Na+ + HCO3-
Most chemical reactions are performed in solution. Therefore we need a method of stating concentrations and there are several ways to do this.
I. Molarity
One way would be to give the number of moles of compound per liter of solution and to do this you would divide the number of moles by the number of liters so that you would have the units of moles/liter. This is called molarity and given the symbol M where:
M = (moles of compound)/(liters of solution)
Note that this liters of solution, not liters of solvent. The solute also occupies volume and we can't ignore it. So let's work an example. What is the molar concentration (or what is the molarity) of a 200. ml solution of 26.3 g of NaHCO3 and H2O? The volume of the solution is 0.200 L so we need to know the moles of the compound. The solute has a molecular mass of 84.0 g/mol so the number of moles is:
26.3 g NaHCO3 * (1 mol/84.0 g) = 0.313 mol NaHCO3
The molarity of the solution is then
M = 0.313 mol NaHCO3/(0.200 L) = 1.57 M or we say that the solution is 1.57 M.
We can also state the ion concentrations. Since NaHCO3 is an ionic compound, it dissociates in water into it's constituent ions:
NaHCO3 Na+ + HCO3-
The solution is then 1.57 M in Na+ and it is 1.57 M in HCO3-.
What are the ion concentrations in a 0.50 M solution of (NH4)2SO4?
Well,
(NH4)2SO4 2 NH4+ + SO4>-2
So the solution is 1.0 M in NH4+ and it is 0.50 M in SO4-2.
Now let's work a problem from the opposite direction. How many grams of Na2CO3 are needed to make 2.0 L of a 1.5 M solution of Na2CO3? The first thing to do is to determine the number of moles of Na2CO3 which would be in such a solution:
2.0 L of solution * (1.5 mol Na2CO3)/(1.0 L of solution)
= 3.0 mol Na2CO3
So now we need to know how many grams is needed to make 3.0 moles of Na2CO3:
3.0 mol * (106 g/mol) = 3.2 x 102 Na2CO3
Fini!
Work another example. Your boss asks you to prepare 500. ml of a 0.200 M solution of KMnO4. How do you do it and keep your job (do it correctly)?
First realize from the definition of molarity (M = # moles/L solution) we can solve for the number of moles of the compound needed: # moles = M * L solution so for this problem
# moles = .200 moles/(liter of solution) * 0.500 liters of solution = 0.100 mols of compound
So now we know the moles of KMnO4 needed so we can calculate the number of grams needed:
0.100 mol KMnO4 * (158 g/mol) = 15.8 g of KMnO4
Dilution
Laboratories will often store large quantities of common solutions of a stated concentration and you need to make up a small volume of the solution to a different concentration. So it is important for you to know how to do this. The principle used is that the number of moles of the solute will be the same in both containers. And from our definition of molarity, how do we obtain the number of moles? # moles = M * V where V is the volume of the solution and M is the molarity. So for two solutions containing the same number of moles of solute:
# moles of solute in solution 1 = # moles of solute in solution 2 so
M1 * V1 = M2 * V2
So the above is our working formula for dilution problems.
Let's work an example. You need 300. ml of a 1.00 M NaOH solution and your lab has a large bottle of 3.00 M NaOH. How do you obtain what you need? Both solutions must contain the same number of moles of NaOH so
M1 * V1 = M2 * V2 is our working formula where M1 = 1.00, V1 = 0.300 L, M2 = 3.00, and V2 is our unknown (what volume of the 0.3 M solution has the same number of moles of NaOH as 300 ml of a 1.00 M solution). Solving for V2 gives
V2 = M1 * V1/M2
= 1.00 * 0.300/3.00 = 0.100 liters
So we need to take 100 ml of the 3.00 M solution(this will give us the number of moles of NaOH needed) and add 200 ml of water and you will have 300 mol of a 1.00 M NaOH solution. You keep your job!
Practice, practice, practice.
Here is an excellent practice site.
This is also an excellent place to practice.
II. Molality
Molality is another way of expressing solution concentrations. The molality (m) is defined as
Molality = m = moles of solute/kilogram of solvent
The numerator is the same as in molarity, but the denominator is in weight instead of volume and it is in terms of solvent rather than solution.
III. Mass percent
Mass percent (sometimes called weight percent) is the percent by mass of the solute in solution:
Mass percent = (mass of solute/mass of solution) * 100
IV. Mole fraction
The mole fraction is the number of moles of one of the solution components divided by the total number of moles of the solution.
Mole fraction of A = moles of A/(total number of moles of solution) or for a binary solution (A and B):
XA = nA/(nA + nB)
Practice Problem:
You make a solution of 50.0 g of CsCl in 50.0 g of water. The total volume of the solution is 63.3 ml. Calculate the mass percent, molarity, molality, and mole fraction of CsCl. The answers are 50.0 %, 4.69 M, 5.94 m, and 0.0968.
Here is a place to go practice and after working there for awhile go to this site for further practice.
| Now take a practice quiz to help you understand if you understand the basic concepts. |
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