Concentrations of Solutions

1. Concentrations of Solutions • The properties and behavior of solutions often depend not only on the type of solute but also on the concentrationof 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 SolutionsInterconverting 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 0.063 mol of NaOH?

8. Concentration of Solutions Interconverting Molarity, Moles, and Volume Example: How many grams of CuSO4 are needed to prepare 250.0 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 250.0 mL of 0.100 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 usingmass • For reactants that are added to a reaction as aqueous solutions, we measure the reactants usingvolume 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 0.215 M phosphoric acid needed to neutralize 125 mL of 0.103 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 calledtitration. • 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 35.50 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)