Chapter 6: Solutions, Acids, And Bases Goss
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Transcript Chapter 6: Solutions, Acids, And Bases Goss
Colligative Properties of Solutions
Colligative properties = physical
properties of solutions that depend
on the # of particles dissolved, not
the kind of particle.
Colligative Properties
Lowering vapor pressure
Raising boiling point
Lowering freezing point
Generating an osmotic pressure
Colligative Properties
Lowering vapor pressure
Raising boiling point
Lowering freezing point
Generating an osmotic pressure
Boiling Point Elevation
a solution that contains a nonvolatile
solute has a higher boiling pt than the
pure solvent; the boiling pt elevation is
proportional to the # of moles of solute
dissolved in a given mass of solvent.
Boiling Point Elevation
Tb
where:
= kb m
Tb = elevation of boiling pt
m = molality of solute
kb = the molal boiling pt elevation constant
kb values are constants; see table 15-4, p. 472
(honors text)
kb for water = 0.52 °C/m
Ex: What is the normal boiling pt of a 2.50
m glucose, C6H12O6, solution?
“normal” implies 1 atm of pressure
Tb = kbm
Tb = (0.52 C/m)(2.50 m)
Tb = 1.3 C
Tb = 100.0 C + 1.3 C = 101.3 C
Ex: How many grams of glucose, C6H12O6,
would need to be dissolved in 535.5 g of water
to produce a solution that boils at 101.5°C?
Tb = kbm
1.5 C= (0.52 C/m)(m)
m = 2.885
x
2.885 m
0.5355 kg
x 1.5449 mol 180g 278.1 g 280g
1 mol
Freezing/Melting Point Depression
The freezing point of a solution is
always lower than that of the pure
solvent.
Freezing/Melting Point
Depression
Tf
= kfm
where:
Tf = lowering of freezing point
m = molality of solute
kf = the freezing pt depression constant
kf for water = 1.86 °C/m
kf values are constants;
see table 15-5, p. 474 (honors text)
Ex: Calculate the freezing pt of a
2.50 m glucose solution.
Tf = kfm
Tf = (1.86 C/m)(2.50 m)
Tf = 4.65 C
Tf = 0.00C - 4.65 C = -4.65C
Ex: When 15.0 g of ethyl alcohol, C2H5OH, is
dissolved in 750 grams of formic acid, the freezing
pt of the solution is 7.20°C. The freezing pt of pure
formic acid is 8.40°C. Determine Kf for formic acid.
15.0 g C2H5OH 1mol
0.3261mol
46 g
0.3261mol
0.4348m
0.75 kg
Tf = kfm
1.20 C= (kf)( 0.4348 m)
kf = 2.8 C/m
Ex: An antifreeze solution is prepared containing 50.0 cm3
of ethylene glycol, C2H6O2, (d = 1.12 g/cm3), in 50.0 g
water. Calculate the freezing point of this 50-50 mixture.
Would this antifreeze protect a car in Chicago on a day
when the temperature gets as low as –10° F?
(-10 °F = -23.3° C)
50.0 cm3 C2H6O2
1.12 g
1 mol
0.90323 mol
3
62.0 g
cm
0.90323 mol
18.06 m
0.050 kg
Tf = kfm
Tf = (1.86C/m)(18.06 m)
Tf = 33.6 C
Tf = 0 C – 33.6 C = -33.6 C
YES!
Electrolytes and Colligative
Properties
• Colligative properties depend on the # of
particles present in solution.
• Because ionic solutes dissociate into ions,
they have a greater effect on freezing pt and
boiling pt than molecular solids of the same
molal conc.
Electrolytes and Colligative
Properties
For example, the freezing pt of water is
lowered by 1.86°C with the addition of any
molecular solute at a concentration of 1 m.
– Such as C6H12O6, or any other covalent
compound
However, a 1 m NaCl solution contains 2
molal conc. of IONS. Thus, the freezing pt
depression for NaCl is 3.72°C…double that of
a molecular solute.
– NaCl Na+ + Cl- (2 particles)
Electrolytes - Boiling Point Elevation and
Freezing Point Depression
The relationships are given by the following equations:
Tf = kf ·m·n
or
Tb = kb·m·n
Tf/b = f.p. depression/elevation of b.p.
m = molality of solute
kf/b = b.p. elevation/f.p depression constant
n = # particles formed from the dissociation of
each formula unit of the solute
Ex: What is the freezing pt of:
a) a 1.15 m sodium chloride solution?
NaCl Na+ + Cl-
n=2
Tf = kf·m·n
Tf = (1.86 C/m)(1.15 m)(2)
Tf = 4.28 C
Tf = 0.00C - 4.28 C = -4.28C
Ex: What is the freezing pt of:
b) a 1.15 m calcium chloride solution?
CaCl2 Ca2+ + 2Cl-
n=3
Tf = kf·m·n
Tf = (1.86 C/m)(1.15 m)(3)
Tf = 6.42 C
Tf = 0.00C – 6.42 C = -6.42C
Ex: What is the freezing pt of:
c) a 1.15 m calcium phosphate solution?
Ca3(PO4)2 3Ca2+ + 2PO43n=5
Tf = kf·m·n
Tf = (1.86 C/m)(1.15 m)(5)
Tf = 10.7 C
Tf = 0.0C – 10.7 C = -10.7C
Determining Molecular Weights
by Freezing Point Depression
Ex: A 1.20 g sample of an unknown molecular compound is dissolved in
50.0 g of benzene. The solution freezes at 4.92°C. Determine the
molecular weight of the compound. The freezing pt of pure benzene
is 5.48°C and the Kf for benzene is 5.12°C/m.
Tf = 0.56°C
Tf = kf·m
0.56°C = (5.12°C/m)(m)
m = 0.1094
x mol
0.1 094m
0.050 kg
x 0.00547mol
1.20g
Molar Mass
219g mol
0.00547mol
Ex: A 37.0 g sample of a new covalent compound was dissolved in 200.0
g of water. The resulting solution froze at –5.58°C. What is the
molecular weight of the compound?
Tf = 5.58°C
Tf = kf·m
5.58°C = (1.86°C/m)(m)
m = 3.00 m
x mol
3.00m
0.200 kg
x 0.60mol
37.0g
Molar Mass
61.7 g mol
0.60mol