1. CHAPTER 16 Solutions & Colligative Properties

2. Solutions Particles less than 1 nm in size. Homogeneous mixtures Particles do not settle and cannot be separated by filtration. Do not exhibit Tyndall effect (don’t scatter light)

3. Components of a Solution Solutions consist of two parts:  Solute = substance being dissolved  Solvent = dissolving medium The solutes and solvents in a solution may exist as gases, liquids, or solids. Solute Solvent Example Gas Air Gas Liquid CO 2 in water Liquid Alcohol in H 2 O Solid LiquidSugar in H 2 O Solid Brass, 14 k gold

4. Solutions The intermolecular forces between solute and solvent particles must be strong enough to compete with those between solute particles and those between solvent particles. Woah wait, what does this mean?!?!?

5. How Does a Solution Form? As a solution forms, the solvent pulls solute particles apart and surrounds, or solvates, them.

6. How Fast is a Solution Formed? Influenced by 3 Factors: 1. Stirring (agitation) 2. Temperature 3. Surface Area of the Dissolving Particles

7. Solubility There is a limit to the amount of solute that can be dissolved in a solvent. The point at which this limit is reached for any solute-solvent combination depends on the nature of the solute (what’s being dissolved in what solvent) the amount of the solvent the temperature

8. Saturated Versus Unsaturated Solutions A solution that contains the maximum amount of dissolved solute is described as a saturated solution.  If more solute is added to a saturated solution, it falls to the bottom of the container and does not dissolve. A solution that contains less solute than a saturated solution under the same conditions is an unsaturated solution.

9. Supersaturated Solutions PUT VIDEO HERE Usually when a saturated solution is cooled, the excess solute usually comes out of solution, leaving the solution saturated at the lower temperature. But sometimes the excess solute does not separate and forms a supersaturated solution.  solution that contains more dissolved solute than a saturated solution contains under the same conditions.  application: rock candy

10. Supersaturated Solutions

11. STOP! Do POGIL

12. Solubility Values The solubility of a substance is the amount of that substance required to form a saturated solution with a specific amount of solvent at a specified temperature.  example: The solubility of sodium chloride is 36 g per 100 g of water at 20°C. Solubilities vary widely and are determined experimentally.  They can be found in chemical handbooks and are usually given as grams of solute per 100 g of solvent at a given temperature.

13. Temperature and Solubility 1. Increasing the temperature usually increases solubility of solids in liquids.  The effect of temperature on solubility for a given solute is difficult to predict.  The solubilities of some solutes vary greatly over different temperatures, and those for other solutes hardly change at all.  A few solid solutes are actually less soluble at higher temperatures. 2. Increasing the temperature usually decreases the solubility of a gas in liquid.

15. 16.2 Concentration of a Solution The concentration of a solution is a measure of the amount of solute in a given amount of solvent or solution. Chemists commonly use several different ways to quantitatively express the concentration of a solution. These include:  Percent Composition by mass  Parts per million / Part per billion  Molarity (M)  Molality (m)

16. Percent Composition by Mass Sometimes the amount of solute present in the solution is expressed as a percent by mass. grams of solute % Composition = ------------------------- x 100 grams of solution Remember: solution is the solute and solvent together

17. Percent Composition Sample Problems What is the percent composition of a solution that contains 115 g NaCl dissolved in 500 g of H 2 O? How many grams of KMnO 4 are needed to make 500 g of a 12.0% KMnO 4 solution?

18. Parts per Million (ppm) and Parts per Billion (ppb) ppm = mass of solute in solution total mass of solution  10 6 Parts per Million (ppm) Parts per Billion (ppb) ppb = mass of solute in solution total mass of solution  10 9

19. What Do You Think? A 900.0 g sample of sea water is found to contain 0.0067 g Zn. Express this concentration in ppm.

20. Molarity (M) Molarity (M) is the number of moles of solute per liter of solution. moles of solute (mol) Molarity (M) = -------------------------------- liters of solution (L)

21. Molarity Sample Problems What is the molarity of a 0.500 L sample of solution that contains 60.0 g of sodium hydroxide (NaOH)? How many grams of sodium chloride (NaCl) are required to prepare 250.0 mL of a 3.00 M solution?

