1. Solution Concentration

2.  Lesson Objectives  Describe the concept of concentration as it applies to solutions, and explain how concentration can be increased or decreased.  Define molarity and molality, including the most commonly used units for each measurement.  Define weight percent and be able to determine the weight percent of a solute in a given solution.  Understand how a dilution changes the concentration of a solution, and be able to perform calculations related to this process.

3. Introduction  A solution is comprised of at least two components - the solvent and one or more solutes. Although many different substances (including solids, liquids, and gases) can act as a solvent, we are now going to focus primarily on aqueous solutions, in which water acts as the solvent. Water is the most common solvent that we encounter in our daily lives. In this section, we will explain how to determine the concentration for a given solution and how to create solutions of known concentrations.

4. Solution Concentration  In its most general form, concentration describes the number of items in a given area or volume. The units generally depend on the types of items being counted. For example, if we were to calculate the concentration of people that live in a city, we would divide the total number of people by the area. If we measured the area in square miles, the units of concentration for this measurement would be people per square mile. Or, we could determine the concentration of fish in a lake by dividing the total number of fish by the volume of the lake. If we measure the volume in cubic meters, the concentration would have units of fish per cubic meter.

5.  To express the concentration of a solution, we can perform a similar calculation. The amount of solute is commonly measured in terms of moles, but it can also be measured by mass or by total number of particles. We can then divide this value by either the amount of solvent or the total amount of solution. These values may have units of mass, volume, moles, or number of particles. Depending on how each component is measured, we get different ways to measure concentration.

6.  In this lesson, we will learn about four different ways to describe the concentration of a solution: Molarity - moles of solute divided by volume (in liters) of solution. Molality - moles of solute divided by mass (in kilograms) of solvent. Weight percent - mass of solute divided by mass of solution. Parts per million (or parts per billion) - particles of solute divided by particles of solution.  Each of these ways of describing concentration will be discussed further below.

7. Molarity  The most common way to express the concentration of a solution is by determining its molarity. The molarity of a solution tells us how many moles of solute are present in each liter of solution. It can be calculated as follows:  Molarity = mol solute/ L solution  Molarity has units of moles per liter (mol/L). Moles per liter is also given the abbreviated name molar (M). For example, a solution that contains 2 moles of solute in each liter of solution would be a 2 molar (2 M) solution.

8.  What is the molarity (M) of a 3.4 liter sample of a solution that contains 0.32 moles of NaCl?  Answer:  M NaCl = 0.32 mol NaCl/ 3.4 L= 0.094 M  When indicating solution concentration in molarity, a bracket notation is often used. For instance, from the previous example we could write:  NaCl =0.094  This indicates a 0.094 M solution of sodium chloride.

9. Making Solutions of a Specific Molarity  To make a solution with a particular concentration of a given solute, the following procedure can be used:  Calculate the moles of solute that would be present in the entire desired solution, and then use the molar mass of the solute to calculate the mass that you will need.  Weigh out the appropriate amount of solute, and place it in a volumetric flask.  Fill the flask about halfway with solvent, and swirl until the solute is completely dissolved.  Finally, add enough solvent so that the meniscus of the solution lines up with the calibration mark on the flask.

10.  You want to make a 0.154 M solution of sodium chloride (the approximate concentration of a standard saline solution). Describe how you would prepare 1.00 L of this solution.

11.  Step 1: Calculate  The molarity of our solution is going to be 0.154 M, and the total volume will be 1.00 L. Plugging these values into the definition of molarity will allow us to calculate the necessary moles of solute.  M = mol/ L mol  = M×L = (0.154 M)×(1.00 L)  = 0.154 mol NaCl  Because we will be measuring out NaCl by mass, we need to determine the mass of 0.154 moles of NaCl using its molar mass (58.44 g/ mol).  0.154 mol NaCl × 58.44 g NaCl / 1 mol NaCl = 9.00 g NaCl

12.  Step 2: Weigh out 9.00 g of NaCl and place it in a 1 L volumetric flask.  Step 3: Fill the flask about halfway with water, and swirl until the NaCl is completely dissolved.  Step 4: Add more water until the meniscus of the solution lines up with the calibration mark. Mix well.

13. Molality  Another way to express the concentration of a solution is by determining its molality. The molality (m) of a solution tells us how many moles of solute are combined with each kilogram of solvent. Note that there are two differences between molarity and molality. Molality uses mass instead of volume, and we are looking at the amount of solvent instead of the total amount of solution. It can be calculated as follows:  molality = moles solute/ kg solvent  Molality is primarily useful for calculating certain physical properties of solutions, such as freezing point, boiling point, and vapor pressure.

14.  What would be the molality of a sugar solution in which 4.00 g C 6 H 12 O 6 was dissolved in 1.00 L of water? Water has a density of 1.00 g/mL.

