# Section One

 Chapter Six, Section One

OBJECTIVES

1. Learn some basic definitions about solvents.
2. Learn the concentration units of molarity, molality, percent, and mole fraction and how to interconvert among them.
3. Learn how to calculate the resulting concentration upon dilution.

I. Introduction

A solution is a homogeneous mixture. The substance present in the larger(est)st amount is the solvent and the substance(s) in the smaller(est) amount is/are the solute(s). So if you have a sugar/water solution you have water as the solvent and sugar as the solute. A water/salt solution would have water as the solvent and the salt as the solvent.

II.Solubility

The solubility of a solute in a solvent is the maximum amount of solute that can be dissolved in the solvent at a particular temperature so that the mixture remains homogeneous. Such a mixture is said to be saturated. The solubility of a solid in a liquid generally increases with increasing temperature, whereas the solubility of a gas in a liquid generally decreases with increasing temperature. When a liquid is soluble in another liquid we say that the two liquids are miscible in each other. When the amount of solute in a solvent is above the saturated level, the mixture is said to be supersaturated. A supersaturated solution is unstable and when it is disturbed in some manner, the extra solute will precipitate out of the solution. Shaking the supersaturated solution or seeding it (adding a crystal that allows the extra solute to begin crystallizing around the added crystal) will cause the supersaturated solution to become a saturated solution.

To determine what solutes will be soluble in what solvent, we say that "like dissolves like" and we are referring to polar and nonpolar with this rule. Polar solvents will dissolve polar solutes and nonpolar solvents will dissolve nonpolar solvents. Polar solvents will not dissolve nonpolar solutes and nonpolar solvents will not dissolve polar solutes.

Usually the solubility of a gas in a liquid decreases as the pressure decreases. Thus when you open a can of a carbonated drink (one which has carbon dioxide dissolved in it), there is a fiz as the pressure is reduced and the gas solute is then released from solution.

III. Concentration Units

Most chemical reactions are performed in solution. Therefore we need a method of stating concentrations and there are several ways to do this.

A. 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 = .0200 moles/(liter of solution) * 0.500 liters of solution = 0.0100 mols of compound

So now we know the moles of KMnO4 needed so we can calculate the number of grams needed:

0.0100 mol KMnO4 * (158 g/mol) = 1.58 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.

B. 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

C. 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)

ASSIGNMENTS

• 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.

QUIZ ONE

After you have studied this material and practiced some problems, take quiz one. If you score at least 80 on the test then you are ready to continue to the next section.

Web Author: Dr. Leon L. Combs