2. Chapter 13 Properties of Solutions
Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Chapter 13 Properties of Solutions
Adapted by SA Green from:
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.

3. Solutions
Solutions are homogeneous mixtures of two or more pure substances.
In a solution, the solute is dispersed uniformly throughout the solvent.

4. Solutions How does a solid dissolve into a liquid?
What ‘drives’ the dissolution process?
What are the energetics of dissolution?

5. How Does a Solution Form?
Solvent molecules attracted to surface ions.
Each ion is surrounded by solvent molecules.
Enthalpy (DH) changes with each interaction broken or formed.
Ionic solid dissolving in water

6. How Does a Solution Form?
Solvent molecules attracted to surface ions.
Each ion is surrounded by solvent molecules.
Enthalpy (DH) changes with each interaction broken or formed.

7. How Does a Solution Form
The ions are solvated (surrounded by solvent).
If the solvent is water, the ions are hydrated.
The intermolecular force here is ion-dipole.

8. Energy Changes in Solution
To determine the enthalpy change, we divide the process into 3 steps.
Separation of solute particles.
Separation of solvent particles to make ‘holes’.
Formation of new interactions between solute and solvent.

9. Enthalpy Changes in Solution
The enthalpy change of the overall process depends on H for each of these steps.
Start
End
Start
End

10. Enthalpy changes during dissolution
DHsoln = DH1 + DH2 + DH3
The enthalpy of solution, DHsoln, can be either positive or negative.
DHsoln (MgSO4)= kJ/mol --> exothermic
DHsoln (NH4NO3)= 26.4 kJ/mol --> endothermic

11. Why do endothermic processes sometimes occur spontaneously?
Some processes, like the dissolution of NH4NO3 in water, are spontaneous at room temperature even though heat is absorbed, not released.

12. Enthalpy Is Only Part of the Picture
Entropy is a measure of:
Dispersal of energy in the system.
Number of microstates (arrangements) in the system.
b. has greater entropy,  is the favored state
(more on this in chap 19)

13. Entropy changes during dissolution
Each step also involves a change in entropy.
Separation of solute particles.
Separation of solvent particles to make ‘holes’.
Formation of new interactions between solute and solvent.

14. Does the entropy of the system increase or decrease?
SAMPLE EXERCISE 13.1 Assessing Entropy Change
In the process illustrated below, water vapor reacts with excess solid sodium sulfate to form the hydrated form of the salt. The chemical reaction is
Does the entropy of the system increase or decrease?

15. Dissolution vs reaction
dry
Ni(s) + HCl(aq)
NiCl2(aq) + H2(g)
NiCl2(s)
Dissolution is a physical change—you can get back the original solute by evaporating the solvent.
If you can’t, the substance didn’t dissolve, it reacted.

16. Degree of saturation Saturated solution
Solvent holds as much solute as is possible at that temperature.
Undissolved solid remains in flask.
Dissolved solute is in dynamic equilibrium with solid solute particles.

17. Degree of saturation Unsaturated Solution
Less than the maximum amount of solute for that temperature is dissolved in the solvent.
No solid remains in flask.

18. Degree of saturation Supersaturated
Solvent holds more solute than is normally possible at that temperature.
These solutions are unstable; crystallization can often be stimulated by adding a “seed crystal” or scratching the side of the flask.

19. Degree of saturation  How much solute can be dissolved in a solution?
Unsaturated, Saturated or Supersaturated?
 How much solute can be dissolved in a solution?
More on this in Chap 17
(solubility products, p 739)

20. Factors Affecting Solubility
Chemists use the axiom “like dissolves like”:
Polar substances tend to dissolve in polar solvents.
Nonpolar substances tend to dissolve in nonpolar solvents.

21. Factors Affecting Solubility
The stronger the intermolecular attractions between solute and solvent, the more likely the solute will dissolve.
Example: ethanol in water
Ethanol = CH3CH2OH
Intermolecular forces = H-bonds; dipole-dipole; dispersion
Ions in water also have ion-dipole forces.

22. Factors Affecting Solubility
Glucose (which has hydrogen bonding) is very soluble in water.
Cyclohexane (which only has dispersion forces) is not water-soluble.

23. Factors Affecting Solubility
Vitamin A is soluble in nonpolar compounds (like fats).
Vitamin C is soluble in water.

24. Which vitamin is water-soluble and which is fat-soluble?
Figure: UNE13.06
Title:
Exercise 13.6
Caption:
Vitamins E and B6.

25. Gases in Solution
In general, the solubility of gases in water increases with increasing mass.
Why?
Larger molecules have stronger dispersion forces.

26. Gases in Solution

27. Gases in Solution
The solubility of liquids and solids does not change appreciably with pressure.
But, the solubility of a gas in a liquid is directly proportional to its pressure.
Increasing pressure above solution forces more gas to dissolve.

28. Henry’s Law Sg = kPg where Sg is the solubility of the gas;
k is the Henry’s law constant for that gas in that solvent;
Pg is the partial pressure of the gas above the liquid.

