1. Solution Concentration

2.  So far, you have studied how solutions can form and the limits to the amount of solute that can possibly dissolve in a solvent at a given temperature and pressure. In this lesson, you will learn how to quantitatively express the amount of solute present in any solution.

3. Percent Solutions  There are several ways to express the amount of solute present in a solution. The concentration of a solution is a measure of the amount of solute that has been dissolved in a given amount of solvent or solution. A concentrated solution is one that has a relatively large amount of dissolved solute. A dilute solution is one that has a relatively small amount of dissolved solute.

4.  However, these terms are relative, and we need to be able to express concentration in a more exact, quantitative manner. Still, concentrated and dilute are useful as terms to compare one solution to another. Also, be aware that the terms “concentrate” and “dilute” can be used as verbs. If you were to heat a solution, causing the solvent to evaporate, you would be concentrating it, because the ratio of solute to solvent would be increasing. If you were to add more water to an aqueous solution, you would be diluting it because the ratio of solute to solvent would be decreasing.

5.  One way to describe the concentration of a solution is by the percent of the solution that is composed of the solute.  This percentage can be determined in one of two ways: (1) the mass of the solute divided by the mass of the solution, or (2) the volume of the solute divided by the volume of the solution. Because these methods generally result in slightly different values, it is important to always indicate whether a given percentage was calculated "by mass" or "by volume."

6. Mass Percent  When the solute in a solution is a solid, a convenient way to express the concentration is a mass percent (mass/mass), which is the grams of solute per 100 g of solution.  Percent by mass = mass of solute/mass of solution X 100%  Suppose that a solution was prepared by dissolving 25.0 g of sugar into 100 g of water. The percent by mass would be calculated as follows:  Percent by mass = 25 g sugar/125 g solution X 100% = 20% sugar

7.  Sometimes, you may want to make a particular amount of solution with a certain percent by mass and will need to calculate what mass of the solute is needed.  For example, let’s say you need to make 3000 g of a sodium chloride solution that is 5% by mass. You can rearrange and solve for the mass of solute.  mass of solute = percent by mass/100% X mass of solution  = 5%/100% X 3000 g  = 150 g NaCl  You would need to weigh out 150 g of NaCl and add it to 2850 g of water. Notice that it was necessary to subtract the mass of the NaCl (150 g) from the mass of solution (3000 g) to calculate the mass of the water that would need to be added.

8. Volume Percent  The percentage of solute in a solution can more easily be determined by volume when the solute and solvent are both liquids. The volume of the solute divided by the volume of solution expressed as a percent, yields the percent by volume (volume/volume) of the solution. If a solution is made by taking 40. mL of ethanol and adding enough water to make 240. mL of solution, the percent by volume is:  Percent by volume = volume of solute/volume of solution X 100%  = 40 mL ethanol/240 mL solution X 100%  = 16.7% ethanol

9.  It should be noted that, unlike in the case of mass, you cannot simply add together the volumes of solute and solvent to get the final solution volume. When adding a solute and solvent together, mass is conserved, but volume is not.  In the example above, a solution was made by starting with 40 mL of ethanol and adding enough water to make 240mL of solution. Simply mixing 40 mL of ethanol and 200 mL of water would not give you the same result, as the final volume would probably not be exactly 240 mL.

10. Molarity  Chemists primarily need the concentration of solutions to be expressed in a way that accounts for the number of particles present that could react according to a particular chemical equation. Since percentage measurements are based on either mass or volume, they are generally not useful for chemical reactions. A concentration unit based on moles is preferable. The molarity (M) of a solution is the number of moles of solute dissolved in one liter of solution.

11.  To calculate the molarity of a solution, you divide the moles of solute by the volume of the solution expressed in liters.  Molarity (M) = moles of solute/liters of solution = mol/L  Note that the volume is in liters of solution and not liters of solvent. When a molarity is reported, the unit is the symbol M, which is read as “molar”. For example, a solution labeled as 1.5 M NH 3 is a “1.5 molar solution of ammonia.”

