Along with the online lecture, read chapter 7 in Basic Chemistry for Water and Wastewater Operators.





Lecture

Bonding in Solutions

Introduction

As you will remember from Lesson 4, a solution is a mixture of at least two substances - the solvent and one or more solutes.  By now, you have dealt with several solutions in lab.  For example, you titrated with a sulfuric acid solution during the alkalinity lab and with an EDTA solution in the hardness lab. 

Not every combination of two substances becomes a solution.  When you pour milk on your cereal in the morning, the resulting mixture is not a solution because the cereal does not become dissolved in the milk.  Similarly, when mud in water causes turbidity, the mud is not part of a solution.  We use the term insoluble to describe two substances which do not form a solution when they are mixed together. 

A solution is only formed when two substances mix homogeneously, meaning that any portion of the solution will contain a specific amount of both the solute and the solvent.  In the cases of cereal and milk or of muddy water, the two substances in the mixture will settle apart into two layers over time, so they are not homogeneously mixed.  When you are dealing with a true solution, the two (or more) substances found in the solution will never settle apart.

The homogeneous mixture found in a solution is the result of a chemical interaction between the solute and the solvent.  In a solution, chemical bonds between the solute and solvent hold the solute in the solution and prevent it from settling out.  The chemical bonds found between solute and solvent typically include the weak intermolecular forces introduced in lesson 5 - van der Waal's forces and hydrogen bonds.  For example, when acetone is added to water, hydrogen bonds form between the two substances as shown below:

These bonds keep the acetone dissolved in the solution.  Once the acetone is fully dissolved and mixed into the water, each milliliter of the solution will contain the same amount of acetone and the same amount of water. 



Solutes and Solvents

Although most of the solutions we will deal with in this course are liquids, you should be aware that solutions can be made up of any combination of gases, liquids, and solids.  For example, the air you breathe is an example of a gaseous solution consisting primarily of nitrogen and oxygen while steel is a solid solution in which iron is the solvent and carbon and manganese are the solutes.  

In each solution, the solvent is the substance which determines the state of the finished solution.  For example, when you add salt (a solid) to water (a liquid), you can tell that the water is the solvent since the resulting solution is a liquid.  If both the solute and the solvent have the same state, the solvent is typically the part of the solution which is present in the highest concentration. 

Another distinction between solutes and solvents is that solutes are sometimes changed when they become part of a solution while solvents are typically unchanged.  For example, ionic compounds such as calcium carbonate and table salt break apart into their constituent ions when they become part of a solution. 



Like Dissolves Like

Not every solvent will be able to dissolve every type of solute.  Chemists use the phrase "like dissolves like" to summarize the idea that solvents are best able to dissolve solutes which have a similar chemical composition and which form the same types of bonds. 

Although there are many characteristics which may affect how a solute and solvent relate, the simplest distinction is whether the molecule can form hydrogen bonds.  Molecules such as water, alcohols, and acetone which can form hydrogen bonds tend to be soluble in each other.  On the other hand, molecules such as oil, gasoline, and grease which cannot form hydrogen bonds and instead are attracted to each other only by van der Waal's forces tend to be soluble in each other but are not soluble in water. 

When insoluble substances are mixed together, they settle out into two separate layers, as shown above.

Universal Solvents

Water is sometimes called the universal solvent because of its ability to dissolve a diverse array of substances.  Acetone is also occasionally referred to as a universal solvent since it is able to dissolve oils, alcohols, and water.  In truth, however, there is no such thing as a universal solvent because no one substance is capable of dissolving every possible solute. 

Even though there is no true universal solvent, you use one very good solvent every day - soap.  Soap is made up of molecules which act like an oil on one end and like an alcohol on the other, so they are soluble in oils, in water, and in alcohol. 

When you add soap to your dishwater, the "head" of the soap molecule makes the soap dissolve easily in the water.  The "tail" of the soap molecule attracts oils and grease which are not usually soluble in water and which would not be removed by washing with water alone.  When you rinse the soapy water off your dishes, the oils and greases are washed away with the soap. 

 

Solubility and Saturation

Solubility

As you will remember from Lesson 4, a variety of units can be used to measure the concentration of solute in a solution.  In general, a concentrated solution is one where the amount of solute in a given volume of solvent is great.  A dilute solution is one where the amount of solute in a given volume of solvent is relatively small.  But can we mix up a solution of any concentration?  For example, can we produce a 90% solution of table salt in water?  How about a 90% solution of table salt in gasoline?

