1. Chapter 13 Properties of Solutions

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

3. 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.

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

5. How Does a Solution Form?
If an ionic salt is soluble in water, it is because the ion–dipole interactions are strong enough to overcome the lattice energy of the salt crystal.

6. Energy Changes in Solution
Simply put, three processes affect the energetics of solution:
Separation of solute particles,
Separation of solvent particles,
New interactions between solute and solvent.

7. Energy Changes in Solution
The enthalpy change of the overall process depends on H for each of these steps.

8. 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.

9. Why Do Endothermic Processes Occur?
Yet we know that in some processes, like the dissolution of NH4NO3 in water, heat is absorbed, not released.

10. Enthalpy Is Only Part of the Picture
The reason is that increasing the disorder or randomness (known as entropy) of a system tends to lower the energy of the system.

11. 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.

12. Student, Beware!
Just because a substance disappears when it comes in contact with a solvent, it doesn’t mean the substance dissolved. It may have reacted.

13. Student, Beware!
Dissolution is a physical change—you can get back the original solute by evaporating the solvent.
If you can’t get it back, the substance didn’t dissolve, it reacted.

14. Types of Solutions Saturated
In a saturated solution, the solvent holds as much solute as is possible at that temperature.
Dissolved solute is in dynamic equilibrium with solid solute particles.

15. Types of Solutions Unsaturated
If a solution is unsaturated, less solute than can dissolve in the solvent at that temperature is dissolved in the solvent.

16. Types of Solutions Supersaturated
In supersaturated solutions, the solvent holds more solute than is normally possible at that temperature.
These solutions are unstable; crystallization can usually be stimulated by adding a “seed crystal” or scratching the side of the flask.

17. 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.

18. Factors Affecting Solubility
The more similar the intermolecular attractions, the more likely one substance is to be soluble in another.

19. Factors Affecting Solubility
Glucose (which has hydrogen bonding) is very soluble in water, while cyclohexane (which only has dispersion forces) is not.

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

21. Sample Exercise 13.1 Predicting Solubility Patterns
Predict whether each of the following substances is more likely to dissolve in the nonpolar solvent carbon tetrachloride (CCl4) or in water: C7H16, Na2SO4, HCl, and I2.
Solve:
C7H16 is nonpolar.
Na2SO4, is ionic.
HCl, is polar.
I2, is nonpolar.
We would therefore predict that C7H16 and I2 (the nonpolar solutes) would be more soluble in the nonpolar CCl4 than in polar H2O, whereas water would be the better solvent for Na2SO4 and HCl (the ionic and polar covalent solutes).

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

23. 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.
Start here 1/21/10--8:30 & 10

24. 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, and
Pg is the partial pressure of the gas above the liquid.

25. SCO2 = kPCO2 = (3.4  102 mol/L-atm)(4.0 atm) = 0.14 mol/L = 0.14 M
Sample Exercise 13.2 A Henry’s Law Calculation
Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4.0 atm over the liquid at 25 C. The Henry’s law constant for CO2 in water at this temperature is 3.4  102 mol/L-atm.
Solution
SCO2 = kPCO2 = (3.4  102 mol/L-atm)(4.0 atm) = 0.14 mol/L = 0.14 M

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

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

28. Ways of Expressing Concentrations of Solutions

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

30. 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

31. Sample Exercise 13.3 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:

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

33. Molarity (M ) moles of solute M = liters of solution
You will recall this concentration measure from Chapter 4.
Since volume is temperature-dependent, molarity can change with temperature.

34. Molality (m) moles of solute m = kilograms of solvent
Since both moles and mass do not change with temperature, molality (unlike molarity) is not temperature-dependent.

35. Sample Exercise 13.4 Calculation of Molality
A solution is made by dissolving 4.35 g glucose (C6H12O6) in 25.0 mL of water at 25 C. Calculate the molality of glucose in the solution. Water has a density of 1.00 g/mL.
Solution:
Because water has a density of 1.00 g/mL, the mass of the solvent is: (25.0 mL)(1.00 g/mL) = 25.0 g = kg

36. An aqueous solution of hydrochloric acid contains 36% HCl by
Sample Exercise 13.5 Calculation of Mole Fraction and Molarity
An aqueous solution of hydrochloric acid contains 36% HCl by
mass. (a) Calculate the mole fraction of HCl in the solution.
(b) Calculate the molality of HCl in the solution.

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

38. A solution with a density of 0.876 g/mL contains 5.0 g of toluene
Sample Exercise 13.6 Calculation of Molarity Using the Density of the Solution
A solution with a density of g/mL contains 5.0 g of toluene
(C7H8) and 225 g of benzene. Calculate the molarity of the
solution.
Solution:

39. Colligative Properties
Changes in 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

40. Vapor Pressure
Because of solute–solvent intermolecular attraction, higher concentrations of nonvolatile solutes make it harder for solvent to escape to the vapor phase.

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

42. Raoult’s Law PA = XAPA where
XA is the mole fraction of compound A, and
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.

43. PA = XA P A
Ideal Solution
PB = XB P B
PT = PA + PB
PT = XA P A
+ XB P B

44. < & > & PT is greater than predicted by Raoults’s law
PT is less than
predicted by Raoults’s law
Force
A-B
A-A
B-B
<
&
Force
A-B
A-A
B-B
>
&

45. Glycerin (C3H8O3) is a nonvolatile nonelectrolyte with a density of
Sample Exercise 13.7 Calculation of Vapor Pressure of a Solution
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), and its density is 1.00 g/mL.