12.  Sample Problem 16.2: Calculating Molarity  A solution is prepared by dissolving 42.23 g of NH 4 Cl into enough water to make 500.0 mL of solution. Calculate  its molarity.  Step 1: List the known quantities and plan the problem.  Known  mass of NH 4 Cl = 42.23 g  molar mass of NH4Cl = 53.50 g/mol  volume of solution = 500.0 mL = 0.5000 L  Unknown  molarity = ? M

13.  The mass of the ammonium chloride is first converted to moles. Then, the molarity is calculated by dividing by liters. Note that the given volume has been converted to liters.  Step 2: Solve.  42.23 g NH 4 Cl X 1 mol NH 4 Cl/53.50 g NH 4 Cl = 0.7893 mol NH 4 Cl  0.7893 mol NH 4 Cl/0.5000 L = 1.579 M  Step 3: Think about your result.  The molarity is 1.579 M, meaning that a liter of the solution would contain 1.579 moles of NH 4 Cl. Having four  significant figures is appropriate.

14.  Practice Problem  1. What is the molarity of a solution for which 250 mL of the solution contains 10.0 g of Pb(NO 3 ) 2 ?

15.  In a laboratory situation, a chemist must frequently prepare a given volume of a solution with a specific molarity.  The first task is to calculate the mass of the solute that is necessary. The molarity equation can be rearranged to solve for moles, which can then be converted to grams.  Sample Problem 16.3: Finding the Necessary Mass of Solute  A chemist needs to prepare 3.00 L of a 0.250 M solution of potassium permanganate (KMnO 4 ). What mass of  KMnO 4 does she need to make the solution?

16.  Step 1: List the known quantities and plan the problem.  Known molarity = 0.250 M volume = 3.00 L molar mass of KMnO 4 = 158.04 g/mol  Unknown mass of KMnO 4 = ? g  Moles of solute is calculated by multiplying molarity by liters. Then, moles is converted to grams.

17.  Step 2: Solve.  0.250 M KMnO 4 X 3.00 L solution = 0.750 mol KMnO 4  0.750 mol KMnO 4 X 158.04 g KMnO 4 /1 mol KMnO 4  = 119 g KMnO 4  Step 3: Think about your result.  When 119 g of potassium permanganate is dissolved in enough water to make 3.00 L of solution, the molarity is 0.250 M.

18.  Practice Problem  2. What mass of CaCl 2 is needed to make 600. mL of a 0.380 M solution?

19. Preparing Solutions  If you are attempting to prepare 1.00 L of a 1.00 M solution of NaCl, you would obtain 58.44 g of sodium chloride.  However you cannot simply add the sodium chloride to 1.00 L of water. After the solute dissolves, the volume of the solution will be slightly greater than a liter because the hydrated sodium and chloride ions take up space in the solution. Instead, a volumetric flask needs to be used. Volumetric flasks come in a variety of sizes and are designed so that a chemist can precisely and accurately prepare a solution of one specific volume.

20.  In other words, you cannot use a 1-liter volumetric flask to make 500 mL of a solution. It can only be used to prepare 1 liter of a solution. The steps to follow when preparing a solution with a 1-liter volumetric flask are outlined.  1. The appropriate mass of solute is weighed out and added to a volumetric flask that has been about half-filled with distilled water.  2. The solution is swirled until all of the solute dissolves.  3. More distilled water is carefully added up to the line etched on the neck of the flask.  4. The flask is capped and inverted several times to completely mix the solution.

21. Dilutions  When additional water is added to an aqueous solution, the concentration of that solution decreases. This is because the number of moles of the solute does not change, but the total volume of the solution increases. We can set up an equality between the moles of the solute before the dilution (1) and the moles of the solute after the dilution (2).

22.  Since the moles of solute in a solution is equal to the molarity multiplied by the volume in liters, we can set those equal.  M 1 X L 1 = M 2 X L 2  Finally, because the two sides of the equation are set equal to one another, the volume can be in any units we choose, as long as that unit is the same on both sides. Our equation for calculating the molarity of a diluted solution becomes:  M 1 X V 1 = M 2 X V 2

23.  Suppose that you have 100. mL of a 2.0 M solution of HCl. You dilute the solution by adding enough water to make the solution volume 500. mL. The new molarity can easily be calculated by using the above equation and solving for M 2.  M 2 = (M 1 X V 1 )/V 2 = (2.0M X 100mL)/500mL = 0.40 M HCl  The solution has been diluted by a factor of five, since the new volume is five times as great as the original volume.  Consequently, the molarity is one-fifth of its original value.  Another common dilution problem involves deciding how much of a highly concentrated solution is required to make a desired quantity of solution with a lower concentration. The highly concentrated solution is typically referred to as the stock solution.