29. Henry’s Law
k for N2 at 25°
=6.8 x 10-4 mol/L atm
Sg = kPg

30. Temperature
Generally, the solubility of solid solutes in liquid solvents increases with increasing temperature.

31. Temperature
The opposite is true of gases. Higher temperature drives gases out of solution.
Carbonated soft drinks are more “bubbly” if stored in the refrigerator.
Warm lakes have less O2 dissolved in them than cool lakes.

32. Chap 13: Ways of Expressing Concentrations of Solutions

33. Mass Percentage mass of A in solution Mass % of A =
total mass of solution
Mass % of A =
 100

34. Parts per Million and Parts per Billion
Parts per Million (ppm)
mass of A in solution
total mass of solution
ppm =
 106
Parts per Billion (ppb)
mass of A in solution
total mass of solution
ppb =
 109

35. total moles in solution
Mole Fraction (X)
moles of A
total moles in solution
XA =
In some applications, one needs the mole fraction of solvent, not solute—make sure you find the quantity you need!

36. Molarity (M) mol of solute M = L of solution
You will recall this concentration measure from Chapter 4.
Because volume is temperature dependent, molarity can change with temperature.

37. Molality (m) mol of solute m = kg of solvent
Because neither moles nor mass change with temperature, molality (unlike molarity) is not temperature dependent.

38. Mass/Mass
Moles/Moles
Moles/L
Moles/Mass

39. Changing Molarity to Molality
If we know the density of the solution, we can calculate the molality from the molarity, and vice versa.

40. SAMPLE EXERCISE 13.4 Calculation of Mass-Related Concentrations
(a) A solution is made by dissolving 13.5 g of glucose (C6H12O6) in kg of water. What is the mass percentage of solute in this solution? (b) A 2.5-g sample of groundwater was found to contain 5.4g of Zn2+ What is the concentration of Zn2+ in parts per million?
PRACTICE EXERCISE
(a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. (b) A commercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution?
PRACTICE EXERCISE
A commercial bleach solution contains 3.62 mass % NaOCl in water. Calculate (a) the molality and (b) the mole fraction of NaOCl in the solution.

41. SAMPLE EXERCISE 13.4 Calculation of Mass-Related Concentrations
(a) A solution is made by dissolving 13.5 g of glucose (C6H12O6) in kg of water. What is the mass percentage of solute in this solution? (b) A 2.5-g sample of groundwater was found to contain 5.4g of Zn2+ What is the concentration of Zn2+ in parts per million?
Solution
Analyze: We are given the number of grams of solute (13.5 g) and the number of grams of solvent
(0.100 kg = 100 g). From this we must calculate the mass percentage of solute.
Plan: We can calculate the mass percentage by using Equation The mass of the solution is the sum of the mass of solute (glucose) and the mass of solvent (water).
Solve: 
Comment: The mass percentage of water in this solution is (100 – 11.9)% = 88.1%.
(b) Analyze: In this case we are given the number of micrograms of solute. Because 1g is 1  10–6 g,
5.4g = 5.4  10–6 g.
Plan: We calculate the parts per million using Equation 13.6.
Solve:  
PRACTICE EXERCISE
(a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. (b) A commercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution?
Answers: (a) 2.91%, (b) 90.5 g of NaOCl
Answers: (a) m, (b) 9.00  10–3

42. Colligative Properties
Colligative properties depend only on the number of solute particles present, not on the identity of the solute particles.
Among colligative properties are
Vapor pressure lowering
Boiling point elevation
Melting point depression
Osmotic pressure

43. Vapor Pressure
As solute molecules are added to a solution, the solvent become less volatile (=decreased vapor pressure).
Solute-solvent interactions contribute to this effect.

44. Vapor Pressure
Therefore, the vapor pressure of a solution is lower than that of the pure solvent.

45. Raoult’s Law PA = XAPA where XA is the mole fraction of compound A
PA is the normal vapor pressure of A at that temperature
NOTE: This is one of those times when you want to make sure you have the vapor pressure of the solvent.

46. SAMPLE EXERCISE 13.8 Calculation of Vapor-Pressure Lowering
Glycerin (C3H8O3) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25°C. Calculate the vapor pressure at 25°C of a solution made by adding 50.0 mL of glycerin to mL of water. The vapor pressure of pure water at 25°C is 23.8 torr (Appendix B).
PRACTICE EXERCISE
The vapor pressure of pure water at 110°C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110°C. Assuming that Raoult’s law is obeyed, what is the mole fraction of ethylene glycol in the solution?

