1. Water and Aqueous Systems
Chapter 17

2. Objectives Describe the hydrogen bonding that occurs in water
Explain the high surface tension and low vapor pressure of water in terms of hydrogen bonding
Account for the high heat of vaporization and the high boiling point of water in terms of hydrogen bonding
Explain why ice floats on water

3. The Water Molecule O-H bonds are highly polar covalent bonds.
the O atom has a slightly negative charge. H has a slightly positive charge.
Hydrogen bonding- occurs when hydrogen is attracted to a very electronegative element (N, O, and F).

4. Hydrogen bonding in water

5. The Unique Properties of Water
High surface tension
Low vapor pressure
High specific heat capacity
High heat of vaporization
High boiling point
Low density of ice

6. Surface properties
Surface tension: the inward pull that tends to minimize the
surface area of a liquid.
Surfactant: surface active agent- surface tension is decreased by detergents (interferes with hydrogen bonding)

7. Why are water droplets spherical as they fall?

8. Surface tension

9. Surface tension

10. Water on your car….

11. Vaporization and Vapor Pressure
Hydrogen bonding allows few water particles to vaporize, resulting in low vapor pressure above the surface of water

12. Water’s low vapor pressure

13. Specific Heat Capacity
Water’s high specific heat capacity helps to moderate daily air temperatures around large bodies of water.
(Water’s specific heat
is more than 4 times
that of most metals.)
This is due to the
hydrogen bonding

14. Water’s high heat of vaporization
Because of hydrogen bonding,
water absorbs large amounts
of heat as it vaporizes.
Water releases large
amounts of heat as it
condenses.

15. Water’s high boiling point
Hydrogen bonding causes water’s high boiling point (Stronger IMFs to overcome).

16. Ice
Hydrogen bonding causes water molecules to freeze in a certain arrangement
Ice has an open framework structure, causing space to be trapped between the molecules
The density of ice is lower than that of water.
This is why ice floats.

17. Objectives
Explain the significance of the statement “like dissolves like”
Distinguish between strong electrolytes, weak electrolytes, and nonelectrolytes, giving examples of each

18. Solution – homogeneous mixture
What is a solution?
Solution – homogeneous mixture
Solvent – substance present in largest amount (the dissolving medium)
Solutes – other substances in the solution
Aqueous solution – water is the solvent

20. Solvation… the process that occurs as a solute dissolves.
There are two types of solvation:
Ionic compounds dissolve by dissociation.
Covalent compounds dissolve by molecular solvation.

21. Solubility of Ionic Substances: Dissociation
The positive and negative ions of a salt come apart (dissociate) as a salt dissolves.

22. Dissociation vs. Molecular Solvation
NaCl(s)  Na +(aq) + Cl -(aq)
Note: the ions come apart from each other
(electrolytes)
Molecular solvation:
C6H12O6(s)  C6H12O6(aq)
Note: no dissociation occurs
(nonelectrolytes)

23. “Like dissolves like”
Polar solutes dissolve in polar solvents (water is polar, so it dissolves polar substances, either ionic or molecular).
Nonpolar solutes dissolve in nonpolar solvents (oil dissolves in kerosene).

24. Solubility of Polar Substances
Ethanol is soluble in water because of the polar OH bond.

25. Why is solid sucrose (C12H22O11), table sugar, soluble in water?

26. Substances Insoluble in Water
Nonpolar oil does not interact with polar water.

27. Electrolytes
Electrolytes- conduct an electric current in solution or in molten state.
(ionic compounds)
Nonelectrolytes- do not conduct an electric current in solution or in molten state.
(molecular compounds)

28. Electrolytes
Strong electrolytes: solute completely breaks apart in solution
[includes soluble salts (such as KCl), inorganic acids (such as HNO3), inorganic bases (such as NaOH)]

29. Electrolytes
Weak electrolytes: only a fraction of the solute breaks apart into solution
[includes poorly soluble salts (such as PbCl2), organic acids (such as HC2H3O2), and organic bases (such as NH3)]

30. Electrolytes
Nonelectrolyte – does not conduct when in solution- does not break apart into ions.
[includes most organic compounds (such as glucose)]

31. Water of hydration
The water in a crystal is called water of hydration.
A hydrate is a compound that includes water of hydration.

32. Water of hydration
A hygroscopic substance removes water from the air. These substances are called dessicants.

33. Heterogeneous Aqueous Systems
Suspensions are mixtures from which particles settle out upon standing. (The particles are much larger than those in a solution.)
Colloids are heterogeneous mixtures that contain particles that are intermediate in size between suspensions and true solutions

34. Colloidal Systems
Colloids exhibit the Tyndall effect, which is the scattering of light in all directions.
Colloids include milk, mayonnaise, marshmallows, egg white, blood, and paint.