24.  Sample Problem 16.4: Dilution of a Stock Solution  Nitric acid (HNO 3 ) is a powerful and corrosive acid. When ordered from a chemical supply company, its molarity is 16 M. How much of the stock solution of nitric acid needs to be used to make 8.00 L of a 0.50 M solution?  Step 1: List the known quantities and plan the problem.  Known  stock HNO 3 (M 1 ) = 16 M  V 2 = 8.00 L  M 2 = 0.50 M  Unknown  volume of stock HNO 3 (V 1 ) = ? L  The unknown in the equation is V1, the necessary volume of the concentrated stock solution.

25.  Step 2: Solve.  V1 = (M 2 X V 2 )/M1 = (0.50 M X 8.00 L)/16 M = 0.25 L = 250 mL  Step 3: Think about your result.  250 mL of the stock HNO 3 solution needs to be diluted with water to a final volume of 8.00 L. The dilution from 16 M to 0.5 M is a factor of 32.

26.  Practice Problems  3. 125 mL of a 2.55 M solution of Zn(NO 3 ) 2 is diluted with water to make the final volume 197 mL. Calculate the new molarity.  4. What volume of a 1.50 M solution of KCl needs to be diluted in order to prepare 2.40 L of a 0.0750 M solution?

27.  Dilutions can be performed in the laboratory with various tools, depending on the volumes required and the desired accuracy. The image below illustrates the use of two different types of pipettes. The use of a calibrated pipette instead of a graduated cylinder improves accuracy. In the figure on the right, the student is using a micropipette, which is designed to quickly and accurately dispense very small volumes. Micropipettes are adjustable and come in a variety of sizes.

28. Molality  A final way to express the concentration of a solution is by its molality. The molality ( m ) of a solution is the moles of solute divided by the kilograms of solvent. A solution that contains 1.0 mol of NaCl dissolved in 1.0 kg of water is a “one molal” solution of sodium chloride. The symbol for molality is a lower-case m written in italics.  Molality (m) = moles of solute/kilograms of solvent = mol/kg

29.  Molality differs from molarity only in the denominator. While molarity is based on the liters of solution, molality is based on the kilograms of solvent. Concentrations expressed in molality are used when studying properties of solutions related to vapor pressure and temperature changes. Molality is used because its value does not change with changes in temperature. The volume of a solution, on the other hand, is slightly dependent upon temperature.

30.  Sample Problem 16.5: Calculating Molality  Determine the molality of a solution prepared by dissolving 28.60 g of glucose (C6H12O6) in 250. g of water.  Step 1: List the known quantities and plan the problem.  Known  mass of solute = 28.60 g C6H12O6  mass of solvent = 250. g = 0.250 kg  molar mass of C6H12O6 = 180.18 g/mol  Unknown  molality = ? m  Convert grams of glucose to moles and convert the mass of the solvent from grams to kilograms.

31.  Step 2: Solve.  28.60 g C 6 H 12 O 6 X (1 mol C 6 H 12 O 6 /180.18 g C 6 H 12 O 6 ) = 0.1587 mol C 6 H 12 O 6  0.1587 mol C 6 H 12 O 6 /0.250 kg H 2 O = 0.635 m C 6 H 12 O 6  Step 3: Think about your result.  The answer represents the moles of glucose per kilogram of water and has three significant figures.

32.  Practice Problem  5. Calculate the molality of a solution containing 2.25 g of LiNO 3 dissolved in 600. g of water.

33.  Molality and molarity have very similar values for dilute aqueous solutions because the density of those solutions is relatively close to 1.0 g/mL. This means that 1.0 L of solution has a mass of just about 1.0 kg. As the solution becomes more concentrated, its density will not be as close to 1.0 g/ml, and the molality value will be different than the molarity. For solutions with solvents other than water, the molality will be very different than the molarity. Make sure that you are paying attention to which quantity is being used in a given problem.