Transcript Solutions Chapter 14

Chemistry and Chemical Reactivity
6th Edition
1
John C. Kotz
Paul M. Treichel
Gabriela C. Weaver
CHAPTER 14
Solutions and Their Behavior
Lectures written by John Kotz
©2006
2006
Brooks/Cole
Thomson
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Brooks/Cole
- Thomson
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Solutions
Chapter 14
Why does a raw egg swell or shrink when
placed in different solutions?
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Some Definitions
A solution is a
One constituent is usually
HOMOGENEOUS
regarded as the SOLVENT
mixture of 2 or more
and the others as
substances in a single SOLUTES.
phase.
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Definitions
Solutions can be
classified as
saturated or
unsaturated.
A saturated solution
contains the maximum
quantity of solute that
dissolves at that
temperature.
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Definitions
Solutions can be
classified as
unsaturated or
saturated.
A saturated solution
contains the maximum
quantity of solute that
dissovles at that
temperature.
SUPERSATURATED
SOLUTIONS contain
more than is possible
and are unstable.
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Dissolving An Ionic Solid
Active Figure 14.9
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Energetics
of the Solution Process
Figure 14.8
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Energetics
of the Solution Process
If the enthalpy of
formation of the
solution is more
negative that that of
the solvent and
solute, the enthalpy
of solution is
negative.
The solution process
is exothermic!
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Supersaturated
Sodium Acetate
• One application of a
supersaturated
solution is the
sodium acetate
“heat pack.”
• Sodium acetate has
an ENDOthermic
heat of solution.
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Supersaturated
Sodium
Acetate
Sodium acetate has an ENDOthermic heat of
solution.
NaCH3CO2 (s) + heat ---->
Na+(aq) + CH3CO2-(aq)
Therefore, formation of solid sodium acetate
from its ions is EXOTHERMIC.
Na+(aq) + CH3CO2-(aq) --->
NaCH3CO2 (s) + heat
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Colligative Properties
On adding a solute to a solvent, the props. of
the solvent are modified.
• Vapor pressure
decreases
• Melting point
decreases
• Boiling point
increases
• Osmosis is possible (osmotic pressure)
These changes are called COLLIGATIVE
PROPERTIES.
They depend only on the NUMBER of solute
particles relative to solvent particles, not on
the KIND of solute particles.
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Concentration Units
An IDEAL SOLUTION is
one where the properties
depend only on the
concentration of solute.
Need conc. units to tell us the
number of solute particles
per solvent particle.
The unit “molarity” does not
do this!
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Concentration Units
MOLE FRACTION, X
For a mixture of A, B, and C
mol A
X A  mol fraction A =
mol A + mol B + mol C
MOLALITY, m
mol solute
m of solute =
kilograms solvent
WEIGHT % = grams solute per 100 g solution
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Calculating Concentrations
Dissolve 62.1 g (1.00 mol) of ethylene glycol
in 250. g of H2O. Calculate mol fraction,
molality, and weight % of glycol.
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Calculating Concentrations
Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g
of H2O. Calculate X, m, and % of glycol.
250. g H2O = 13.9 mol
1.00 mol glycol
X glycol =
1.00 mol glycol + 13.9 mol H2O
X glycol = 0.0672
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Calculating Concentrations
Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g
of H2O. Calculate X, m, and % of glycol.
Calculate molality
1.00 mol glycol
conc (molality) =
 4.00 molal
0.250 kg H2O
Calculate weight %
62.1 g
%glycol =
x 100% = 19.9%
62.1 g + 250. g
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Dissolving Gases &
Henry’s Law
Gas solubility (mol/L) = kH • Pgas
kH for O2 = 1.66 x 10-6 M/mmHg
When Pgas drops, solubility drops.
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Understanding
Colligative Properties
To understand colligative properties, study
the LIQUID-VAPOR EQUILIBRIUM for a
solution.
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Understanding
Colligative Properties
To understand
colligative
properties,
study the
LIQUID-VAPOR
EQUILIBRIUM
for a solution.
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Understanding
Colligative Properties
VP of H2O over a solution depends on the
number of H2O molecules per solute
molecule.
Psolvent proportional to Xsolvent
Psolvent = Xsolvent • Posolvent
VP of solvent over solution
= (Mol frac solvent)•(VP pure solvent)
RAOULT’S LAW
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Raoult’s Law
An ideal solution is one that obeys Raoult’s
law.
PA = XA • PoA
Because mole fraction of solvent, XA, is always
less than 1, then PA is always less than PoA.
The vapor pressure of solvent over a solution is
always LOWERED!
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Raoult’s Law
Assume the solution containing 62.1 g of
glycol in 250. g of water is ideal. What is the
vapor pressure of water over the solution at
30 oC? (The VP of pure H2O is 31.8 mm Hg; see
App. E.)
Solution
Xglycol = 0.0672 and so Xwater = ?
Because Xglycol + Xwater = 1
Xwater = 1.000 - 0.0672 = 0.9328
Pwater = Xwater • Powater = (0.9382)(31.8 mm Hg)
Pwater = 29.7 mm Hg
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Raoult’s Law
For a 2-component system where A is the
solvent and B is the solute
∆PA = VP lowering = XBPoA
VP lowering is proportional to mol frac solute!
