1. Thursday, March 6, 2008 Discussion of Molarity Lab Results Introduce Section 15.2b -- Dilutions Homework: Pg. 555, #31a-d, 32, 33, 34

2. Concentrated Solution To save time and space, usually buy solutions that are concentrated- stock solutions Add solvent to get the molarity you want Dilution- process of adding more solvent to a solution

3. Dilutions Only water is added to the solutions Figure out how much water needs to be added Moles of solute after dilution = moles of solute before dilution

4. What happens? Moles of solute stays the same Volume of water increases So Molarity decreases

5. Example Prepare 500. mL of 1.00M acetic acetic acid, HC 2 H 3 O 2 from a 17.5 M stock solution of acetic acid. What volume of stock solution is required? 1 st - find the number of moles of acetic acid needed in the final solution V dilute x M dilute = moles solute

6. Example (continued) Only source of acetic acid is the stock solution So moles of solute in dilute solution must equal moles in concentrated solution This is always true!

7. Example (continued) Now we need to find the volume of 17.5 M acetic acid that contains 0.500 mol of HC 2 H 3 O 2 This is the unknown volume V

8. Volume x molarity = moles Solving for V gives V = 0.0286 L or 28.6 mL of solution To make 500 mL of 1.00M acetic acid solution, we take 28.6 mL of 17.5 M acetic acid and dilute it to a total volume of 500 mL

9. Because moles of solute remain the same before and after dilution, M 1 x V 1 = moles of solute = M 2 x V 2 1 is initial conditions, 2 is final conditions

10. Preparing a dilution

11. Example What volume of 18 M H 2 SO 4 would be needed to prepare 1.5 L of 0.10 M H 2 SO 4 ? V x 18 M = 1.5 L x 0.10 M V= 8.3 mL

12. Common Stock Solutions Stock Solutions Sulfuric acid 18 M Nitric acid16 M HCl12 M

13. Practice Problem Exercise 15.8 What volume of 12M HCl must be taken to prepare 0.75L of 0.25M HCl? 16 mL