35. Solutions and their Behavior
Chapter 18

36. Identify factors that determine the rate at which a solute dissolves
Objectives
Identify factors that determine the rate at which a solute dissolves
Identify factors that affect the solubility of a solute in solution
Calculate the solubility of a gas in a liquid under various pressure conditions

37. Factors affecting the rate of dissolving
How could you speed up the dissolving of sugar in a glass of iced tea?
Which dissolves faster, table salt or rock salt?

38. Factors affecting the rate of dissolving
Temperature – increasing the temperature speeds up the rate of dissolving
Agitation – stirring speeds up the rate of dissolving
Particle size – smaller particles dissolve faster than large particles (surface area)

39. Solubility
The solubility- the amount that dissolves in a given quantity of solvent at a given temperature.

40. Solubility
A saturated solution contains the maximum amount of solute at a constant temperature.

41. Solubility
An unsaturated solution contains less solute than a saturated solution.
A supersaturated solution contains more solute than a saturated solution. (This occurs when a solution is saturated and then allowed to cool but all of the solid remains dissolved. It is an unstable solution, adding a crystal causes precipitation.)

42. What would happen…
…if you added more sugar to a saturated sugar solution and stirred?
…if you added more sugar to an unsaturated sugar solution and stirred?

43. Factors affecting the solubility of a substance
How could you increase the amount of sugar that would eventually dissolve in a glass of tea?

44. Factors affecting the solubility of a substance
Only two factors affect the amount of solute that can dissolve.
Temperature affects solubility of both solids and gases in liquid solvents.
Pressure affects solubility of gases in liquid solvents.

45. Factors affecting the solubility of a substance
The solubility of most solid substances increases as the temperature of the solvent increases.
For a few substances, the reverse occurs.

46. The Effect of Temperature on the Solubility of Solids

47. The Effect of Temperature on Gas Solubility
Increasing the temperature of a dissolved gas solution decreases the concentration of the gas.
Have you ever tried a hot Dr. Pepper? Heat it in a pan on the stove, pour a cup, and it has no bubbles!
Thermal pollution occurs when hot water is added to a lake, the dissolved oxygen levels fall in the water and it kills the fish.

48. The Effect of Temperature on Gas Solubility

49. The Effect of Pressure on Gas Solubility
Increasing the pressure of a gas over the surface of a solvent increases the solubility of the gas in the solvent.
In a bottled soda, the pressure of CO2 over the liquid is high and when the cap is opened, the pressure is reduced and bubbles begin to come out of the solution.

50. The Effect of Pressure on Gas Solubility
Henry’s Law: 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.
S1/P1 = S2/P2

51. Question:
If the solubility of a gas in water is 0.77 g/L at 3.5 atm pressure, what is the solubility (in g/L) at 1.0 atm?
(The temperature is held constant at 25oC.)
Answer: S1/P1 = S2/P2
0.77 g/L / 3.5 atm = S2 / 1.0 atm
S2 = 0.22 g/L

52. Solve problems involving the molarity of a solution
Objectives
Solve problems involving the molarity of a solution
Describe how to prepare dilute solutions from concentrated solutions of known molarity.

53. Solution Composition: Molarity
Concentration of a solution is the amount of solute in a given volume of solution.
A concentrated solution has a high concentration of solute.
A diluted solution has only a low concentration of solute.

54. A unit of solution concentration
A common unit of concentration:
Molarity (M)
(Notice that the ratio includes liters of total solution,
not liters of solvent.)

55. Solution Composition: Molarity
Consider both the moles of solute and the volume of solution to find concentration.

56. Solution Composition: Molarity
To find the moles of solute in a given volume of solution of known molarity use the definition of molarity.

57. Solution Preparation: Molarity
To make a molar solution:
Weigh out a sample of solute that contains the necessary moles of solute.
Transfer to a volumetric flask.
Add only enough solvent to mark on flask to get total solution volume.

58. Problem:
Calculate the molarity when 3.0 mol of solute is dissolved in enough water to make 2.0 L of solution.
Answer:
mol solute/L solution = molarity
3.0 mol / 2.0 L = 1.5M

59. Problem
A solution contains 17.0 grams of sodium chloride in 200. mL of solution. What is the molarity of the solution?
Answer: (convert grams to moles)
17.0 grams . (1 mol/58.5 g) = mol
mol solute/L solution = molarity
mol / L = molar NaCl

60. Problem
How many grams of sodium chloride are needed to prepare 275 mL of a 0.200M solution of NaCl?
Answer:
molarity . L solution = mol solute
0.200M L = moles of NaCl
mol . (58.5 g/1 mol) =3.22 grams NaCl

61. Another unit of solution concentration
Another common unit of concentration:
Molality (m)
(Notice that the ratio includes liters of solvent,
not liters of total solution.)

