(black)("g"))) * "1 mole PbCl"_2/(278.1color(red)(cancel(color(black)("g")))) = "0.05394 moles PbCl"_2

As you know, the molarity of the solution tells you the number of moles of solute present per liter of solution. Since our sample has a volume of $\text{1.0 L}$, you can say that its molarity will be equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molarity = 0.054 mol L}}^{- 1}}}}$

Now, to calculate the solubility of the salt, you need to know its solubility product constant, ${K}_{s p}$

${K}_{s p} = 1.7 \cdot {10}^{- 4}$

When lead(II) chloride is dissolved in water, an equilibrium is established between the undissolved solid and the dissolved ions

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

By definition, the solubility product constant will be equal to

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

If you take $s$ to be the molar solubility of the salt, i.e. the maximum concentration of lead(II) chloride that dissolves in water to produce a saturated solution, you can say that you'll have

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

This means that you have

$s = \sqrt[3]{{K}_{s p} / 4}$

which, in your case, is equal to

$s = \sqrt[3]{\frac{1.7 \cdot {10}^{- 4}}{4}} = 0.035$

Therefore, you can say that the salt has a molar solubility equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molar solubility = 0.035 mol L}}^{- 1}}}}$

Both values are rounded to two sig figs, the number of sig figs you have for the concentration of the solution.

Now, you should notice something interesting here. The molar solubility of the salt is lower than the concentration of the $\text{1.5% m/v}$ solution. This should let you know that you cannot have a $\text{1.5% m/v}$ solution of lead(II) chloride in $\text{1 L}$ of solution at room temperature.

A saturated lead(II) chloride solution will contain

0.035 color(red)(cancel(color(black)("mol")))/"L" * "278.1 g"/(1 color(red)(cancel(color(black)("mole PbCl"_2)))) = "9.7 g L"^(-1)

This means that $\text{100 mL}$ of solution will contain

100 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "9.7 g PbCl"_2/(1color(red)(cancel(color(black)("L")))) = "0.97 g PbCl"_2#

Therefore, the mass by volume percent concentration of a saturated lead(II) chloride solution at room temperature is

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{% PbCl"_2 = "0.97% m/v}}}}$

In other words, you can only hope to dissolve about $\text{0.97 g}$ of lead(II) chloride per $\text{1 L}$ of solution at room temperature.