# #"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?

##### 1 Answer

#### 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

#"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

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

#"CaCl"_ (2(aq)) -> "Ca"_ ((aq))^(2+) + color(red)(2)"Cl"_ ((aq))^(-)#

This tells you that an aqueous solution of calcium chloride has

#["Cl"^(-)] = color(red)(2) * ["CaCl"_2]#

This is the case becauseevery moleof calcium chloride that dissociates produces#color(red)(2)# .molesof 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)# #"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)# #"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# #"g"#

So, you know that

#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

#color(darkgreen)(ul(color(black)("ppm concnetration = 0.532 ppm Cl"^(-))))#

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