1. Solutions

2. Solutions: Basic Definitions Solute – substance that is being dissolved Solvent – substance that dissolves the solute Solution – a mixture of substances that has a uniform composition; a homogeneous mixture

3. Solutions: Basic Definitions Soluble – when a substance will dissolve in another substance (salt & water) Insoluble – when a substance will not dissolve in another substance (sand & water)

4. Solutions: Basic Definitions Miscible – when two liquids are soluble in each other (alcohol & water) Immiscible – when two liquids are not soluble in each other (oil & water) Aqueous – dissolved in water

5. Solutions: Basic Definitions unsaturated solution - If the amount of solute dissolved is less than the maximum that could be dissolved saturated solution - solution which holds the maximum amount of solute per amount of the solution under the given conditions supersaturated solution - solutions that contain more solute than the usual maximum amount and are unstable.

6. Suspension – mixture containing particles that will settle out is left undisturbed (cornstarch & water) Colloid – heterogeneous mixture that will not settle out if left alone (blood) Emulsion – colloid in which a liquid is suspended in another liquid (mayo) Solutions: Basic Definitions

7. Electrolyte – solution that conducts an electric current Non electrolyte – solution that does not conduct an electric current Solutions: Basic Definitions

8. Supersaturated Solutions They cannot permanently hold the excess solute in solution and may release it suddenly. Supersaturated solutions, as you might imagine, have to be prepared carefully. Generally, this is done by dissolving a solute in the solution at an elevated temperature, at which solubility is higher than at room temperature, and then slowly cooling the solution.

9. Reading Solubility Graphs According to the graph above, about how many grams of KBr are needed to make a saturated solution in 100 g of water at 30° C? 70 grams KBr

10. Reading Solubility Graphs According to the graph above, what kind of solution would you have if you dissolved 10 g of KCl in 100 g of water at 0°C? Unsaturated

11. Solubility Solvation – process of surrounding solute particles with solvent particles to form a solution The rule for dissolving solutions is “like dissolves like” Polar substances will dissolve in polar solvents Non polar substances will dissolve in non polar solvents Non polar will NOT dissolve in polar and vice versa

12. Increasing the Rate of Solution 1.Agitation 2.Increasing Temperature 3.Increasing Surface Area

13. Agitation Increases the speed of the particles –speeds up the dissolving process in solids.

14. Increasing Temperature More collisions of particles as temperature increases.

15. Particle Size (Increasing Surface Area) Smaller particles dissolve faster than larger particles. –more surface area –Sugar cube vs. ½ teaspoon sugar –Teaspoon will dissolve faster

16. Solubility of a gas Two main factors that affect the solubility of a gas in a liquid 1.Temperature –Normally, the higher the temperature, the faster a solute will dissolve…NOT with a gas! –In a gas, the cooler the temperature, the faster the gas will dissolve –Think of drinking a coke…what’s fizzier, a cold coke or a hot coke?

17. Solubility of a gas The second factor affecting the solubility of a gas is pressure 2.Pressure –The higher the pressure, the more gas that will dissolve –Think of a coke bottle…What will happen if you leave the lid off?

18. Henry’s Law The solubility of a gas is directly proportional to the pressure The higher the pressure, the more gas will dissolve S 1 = S 2 P 1 P 2 S = solubility (g/L) P = pressure

19. Example If 0.85 g of a gas at 4.0 atm of pressure dissolves in 1.0 L of water at 25°C, how much will dissolve in 1.0 L of water at 1.0 atm of pressure at the same temperature?

20. Example S 1 = 0.85 g/L P 1 = 4.0 atm S 2 = ? P 2 = 1.0 atm S 1 = S 2 P 1 P 2 0.85 = S 2 4.0 1.0 S 2 = 0.21 g/L

21. Another Example The solubility of a gas is 2.0 g/L at 50.0 kPa. How much gas will dissolve in 1.5 L at 10.0 kPa?

22. Another Example S 1 = 2.0 g/L P 1 = 50.0 kPa S 2 = ? P 2 = 10.0 kPa V 2 = 1.5 L S 1 = S 2 P 1 P 2 2.0 = S 2 50.0 10.0 S 2 = 0.40 g/L S 2 = 0.40 g/L (1.5 L) 0.60 g

23. Concentration Concentration units can vary greatly. They express a ratio that compares an amount of the solute with an amount of the solution or the solvent. For chemistry applications, the concentration term molarity is generally the most useful.

24. % by Mass Remember … % = part x 100 whole % by mass = mass solute x 100 mass solution

25. Example What is the % by mass of a solution with 3.6 g of NaCl dissolved in 100.0 g of water? % = (3.6 / 103.6) x 100 = 3.5% NaCl

26. % by Volume Remember … % = part x 100 whole % by volume = volume solute x 100 volume solution

27. Example What is the % by volume of 75.0 ml of ethanol dissolved in 200.0 ml of water? % = (75.0 / 275.0) x 100 = 27.3%

28. Molarity Molarity is defined as the number of moles of solute per liter of solution. Molarity = moles of solute/liter of solution Note that the volume is the total solution volume that results, not the volume of solvent alone.

29. Molarity Examples Calculate the molarity of a solution made by dissolving 23.4 g of sodium sulfate in 125 ml of solution 23.4 g Na 2 SO 4  0.165 mol 125 ml  0.125 L M = mol / L M = 0.165mol / 0.125 L M = 1.32 M

30. Molarity Examples Calculate the molarity of a solution made by dissolving 5.00 g of C 6 H 12 O 6 in enough water to make 100.0 ml of solution 0.280 M

31. Molarity Examples How many grams of Na 2 SO 4 are required to make 0.350 L of a 0.500 M solution of Na 2 SO 4 ? 24.9 g Na 2 SO 4

32. Dilution When chemists purchase solutions, they generally purchase “stock solutions” which are extremely concentrated solutions This way a chemist can dilute the strong solution to any concentration that they wish. This stops the chemist from having to buy several concentrations

33. Dilution Equation M 1 V 1 = M 2 V 2 M 1 = initial molarity V 1 = initial volume M 2 = final molarity V 2 = final volume The units for V 1 & V 2 do not matter as long as they are the same M 1 & M 2 MUST be in molarity

34. Dilution Problems Suppose we want to make 250 ml of a 0.10 M solution of CuSO4 and we have a stock solution of 1.0 M CuSO4. How would we prepare the solution? First do the math M 1 V 1 = M 2 V 2 (0.10M)(250ml) = (1.0)(V 2 ) V 2 = 25 ml

35. More Dilution Problems How many ml of 3.0 M H 2 SO 4 are required to make 450 ml of a 1.0 M solution? 150 mL