46. Boiling-Point Elevation and Freezing-Point Depression
Nonvolatile solute–solvent interactions also cause solutions to have higher boiling points and lower freezing points than the pure solvent.

47. Freezing-Point Depression
Tf = Kf  m
Here Kf is the molal freezing-point depression constant of the solvent (0C/m).
DTf = T f – Tf
T f is the freezing point of
the pure solvent
T f is the freezing point of
the solution
T f > Tf
DTf > 0
Tf is subtracted from the normal freezing point of the solvent.
m is the molality of the solution

48. Boiling-Point Elevation
Tb = Kb  m
where Kb is the molal boiling-point elevation constant (0C/m).
DTb = Tb – T b
T b is the boiling point of
the pure solvent.
T b is the boiling point of
the solution.
Tb > T b
DTb > 0
Tb is added to the normal boiling point of the solvent.
m is the molality of the solution

49. Boiling-Point Elevation and Freezing-Point Depression
Tb = Kb  m
Tf = Kf  m
Note that in both equations, T does not depend on what the solute is, but only on how many particles are dissolved.
Start here 1/26/ AM

50. Tb = Kbm = (0.51 C/m)(5.37 m) = 2.7 C
Sample Exercise 13.8 Calculating Of Boiling-Point Elevation and Freezing-Point Depression
Automotive antifreeze consists of ethylene glycol, CH2(OH)CH2(OH),
a nonvolatile nonelectrolyte. Calculate the boiling point and freezing point of a 25.0 mass % solution of ethylene glycol in water.
Tb = Kbm = (0.51 C/m)(5.37 m) = 2.7 C
Tf = Kfm = (1.86 C/m)(5.37 m) = 10.0 C
Boiling point = (normal bp of solvent) + Tb
= C C = C
Freezing point = (normal fp of solvent) Tf
= 0.0 C  10.0 C = 10.0 C

51. Colligative Properties of Electrolytes
Since the colligative properties of electrolytes depend on the number of particles dissolved, solutions of electrolytes (which dissociate in solution) should show greater changes than those of nonelectrolytes.

52. Colligative Properties of Electrolytes
However, a 1M solution of NaCl does not show twice the change in freezing point that a 1M solution of methanol does.

53. van’t Hoff Factor
One mole of NaCl in water does not really give rise to two moles of ions.
Some Na+ and Cl reassociate for a short time, so the true concentration of particles is somewhat less than two times the concentration of NaCl.

54. van’t Hoff Factor
Reassociation is more likely at higher concentration.
Therefore, the number of particles present is concentration-dependent.
We modify the previous equations by multiplying by the van’t Hoff factor, i: Tf = Kf  m  i

55. List the following aqueous solutions in order of their expected
Sample Exercise 13.9 Freezing-Point Depression in Aqueous Solutions
List the following aqueous solutions in order of their expected
freezing point: m CaCl2, 0.15 m NaCl, 0.10 m HCl,
0.050 m CH3COOH, 0.10 m C12H22O11.
Because the freezing points depend on the total molality of particles in solution, the expected ordering is 0.15 m
NaCl (lowest freezing point), 0.10 m HCl, m CaCl2, 0.10 m C12H22O11, and m CH3COOH (highest freezing point).

56. Osmosis
Some substances form semipermeable membranes, allowing some smaller particles to pass through, but blocking other larger particles.
In biological systems, most semipermeable membranes allow water to pass through, but solutes are not free to do so.

57. Osmosis
In osmosis, there is net movement of solvent from the area of higher solvent concentration (lower solute concentration) to the area of lower solvent concentration (higher solute concentration).

58. Osmotic Pressure n P = ( )RT = MRT V
The pressure required to stop osmosis, known as osmotic pressure, P, is
P = ( )RT = MRT n V
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.

59. 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.

60. 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.

61. The average osmotic pressure of blood is 7.7 atm at 25 C. What
Sample Exercise Calculating Involving Osmotic Pressure
The average osmotic pressure of blood is 7.7 atm at 25 C. What
molarity of glucose (C6H12O6) will be isotonic with blood?
Solution:

62. A solution of an unknown nonvolatile nonelectrolyte was prepared
Sample Exercise Molar Mass from Freezing-Point Depression
A solution of an unknown nonvolatile nonelectrolyte was prepared
by dissolving g of the substance in 40.0 g of CCl4. The
boiling point of the resultant solution was C higher Than
that of the pure solvent. Calculate the molar mass of the solute.

63. The osmotic pressure of an aqueous solution of a certain protein
Sample Exercise Molar Mass from Osmotic Pressure
The osmotic pressure of an aqueous solution of a certain protein
was measured to determine the protein’s molar mass. The
solution contained 3.50 mg of protein dissolved in sufficient water
to form 5.00 mL of solution. The osmotic pressure of the solution
at 25 C was found to be 1.54 torr. Treating the protein as
a nonelectrolyte, calculate its molar mass.
Solution:

64. Colligative Properties of Ionic Solutions
For ionic solutions we must take into account the number of ions present
i = van’t Hoff factor or the number of ions present
For vapor pressure lowering: P = i XsoluteP 0solvent
For boiling point elevation: Tb = i Kb m
For freezing point depression: Tf = i Kf m
For osmotic pressure:  = i MRT
im = conc. of particles

65. Colloids
Suspensions of particles larger than individual ions or molecules, but too small to be settled out by gravity, are called colloids.

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

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

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