1. Chapter 13 – Properties of Solutions Homework: 33, 34, 35, 38, 39, 41, 44, 46, 47, 50, 55, 59, 63, 65, 66, 67, 69, 72, 74

2. 13.4 – Ways of Expressing Concentration Concentration of a solution can be expressed qualitatively or quantitatively Terms dilute and concentrated are qualitative statements Dilute meaning a relatively small concentration of solute Concentrated meaning a relatively large concentration of solute Our focus will be on the quantitative descriptions of concentration

3. Mass Percentage One of the simplest quantitative methods of expressing concentration Given by So if I had a solution of 36% HCl by mass It would contain 36 g of HCl for each 100 g of solution Note: NOT per 100g of solvent

4. ppm Oftentimes, very dilute solutions are given in terms of parts per million (ppm) ppm is given by What this means If I have a solution whose concentration is 1 ppm It has 1 g of solute for every million (10 6 ) g of solution Or 1 mg of solute per kg of solution

5. Because density of water is 1 g/mL 1 kg of a very dilute aqueous solution will have a volume very close to 1 L So, 1 ppm also means 1 mg/L of solution ppm often used to express maximum concentration of toxic or carcinogenic substances in the environment In US, maximum arsenic in water is 0.010 ppm

6. ppb A solution that is VERY dilute is often measured in terms of parts per billion (ppb) 1 ppb means 1 g of solute per billion (10 9 ) g of solution Or 1 microgram (μg) per liter of solution So the allowable concentration of arsenic in water could also be 10 ppb.

7. Mole Fraction Concentration can also be expressed by the number of moles of components of the solution. Mole fraction we’ve talked about No units, just a ratio The sum of all of the mole fractions of the components adds up to 1

8. Molarity The big one Molarity (M) relates moles of solute to volume of solution Very useful when doing stoichiometry, because of the easy mole relationship

9. Molality Molality (m) is the concentration unit that looks at the number of moles of solute per kilogram of solvent Don’t get it confused with molarity Molarity is moles/L Molality is moles/kg Molality doesn’t change with temperature, while molarity does Molality often used when the solution is used over a range of temperatures

10. Conversion of Concentration Units Sometimes the concentration needs to be known in several different units It is possible to convert concentration units

11. Example An aqueous solution of hydrochloric acid contains 36% HCl by mass. Find the mole fraction of HCl in the solution. Find the molality of HCl in the solution

12. Working Mole fraction When dealing with percentages, assume 100g of the substance So 36% = 36g per 100 g of the substance So if HCl is 36g, that means that water would be 64g To do mole fraction, we now convert the g of each of the components to moles

14. Converting to molality Need moles of HCl and mass of water Luckily, we got those from doing the mole fraction 0.99 mol HCl and 0.064 kg H 2 O

15. Molality ↔ Molarity To convert between molality and molarity we need to know the density of the solution

17. Example A solution contains 5.0g of toluene (C 7 H 8 ) and 225 g of benzene and has a density of 0.876 g/mL. Calculate the molarity of the solution.

18. To find molarity, we need mol of toluene and volume of solution We find volume by using the density of the solution But first, we need total mass of solution

19. To find volume, we use density Note: We actually have 230 g of substance 225 g benzene and 5 g of toluene Finally, to find molarity, we just divide moles of toluene by liters of solution

20. Final Step – Almost There! Note: To solve this problem we needed the following Density (allows us to move from mass of solution to volume of solution) Mass of solute (to find moles of solute)

21. 13.5 – Colligative Properties The physical properties of many solutions differ from those of the pure solvent We add ethylene glycol to water in radiators to lower the freezing point (antifreeze) and to raise the boiling point (so that the water won’t boil in the radiator)

22. This lowering of the freezing point and raising of the boiling point are physical properties These properties (in a solution) depend on the quantity (concentration) but not the identity of the solute These types of properties are called colligative properties. Colligative means “depending on the collection” So these properties depend on the collective effect of the number of solute particles

23. Common Colligative Properties The common (the one’s we’re going to look at) colligative properties include: Lowering freezing point Raising boiling point Vapor-pressure reduction Osmotic pressure For each of these, notice how the concentration of the solute affects the property we’re looking at (relative to the pure substance)

24. Lowering the Vapor Pressure We know that a liquid in a closed container will create a dynamic equilibrium with its vapor At equilibrium, the pressure exerted by the vapor is called the vapor pressure Remember, a substance that has no measurable vapor pressure is called nonvolatile And a substance that has a vapor pressure is called volatile

25. We find that the vapor pressure of solutions (if we have added a nonvolatile solute) will be lower than the vapor pressure of the solvent The amount the vapor pressure is lowered is directly proportional to the concentration of the solute This relationship is given by Raoult’s law

26. Raoult’s Law P A = partial pressure of the vapor from the solution In other words: vapor pressure of the solution X A = mole fraction of the solvent Pº A = vapor pressure of the pure solvent

27. Example The vapor pressure of water is 17.5 torr at 20ºC. If we add glucose to the water, so that the mole fraction of water is 0.800, then the new vapor pressure of water will be P H2O = (0.800)(17.5 torr) = 14.0 torr Doesn’t matter which pressure units we use! Just be consistent

28. Raoult’s law assumes that the solutes are non-volatile, molecular compounds. Not always the case Will deal with ionic compounds (and their effects) when we discussion freezing and boiling points

