1. II III I C. Johannesson I. The Nature of Solutions Ch. 14 - Solutions

2. C. Johannesson Definitions  Solution -  Solution - homogeneous mixture Solvent Solvent - present in greater amount Solute Solute - substance being dissolved

3. C. Johannesson Definitions Solute Solute - KMnO 4 Solvent Solvent - H 2 O

4. C. Johannesson Solvation  Solvation –  Solvation – the process of dissolving solute particles are separated and pulled into solution solute particles are surrounded by solvent particles

5. C. Johannesson Solvation Strong Electrolyte Non- Electrolyte solute exists as ions only - + salt - + sugar solute exists as molecules only - + acetic acid Weak Electrolyte solute exists as ions and molecules DISSOCIATIONIONIZATION View animation online.animation

6. C. Johannesson Solvation  Dissociation separation of an ionic solid into aqueous ions NaCl(s)  Na + (aq) + Cl – (aq)

7. C. Johannesson Solvation  Ionization breaking apart of some polar molecules into aqueous ions HNO 3 (aq) + H 2 O(l)  H 3 O + (aq) + NO 3 – (aq)

8. C. Johannesson Solvation  Molecular Solvation molecules stay intact C 6 H 12 O 6 (s)  C 6 H 12 O 6 (aq)

9. C. Johannesson Solvation NONPOLAR POLAR “Like Dissolves Like”

10. C. Johannesson Solvation  Soap/Detergent polar “head” with long nonpolar “tail” dissolves nonpolar grease in polar water

11. C. Johannesson Solubility SATURATED SOLUTION no more solute dissolves UNSATURATED SOLUTION more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form concentration

12. C. Johannesson Solubility  Solubility maximum grams of solute that will dissolve in 100 g of solvent at a given temperature varies with temp based on a saturated soln

13. C. Johannesson Solubility  Solubility Curve shows the dependence of solubility on temperature

14. C. Johannesson Solubility  Solids are more soluble at... high temperatures.  Gases are more soluble at... low temperatures & high pressures (Henry’s Law). EX: nitrogen narcosis, the “bends,” soda

15. II III I C. Johannesson II. Concentration Ch. 14 - Solutions

16. C. Johannesson Concentration  The amount of solute in a solution.  Describing Concentration % by mass - medicated creams % by volume - rubbing alcohol ppm, ppb - water contaminants molarity - used by chemists molality - used by chemists

17. C. Johannesson Concentration SAWS Water Quality Report - June 2000

18. C. Johannesson Molarity Must be in liters 1000 mL = 1 L

19. C. Johannesson Molality mass of solvent only 1 kg water = 1 L water

20. C. Johannesson Molality  Find the molality of a solution containing 75 g of MgCl 2 in 250 mL of water. 75 g MgCl 2 1 mol MgCl 2 95.21 g MgCl 2 = 3.2 m MgCl 2 0.25 kg water

21. C. Johannesson Molality  How many grams of NaCl are req’d to make a 1.54m solution using 0.500 kg of water? 0.500 kg water1.54 mol NaCl 1 kg water = 45.0 g NaCl 58.44 g NaCl 1 mol NaCl

22. C. Johannesson Dilution  Preparation of a desired solution by adding water to a concentrate.  Moles of solute remain the same.

23. C. Johannesson Dilution  What volume of 15.8M HNO 3 is required to make 250 mL of a 6.0M solution? GIVEN: M 1 = 15.8M V 1 = ? M 2 = 6.0M V 2 = 250 mL WORK: M 1 V 1 = M 2 V 2 (15.8M) V 1 = (6.0M)(250mL) V 1 = 95 mL of 15.8M HNO 3

24. C. Johannesson Preparing Solutions  500 mL of 1.54M NaCl 500 mL water 45.0 g NaCl mass 45.0 g of NaCl add water until total volume is 500 mL mass 45.0 g of NaCl add 0.500 kg of water 500 mL mark 500 mL volumetric flask  1.54m NaCl in 0.500 kg of water

25. C. Johannesson Preparing Solutions Copyright © 1995-1996 NT Curriculum Project, UW-Madison (above: “Filling the volumetric flask”)

26. C. Johannesson Preparing Solutions Copyright © 1995-1996 NT Curriculum Project, UW-Madison (above: “Using your hand as a stopper”)

27. C. Johannesson Preparing Solutions  250 mL of 6.0M HNO 3 by dilution measure 95 mL of 15.8M HNO 3 95 mL of 15.8M HNO 3 water for safety 250 mL mark combine with water until total volume is 250 mL Safety: “Do as you oughtta, add the acid to the watta!”

28. C. Johannesson Solution Preparation Lab  Turn in one paper per team.  Complete the following steps: A) Show the necessary calculations. B) Write out directions for preparing the solution. C) Prepare the solution.  For each of the following solutions: 1) 100.0 mL of 0.50M NaCl 2) 0.25m NaCl in 100.0 mL of water 3) 100.0 mL of 3.0M HCl from 12.1M concentrate.

29. II III I C. Johannesson III. Colligative Properties Ch. 14 - Solutions

30. C. Johannesson Definition  Colligative Property property that depends on the concentration of solute particles, not their identity

31. C. Johannesson Types  Freezing Point Depression  Freezing Point Depression (  t f ) f.p. of a solution is lower than f.p. of the pure solvent  Boiling Point Elevation  Boiling Point Elevation (  t b ) b.p. of a solution is higher than b.p. of the pure solvent

32. C. Johannesson Types View Flash animation.Flash animation Freezing Point Depression

33. C. Johannesson Types Solute particles weaken IMF in the solvent. Boiling Point Elevation

34. C. Johannesson Types  Applications salting icy roads making ice cream antifreeze cars (-64°C to 136°C) fish & insects

35. C. Johannesson Calculations  t :change in temperature (° C ) k :constant based on the solvent (° C·kg/mol ) m :molality ( m ) n :# of particles  t = k · m · n

36. C. Johannesson Calculations  # of Particles Nonelectrolytes (covalent) remain intact when dissolved 1 particle Electrolytes (ionic) dissociate into ions when dissolved 2 or more particles

37. C. Johannesson Calculations  At what temperature will a solution that is composed of 0.73 moles of glucose in 225 g of phenol boil? m = 3.2m n = 1  t b = k b · m · n WORK: m = 0.73mol ÷ 0.225kg GIVEN: b.p. = ?  t b = ? k b = 3.60°C·kg/mol  t b = (3.60°C·kg/mol)(3.2m)(1)  t b = 12°C b.p. = 181.8°C + 12°C b.p. = 194°C

38. C. Johannesson Calculations  Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. mL water. m = 4.8m n = 2  t f = k f · m · n WORK: m = 0.48mol ÷ 0.100kg GIVEN: f.p. = ?  t f = ? k f = 1.86°C·kg/mol  t f = (1.86°C·kg/mol)(4.8m)(2)  t f = 18°C f.p. = 0.00°C - 18°C f.p. = -18°C