62. Molality
Calculate the molality of a solution prepared by dissolving 45.0 grams of KCl in 500. grams of water.
Answer:
45.0 grams KCl . (74.55 g/1 mol) =0.604 mol
m = mol solute/kg solvent
0.604 mol KCl/ .500 kg water = 1.21 molal

63. Molality
How many grams of sodium chloride must be dissolved in grams of water to prepare a 0.75 molal NaCl solution?
Answer:
mol solute = m . kg solvent
mol = 0.75m kgwater = 0.19 mol NaCl
0.19 mol NaCl . (58.5 g/1 mol) = 11 grams NaCl

64. Dilution
When diluting a solution to a lower concentration, only water is added in the dilution – the amount of solute remains the same. (remember: mol solute = molarity . L solution)
(mol solute)conc = (mol solute)diluted
Mconc . Lconc = Mdiluted . Ldiluted
Mconc . Vconc = Mdiluted . Vdiluted
M1. V1 = M2. V2

65. Dilution Diluting a solution
Transfer a measured amount of original solution to a flask containing some water.
Add water to the flask to the mark (with swirling) and mix by inverting the flask.
M1 . V1 = M2 . V2

66. Problem
How many mL of a concentrated stock solution of 5.0M HCl is needed to prepare 250. mL of 1.75M HCl?
Answer:
M1 . V1 = M2 . V2
5.0M . V1 = 1.75M mL
V1 = 87.5 mL
(notice: you do not have to convert to liters in dilution problems)

67. Calculate the molality and mole fraction of a solution
Objectives
Explain on a particle basis why a solution has a lower vapor pressure than the pure solvent of that solution
Explain on a particle basis why a solution has an elevated boiling point and a depressed freezing point compared to the pure solvent
Calculate the molality and mole fraction of a solution
Calculate the molar mass of a molecular compound from the freezing point depression or boiling point elevation of a solution of the compound

68. Colligative Properties
Colligative property – a solution property that depends on the number, but not the type, of solute particles present
There are three colligative properties of solutions:
Vapor pressure lowering
Boiling point elevation
Freezing point depression

69. Vapor Pressure Lowering
When solute particles are added to a solvent, the solute forms associations with the solvent. Fewer solvent particles are free to vaporize, so the vapor pressure is lower over a solution than over the pure solvent.

70. Colligative Properties
Adding solute particles causes the liquid range to become wider. The lowering of vapor pressure causes the boiling point to rise and the freezing point to fall. ( interferes with crystal formation)
Boiling point increases
Freezing point decreases

71. Colligative Properties
The higher the number of particles (ions, molecules, etc.) the greater the change in the colligative property.
Which would have the greatest elevation in boiling point?
1.0M C6H12O6
1.0M NaCl
1.0M CaCl2
1.0M AlCl3

72. Answer:
All 4 solutions have the same concentration, but the number of particles that form is not the same:
C6H12O6(s)  C6H12O6(aq)
(1 particle)
NaCl(s)  Na+(aq) + Cl-(aq)
(2 particles)
CaCl2(s)  Ca2+(aq) + 2Cl-(aq)
(3 particles)
AlCl3(s)  Al3+ (aq)+ 3Cl-(aq)
(4 particles, greatest effect)
(Remember: ionic compounds dissociate, breaking down into multiple particles, molecular compounds do not.)

73. Calculating Solution Boiling Point
The boiling point is raised by the addition of a nonvolatile solute. To calculate the change in boiling point, you must consider the concentration of the particles in solution.
Remember: water normally boils at 100oC.
DT = Kb i m
change in boiling point =
(boiling point constant) (number if ions) (molality)

74. Problem
What is the boiling point of 2.75m NaCl(aq) solution? (Kb for water is 0.512oC/m)
Answer:
DT = Kb i m
DT = (0.512oC/m) (2) (2.75m)
DT = 2.82oC (this is the change in boiling point)
Boiling point = 100oC oC = oC

75. Calculating Solution Freezing Point
You calculate the freezing point of a solution in a similar way.
Remember: water normally freezes at 0.0oC.
DT = Kf i m
change in freezing point =
(freezing point constant) (number if ions) (molality)

76. Problem
What is the freezing point of a solution that contains 2.50 mol of CaCl2 in 1250 g of water?
(Kf for water is 1.86oC/m)
Answer:
DT = Kf i m
DT = (1.86oC/m) (3) (2.50 mol/1.250 kg)
= 11.2oC (this is the freezing point change)
Freezing point = 0.0oC oC = -11.2oC

77. Determining Molar Mass
You can use the change in boiling point or freezing point to calculate the molar mass of an unknown solute.
If you are given a mass of an unknown solute, you can use the boiling point elevation to determine the molality of the solution, then use the volume of solvent to determine the moles of solute. Use the given mass and the moles of solute to determine the molar mass of the solute.

78. Problem:
The boiling point of water is raised to 102.5oC when grams of a nonvolatile molecular solute is dissolved in 500. g of water. Calculate the molar mass of the solute.
Answer:
DT = Kf i m
2.5oC = (.512oC/m) (1) (mol solute/.500 kg)
mol solute = 2.44 mol
65.50 grams/2.44 mol = 26.8 g/mol