A few substances are infinitely soluble, or miscible, which means that they can be mixed together in any proportion.  Ethyl alcohol and water are an example of miscible substances since they can be mixed together in concentrations ranging from pure alcohol to pure water.  However, most substances are not infinitely soluble. 

Solubility is a term used to describe the amount of solute which can be dissolved in a solvent.  The solubility of two substances will depend on the similarity of the two substances.  As mentioned on the last page, some substances (such as oil and water) have a solubility of close to zero.  Most substances, however, will have a solubility somewhere between complete insolubility and complete miscibility. 



Saturation

Once a solution has reached the limit of the solute's and solvent's solubility, the solution is said to be saturated, meaning that it can hold no more solute.  If additional solute is added to a saturated solution, the extra solute will settle out, forming a separate layer like the kind you would see when two substances are insoluble. 

A saturated solution of table salt and water.

You can form your own saturated solution of table salt and water as follows.  Add salt to water, stirring constantly until the salt dissolves.  At first, the salt will completely dissolve in the water, discoloring the water slightly but leaving no visible solid residue.  However, once you have added a certain amount of salt to the water, the solution becomes saturated.  When you add more salt past the saturation point, the salt will not dissolve into the water no matter how long you mix the solution.  Instead, the extra salt will settle out in a layer at the bottom of the solution as shown above. 

Factors Influencing Solubility

The solubility of two substances depends on several factors in addition to the identity of the substances.  These factors can include characteristics of the environment and the state of the solute and solvent.  In every case, these factors will influence the amount of solute which can be dissolved in a solvent.

Environmental factors which influence solubility include temperature and pressure.  For example, warm air is able to dissolve much more water vapor than cold air can.  You may  have noticed that the air is humid (full of water vapor) only on warm days.  If very humid air cools suddenly, the extra water vapor will fall out of solution and will turn into liquid water.  The liquid water in the air forms clouds or fog.

The solubility of a solute can also depend on its oxidation state, which refers to the amount of electrons found in the substance.   For example, iron and manganese tend to enter water in a low oxidation state, meaning that they have a relatively large number of electrons.  But when these metals are exposed to air, they become oxidized (lose electrons) and are no longer very soluble in water.  So the oxidized metals drop out of solution and cause red or black water complaints as well as stains on sinks and clothes. 

Supersaturation

Whenever a solution contains more solute than it can hold, it is said to be supersaturated.  We have mentioned one supersaturated solution already in this lesson - the warm air which cooled and thus contained more water vapor than it could hold. 

Solutions can become supersaturated in a variety of ways, but in every case the supersaturated solution is unstable.  If more of the solute is added or if the conditions change in any way, the extra solute will settle out of the supersaturated solution.  In the water vapor and air solution, dust particles in the air provided the slight change which caused water vapor to settle out, forming clouds and rain. 
 
If you've ever made sweetened iced tea, you will have taken advantage of the characteristics of a supersaturated solution.  By adding sugar to hot tea, you were able to dissolve much more sugar into the water than you would have been able to dissolve into cold tea.  When the tea cooled, the additional sugar remained dissolved in the tea as a supersaturated solution.  If you tried to add more sugar to the cooled tea, however, the excess solute would drop out of solution and the tea would become less sweet. 



Layering

The layering seen in a solution which has passed its saturation point resembles the layering seen when two insoluble substances are mixed.  But, as you can see in the illustration below, these two situations are actually quite different.  One of the layers in the saturated solution contains both solute and solvent while the layers in the insoluble mixture contain only one substance per layer. 

The extra solute added to a saturated solution may settle either to the top or to the bottom of the solution.  The location of the extra solute depends on its density, a concept we will discuss in a later lesson. 

 

 

Concentration

Solutions and Concentration

In lab, we will often deal with solutions.  This lesson introduces two related concepts - the concentration of a solution and the process of diluting a solution. 

A solution consists of a liquid (the solvent) with a substance (the solute) dissolved in it.  You are probably familiar with many solutions from your everyday life.  Milk is a solution consisting of water (the solvent) with lactose and salts dissolved in it.  Ocean water is another type of solution.  Both of the examples given above, along with many of the solutions we work with in lab, are known as aqueous solutions because the solvent is water. 

