1. Solutions Chapter 12

3. Types of Mixtures Solutions –Solvent dissolves a solute Suspensions –The solute particles are too large to be dissolved. They settle out unless the mixture is constantly stirred. Colloids –Particles intermediate in size between solutions and suspensions.

4. Solutions Solvent –The component of a solution that does the dissolving. Solute –The component of a solution that is dissolved. Soluble –Capable of being dissolved. Insoluble –Incapable of being dissolved.

5. Solutions Solute StateSolvent Stateexample Gas Air GasLiquidCO 2 in water Liquid Alcohol in water

6. Solutions Solute StateSolvent StateExample LiquidSolidHg in Ag and Sn (fillings) SolidLiquidSugar in water Solid Cu in Ni

7. Examples of Solutions Sugar Water –Sugar in water (aqueous solution) Salt Water –Salt in water (aqueous solution) Alloys –Brass (zinc and copper) –Sterling Silver (silver and copper) –Stainless Steel (iron, chromium, and nickel) Air –Oxygen in nitrogen

8. Suspensions Oil and vinegar –Nonpolar and polar Oil and water –Nonpolar and polar Muddy water –Large particles

9. Colloids Paint Gelatin Milk Mayonnaise Shaving and whipped cream Fog Cheese butter

10. Tyndall Effect Light can be scattered by colloidal particles. Used to distinguish a solution from a colloid. –T–That is why it is hard to see in the fog, because the light is scattering and is not focused in one direction.

11. Electrolytes Substances that dissolve in water that can conduct current. -Ionic Compounds- NaCl, CaCl 2, MgCl 2, KCl We need electrolytes in our bodies to keep our nervous system and muscular systems functioning properly. Video The Assault on Salt Salt vs. Sea Salt

12. Solubility Unsaturated solution –A solution in which more solute can still dissolve. Saturated Solution –A solution in which no more solute can dissolve. –**A solution can be saturated when at a warmer temperature and then when cooled, form a supersaturated solution in which crystals form (rock candy) Solubility –The amount of substance required to form a saturated solution with a specific amount of solvent at a specific temperature.

13. Solubility curves Solubility curves Tutorial Y axis- grams of solute/100g water X axis temperature Positive slopes- the higher the temperature, the more solute that will dissolve. Negative slopes- the higher the temperature, the less solute that will dissolve.

14. What influences strength of a solution? SampleVolume of waterMass of sucroseMass of salt A50 mL5 g0 B50 mL05g C50 mL10 g0 D50 mL010g

15. Concentrations Of Solutions Molarity –The number of moles of solute in 1 L of solution. –Mol –L –Represented by M Molality –The number of moles of a solute per kg of solvent. –Mol –Kg –Represented by m

16. What influences strength of a solution? SampleMass of sucrose Mass of saltMoles of solute Molarity of solution A5 g00.015.30 M B05g0.0861.72 C10 g00.029.58 M D010g0.1713.42

17. Practice Molarity You have 3.50 L of solution that contains 90.0 g of sodium chloride, NaCl. What is the molarity of the solution? M= mol L 90.0 g NaCl x 1 mol NaCl = 1.54 mol 58.44 g 1.54 mol NaCl = 0.440 M NaCl 3.50 L solution

18. Practice Molarity You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain? How many grams of HCl would that be? M = mol L 0.5 = mol 0.8 Mol = 0.4 mol NaCl x 36.46 g/mol = 14.58 g HCl

19. Practice Molarity To produce 40.0 g of Ag 2 CrO 4 you will need at least 23.4 g of K 2 CrO 4 in solution as a reactant. All you have on hand is 5 L of a 6.0 M K 2 CrO 4 solution. What volume of the solution is needed to give you the 23.4 g K 2 CrO 4 needed for the reaction? 1 mol K 2 CrO 4 = 194.2 g You need at least 23.4 g x 1 mol = 0.120 mol K 2 CrO 4 194.2 g M= mol therefore, 6.0 M = 0.120 L = 0.020 L K 2 CrO 4 solution L L

20. Practice molality A solution was prepared by dissolving 17.1 g of C 12 H 22 O 11, sucrose, in 125 g of water. Find the molal concentration of the solution. M = mol solute kg of solvent 17.1 g C 12 H22O 11 = 1 mol = 0.0500 mol C 12 H 22 O 11 342.34 g 0.0500 mol = 0.400m C 12 H 22 O 11 0.125 kg H2O

21. Practice molality A solution of Iodine, I 2, in CCl 4 is used for chemical tests. How much iodine must be added to prepare a 0.480m solution of iodine in CCl 4 if 100.0 g of CCl 4 is used? 100.0 g =.100 kg CCl4 0.480 m = x mol = 0.0480 mol I2 0.1 kg H 2 O 0.0480 mol I 2 x 253.8 g I 2 = 12.2 g I 2 1 mol