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Solution - a homogeneous mixture of two or
more chemical substances
Solvent - the major component of a solution
( or sometimes the only liquid component)
Solute - a minor component of a solution
(the stuff which is dissolved in the solution)
Solutions: Concentration Expressions
Molarity:
Moles of solute per liter of solution
Molality:
Moles of solute per Kg of solvent
Mole Fraction:
Moles of solute divided by total
moles of solute and solvent.
Thermodynamics of Solubility
What is DH° of solution?
Thermodynamics of Solubility
+
NaCl(s)
Na (g)
+
H2O(l) + Na (g)
+
+
-
-
o
1
DH
Cl (g)
= 786 kJ/mol
+
Cl (g)
Na (aq) + Cl-(aq)
DHohyd =
DHo2 + DHo3 = -783 kJ/mol
DH° = +3 kJ/mol
So why do salts with DH positive dissolve?
Entropy
DG = DH - TDS
NaCl
DS° = 46 J/Kmol
But some DS° values are negative!
Hydration of a salt - MX
H
O H
H
H O
H O
H
H
H
M
O
H
O HH
O
H
H
H
O H
X
H
H
H
O
O
H
H O
The higher the charge on the ion and the smaller
the size of the ion, the greater the interaction. This
leads to a greater ordering of the water.
The solvation of the ions leads to an
ordering of the water solution. Thus an
decrease in the entropy of the water.
Chemical species that interact favorably
with water are call hydrophilic.
Chemical species that do not interact favorably
with water are called hydrophobic.
Non polar molecules such as hexane do
not dissolve in water.
Oil floats on the top of water.
The enthalpy for dissolving hexane in water
is actually negative. So why doesn’t it dissolve?
When a non polar molecule is dissolved in water the
water molecules form a highly ordered clatharate structure
H
H
H
H
O
H
O
H
H
C H
H
H
O
O
H
O
H
O
H
H
H
The structure here
is a two dimensional
representation of the
true structure which
three dimensional.
This ordering of the water
molecules leads to a
negative DS.
H
This is the basis of the hydrophobic effect
Solubilities of alcohols in water
Methanol
infinite
Ethanol
infinite
1-propanol
infinite
1-butanol
80 g/L
1-pentanol
22 g/L
1-hexanol
5 g/L
Surfactants
Na
O
S
O
O
The amount of gas dissolved in a liquid solution
is directly proportional to the pressure.
Partial Pressure = P = kHC
kH (atm) for water
CH4 4.13 x 102
O2
4.34 x 104
N2
8.57 x 104
Henry’s Law
In the atmosphere how much O2 will dissolve in H2O?
P = k HC
kH O2 4.34 x 104 atm
P(O2) in atm .21atm
.21atm = 4.34 x 104 atm x C
C = 4.84 x 10-6
mol fraction of O2
1000 g H2O/18 g/mol = 55.6 mol H2O
4.84 x 10-6 x 55.6 = 2.69 x 10-4 mol O2
2.69 x 10-4 mol x 32 g/mol = .0086 g O2
.0086 g O2 / liter H2O
at normal atmospheric pressure
In the actual atmosphere
.21 x 32.0 g / 22.4 =
.30 g O2 / liter air
Aren’t you glad you are not a fish!
What will happen if you have two beakers in a
closed system? One contains 1 liter of 1M NaCl,
the other 1 liter of pure H2O.
Raoult’s Law
Vapor pressure of
solution with a
nonvolatile solute.
Pobs = CsolvPosolv
Raoult’s Law for a mixture of two volatile liquids.
Ptotal = PA + PB = CAPoA + CBPoB
Raoult’s says that the presence of a solute
will lower the vapor pressure of a solution as
compared to the pure solvent.
What will this do to the freezing point and
boiling point of the solution?
Freezing Point Depression
DT = Kf msolute
Boiling Point Elevation
DT = Kb msolute
Calculate the boiling point of a solution containing
6.16 g of sucrose, C12H22O11, dissolved in 30.g of H2O.
DT = Kb msolute
Kb = 0.51 C°kg/mol
m.wt. sucrose 342 g/mol
mol = 6.16 g/(342g/mol) = .018 mol
m = .018 mol / .030 kg
= .60 mol/kg
DT = 0.51 C°kg/mol x .60 mol/kg =.31 C°
Boiling point of solution = 100.31 C°
A solution of .90g of glucose in 20.0g of
water hasa freezing point of -0.465 C°.
What is mwt. of glucose?
Kf = 1.86 C°kg/mol
m = DT/Kf = .465 C° / (1.86 C°kg/mol) =.25 m
.25 mol/kg = X mol/.020kg
X mol of glucose = .005 mol
.90 g/ .005 mol = mwt. / 1mol = 180 g / mol
Osmosis
Osmotic Pressure is
the pressure that just
stops the osmosis.
p = MRT
p = MRT
A 2.2 gram sample of polyethylene was disolved in
toluene to give 100ml solution. The osmotic
pressure at 25°C was measured to be 1.10 x 10-2atm.
What is the mwt. of the polyethylene?
p = MRT
M = p / RT
M = 1.10 x 10-2 / (.082 x 298) = 4.5 x 10-4 M
.10 L x 4.5 x 10-4 mol/L = 4.5 x 10-5 mol
2.2 g / 4.5
x 10-5 mol = mwt = 49,000 g/mol
CH2 CH2
n
49,000/ 28 =
1750 units
Reverse
Osmosis