1. Chapter 16 Properties of Solutions 1

2. Solution Formation Solutions are homogeneous mixtures that may be solid, liquid, or gaseous. The compositions of the solvent and the solute determine whether a substance will dissolve. Stirring (agitation), temperature, and the surface area of the dissolving particles determine how fast the substance will dissolve. 2

3. Stirring affects only the rate at which a solid solute dissolves. It does not influence the amount of solute that will dissolve. An insoluble substance remains undissolved regardless of how vigorously or for how long the solvent/solute system is agitated. 3

4. Temperature & Solution Formation At higher temperatures, the kinetic energy of the solvent molecules is greater than at lower temperatures so they move faster. 4

5. Particle Size & Solution Formation The more surface of the solute that is exposed, the faster the rate of dissolving. A spoonful of granulated sugar dissolves more quickly than a sugar cube because the smaller particles in granulated sugar expose a much greater surface area to the colliding solvent molecules. 5

6. Solubility Solubility Solubility of a substance is the amount of solute that dissolves in a given quantity of a solvent at a specified temperature and pressure to produce a saturated solution. Saturated solution Saturated solution – contains the maximum amount of solute for a given quantity of solvent at a constant temperature and pressure. 6

7. Unsaturated solution Unsaturated solution – a solution that contains less solute than a saturated solution at a given temperature and pressure. If additional solute is added to an unsaturated solution, it will dissolve until the solution is saturated. Some liquids are infinitely soluble in each other. Any amount will dissolve in a given volume. miscible Two liquids are miscible if they dissolve in each other in all proportions (water and ethanol) 7

8. Factors Affecting Solubility Temperature affects the solubility of a solid, liquid and gaseous solutes in a solvent. Both temperature and pressure affect the solubility of gaseous solutes. The solubility of most solid substances increases as the temperature of the solvent increases. Supersaturated solution Supersaturated solution – contains more solute than it can theoretically hold at a given temperature. 8

9. Pressure and Solubility Changes in pressure have little affect on the solubility of solids and liquids, but pressure strongly influences the solubility of gases. Carbonated beverages contain large amounts of carbon dioxide dissolved in water. Dissolved CO 2 makes the drink fizz. The drinks are bottle under higher pressure of CO 2 gas, which forces large amounts of the gas into solution. When opened, the partial pressure of CO 2 above the liquid decreases. 9

10. Pressure and Solubility Immediately, bubbles of CO 2 form in the liquid and escape from the bottle and the concentration of dissolved CO 2 decrease. If the drink is left open, it becomes “flat” as it loses its CO 2. Henry’s Law Henry’s Law – stated that at a given temperature, the solubility (S) of a gas in a liquid is directly proportional to the pressure (P) of the gas above the liquid. As the pressure of the gas above the liquid increases, the solubility of the gas increases. 10

11. Pressure and Solubility Henry’s Law S 1 = S 2 P 1 P 2 11

12. Question The solubility of a gas in water is 0.16 g/L at 104 kPa. What is the solubility when the pressure of the gas is increased to 288 kPa. Assume the temperature remains constant. S 1 = S 2 P 1 P 2 (288 kPa) ( 0.16g/L) = 0.44 g/L (104 kPa) 12

13. End of Section 16.1 13

14. Concentration Concentration Concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent. Dilute solution Dilute solution is one that contains a small amount of solute. Concentrated solution Concentrated solution – contains a large amount of solute. In chemistry the most important unit of concentration is molarity. 14

15. Molarity Molarity Molarity (M) is the number of moles of solute dissolved in one liter of solution Note that the volume involved is the total volume of the resulting solution, not the volume of the solvent alone. 3 M NaCl is read as “three molar sodium chloride” 15

16. Molarity Questions A solution has a volume of 2.0 L and contains 36.0 g of glucose (C 6 H 12 O 6 ). If the molar mass of glucose is 180 g/mol, what is the molarity of the solution? M = moles of solute volume of solution No. of moles = mass= 36 = 0.2 mol molar mass 180 M = 0.2 2 M = 0.1 mol/L or 0.1M C 6 H 12 O 6 16

17. Molarity Questions A solution has a volume of 250 mL and contains 0.70 mol NaCl. What is its molarity? M = moles of solute volume of solution M = 0.70 mol NaCl 0.250L solution (convert to L) M = 2.8 mol/L or 2.8M NaCl 17

18. Molarity Sometimes you may need to determine the number of moles of solute dissolved in a given volume of solution. How many moles are in 2.00 L of 2.5M lithium chloride (LiCl)? Moles of solute = molarity (M) x liters of solution (V) Moles of solute = (2.5 moles/L) ( 2.00L) Moles of solute = 5.0 mol 18

19. Molarity Questions How many moles of ammonium nitrate are in 335 mL of 0.425 M NH 4 NO 3 ? mol NH 4 NO 3 = M x L of solution = (0.425 mol/L) (0.335L) = 0.142 mol NH 4 NO 3 How many moles of solute are in 250 mL of 2.0M CaCl 2 ? How many grams of CaCl 2 is that? mol CaCl 2 = (2.0 mole/L) (0.250 L) = 0.5 mol 0.50 mol CaCl 2 X 111.11 g CaCl 2 = 56 g 19

20. Making Dilutions Diluting Diluting - To make less concentrated by adding solvent. Diluting a solution reduces the number of moles of solute per unit volume, but the total number of moles of solute in solution does not change. Moles of solute before dilution = moles of solute after dilution moles of solute = M x L of solution and total number of moles of solute remains unchanged upon dilution. M 1 V 1 = M 2 V 2 20

21. Making Dilutions M 1 V 1 = M 2 V 2 molarity & volumemolarity and volume of original solutionof diluted solution Volumes can be L or mL as long as the same units are used for both V 1 and V 2 21

22. Making Dilutions A student is preparing a 100 mL of 0.40M MgSO 4 from a stock solution of 2.0 M MgSO 4. How would she do this? M 2 V 2 M 1 V 2 = 20 ml – She would measure 20 mL of the stock solution (2.0 M MgSO 4 ) and transfer it to a volumetric flask. Then she would add water to the flask to make 100 mL of solution. Try the class activity on page 482 = V1V1 22

23. Questions How many milliliters of a solution of 4.0 M KI are needed to prepare a 250.0 mL of 0.760 M KI? V 1 (0.760M)(250.0 mL) (47.5 mL) (4.0 M) How could you prepare 250 mL of 0.20M NaCl using on a solution of 1.0M NaCl and water? V 1 = (0.20 M) ( 250 mL) = 50 mL ( 1.0 M) Use a pipet to transfer 50 mL of the 1.0M solution to a 250 mL flask. Then add distilled water up to the mark. = = 23

24. End of Section 16.2 24