1. occurs when neutral combinations of particles separate into ions while in aqueous solution.
Dissociation
sodium chloride
NaCl  Na Cl1–
sodium hydroxide
NaOH  Na OH1–
hydrochloric acid
HCl  H Cl1–
sulfuric acid
H2SO4  2 H SO42–
acetic acid
CH3COOH  CH3COO1– H1+
In general, acids yield hydrogen ions
(H1+) ?
in aqueous solution; bases yield hydroxide ions.
(OH1–) ?

2. Strong electrolytes exhibit nearly 100% dissociation.
NaCl Na Cl1–
NOT in water:
in aq. solution:
Weak electrolytes exhibit little dissociation.
CH3COOH CH3COO1– H1+
NOT in water:
in aq. solution:
“Strong” or “weak” is a property of the substance.
We can’t change one into the other.

3. electrolytes: solutes that dissociate in solution
-- conduct electric current because of free-moving ions
e.g., acids, bases, most ionic compounds
-- are crucial for many cellular processes
-- obtained in a healthy diet
-- For sustained exercise or a bout of the flu, sports drinks
ensure adequate electrolytes.
nonelectrolytes: solutes that DO NOT dissociate
-- DO NOT conduct electric current (not enough ions)
e.g., any type of sugar

4. Colligative Properties  depend on concentration of a solution
Compared to solvent’s… a solution w/that solvent has a…
…normal freezing point (NFP) …lower FP
FREEZING PT.
DEPRESSION
…normal boiling point (NBP) …higher BP
BOILING PT.
ELEVATION

5. Applications (NOTE: Data are fictitious.)
1. salting roads in winter FP BP
water
0oC (NFP)
100oC (NBP)
water + a little salt
–11oC
103oC
water + more salt
–18oC
105oC
2. antifreeze (AF) /coolant FP BP
water
0oC (NFP)
100oC (NBP)
water + a little AF
–10oC
110oC
50% water + 50% AF
–35oC
130oC
Why do you think some towns use calcium chloride on roads in the winter versus sodium chloride?
CaCl2 yields three ions while NaCl yields only two ions. Calcium chloride will work in colder weather. Calcium chloride will work better than sodium chloride. Secondly, calcium chloride doesn’t kill grass like sodium chloride.

6. 3. law enforcement white powder starts melting at… finishes
penalty, if
convicted A
109oC
175oC
comm. service B
150oC
180oC
2 years C
194oC
196oC
20 years
Drugs, especially pharmaceutical drugs, are tested for purity. A simple test is to measure the melting point range of the substance. A pure sample will have a very narrow range for its melting point. As impurities are introduced into the sample, an increase in the melting point range is observed.

7. Effect of Pressure on Boiling Point
Boiling Point of Water at Various Locations
Location
Feet above sea level
Patm
(kPa)
Boiling Point
(C)
Top of Mt. Everest, Tibet
29,028 32 70
Top of Mt. Denali, Alaska
20,320
45.3 79
Top of Mt. Whitney, California
14,494
57.3 85
Leadville, Colorado
10,150 68 89
Top of Mt. Washington, N.H.
6,293
78.6 93
Boulder, Colorado
5,430
81.3 94
Madison, Wisconsin
900
97.3 99
New York City, New York 10
101.3
100
Death Valley, California
-282
102.6
100.3

8. Calculations for Colligative Properties
The change in FP or BP is found using… DTx = Kx m i
DTx = change in To (below NFP or above NBP)
Kx = constant depending on… (A) solvent
(B) freezing or boiling
m = molality of solute = mol solute / kg solvent
i = integer that accounts for any solute dissociation
any sugar (all nonelectrolytes)……………...i = 1
table salt, NaCl  Na1+ + Cl1–………………i = 2
barium bromide, BaBr2  Ba Br1–……i = 3

9. Freezing Point Depression Boiling Point Elevation
DTf = Kf m i DTb = Kb m i
Then use these in conjunction with the NFP and NBP to
find the FP and BP of the mixture.
The normal boiling point of a substance is the temperature at which the vapor pressure equals 1 atm.
Because the vapor pressure of the solution at a given temperature is lower than the vapor pressure of the pure solvent, achieving a vapor pressure of 1 atm for the solution requires a higher temperature than the normal boiling point of the solvent.
The boiling point of a solution is always higher than that of the pure solvent.
The magnitude of the increase in the boiling point is related to the magnitude of the decrease in the vapor pressure; the decrease in the vapor pressure is proportional to the concentration of the solute in solution.
The boiling point elevation (Tb) is defined as the difference between the boiling points of the solution and the pure solvent:
Tb = Tb – T0b
where Tb is the boiling point of the solution and T0b is the boiling point of the pure solvent.
The relationship between Tb and concentration can be expressed as
Tb = mKb
where m is the concentration of the solute expressed in molality, and Kb is the molal boiling-point elevation constant of the solvent, which has units of ºC/m.
Concentration of the solute is expressed as molality rather than mole fraction or molarity for two reasons:
1. Because the density of a solution changes with temperature, the value of molarity also varies with temperature, and if the boiling point depends on the solute concentration, then the system is not maintained at a constant temperature.
2. Molality and mole fraction are proportional for relatively dilute solutions, but molality has a larger numerical value and eliminates nonsignificant zeros.
• Boiling-point elevation depends on the total number of dissolved nonvolatile solute particles.
Dissolving a nonvolatile solute in water not only raises the boiling point of the water but also lowers its freezing point.
The freezing-point depression (Tf) is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution:
Tf = T0f – Tf
where T0f is the freezing point of the pure solvent and Tf is the freezing point of the solution:
The relationship between Tf and the solute concentration is
Tf = mKf
where m is the molality of the solution and Kf is the molal freezing-point depression constant for the solvent in units of ºC/m.
Freezing-point depression depends on the total number of dissolved nonvolatile solute particles.
The molar mass of an unknown compound can be determined by measuring the freezing point of a solution that contains a known mass of solute, which is accurate for dilute solutions.
Changes in the boiling point are smaller than changes in the freezing point for a given solvent, so boiling point elevations are difficult to measure and are not used to determine molar mass.
(Kf = cryoscopic constant, which is 1.86 K kg/mol for the freezing point of water)
(Kb = ebullioscopic constant, which is 0.51 K kg/mol for the boiling point of water)

10. 168 g glucose (C6H12O6) are mixed w/2.50 kg H2O.
Find BP and FP of mixture. For H2O, Kb = 0.512, Kf = –1.86.
i = 1
(NONELECTROLYTE)
DTb = Kb m i = (0.373) (1) = 0.19oC
BP = ( )oC = oC
DTf = Kf m i = –1.86 (0.373) (1) = –0.69oC
FP = (0 + –0.69)oC = –0.69oC

11. 168 g cesium bromide are mixed w/2. 50 kg H2O
168 g cesium bromide are mixed w/2.50 kg H2O. Find BP and FP of mixture.
For H2O, Kb = 0.512, Kf = –1.86.
Cs1+ Br1– i = 2
CsBr  Cs Br1–
DTb = Kb m i = (0.316) (2) = 0.32oC
BP = ( )oC = oC
DTf = Kf m i = –1.86 (0.316) (2) = –1.18oC
FP = (0 + –1.18)oC = –1.18oC