47. SAMPLE EXERCISE 13.8 Calculation of Vapor-Pressure Lowering
Glycerin (C3H8O3) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25°C. Calculate the vapor pressure at 25°C of a solution made by adding 50.0 mL of glycerin to mL of water. The vapor pressure of pure water at 25°C is 23.8 torr (Appendix B).
Solution
Analyze: Our goal is to calculate the vapor pressure of a solution, given the volumes of solute and solvent and the density of the solute.
Plan: We can use Raoult’s law (Equation 13.10) to calculate the vapor pressure of a solution. The mole fraction of the solvent in the solution, XA, is the ratio of the number of moles of solvent (H2O) to total solution (moles C3H8O3 + moles H2O).
Solve: To calculate the mole fraction of water in the solution, we must determine the number of moles of C3H8O3 and H2O:

48. SAMPLE EXERCISE 13.8 continued
The vapor pressure of the solution has been lowered by 0.6 torr relative to that of pure water.
We now use Raoult’s law to calculate the vapor pressure of water for the solution:
PRACTICE EXERCISE
The vapor pressure of pure water at 110°C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110°C. Assuming that Raoult’s law is obeyed, what is the mole fraction of ethylene glycol in the solution?
Answer: 0.290

49. Boiling Point Elevation and Freezing Point Depression
Solute-solvent interactions also cause solutions to have higher boiling points and lower freezing points than the pure solvent.

50. Boiling Point Elevation
The change in boiling point is proportional to the molality of the solution:
Tb = Kb  m
where Kb is the molal boiling point elevation constant, a property of the solvent.
Tb is added to the normal boiling point of the solvent.

51. Freezing Point Depression
The change in freezing point can be found similarly:
Tf = Kf  m
Here Kf is the molal freezing point depression constant of the solvent.
Tf is subtracted from the normal freezing point of the solvent.

52. Boiling Point Elevation and Freezing Point Depression
In both equations, T does not depend on what the solute is, but only on how many particles are dissolved.
Tb = Kb  m
Tf = Kf  m

53. Colligative Properties of Electrolytes
Because these properties depend on the number of particles dissolved, solutions of electrolytes (which dissociate in solution) show greater changes than those of nonelectrolytes.
e.g. NaCl dissociates to form 2 ion particles; its limiting van’t Hoff factor is 2.

54. Colligative Properties of Electrolytes
However, a 1 M solution of NaCl does not show twice the change in freezing point that a 1 M solution of methanol does.
It doesn’t act like there are really 2 particles.

55. van’t Hoff Factor
One mole of NaCl in water does not really give rise to two moles of ions.

56. van’t Hoff Factor
Some Na+ and Cl− reassociate as hydrated ion pairs, so the true concentration of particles is somewhat less than two times the concentration of NaCl.

57. The van’t Hoff Factor
Reassociation is more likely at higher concentration.
Therefore, the number of particles present is concentration dependent.

58. The van’t Hoff Factor Tf = Kf  m  i
We modify the previous equations by multiplying by the van’t Hoff factor, i
Tf = Kf  m  i
i = 1 for non-elecrtolytes

59. Osmosis
Semipermeable membranes allow some particles to pass through while blocking others.
In biological systems, most semipermeable membranes (such as cell walls) allow water to pass through, but block solutes.

60. Osmosis
In osmosis, there is net movement of solvent from the area of higher solvent concentration (lower solute concentration) to the are of lower solvent concentration (higher solute concentration).
Water tries to equalize the concentration on both sides until pressure is too high.

61. Osmotic Pressure n  = ( )RT = MRT V
The pressure required to stop osmosis, known as osmotic pressure, , is n V
 = ( )RT = MRT
where M is the molarity of the solution
If the osmotic pressure is the same on both sides of a membrane (i.e., the concentrations are the same), the solutions are isotonic.

62. Osmosis in Blood Cells
If the solute concentration outside the cell is greater than that inside the cell, the solution is hypertonic.
Water will flow out of the cell, and crenation results.

63. Osmosis in Cells
If the solute concentration outside the cell is less than that inside the cell, the solution is hypotonic.
Water will flow into the cell, and hemolysis results.

65. Molar Mass from Colligative Properties
We can use the effects of a colligative property such as osmotic pressure to determine the molar mass of a compound.

66. Colloids:
Suspensions of particles larger than individual ions or molecules, but too small to be settled out by gravity.

67. Tyndall Effect Colloidal suspensions can scatter rays of light.
This phenomenon is known as the Tyndall effect.

68. Colloids in Biological Systems
Some molecules have a polar, hydrophilic (water-loving) end and a nonpolar, hydrophobic (water-hating) end.

69. Colloids in Biological Systems
Sodium stearate is one example of such a molecule.

70. Colloids in Biological Systems
These molecules can aid in the emulsification of fats and oils in aqueous solutions.

71. END Chap 13

72. Why Do Endothermic Processes Occur?
Things do not tend to occur spontaneously (i.e., without outside intervention) unless the energy of the system is lowered.

73. Enthalpy Is Only Part of the Picture
So even though enthalpy may increase, the overall energy of the system can still decrease if the system becomes more disordered.

74. Student, Beware!
Just because a substance disappears when it comes in contact with a solvent, it doesn’t mean the substance dissolved.