For very dilute solutions, ∆PA = K•molalityB
where K is a proportionality constant.
This helps explain changes in melting and
boiling points.
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Changes in Freezing and Boiling
Points of Solvent
VP Pure solvent
1 atm
VP solvent
after adding
solute
P
BP solution
BP pure
solvent
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See Figure 14.14
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Vapor
Pressure
Lowering
Figure 14.14
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The boiling point of a
solution is higher than that
of the pure solvent.
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Elevation of Boiling Point
Elevation in BP = ∆TBP = KBP•m
(where KBP is characteristic of solvent)
VP Pure solvent
1 atm
VP solvent
after adding
solute
P
BP solution
BP pure
solvent
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Change in Boiling Point
Dissolve 62.1 g of glycol (1.00 mol) in 250. g
of water. What is the BP of the solution?
KBP = +0.512 oC/molal for water (see Table
14.3).
Solution
1.
Calculate solution molality = 4.00 m
2.
∆TBP = KBP • m
∆TBP = +0.512 oC/molal (4.00 molal)
∆TBP = +2.05 oC
BP = 102.05 oC
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Change in Freezing Point
Pure water
Ethylene glycol/water
solution
The freezing point of a solution is
LOWER than that of the pure solvent.
FP depression = ∆TFP = KFP•m
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Lowering the Freezing Point
Water with and without antifreeze
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When a solution freezes, the
solid phase is pure water. The
solution becomes more
concentrated.
Freezing Point Depression
Calculate the FP of a 4.00 molal glycol/water
solution.
KFP = -1.86 oC/molal (Table 14.4)
Solution
∆TFP = KFP • m
= (-1.86 oC/molal)(4.00 m)
∆TFP = -7.44 oC
Recall that ∆TBP = +2.05 ˚C for this solution.
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Freezing Point Depression
How much NaCl must be dissolved in 4.00
kg of water to lower FP to -10.00 oC?.
Solution
Calc. required molality
∆TFP = KFP • m
-10.00 oC = (-1.86 oC/molal) • Conc
Conc = 5.38 molal
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Freezing Point Depression
How much NaCl must be dissolved in 4.00 kg of water to lower
FP to -10.00 oC?.
Solution
Conc req’d = 5.38 molal
This means we need 5.38 mol of dissolved
particles per kg of solvent.
Recognize that m represents the
total concentration of all dissolved
particles.
Recall that 1 mol NaCl(aq)
--> 1 mol Na+(aq) + 1 mol Cl-(aq)
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Freezing Point Depression
How much NaCl must be dissolved in 4.00 kg of water to lower
FP to -10.00 oC?.
Solution
Conc req’d = 5.38 molal
We need 5.38 mol of dissolved particles per
kg of solvent.
NaCl(aq) --> Na+(aq) + Cl-(aq)
To get 5.38 mol/kg of particles we need
5.38 mol / 2 = 2.69 mol NaCl / kg
2.69 mol NaCl / kg ---> 157 g NaCl / kg
(157 g NaCl / kg)•(4.00 kg) = 629 g NaCl
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Boiling Point Elevation and
Freezing Point Depression
∆T = K•m•i
A generally useful equation
i = van’t Hoff factor = number of particles
produced per formula unit.
Compound
Theoretical Value of i
glycol
1
NaCl
2
CaCl2
3
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Osmosis
Dissolving the
shell in vinegar
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Egg in pure water
Egg in corn syrup
Osmosis
The semipermeable membrane allows only the
movement of solvent molecules.
Solvent molecules move from pure solvent to
solution in an attempt to make both have the
same concentration of solute.
Driving force is entropy
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Process of Osmosis
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Osmosic Pressure, ∏
Equilibrium is reached when
pressure — the
OSMOTIC PRESSURE,
∏ — produced by extra
Osmotic
pressure
solution counterbalances
pressure of solvent
molecules moving thru the
membrane.
∏ = cRT
(c is conc. in mol/L)
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Osmosis
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Osmosis
at the Particulate Level
Figure 14.17
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Osmosis
• Osmosis of solvent
from one solution to
another can continue
until the solutions are
ISOTONIC — they
have the same
concentration.
• Osmotic pressure in
living systems:
FIGURE 14.18
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Osmosis and Living Cells
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Reverse Osmosis
Water Desalination
Water desalination plant in Tampa
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Osmosis
Calculating a Molar Mass
Dissolve 35.0 g of hemoglobin in enough water
to make 1.00 L of solution. ∏ measured to
be 10.0 mm Hg at 25 ˚C. Calc. molar mass of
hemoglobin.
Solution
(a)
Calc. ∏ in atmospheres
∏ = 10.0 mmHg • (1 atm / 760 mmHg)
= 0.0132 atm
(b) Calc. concentration
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Osmosis
Calculating a Molar Mass
Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L
of solution. ∏ measured to be 10.0 mm Hg at 25 ˚C. Calc.
molar mass of hemoglobin.
Solution
(b) Calc. concentration from ∏ = cRT
0.0132 atm
Conc =
(0.0821 L • atm/K • mol)(298K)
(c)
Conc = 5.39 x 10-4 mol/L
Calc. molar mass
Molar mass = 35.0 g / 5.39 x 10-4 mol/L
Molar mass = 65,100 g/mol
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