2. CONCENTRATION OF SOLUTIONS Day 4

3. CONCENTRATION  A measurement of the amount of solute in a given amount of solvent or solution (unit of measurement = molarity )  Kind of like the “strength” of the solution

4. MOLARITY Molarity (M)= mol solute L solution

5. MOLARITY Example #1: If we put 0.500 mol of sodium hydroxide into 1.00 L of solution, what is the molarity of the solution?

6. Molarity= mol solute/L sol’n M= 1.00 L 0.500 mol NaOH = 0.500 M NaOH

7. Example #2: What is the molarity if we have 80.0 g NaOH in 1.00 L of solution?

8. Molarity= mol solute/L sol’n M= 1.00 L 2.00 mol NaOH = 2.00 M NaOH 80.0 g NaOH 40.0 g NaOH 1 mol NaOH =2.00 mol

9. MOLARITY  Example #3: How many moles of KOH are in 0.500 L of 0.100 M KOH?

10. Molarity= mol solute cross multiply! 1 L sol’n mol KOH= (0.500 L) (0.100 M KOH) = 0.0500 mol KOH

11. MOLARITY  Ex 4: How many grams of KOH are in 2.75 L of 0.25 M KOH?

12. Molarity = moles cross multiply! L Moles = Molarity * L = 0.25 mol/L * 2.75L = 0.69 moles KOH 0.69 mol KOH 56.097 g KOH = 38.7 g KOH 1 mole KOH (39)

13. MOLARITY - DILUTION Some chemicals are sold as pre-prepared concentrated solutions (stock solutions). To be used, stock solutions usually must be diluted.

14. A LWAYS A DD A CID

15. 2.) MOLARITY - DILUTION M S V S = M D V D M S – Molarity of stock solution V S – Volume of stock solution (L or mL) M D – Molarity of dilute solution V D – Volume of dilute solution (L or mL)

16. *In dilution calculations, the units for volume must be the same.

17. 2.) MOLARITY - DILUTION  Example #1: How much 12.0 M HCl is required to make 2.50 L of a 0.500 M solution?  M s * V s = M D * V D

18. M S V S = M D V D OR M 1 V 1 = M 2 V 2 M S = 12.0 M HCl M D = 0.500 M HCl V S = ?? V D = 2.50 L V S = (0.500 M HCl) (2.50 L) (12.0 M HCl) = 0.104 L HCl

19. 2.) MOLARITY - DILUTION  Example #2: What volume of a 1.50 M solution can be made using 0.0250 L of 18.0 M H 2 SO 4 ?  M S V S = M D V D

20. M S V S = M D V D M S = 18.0 M H 2 SO 4 M D = 1.50 M H 2 SO 4 V S =.0250 L V D = ???? V D = (18.0 M ) (18.0 M H 2 SO 4 ) (.0250 L) (1.50 M ) (1.50 M H 2 SO 4 ) = 0.300 L 0.300 L H 2 SO 4