Lesson 8:
The Chemistry of Solutions

Dilution

Introduction

In lab, you will often be given a stock solution which you will need to dilute to a given concentration for use in a lab exercise.  Dilution consists of adding more solvent to a solution so that the concentration of the solute becomes lower.  The total number of solutes in the solution remains the same after dilution, but the volume of the solution becomes greater, resulting in a lower molarity, ppm, mg/L, or % concentration.

In the picture above, I've shown the solute as yellow dots and the solvent as solid blue.  The 1 L beaker on the left shows the initial concentration, which we might represent as 13 dots/L.  The beaker on the right is the result of dilution of the left beaker.  We added more solvent so that the solution's total volume was 3 L.  As a result, the concentration of the diluted beaker is (13 dots)/(3 L), or 4.3 dots/L.

Calculating Dilution

Dilution calculations are simplified by using the following equation:

M1V1 = M2V2

Where:
M1 = concentration of the first solution
V1 = volume of the first solution
M2 = concentration of the second solution
V2 = volume of the second solution

Concentration and volume in the equation above can have any units as long as the units are the same for the two solutions.

As long as you know three of the four values from the equation above, you can calculate the fourth.  Let's consider a sample problem:

You have 1 L of a 0.125 M aqueous solution of table sugar.  You want to dilute the solution to 0.05 M.  What do you do?

To solve the problem, you simply plug in the three numbers you know:

(0.125 M) (1 L) = (0.05 M) V2

2.5 L = V2

Using the equation, you determine that the volume of the diluted solution should be 2.5 L.  So we simply add enough water to the first solution so that the solution's volume becomes 2.5 L.

More Complicated Calculations

Dilution calculations run the gamut from very simple to very complicated depending on the situation.  Let's consider a more complicated situation so that you'll be prepared for any dilution you may be asked to perform in lab.

Your plant feeds 50 ppm chlorine using a solution feeder.  You need to fill the 10 gallon tank of the solution feeder, but all you have on hand is a stock solution of 60% sodium hypochlorite.  How much of the sodium hypochlorite do you need to mix with water to fill the tank?

This problem is more complicated because it contains two different units of concentration - ppm and percent.  The first step in the problem is to convert them both to the same measurement of concentration.  We'll convert the 60% sodium hypochlorite into ppm, using one of the equations from the last page:

Now we can plug our known values into the equation:

(600,000 ppm) V1 = (50 ppm) (10 gal)

V1 = 0.0008 gal

Finally, we have to convert our answer into a manageable form since it would be impossible to measure out 0.0008 gal.  Let's convert it to mL:

So, in order to achieve a concentration of 50 ppm of chlorine in our solution feeder's tank, we must add 3 mL of the 60% sodium hypochlorite and then fill the rest of the tank with water.

A Few More Notes

Since this lesson is concerned nearly exclusively with math, I'll give you a few more hints to make sure you solve your math problems correctly:

• Remember that a solution consists of the solute and the solvent.  Make sure that you know whether a formula calls for the amount of solution, solute, or solvent.  If the formula calls for the amount of solution, you may have to add together the amount of solute and solvent to find the amount of solution.  For example:

You add 3 g of salt to 20 grams of water.  What is the percent concentration of the resulting solution?

This problem has given you the amount of solute (3 g) and the amount of solvent (20 g), but you need to know the amount of solution in order to solve the problem.  So you add them together to get 23 g.

Now the problem is easy to solve.  You can just plug the numbers into the correct equation and get 13%.

• When calculating dilutions, don't get your two solutions mixed up.  Let's consider a couple more examples:

You have 4 L of an 90% aqueous solution of hydrochloric acid.  You want 1 L of a 25% solution.  How much of the stock solution do you need to use?

The first solution is the 90% stock solution.  The "4L" is a red herring - you aren't going to use the entire 4L of the first solution.  So:

M1 = 90%
V1 = ?

The second solution will be the solution you make after dilution.  You know:

M2 = 25%
V2 = 1 L

Now you can easily solve the problem by plugging numbers into your equation.  Try it - if you get 0.3 L, then you solved the problem correctly.

Water from your well contains 4 ppm iron.  You take a 5 mL sample of well water and dilute it with distilled water until it reaches a volume of 10 mL.  What is the iron concentration of the diluted solution?

Think of the solutions as the "before" solution and the "after" solution.  The "before solution" is the 5 mL sample which contains 4 ppm iron, so:

M1 = 4 ppm
V1 = 5 mL

The "after" solution will be the solution you make after dilution.  You know:

M2 = ?
V2 = 10 mL

Now you can solve the problem easily, getting 2 ppm as your answer.

• When you're working with percent concentrations, remember that the pure chemical has a concentration of 100%.  You can never have concentrations greater than 100%.
• Remember that all of the conversions between different types of concentration (M, mg/L, ppm, and %) assume that you are dealing with aqueous solutions.  If your solution has a solvent other than water, you should not use these conversions.
• And, as always, pay attention to significant figures and units.  Remember to round to the correct number of significant figures after every step.

Review

A solution consists of a homogeneous mixture of two or more substances.  Intermolecular forces such as van der Waal's forces or hydrogen bonds hold the solute and solvent together.  The similarities of structure and bonding between the solute and solvent determine the degree of solubility of the pair.

A few substances are completely miscible, so they can be mixed in any proportion.  In contrast, most solutions will become saturated once the solubility of the solute in the solvent has been reached.  Solutions can become supersaturated when environmental or physical characteristics change.

Concentration is the amount of solute in a solution, measured as molarity, ppm, mg/L, or percent.  Dilution is a process used to lower the concentration of a solution by increasing its volume.

Sources

Olmsted, J., and G.M. Williams.  1997.  Chemistry: The Molecular Science.  Wm. C. Brown Publishers: Boston.

"Concentration."  Basic Concepts in Environmental Sciences.  30 September 2002.  U.S. Environmental Protection Agency.

Langley, J.  "Concentration."  Units of Measurement - Manipulation.  1998.  Safetyline Institute.  State of Western Australia.

Olmsted, J., and G.M. Williams.  1997.  Chemistry: The Molecular Science.  Wm. C. Brown Publishers: Boston.

New Formulas Used

To calculate molarity :

To calculate parts per million:

To calculate milligrams per liter:

To calculate percent concentration:

Conversions between types of concentration:

1 mg/L = 1 ppm

1,000,000 ppm = 100%

1,000,000 mg/L = 100%

To calculate dilutions:

M1V1 = M2V2

Assignments

Complete the questions for Assignment 8. When you have gotten all the answers correct, print the page and either mail or fax it to the instructor. You may also take the quiz online and directly submit it into the database for a grade.

Lab

There are no labs associated with this lesson.

Quiz

Answer the questions in Quiz 8 When you have gotten all the answers correct, print the page and either mail or fax it to the instructor. You may also take the quiz online and directly submit it into the database for a grade.