Scott 1
Concentrations of Solutions Tutorial
David Scott
Recall that solutions are homogeneous mixtures of a solute and solvent. The solute is the substance that is dissolved. The solvent does the dissolving. For example, when table salt (NaCl) is mixed with water, it dissolves. In this example, sodium chloride is the solute and water is the solvent.
Concentration is a way to expressing how much solute is dissolved in a given amount of solvent. The more solute there is per amount of solvent, the greater the concentration. A solution of 25 grams of NaCl in one liter of water is more concentrated than a solution of 10 grams in one liter. The opposite of concentrated is dilute. A dilute solution contains less solute per unit of volume relative to a concentrated one.
Molarity
Molarity is one way of describing concentration of solutions. It is probably the most common unit of concentration used in beginning chemistry. Molarity, or molar concentration, tells how many moles of solute are dissolved in one liter of solution. The symbol for molarity is M and is calculated as follows.
Molarity (M) = moles of solute/liters of solution
Practice calculating molarity
When someone receives intravenous fluids at a hospital, the solution contains 0.90 g of NaCl per 100 mL of solution. What is the molarity of the solution?
Because molarity is moles of solute per liter of solution, you must determine the moles of solute present and divide it by the liters of solution.
1. Determine moles of solute
0.90 g NaCl x (1 mole NaCl/58.5 g NaCl) = 0.015 moles NaCl
2. Determine liters of solution
100 mL x (1 L/1000 mL) = 0.1 L
3. Divide moles of solute by liters of solution.
0.015 mol NaCl/0.1 L solution = 0.15 moles/L = 0.15 M
This type of problem essentially is converting grams to moles and mL to liters.
Preparing molar concentration solutions
It is very important that you notice the denominator. The denominator is liters of solution, not liters of solvent! The solute also has volume. If one mole of NaCl was added to one liter of water, the total volume would be more than one liter of solution.
This has important effects upon how you prepare molar solutions. In order to prepare one liter of a 1.0 M solution of NaCl you would follow these steps.
1. Measure 58.5g (1 mole) of NaCl
2. Fill a 1.00 L volumetric flask half way with distilled water. A volumetric flask is a specialized flask that is highly accurate and measures only one volume. See the figure to the right.[1]
3. Add the solute (NaCl) to the water.
4. Mix/stir/swirl to dissolve the solute.
5. Add the remaining volume of water up to the total (final) volume of 1.00 L.
When following this procedure, the volume of the solute is included in the total (final) volume of the solution.
Practice
Solve the molarity problems on the next few pages. The answers are given. Show your work for all calculations. For example,
What is the molarity of a solution of 25.0 g KCl dissolved in 250 mL of solution?
How many grams of LiCl are needed to make 150 mL of 0.5 M solution?
What volume of 2.5 M solution can be made using 300 g of KH2PO4?
Name ______Date ______Per_____
Molarity Practice Problems[2]
1. Calculate the molarity of a solution prepared by dissolving 10.8 g NaCl in distilled water to a final volume of 500.0 mL. (0.369 M)
2. What is the concentration of 450 mL of solution that contains 200 grams of iron (III) chloride, FeCl3? (2.74 M)
3. What is the concentration of a solution that has a volume of 2.5 L and contains 660 grams of calcium phosphate, Ca3(PO4)2? (0.85 M)
4. What is the concentration of a solution having a volume of 660 mL in which 33.4 grams of aluminum acetate Al(C2H3O2)3 is dissolved? (0.248 M)
5. What is the concentration of a solution with a volume of 9 mL that contains 2 grams of iron (III) hydroxide, Fe(OH)3? (2.08 M)
6. What is the concentration of a solution having a volume of 3.3 mL and containing 12 grams of ammonium sulfite (NH4)2SO3? (31.3 M)
Name ______Date ______Per_____
7. How many grams of potassium carbonate, K2CO3, are needed to make 200 mL of a 2.5 M solution? (69.1 g)
8. How many grams of ammonium sulfate (NH4)2SO4 are needed to make a 0.25 L of solution at a concentration of 6 M? (198 g)
9. How many grams of copper (II) fluoride CuF2 are needed to make 6.7 liters of a 1.2 M solution? (816 g)
10. How many liters of 4 M solution can be made using 100 grams of lithium bromide LiBr? (0.29 L)
11. How many liters of 0.88 M solution can be made with 25.5 grams of lithium fluoride LiF? (1.11 L)
12. What volume of 0.75 M solution can be made using 75 grams of lead (II) oxide PbO? (0.43 L)
13. What volume of 3.4 M solution can be made using 78 grams of isopropanol (C3H8O)? (0.38 L)
Dilutions
In the laboratory, very often you will have concentrated solutions from which you prepare more dilute solutions. These concentrated solutions are called stock solutions. Dilutions are prepared by taking a small volume of a stock solution and adding enough water (solvent) until the desired concentration is achieved.
How much water should you add? Use the following formula to calculate your dilutions.
M1V1 = M2V2
Where M1 and V1 are the initial concentration and volume; M2 and V2 are the final or desired concentration and volume.
Sample Problem
How many mL of 6.0 M HCl are needed to prepare 500 mL of 0.50 M HCl?
In the problem, M1 = 6.0 M, M2 = 0.5 M, V2 = 500 mL, V1 = unknown.
To prepare the dilution, then, 41.7 mL of 6.0 M solution, when diluted to a final volume of 500 mL (i.e. add 458.3 mL of water) yields 0.50 M solution.
Practice: Solve the following dilution problems. Show your work.
1. How many mL of 3.5 M NaCl are needed to prepare one liter of 0.35 M NaCl? (0.10 L)
2. What is the final concentration when 2.8 ml of 18 M phosphoric acid is diluted to a volume of 1.5 L? (0.036 M)
3. If 10.0 mL of a stock solution was diluted to 1000 mL and the resulting concentration was 0.0250 M, what was the concentration of the stock solution? (2.50 M)
Percent solutions
Another way of expressing concentration is by percent. Percent solutions can be either volume % or mass %.
Volume % is the ratio of the volume of solute to volume of solution.
Mass% is the ratio of the mass of solute to mass of solution.
Sample problem:
Calculate the volume % when 45 mL of ethanol is added to 150 mL of water, for a final volume of 195 mL
%V = 45 mL/(45 + 150)mL = 0.23 = 23%
How many grams of NaCl are in 400 g of 0.90% NaCl solution?
400 g solution x (.9 g NaCl/100 g solution) = 3.6 g
Practice: Solve the following problems. Show your work.
1. Calculate the volume percent when 5.5 mL of hexane are added to 55 mL of benzene. (9.1%)
2. How many mL of ethanol are contained in 5 L of wine if the alcohol concentration is 12%? (Ignore any other solutes.) (0.60 L, 600 mL)
3. Calculate the mass percent of a solution of 7.5 g ofNa3PO4 in 550 g of solution. (1.3 %)
4. What mass of solute is present in 25 g of a 0.15% solution? (0.038 g)
David Scott, 2010 Chemistry
[1] http://z.about.com/d/chemistry/1/0/I/b/volumetricflask.jpg
[2] Adapted and corrected from www.chemfiesta.com