15.  Answer:  molality = moles solute/ kg solvent  We need to find the amount of solute (sugar) in moles and the mass of the solvent in kilograms.  moles solute = 4.00 g C 6 H 12 O 6 X 1 mol C 6 H 12 O 6 / 342.30 g C 6 H 12 O 6 = 0.0117 mol C 6 H 12 O 6  kg solvent = 1.00 L H 2 O X 1000 mL H 2 O/ 1 L H 2 O X 1.00 g H 2 O/ 1 mL H 2 O X 1 kg/ 1000 g = 1.00 kg H 2 O  Substituting these values into our molality expression, we get the following:  molality = 0.0117 mol C 6 H 12 O 6 / 1.00 kg H 2 O = 0.0117 m

16. Weight Percent  Another way to express the concentration of a solution is by its weight percent. This is commonly used to describe stock solutions of things like acids and bases. Weight percent can be calculated by dividing the mass of a solute by the mass of the entire solution:  Weight Percent = grams of solute/ grams of solution X 100%

17.  A solution in which 9.3% of the mass is due to NaCl would be referred to as a 9.3% (w/w) NaCl solution. The presence of "(w/w)" indicates that the ratio is between the weight (mass) of the solute and the weight of the total solution. Other ratios in which one or both of the components are measured in terms of volume instead of mass are also common; these percentages are labeled as either weight/volume (w/v) or volume/volume (v/v).

18.  What is the weight percent of a solution that has 9.01 g of NaCl dissolved in 1000. g of water?  Answer:  Grams of solution = grams of NaCl + grams of water = 9.01 g NaCl + 1000 g H 2 O = 1009 g solution Weight Percent = 9.01 g NaCl/ 1009 g solution X 100% = 0.893%  This mixture could be described as a 0.893% (w/w) solution of NaCl in water.

19. Parts Per Million  We can also express concentrations by dividing the number of particles of a certain solute by the total number of particles in a solution. These types of values are commonly used to describe small amounts of a substance in a complex mixture. Common units for this type of concentration are parts per million (ppm) and parts per billion (ppb).

20.  Let’s say we have a sample of water in which Pb 2+ is present at a concentration of 20 ppm. This would mean that if we randomly took a million particles from our solution, 20 of them (on average) would be Pb 2+ ions. The rest would be mostly water molecules, with a possibility of the presence of various other ions. The maximum allowable concentrations of various toxic gases in the air or heavy metal ions in drinking water are often reported in units of ppm or ppb.

21.  If given information on aqueous solution concentration by ppm, these units can be converted to molarity or molality using the following conversion factor:  1 ppm= 1 mg/ L  This conversion is derived from the fact that ppm is really just a ratio between amount of solute and amount of solution. If you had a 1 ppm solution, this would indicate 1 g of solute per 1 × 10 6 g of solution. Assuming an aqueous solution density of 1 g/mL as is standard for water, this indicates 1 g of solute per 1 × 10 6 mL of solution, or 1 mg of solute per 1 L of solution. You can use this conversion unit and a similar conversion method for converting concentration from ppm or ppb.

22. Dilutions  Many chemicals that we use on a daily basis are transported in a concentrated form but used in a more diluted form. For example, concentrated cleaners are often diluted before they are used. To perform a dilution, pure solvent is added to a concentrated solution in order to make a less concentrated (more dilute) solution. The resultant solution will contain the same amount of solute but a greater amount of solvent.

23.  It will therefore have a lower concentration than the original solution. When performing a simple dilution, the concentration and volume of the initial solution are related to the new concentration and volume as follows:  M 1 V 1 = M 2 V 2  M 1 = initial molarity, V 1 = initial volume  M 2 = final molarity, V 2 = final volume  This relationship holds true due to the fact that the moles of solute stays constant through a dilution process.

24.  Example:  50.0 mL of a 0.40 M NaCl solution is diluted to 1000.0 mL. What is the concentration of NaCl in the new solution?  Answer:  Determine the values of each variable in the dilution equation above, and then solve for the unknown variable.  V 1 = 50.0 mL, M 1 = 0.40 M, V 2 = 1000.0 mL, and M 2 = ?  Solving the above equation for M 2 gives us the following:  M 2 = M 1 V 1/ V 2 = (50.0 mL)(0.40 M) / 1000 mL = 0.020 M

25.  Lesson Summary  The concentration of a solution can be expressed as the amount of solute present in a given amount of solvent or solution.  The most common way to express the concentration of a solution is molarity, which is equal to moles of solute per liter of solution.  Another way to express concentration is molality, which is equal to moles of solute per kilogram of solvent.  The ratio of solute mass to solution mass, when expressed as a percentage, is known as weight percent.  The ratio of solute particles to the total number of particles in a solution can be expressed in units of parts per million (ppm) or parts per billion (ppb).  Concentrated solutions can be diluted by adding pure solvent. The concentration and volume of the initial solution are related to the new concentration and new volume by the equation M 1 V 1 = M 2 V 2.