Lesson26:Mixing Compounds with Water to Make Solutions
(See pages 460-467)
Up to this point in our discussion of chemical reactions, we have been assuming all of our reactants were in
the dry form such as powders or crystals. However, some compounds are actually mixed with water or other solvents
to make them more willing to react. Plus, some compounds are safer and easier to handle if they are mixed with
water. However, it's still important to know how much of the compound has been mixed into the solution before we
use the solution in reactions. Like before, we will still use moles as the unit of measure in our reactions, but because
these moles in some reactants will now be mixed into water, we use the measurement called molarity (M).
Before we continue, lets take a moment to learn some names of the parts of a solution. The compound that we
are mixing into the water is called the solute. For example, if we are mixing sugar into our ice tea, the sugar will be
called the solute. The water part of the solution is known as the solvent. In our ice tea example, the tea would be
considered the solvent. So, solute is what is being added or mixed into the solvent (the stuff doing the dissolving). Now,
lets look back at the term molarity.
Molarity tells how many moles of the solute has been mixed to form one liter of solution (the solvent is
usually water). For example, a solution that is labeled a 1 M or 1 Molar solution means that one mole of the
compound has been mixed with water for each 1 liter of solution (remember, a solution is composed of the solute
and the solvent. A 1 M solution is NOT one mole of solvent dissolved in 1 liter of water). A 2 M or 2 Molar solution
means that 2 moles of the compound has been mixed with water for each 1 liter of solution. A 3 M solution means
that 3 moles of the compound has been mixed with water for each 1 liter of solution. Obviously, the greater the
molarity of the solution, the more concentrated the solution would be. A 12 M soluti on would be twice as
concentrated as a 6 M solution.
Consider how these solutions are made. Suppose you needed to prepare a 1 M solution of table salt (sodium
chloride). For example, you can make a 1 molar solution by dissolving one mole of sodium chloride to form 1 liter of
solution. In order to make this solution, you will need to get 1 mole of the sodium chloride. Since we don't have scales
that measure in moles, we need to calculate the number of grams in one mole of sodium chloride. Hopefully, you'll
readily recall that this is a formula weight for the sodium chloride. By doing your calculations, you should find that one
mole of sodium chloride weighs 58.5 grams. A 1 molar solution of sodium chloride would therefore have 58.5 grams of
sodium chloride mixed with water to form one liter of solution.
Consider how many grams of sodium chloride would be required to make a two molar (2 M) solution. If a 1 M
solution requires 58.5 grams, it should be obvious that a 2 M solution would require 2 times as much sodium chloride
or 2 x 58.5 grams = 117 grams of NaCl.
Try this problem: Suppose you were asked to prepare a 5 M solution of potassium hydroxide. How many grams
of potassium hydroxide powder would you need if you want to make 1 liter of solution?
KOH
K: 1 x 39.0g = 39.0 g
O: 1 x 16.0 = 16.0 g
H: 1 x 1.01 = 1.01 g
56.0 g in one mole
Therefore, a 5 M solution would be prepared by mixing 5 times the grams found in one mole or 5 x 56.0 grams =
280 grams of KOH to make 1 liter of solution.
Try another problem: Suppose you were asked to prepare 2 L of a 3 molar calci um chloride solution. Where
do you start? __________________________________
Good, find the formula weight for calcium chloride. Do your work here:
So, the formula weight for calcium chloride is _______________________________________ grams. (If you
didn't get 110 grams, check back over your work!) One mole of calcium chloride weighs 110 grams.
Now, the problem asks you to make a 3 M solution. Therefore, 3 x 110 grams = 330 grams. But, remember
this is how much calcium chloride is needed to be added to make one liter of solution. The problem asks you to
prepare 2 liters of the solution. Since one liter will require 330 grams, it should be intuitive that to prepare 2
liters of solution, it will take 2 x 330 grams or 660 grams of calcium chloride. Therefore, in order to pre pare 2
liters of a 3 M calcium chloride solution, we would need to use 660 grams of cal cium chloride.
Let's look at one final example. Suppose you worked in a college chemistry lab and your boss asked you to prepare 5
liters of a 12 molar barium oxide solution. How many grams of barium oxide would you need to get from your stock
supply?
Where do you begin? Hopefully, you recall to find the formula weight of barium oxide first. Do your work here:
Based upon these calculations, a 1 M solution would require ___________________________ grams.
Therefore, a 12 M solution of barium oxide would require _____________________________ grams.
Since you need five liters of this solution, now you will take 5 times the amount
needed for one liter: 5 x
grams = ________ grams.
Correct answers can be found on the next page.
Correct answers are found on the next page
To review, in this lesson we learned that some compounds are more readily han dled if they are mixed into
solutions. We learned that we still had to be aware of how much of the compound was mixed into the solution.
As in the past lessons, we learned we will still use moles to measure amounts of chemicals and that the amount of
solute mixed into the solvent was called molarity (M). A 1 M solution has one mole of solute mixed with water to form
1 liter of solution. In order to prepare a solution of a desired molarity (M), we found the number of grams per 1 mole of
solvent (formula weight). Finally, if we needed more than one liter of the solution, we learned to multiply the amount
needed in one liter, times the number of liters desired.
Answers for the barium oxide (BaO) example:
Formula weight of barium oxide = 153 grams.
A 12 M solution would require 1840 grams.
And 5 liters of a 12 M solution would require 9200 grams.
Problems: Preparing solutions
1. Tony was asked to prepare a 3 M potassium chloride solution. How many grams of potassium chloride does
Tony need (assume you want to make 1 liter of solution)?
2. Francis needed to prepare a 5.5 M sodium sulfate solution. How many grams of sodium sulfate does she need
(assume you want to make 1 liter of solution)?
3. Harry was told to make a 0.5 M solution of nickel chromate. How many grams does of nickel chromate does
Harry need (assume you want to make 1 liter of solution)?
4. If you needed to prepare 2 liters of a 6 M calcium carbonate solution, how many grams of calcium carbonate do
you need?