1. Concentrations of Solutions
The properties and behavior of solutions often depend not only on the type of solute but also on the concentration of the solute.
Concentration: the amount of solute dissolved in a given quantity of solvent or solution
many different concentration units
(%, ppm, g/L, etc)
often expressed as molarity

2. Concentrations of Solutions
Molarity (M): the number of moles of solute per liter of solution
Molarity = moles of solute
volume of solution in liters
6.0 M HCl
6.0 moles of HCl per liter of solution.
9.0 M HCl
9.0 moles of HCl per liter of solution.

3. Concentration of Solutions
To calculate the molarity of a solution:
Convert amount of solute to moles
Convert volume to liters
Divide amount of solute in moles by volume in liters
Units of molarity are mol/L (M)

4. Concentration of Solutions
Example: What is the molarity of a solution prepared by dissolving 225 g of glucose (C6H12O6) in enough water to make 825 mL of solution?

5. Concentration of Solutions Interconverting Molarity, Moles, and Volume
Molarity can be used as a conversion factor.
The definition of molarity contains 3 quantities:
Molarity = moles of solute
volume of solution in liters
If you know two of these quantities,
you can find the third.

6. Concentration of Solutions Interconverting Molarity, Moles, and Volume
Example: How many moles of HCl are present in 2.5 L of 0.10 M HCl?

7. Concentration of Solutions Interconverting Molarity, Moles, and Volume
Example: What volume (in mL) of a 0.10 M NaOH solution is needed to provide mol of NaOH?

8. Concentration of Solutions Interconverting Molarity, Moles, and Volume
Example: How many grams of CuSO4 are needed to prepare mL of 1.00 M CuSO4?

9. Concentrations of Solutions: Preparing solutions from pure solids
Steps involved in preparing solutions from pure solids

10. Concentrations of Solutions: Preparing solutions from pure solids
Steps involved in preparing solutions from pure solids
Calculate the amount of solid required
Weigh out the solid
Place in an appropriate volumetric flask
Fill flask about half full with water and mix.
Fill to the mark with water and invert to mix.
You should be able to describe this process (including calculating the mass of solid to use) for any solution I specify.

11. Concentration of Solutions Dilution Calculations
Many laboratory chemicals such as acids are purchased as concentrated solutions (stock solutions).
12 M HCl
12 M H2SO4
More dilute solutions are prepared by taking a certain quantity of the stock solution and diluting it with water.

12. Concentration of Solutions Dilution Calculations
A given volume of a stock solution contains a specific number of moles of solute.
25 mL of 6.0 M HCl contains 0.15 mol HCl
(How do you know this???)
If 25 mL of 6.0 M HCl is diluted with 25 mL of water, the number of moles of HCl present does not change.
Still contains 0.15 mol HCl

13. Concentration of Solutions Dilution Calculations
Moles solute = moles solute
before dilution after dilution
Although the number of moles of solute does not change, the volume of solution does change.
The concentration of the solution will change since Molarity = mol solute
Vol soln

14. Concentration of Solutions Dilution Calculations
When a solution is diluted, the concentration of the new solution can be found using:
Mi x Vi = Mf x Vf
where Mi = initial concentration (mol/L)
Vi = initial volume
Mf = final concentration (mol/L)
Vf = final volume

15. Concentration of Solutions Dilution Calculations
Example: What is the concentration of a solution prepared by diluting 25.0 mL of 6.00 M HCl to a total volume of 50.0 mL?

16. Concentration of Solutions Dilution Calculations
Example: What volume of 1.00 M CuSO4 would be needed to prepare mL of M CuSO4?

17. Concentration of Solutions: Preparing dilute solutions from stock solutions
Steps involved in preparing a dilute solution from a more concentrated stock solution.

18. Concentration of Solutions: Preparing dilute solutions from stock solutions
Steps involved in preparing a dilute solution from a more concentrated stock solution.
Calculate the volume of stock solution needed.
Pipet the required amount of stock solution into an appropriate volumetric flask.
Dilute to the mark with DI water. Mix well.
You should be able to describe this process (including calculating the volume of stock solution to use) for any solution I specify.

19. Concentration of Solutions Solution Stoichiometry
Recall that reactions occur on a mole to mole basis.
For pure reactants, we measure reactants using mass
For reactants that are added to a reaction as aqueous solutions, we measure the reactants using volume of solution.

20. Concentration of Solutions Solution Stoichiometry
grams A
Molar
mass
moles A
Molarity A
Vol
Soln A
Molar ratio
Molar
mass
grams B
moles B
Molarity B
Vol Soln B

21. Concentration of Solutions Solution Stoichiometry
Example: If 25.0 mL of 2.5 M NaOH are needed to neutralize (i.e. react completely with) a solution of H3PO4, how many moles of H3PO4 were present in the solution?
3NaOH (aq) + H3PO4 (aq)  Na3PO4 (aq) + 3H2O(l)

22. Concentration of Solutions Solution Stoichiometry
Example: What mass of aluminum hydroxide is needed to neutralize 12.5 mL of 0.50 M sulfuric acid?

23. Concentration of Solutions Solution Stoichiometry
Example: Calculate the volume of M phosphoric acid needed to neutralize 125 mL of M sodium hydroxide.

24. Concentration of Solutions Solution Stoichiometry
Solution stoichiometry can be used to determine the concentration of aqueous solutions used in reactions.
The concentration of an acid can be determined using a process called titration.
reacting a known volume of the acid with a known volume of a standard base solution (i.e. a base whose concentration is known)

25. Concentration of Solutions Solution Stoichiometry
Titration:

26. Concentration of Solutions Solution Stoichiometry
Example: If mL of 2.5 M NaOH are needed to neutralize 50.0 mL of an H3PO4 solution, what is the concentration (molarity) of the H3PO4 solution?
3NaOH (aq) + H3PO4 (aq)  Na3PO4 (aq) + 3H2O(l)