Have you ever mixed up orange juice from concentrate and added too many cans of water?  I'm sure you could taste the difference between the watery orange juice and the properly prepared orange juice.  The amount of solute (orange juice concentrate in the case of this example) in a solution is known as the solution's concentration.  As you will find, solutions with different concentrations act differently in the lab, just as different concentrations of orange juice taste different in your kitchen.  The rest of this page will be devoted to the terminology and math we use to determine the concentration of solutions. 



Molarity

There are many different ways to measure concentration, including molarity, ppm, mg/L, and percent concentration.  Molarity is typically used in very concentrated solutions and will be used in many of the solutions we prepare in lab.  We calculate molarity using the following formula:

As you can see, the unit for molarity - "M" - is equivalent to mol/L.

How many grams of table sugar (C₁₂H₂₂O₁₁) would you need to dissolve in water to produce 0.75 liters of a 0.125 M aqueous solution of table sugar?  The first step in finding the answer to this question is to calculate the number of moles of table sugar which would be needed:

Next, you have to transform the number of moles of sugar into grams of sugar.  We use the technique introduced in the last lesson:

So the answer is that we must dissolve 32 grams of table sugar in enough water to make up 0.75 L of solution.  This will result in a 0.125 M solution.



ppm

While molarity is used for solutions with relatively large concentrations of solute, we use ppm and mg/L to denote much lower concentrations.  This will often be the case when you are measuring solutes in water at a treatment plant, such as the concentration of iron in the water.  We will discuss ppm in this section, then move on to mg/L in the next section. 

Parts per million, or ppm, is just what the name suggests - the number of parts of solute in one million parts of solution.  Concentration in ppm is calculated using the following formula:

Let's consider a simple example:

In order to solve this problem, you first must make the units of the solute the same as the units of the solvent.  So, you will convert from milligrams to grams:

Then you simply plug the numbers into the formula:

So, the concentration of sulfuric acid in the resultant solution is 5.5 ppm. 



mg/L

Milligrams per liter, or mg/L, can be used to denote concentration in similar circumstances to ppm.  The following formula is used to calculate concentration in mg/L:

As you can see, the primary difference between the two calculations is that ppm is a mass per mass calculation while mg/L is a mass per volume calculation.  Due to the special characteristics of water, the concentration of an aqueous solution is the same when calculated in mg/L as it is when calculated in ppm.  So, the 5.5 ppm sulfuric acid aqueous solution discussed in the last section has a concentration of 5.5 mg/L.

Let's consider the following example problem:

You could choose to calculate the concentration as either ppm or mg/L, but mg/L is the better choice since you are given the amount of solution in liters.  We would calculate the answer as follows:

You can state the concentration as either 0.5 mg/L or 0.5 ppm.

Percent

The final unit we use to measure concentration is percent.  Percent concentration is calculated using the following formula:

Notice that I have given no units for the amounts of solute and solution.  That is because you can either calculate weight per weight (w/w) percent concentration or volume per volume (v/v) percent concentration.  Since these two methods can give you different answers, you should always note which method you used.

How does percent concentration relate to concentration in ppm?  In order to figure out the answer, let's consider the same solution we considered in the section on ppm:

We would calculate the percent concentration as follows:

The w/w percent concentration is 0.00055%.  You will notice that this is the same as the ppm concentration divided by 10,000.  W/w concentrations always show this relationship to ppm concentrations since the calculations are identical except for multiplying by one hundred in percent concentration and by one million in ppm concentration.

A concentration of 0.00055% is less understandable than a concentration of 5.5 ppm.  As a result, percent concentration is usually used in situations more like that in which molarity is used, when the solute makes up a larger percentage of the solution. 



Converting Between Types of Concentration

Next, you will be called on to convert between different types of concentration.  For example, if you had a 35.7 ppm solution, what would this be in percent concentration?  If you had a 0.2 M solution, could you convert this to mg/L? 

We've already mentioned a few conversion factors previously.  In aqueous solutions, the following conversion factors are in effect:

Converting to molarity is a little more complicated.  You must use the following formula to convert from mg/L (or ppm) to molarity in an aqueous solution:

We'll give you some practice with making conversions on the next page. 

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.