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The highlights are:
1. Henry's law
2. Raoult's law
3. Colligative properties
1.
Henry’s Law, Freezing Point
Depression, Boiling Point
Elevation and Raoult’s Law
Wow, That is a Mouthful
2.
Henry’s Law
The solubility of a gas is directly proportional to
the gas pressure
Sg= khPg
When the partial pressure of the solute above a
solution drops, the solubility of the gas in the
solution drops as well to maintain the equilibrium.
This can be used to calculate the molar solubility
of a gas.
What is the concentration of O2 in a fresh water
stream in equilibrium with air at 25oC and 1.0 atm.
(Hint there is 21% O2 in the air)
3.
Does this always fit?
As concentrations and partial pressures
increase, deviations from Henry's law
become noticeable. This behavior is very
similar to the behavior of gases, which are
found to deviate from the ideal gas law as
pressures increase and temperatures
decrease. For this reason, solutions which
are found to obey Henry's law are
sometimes called ideal dilute solutions
4.
Colligative Properties
Colligative properties are “properties that
depend only on the relative number of
particles and not on what the actual
substance is”
Remember that
5.
Changing Vapor Pressure
(Raoult’s Law)
Vapor pressure at a given temperature is
the pressure that the vapor exerts when
the rate of molecules leaving the surface is
equal to the rate of them re-condensing.
But what happens when something is now
dissolved in the solvent.
2 things- 1. less solvent molecules at the
surface. 2. different sets of attractive
6.
Lets look at each one individually
1. less solvent molecules at the surface.
Therefore less chance the water leaves, the
vapor pressure is lowered.
This makes sense based on Henry’s law. The
vapor pressure of the solvent will be proportional
to the mole fraction in the liquid.
Psolvent= XsolventK
If we also look at a pure solvent Po,
then Po= XsolventK, Therefore Psolvent= Xsolvent Po
This is Raoult’s Law
7.
Raoult’s Law
Raoult’s law assumes that the solution is ideal.
Therefore, the forces between solute and
solvent molecules must be the same as the
solvent to solvent.
If solvent-solute interactions are stronger than
solvent-solvent, the actual vapor pressure will be
lower than calculated
If solvent-solute interactions are weaker than
solvent-solvent, the actual vapor pressure will be
higher than calculated
8.
Try a problem
Assume you dissolve 10.0g of sugar
(C12H22O11) in 225mL (225g) of water and
warm the water to 60oC. What is the
vapor pressure of the water over this
solution? The normal vapor pressure of
water at 60oC is 149.4 torr.
9.
Raoult’s Law Cont.
Adding a nonvolatile solute to a solvent
will lower the vapor pressure.
∆Psolvent= Psolvent- Posolvent
∆Psolvent= (XsolventPosolvent) – Posolvent = -(1-
Xsolvent+Xsolute = 1
∆Psolvent= -XsolutePosolvent
10.
Why does this matter?
Well remember that vapor pressure
determines the boiling point of a liquid.
If you add solute it will change the
solvent’s vapor pressure, therefore the
boiling point changes.
This is called boiling point elevation!
11.
Boiling Point Elevation
The Boiling point elevation, Δtbp, is directly
proportional to the molality of the solute
Δtbp= Kbpmsolute
Kbp is called the molal boiling point elevation
constant by solvent and is (oC/m)
How many grams of ethylene glycol,
HOCH2CH2OH, do you have to add to 125 g of
water to increase the bp by 1oC? (The KbpWater
= +0.5121 oC/m
12.
What is another use?
Molar mass by boiling point
elevation!
A solution prepared from 1.25 g of oil of
wintergreen (methyl salicylate) in 99.0 g of
benzene has a boiling point of 80.31oC.
Determine the molar mass of the
compound. (Benzene’s normal bp is 80.10
and the Kbp is +2.53 oC/m)
Answer is 150 g/mol
13.
Freezing Point Depression
Very similar to boiling point
Δtfp= Kfpmsolute
The reason for this is very similar in
changes in vapor pressure equilibrium
There are more atoms of pure solvent
going from solid to liquid than from liquid
to solid.
14.
What about for electrolytes?
We would assume that adding NaCl or such to water
would have twice the effect
That is pretty much true.
To see the real effect, we need a van’t Hoff factor
i = Δtfp, measured/ Δtfp calculated
As the Δtfp calculated is if no ionization occurred. The i is
not a perfect for the number of ions, but is closest to it for
dilute solutions due to the intermolecular attractive
forces.
Δtfp, measured= Kfpmsolute i
15.
Calculate the freezing point of 525 g of
water that contains 25.0 g of NaCl.
Assume i is 1.85 for NaCl.
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