# "1.50 mL" of a "0.0500-M" solution of calcium chloride, "CaCl"_2, was diluted with water to a volume of "10.0 L". What is the concentration of "Cl"^(–) ions in the diluted solution in ppm?

Jul 27, 2017

${\text{0.532 ppm Cl}}^{-}$

#### Explanation:

The idea here is that a solution's concentration in parts per million, or ppm, is supposed to tell you the number of grams of solute present for every

$1 , 000 , 000 = {10}^{6}$

grams of solution. Because you'll have a very, very small amount of calcium chloride in the diluted solution, you can assume that the total mass of the solution will be equal to the mass of the solvent, i.e. the mass of water.

This will allow you to use the density of water to find the total mass of the solution, since

$\text{mass of solution " ~~ " mass of water}$

Since the problem doesn't provide a value for the density of water, you can approximate it to be equal to ${\text{1 g mL}}^{- 1}$.

Now, the first thing to do here is to figure out exactly how many grams of chloride anions are present in the diluted solution.

Calcium chloride is soluble in water, which means that you have

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

This tells you that an aqueous solution of calcium chloride has

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

This is the case because every mole of calcium chloride that dissociates produces $\textcolor{red}{2}$ moles of chloride anions.

In your case, you will have

["Cl"^(-)] = color(red)(2) * "0.0500 M" = "0.100 M"

Use the molarity and volume of the initial solution to determine how many moles of chloride anions it contains

1.50 color(red)(cancel(color(black)("mL solution"))) * "0.100 moles Cl"^(-)/(10^3color(red)(cancel(color(black)("mL solution")))) = 1.5 * 10^(-4) ${\text{moles Cl}}^{-}$

As you know, the mass of an ion is approximately equal to the mass of the neutral atom, so use the molar mass of elemental chlorine to determine how many grams of chloride anions are present in the initial solution

1.5 * 10^(-4) color(red)(cancel(color(black)("moles Cl"^(-)))) * "35.453 g"/(1color(red)(cancel(color(black)("mole Cl"^(-))))) = 5.318 * 10^(-3) $\text{g}$

Now, after you dilute this initial solution, the mass of chloride anions remains unchanged--this is the case because you're diluting the solution by adding water.

Use the density of water to find the total mass of the diluted solution

10.0 color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1 g"/(1color(red)(cancel(color(black)("mL")))) = 10^4 $\text{g}$

So, you know that ${10}^{4}$ $\text{g}$ of solution contain $5.318 \cdot {10}^{- 3}$ $\text{g}$ of chloride anions, so all you have to do now is figure out how many grams of chloride anions are present in ${10}^{6}$ $\text{g}$ of this solution

10^6 color(red)(cancel(color(black)("g solution"))) * (5.318 * 10^(-3)color(white)(.)"g Cl"^(-))/(10^4color(red)(cancel(color(black)("g solution")))) = "0.5318 g Cl"^(-)

You can thus say that the concentration of the chloride anions in ppm will be equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{ppm concnetration = 0.532 ppm Cl}}^{-}}}}$

The answer is rounded to three sig figs, the number of sig figs you have for your values.