22. MAKING DILUTIONS! Pay attention future lab workers, this one is for you! Make dilutions from a more concentrated solution: M1V1 = M2V2 (sometimes also seen as C1V1=C2V2)

23. Dilutions practice worksheet

24. 16.3 Colligative Properties & 16.4 Calculations w/ Colligative Properties

25. Another Way to Measure Concentration: Molality (m) Molality is the concentration of a solution expressed in moles of solute per kilogram of solvent. moles of solute (mol) Molality (m) = ----------------------------------- kilogram of solvent (kg)

26. Molality and Mole Fraction  To make a 0.500m solution of NaCl, use a balance to measure 1.000 kg of water and add 0.500 mol (29.3 g) of NaCl. 16.4

27. 16.6

31. Molality Sample Problem A solution was prepared by dissolving 17.1 g of glucose, C 6 H 12 O 6, in 275 g of water. What is the molality (m) of this solution?

32. Molality and Mole Fraction  The mole fraction of a solute in a solution is the ratio of the moles of that solute to the total number of moles of solvent and solute. 16.4

33. Molality and Mole Fraction  In a solution containing n A mol of solute A and n B mol of solvent B (X B ), the mole fraction of solute A (X A ) and the mole fraction of solvent B (X B ) can be expressed as follows. 16.4

34. 16.7

37. Sample Problem 16.7 16.7

38. Colligative Properties These are the effects that a solute has on a solvent. When water has something dissolved in it, its physical properties change. It will no longer boil at 100 o C and it will no longer freeze at 0 o C like pure water.

39. Three Main Effects Lowers the vapor pressure of a solvent.  Lower vapor pressure means that fewer water molecules can escape from the liquid phase into the gas phase at given temperature. Remember, a lower vapor pressure means a higher boiling point! 2) Raises the boiling point of a solvent. 3) Lowers the freezing point of a solvent.

40. Freezing Point Depression

41. Boiling Point Elevation

42. Applications  salting icy roads  making ice cream  antifreeze  cars (-64°C to 136°C)  fish & insects

43. Colligative Properties -THESE DEPEND ONLY ON THE NUMBER OF DISSOLVED PARTICLES -NOT ON WHAT KIND OF PARTICLE GENERAL RULE: THE MORE SOLUTE PARTICLES THAT ARE PRESENT IN A SOLVENT, THE GREATER THE EFFECT.

44. # of Particles  Nonelectrolytes (covalent)  remain intact when dissolved  1 particle  Electrolytes (ionic)  dissociate into ions when dissolved  2 or more particles  Electrolytes have a stronger affect in lowering the freezing point and elevating the boiling point because it puts more particles into the solution.

45. Remember the rule, the more particles, the greater the effect! The Dissociation Factor (d.f.) for an electrolyte is the number of ions a compound dissociates into. NaCl gives Na + ions and Cl - ions, which is 2 particles, therefore d.f. = 2 What is “ d.f. “ for Al(NO 3 ) 3 ? Al(NO 3 ) 3  Al 3+ + 3 NO 3 - = 4 particles

46. Non-electrolytes A compound that does not conduct electricity when dissolved in water. Examples are glucose or any other sugars and alcohols, such as ethanol (CH 3 CH 2 OH) The d.f. = 1 for any non-electrolyte. Why? Covalent molecules do not break apart when they become solvated by water molecules.

47. Calculations  t :change in temperature (° C ) K :constant based on the solvent (° C·kg/mol ) Use K b for boiling point elevation and K f for freezing point depression. Each solvent has it’s own unique factors! m :molality ( m ) = moles solute/Kg solvent d.f. :# of particles  t = m · d.f. · K

48. Example 1: If 48 moles of ethylene glycol, C 2 H 4 (OH) 2, is dissolved in 5.0kg of water. What is the boiling point of the solution? ∆Tbp = m  d.f.  Kb d.f. = 1 ( it is a non-eletrolyte) Kb = 0.52 o C/m (a constant for water) Molality = m = moles solute/kg solvent ∆Tbp = (48 moles) (1) (0.52 o C/molal) (5.0 Kg) (5.0 Kg) ∆Tbp = 5.0 o C New BP = 100 + 5 = 105 o C

49. Example 2: 55.5 grams of CaCl 2 are dissolved in 2.0kg of water. What is the freezing point of this solution? ∆Tfp = m  d.f.  Kf d.f. = 3 (CaCl 2 produces 3 particles) CaCl 2  Ca 2+ + 2 Cl - Kf = 1.86 o C/m (a constant for water) ∆Tfp = (0.25molal) (3) (1.86 o C/molal) Must calculate molality Molality = mass/molar mass Kg Solvent = (55.5 gram/111g/mole) 2.0 kg = 0.25m = 1.4 o C New FP = -1.4 o C