SOLUTIONS

1. SOLUTIONS

2. Solutions (ch.16) • Solution – a homogeneous mixture of pure substances • The SOLVENT is the medium in which the SOLUTESare dissolved. (The solvent is usually the most abundant substance.) • Example: • Solution: Salt Water • Solute: Salt • Solvent: Water

3. The process of dissolution is favored by: • A decrease in the energy of the system (exothermic) 2) An increase in the disorder of the system (entropy)

4. Liquids Dissolving in Liquids • Liquids that are soluble in one another (“mix”) are MISCIBLE. • “LIKE dissolves LIKE” • POLAR liquids are generally soluble in other POLAR liquids. • NONPOLAR liquids are generally soluble in other NONPOLAR liquids. LIKE DISSOLVES LIKE : demo

5. Factors affecting rate of dissolution: think iced tea vs. hot tea & the type of sugar you use: cubes or granulated 1) Surface area / particle size • Greater surface area, faster it dissolves 2) Temperature • Most solids dissolve faster @ higher temps 3) Agitation • Stirring/shaking will speed up dissolution

6. Saturation: a solid solute dissolves in a solvent until the soln is SATURATED • Unsaturated solution – is able to dissolve more solute • Saturated solution – has dissolved the maximum amount of solute • Supersaturated solution – has dissolved excess solute (at a higher temperature). Solid crystals generally form when this solution is cooled.

7. ROCK CANDY, YUM!!

8. Applying Concepts QUESTION • When the crystallization has stopped, will the solution be saturated or unsaturated? •  answer

9. ANSWER: SATURATED • Solution has the maximum amount of solute for a given quantity of solvent at a constant temperature and pressure.

10. SOLUBILITY • Solubility = the amount of solute that will dissolve in a given amount of solvent

11. Factors Affecting Solubility • The nature of the solute and solvent: different substances have different solubilities • Temperature: many solids substances become more soluble as the temp of a solvent increases; however, gases are less soluble in liquids at higher temps. • Pressure: Only affects the solubility of gases. As pressure increases, the solubility of gases increases.

12. Concentration of Solution • Concentration refers to the amount of solute dissolved in a solution.

13. *MOLARITY

14. Example: Describe how you would prepare 2.50 L of 0.665 M Na2SO4 solution starting with:a) solid Na2SO4b) 5.00 M Na2SO4 Dissolve 236 g of Na2SO4 in enough water to create 2.50 L of solution.

15. MOLARITY BY DILUTION • When you dilute a solution, you can use this equation:

16. Example: Describe how you would prepare 2.50 L of 0.665 M Na2SO4 solution starting with:a) 5.00 M Na2SO4 Add 0.333 L of Na2SO4 to 2.17 L of water.

17. MASS PERCENT

18. MASS PERCENT • Example: What is the percent of NaCl in a solution made by dissolving 24 g of NaCl in 152 g of water?

19. *MOLALITY

20. MOLALITY • Example: What is the molality of a solution that contains 12.8 g of C6H12O6 in 187.5 g water?

21. MOLALITY • Example: How many grams of H2O must be used to dissolve 50.0 g of sucrose to prepare a 1.25 m solution of sucrose, C12H22O11?

22. Colligative Properties of Solutions (chapter 16) Colligative properties = physical properties of solutions that depend on the # of particles dissolved, not the kind of particle.

23. Colligative Properties • Lowering vapor pressure • Raising boiling point • Lowering freezing point • Generating an osmotic pressure

24. 2 to focus on… • Lowering vapor pressure • Raising boiling point • Lowering freezing point • Generating an osmotic pressure

25. Boiling Point Elevation • a solution that contains a nonvolatile solute has a higher boiling pt than the pure solvent; the boiling pt elevation is proportional to the # of moles of solute dissolved in a given mass of solvent. Like when adding salt to a pot of boiling water to make pasta 

26. Boiling Point Elevation • Tb = kbm where: Tb = elevation of boiling pt m = molality of solute (mol solute/kg solvent) kb = the molal boiling pt elevation constant • kb values are constants; see table 16.3 pg. 495 • kb for water = 0.52 °C/m

27. Ex: What is the normal boiling pt of a 2.50 m glucose, C6H12O6, solution? • “normal” implies 1 atm of pressure • Tb = kbm • Tb = (0.52 C/m)(2.50 m) • Tb = 1.3 C • Tb = 100.0 C + 1.3 C = 101.3 C

28. Freezing/Melting Point Depression • The freezing point of a solution is always lower than that of the pure solvent. Like when salting roads in snowy places so the roads don’t ice over or when making ice cream 

29. Freezing/Melting Point Depression • Tf = kfm where: Tf = lowering of freezing point m = molality of solute kf = the freezing pt depression constant • kf for water = 1.86 °C/m • kf values are constants; see table 16.2 pg. 494

30. Ex: Calculate the freezing pt of a 2.50 m glucose solution. • Tf = kfm • Tf = (1.86 C/m)(2.50 m) • Tf = 4.65 C • Tf = 0.00C - 4.65 C = -4.65C

31. Electrolytes and Colligative Properties • Colligative properties depend on the # of particles present in solution. • Because ionic solutes dissociate into ions, they have a greater effect on freezing pt and boiling pt than molecular solids of the same molal conc.

32. Electrolytes and Colligative Properties • For example, the freezing pt of water is lowered by 1.86°C with the addition of any molecular solute at a concentration of 1 m. • Such as C6H12O6, or any other covalent compound • However, a 1 m NaCl solution contains 2 molal conc. of IONS. Thus, the freezing pt depression for NaCl is 3.72°C…double that of a molecular solute. • NaCl  Na+ + Cl- (2 particles)

33. Electrolytes - Boiling Point Elevation and Freezing Point Depression The relationships are given by the following equations: • Tf = kf ·m·norTb = kb·m·n Tf/b = f.p. depression/elevation of b.p. m = molality of solute kf/b = b.p. elevation/f.p depression constant n = # particles formed from the dissociation of each formula unit of the solute

34. Ex: What is the freezing pt of a 1.15 m sodium chloride solution? • NaCl  Na+ + Cl- n=2 • Tf = kf·m·n • Tf = (1.86 C/m)(1.15 m)(2) • Tf = 4.28 C • Tf = 0.00C - 4.28 C = -4.28C