1. 1 Chapter 15 Solutions

2. 2 Definitions solution: a homogeneous mixture in which the components are uniformily intermingled solvent: substance of a solution present in the largest amount solute: all other substances of a solution Aqueous solution: solution with water as the solvent (solutions are homogeneous mixtures)

3. 3 Solution composition saturated: a solution which contains the maximum amount of solute dissolved unsaturated: a solution that has NOT reached the maximum amount of solute supersaturated: a solution that has more dissolved solid then it is designed to hold concentrated: a relatively large amt. of solute is dissolved dilute: a relatively small amount of solute is dissolved

4. 4 acids and bases Chap 8: acid-base reactions also known as neutralization reactions. acid = ?? base =?? equivalent: amount needed to furnish 1 mole of acid/base. 1 equivalent of acid is amt. of the acid needed to furnish 1 mole of H +

5. 5 colligative properties colligative property: properties that depend on the concentration of solute particles. freezing point: when a solute is added to a liquid, the freezing point goes down (freezing point depression):  T= K f m boiling point: when a solute is added to a liquid, the boiling point goes up (boiling point elevation):  T = K b m

6. 6 concentration measurements M = molarity (moles/L) m = molality (moles/kg of solvent) Keep in mind that 1kg=1000g

7. 7 Boiling Point and freezing point Adding a solute to a liquid changes the liquids freezing AND boiling points freezing point depression and boiling point elevation  T = K f/b *m

8. 8 Example problems The freezing point of water is 0 o C, what is the new freezing point if 250.0g of C 6 H 12 O 6 is dissolved in 100.0g of water? (K f = 1.853 o C*kg/mol) Answer= -25.7 o C

9. 9 Soln composition: Mass % mass % = mass of solute x 100% mass of solution

10. 10 Soln Composition: Molarity Molarity (M)= moles of solute/ L of solution Let’s say you dissolved 58g of salt in 10 L of water 58g of salt is equal to 1 mole of salt So, the molarity of the solution would be 0.1 M

11. 11 Dilution Many times in chemistry we store a concentrated stock solution and then dilute it to the proper concentration M 1 V 1 =M 2 V 2

12. 12 dilution problem You have a stock solution of acetic acid at a concentration of 5M. You need 25mL of a 1M solution. How do you make it? AnswerV 1 (5M) = (25 mL)(1M) V 1 = 5 mL. That means we need to take 5 mL of the stock solution and dilute it to 25 mL to get the proper concentration.

13. 13 More problems Calculate the number of moles of KOH in 150 mL of a 0.500M solution. If 4.25g of CaBr 2 is dissolved in enough water to make a 125 mL solution, what is the molarity?

14. 14 Solutions and Stoich Sodium chloride will react with silver nitrate to produce silver (I) chloride. What mass of silver (I) chloride can be produced if you have 3.00L of a 3M solution of NaCl and 2.00L of a 2M AgNO 3 solution and what is the molarity of the AgCl?

15. 15 Answer 573.28g AgCl produced Molarity = 0.8M What is the molarity of the remaining ions in solution (Na +,NO 3 -, Ag + ) ?

16. 16 Why does salt melt ice in the winter time? Adding salt to ice lowers the freezing point of water. This means that the freezing point is less than 0 o C. So, for ice to stay ice it is going to have to be even colder than 0 o C. So, at 0 o C water is now going to be a liquid instead of a solid (ice)

17. 17 Why does adding salt to ice make the ice colder? because the freezing point is lower, the ice will not melt