29. Deviation from Prediction An ideal solution obeys Raoult’s law. Real solutions best approximate ideal solutions when solute concentration is low and when solute and solvent have similar molecular sizes and types of intermolecular attractions Understand that this occurs, but assume ideal solution unless told otherwise

30. Boiling Point Elevation Remember, adding solute lowers vapor pressure Also remember, boiling occurs when vapor pressure = atmospheric pressure So we need a higher temperature to give us more vapor pressure when we add a solute So, the boiling point of a solution is higher than that of a pure liquid

31. The increase in boiling point (relative to the pure solvent) is given by the term ΔT b Is directly proportional to the number of solute particles per mole of solvent molecules Molality is the number of moles of solute per 1000 g of solvent, which shows a fixed number of moles of solvent. Thus, ΔT b is directly proportional to molality

32. ΔT b = K b m The magnitude of K b (called the molal boiling-point-elevation constant) depends on the solvent in question Typical solvents listed on pg. 551 For water, K b = 0.51ºC/m So a 1 m aqueous solution of sucrose (sugar) would boil at 0.51ºC higher than pure water

33. The boiling-point elevation is proportional to the concentration of the solute particles Doesn’t matter if they are ions or molecules Note: When an ionic compound dissolves, total molality increases When 1 mol of NaCl dissolves, 2 moles of solute form (1 mol of Na + and 1 mol of Cl - ) So if we have a 1 m aqueous solution of NaCl, we actually have a total of 2 m in solute particles So double the effect of boiling point increase

34. Freezing-Point Depression The freezing point of a solution is the temperature at which the first crystals of pure solvent begin to form In equilibrium with the solution Lower vapor pressure effects freezing point too Lower vapor pressure (we’ve added solute) means the freezing point is lower than that of a pure solvent

35. ΔT f = K f m Like the boiling-point elevation, the decrease in freezing point, ΔT f is directly proportional to the molality of the solute The values of K f, the molal freezing-point- depression constant for several common solvents are also found on pg. 551 For water, K f = 1.86ºC/m So 1m aqueous solution of sucrose will freeze 1.86ºC lower than pure water While the same rule for ionic compounds applies here A 1m aqueous solution of NaCl will freeze 3.72ºC below the freezing point of the pure solvent

36. Example Calculate the freezing point of a solution containing 0.600 kg of CHCl 3 and 42.0g of eucalyptol (C 10 H 18 O), a fragrant substance found in the leaves of eucalyptus trees.

37. Workspace K f = 4.68ºC/m normal freezing point = -63.5ºC Find molality!

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39. Example 2 Which of the following solutes will produce the largest increase in boiling point upon addition to 1 kg of water? 1 mol of Co(NO 3 ) 2 2 mol of KCl 3 mol of ethylene glycol (C 2 H 6 O 2 )

40. Osmosis Certain materials are semipermeable Often in biological systems and synthetic substances When in contact with a solution, allows some molecules to pass through them, but not others Generally allow small solvent molecules (like water) to pas through But blocks larger solute molecules or ions

41. Consequences of Osmosis Consider a membrane that only allows solvent particles to pass If placed between two solutions of different concentration, solvent molecules move in both directions through the membrane However, this process is called osmosis, and the net movement of solvent is always toward the solution with the higher SOLUTE concentration So the less concentrated solution loses solvent, and more concentrated solution gains solvent What does this mean?

43. If we wanted to stop this from happening, we apply pressure on the side of the more concentrated solution The pressure required to prevent osmosis by pure solvent is the osmotic pressure, π

44. The osmotic pressure obeys a law similar in form to the ideal gas law πV=nRT V is the volume of the solution n is the number of moles of solute R is the ideal-gas constant T is temperature in Kelvin scale

45. From this equation, we can write Where M is the molarity of the solution R value determines which units your pressure will be in

46. If two solutions of identical osmotic pressure are separated by a semi-permeable membrane, no osmosis will occur The two solutions are called isotonic If one solution is of lower osmotic pressure, that solution is hypotonic with respect to the more concentrated solution The more concentrated solution is hypertonic with respect to the dilute solution

47. Biological Systems See! I know biology! The membranes of blood cells are semipermeable Placing a red blood cell in solution that is hypertonic relative to the solution inside the cell causes water to come out of the cell Causes cell to shrivel (crenation) Placing a red blood cell in a solution that is hyptotonic relative to the solution inside the cell causes water to move into the cell Causes cell to rupture (hemolysis) When giving IV infusions to people, IV solution must be isotonic with the solution inside the cell

48. Example What is the osmotic pressure at 20ºC of a 0.0020 M sucrose (C 12 H 22 O 11 ) solution.

49. Determination of Molar Mass Any of the colligative properties can be used to determine the molar mass of a solute.

50. Molar Mass from Boiling Point Elevation A solution of unknown nonvolatile non- ionic solute was prepared by dissolving 0.250 g of the substance in 40.0 g of CCl 4. The boiling point of the solution was 0.357ºC higher than that of the pure solvent. What is the molar mass of the solute?

51. To find molar mass, we need grams of solute and moles of solute Step 1: Solve for molality Step 2: Use molality to solve for mol solute Step 3: Use given data and mol solute to find molar mass

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54. Molar Mass from Osmotic Pressure A solution contains 3.50 mg of protein dissolved in enough water to form 5.00 mL of solution. The osmotic pressure of the solution at 25ºC was found to be 1.54 torr. Calculate the molar mass of the protein.

55. Step 1: Solve for molarity Step 2: Using molarity, solve for number of moles Step 3: Divide number of grams given by moles

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