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Solutions Part I: Colligative Properties (loosely from Jespersen Chap. 13 Sec 6 & 7. Dr. C. Yau Spring 2014. What are Solutions?. Quick reminder: Solutions are homogeneous mixtures consisting of... generally one solute dissolved in a solvent . (More than one solute is possible.).
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Dr. C. Yau
Solutions are homogeneous mixtures consisting of...
generally one solute
dissolved in a solvent.
(More than one solute is possible.)
The term "colligative" comes from a Greek word meaning "dependent on the number."
Colligative properties refer to the properties of the solventthat depends on the number of particles (of the solute) and not the nature of the particles. However, with the stipulation that...
These solutes must be nonvolatile (and therefore does not contribute to the vapor pressure of the solvent). 3
Since the properties are NOT dependent on the "nature" of the solute, this means the IMF of the solute plays no part in this.
We will be studying how the concentration of the solute affects the properties of the solvent.
Psoln = Xsolvent Posolvent
Psoln = vp of the solution (only due to the solvent because the solute is nonvolatile)
Xsolvent = mole fraction of solvent
Posolvent = vp of the pure solvent(without solute)
Psoln = Xsolvent Posolvent
Vp of solution is a fraction of the vp of the pure solvent.
This means that the solute has lowered the vp of the solvent.
One way to explain it is to consider the entropy of the system (which will be covered in a later chapter).
Another way is to consider the solute particles at the surface of the solution.
Their presence decreases the number of solvent particles on the surface, thus decreasing the chances of them breaking off into the vapor phase.
solvent alone solute + solvent
VP of solvent is lowered by the solute
What happens to the BP and FP when a solute is added?
Again we are considering only solutes that are nonvolatile…
The BP is raised
and the FP is lowered by the presence of the solute.
These are referred to as…
and FP depression.
Before we begin to study the quantitative effect of solute on the certain physical properties of a solvent, we must consider a special unit of concentration:
which is not to be confused with
Take a moment to compare the two.
Can you see why m is not affected by T?
Is molarity affected by T? Why?
Learn the definitions of these
units of concentration NOW!
The change in freezing point ΔTf is proportional to the molality:
ΔTf = Kfm
and change in boiling point ΔTb is
also proportional to the molality:
ΔTb = Kbm
Table 13.3 p.609 gives these constants for various solvents.
ΔTb = Kb m
ΔTf = Kfm
NOTE: BP is NOT ΔTb
What is the difference between BP and ΔTb?
This is a common mistake!!!
What is the symbol for BP?
Example 13.8 p. 609
Estimate the fp of a solution made from 10.0 g of urea, CO(NH2)2, with MM 60.06 g mol-1, and 125 g of water.
Table 13.3 tells us
Kf for water = 1.86oC m-1 (or oC/m)
Do Practice Exercise 17 & 18 p.610.
Example 13.9 p.611
A solution made by dissolving 5.65 g of an unknown molecular compound in 110.0 g of benzene froze at 4.39oC. What is the molar mass of the solute?
Table 13.3 tells us
Kf for benzene = 5.07 oC/m
Fp for benzene = 5.45 oC
Do Practice Exercise 19 & 20 p.612.
Osmosis is the movement of solvent molecules through a semipermeable membrane from a dilute soln to a more concentrated soln.
Fig. 13.19 p.614
= MRT (Note: M not m)
where = osmotic pressure
M = concentration in molarity
R = gas constant = 0.08206 atm.L.mol-1.K-1)
T = temperature in K
Example 13.10 p.616
A very dilute soln , 0.00100 M sugar in water, is separated from pure water by an osmotic membrane. What osmotic pressure in torr develops at 25oC?
Do Practice Exercises 21 ans 22 p.617.
Example 13.11 p.617
An aqueous solution with a volume of 100. mL and containing 0.122 g of an unknown molecular compound has an osmotic pressure of 16.0 torr at 21.0oC. What is the molar mass of the solute?
Do Practice Exercises 23 & 24 p.618.
When the solute is ionic, such as NaCl the effect is doubled. Why?
When the solute is (NH4)2SO4, the effect is tripled. Why?
What does NaCl do when it dissolves in water?
NaCl(s) Na+ (aq) + Cl- (aq)
One f.u. transforms into TWO particles, so the effect is doubled.
ΔTf = i Kf m
ΔTB = i KB m
= i MRT
i = number of particles into which the ionic solute breaks up.
i = 2 for NaCl (effect is doubled)
i = 3 for (NH4)2SO4 (effect is tripled)
FP of a 0.10 m sucrose (C12H11O22) solution is found to be -0.19C.
FP of a 0.10 m NaCl solution = -0.38C (lowered by twice as much).
Kb of water = 0.51 C/m
What do you expect the BP to be for…
Which do you expect to have the higher BP?
a) 0.1m NaCl or b) 0.05m Na2SO4
Nonelectrolyte: ΔTf = Kf m
(Solute is molecular.)
Electrolyte: ΔTf = i Kf m
(Solute is ionic.)
If we don’t know whether a solute is ionic, we can identify it as such as follows:
Calc ΔTfas a nonelectrolyte.
Compare it with the observed ΔTf .
e.g. If calc ΔTf is 6oC and observed is 12oC, we know i = 2. (Effect is doubled.)
Experimentally, the i value is determined thus:
Example: Experimentally we determined 0.10 m NaCl to have a Tf of 0.38C. We calculate the theoretical
Tf = kf m = (1.86C/m)(0.10m)=0.18C.
As concentration increases, deviation increases.
Dilution leads towards the theoretical i value.
Compare NaCl with MgSO4. Both have theoretical i = 2.
Why does MgSO4 deviate much more than NaCl?
We see that the effect of ionic solute causes the effect to be GREATER than expected.
What kind of solute would cause the effect to be LESS than expected?
Benzoic acid forms a "dimer" in solution. Two molecules are not freely working as two particles, but working as ONE unit. Thus, i = ½ and effect is one half of expected.
We had previously begun with the explanations to why the vapor pressure of a solvent is lowered when a solute is added.
Let us now return to “vapor pressure lowering” and examine the implications.
Posolventis a constant,characteristic of the solvent
Psoln and Xsolvent are variables.
If we rearrange the equation, we get
Psoln = Posolvent Xsolvent
y = m x
If we plot Psoln vs Xsolvent ,
the slope is Posolvent.
Carbon tetrachloride has a vp of 155 torr at 20oC. This solvent can dissolve candle wax, which is essentially nonvolatile. Although candle wax is a mixture, we can take its molecular formula to be C22H46 (MM 311g mol-1).
What is the vapor pressure at 20oC of a solution prepared by dissolving 10.0 g of wax in 40.0 g of CCl4 (MM 154 g mol-1).
Do Practice Exercises 13 & 14 p.605
P = change in vp due to solutes
P = Xsolute P°solv
How do we get this equation?
(See lecture notes. This was derived in class. It is also in your textbook, more or less.)
Let us now consider the case when the solute is VOLATILE.
Applying Dalton's Law of Partial Pressures to a mixture of gases A + B,
PTotal = PA + PB
where PA is partial pressure of A = XAPoA
and PB is partial pressure of B = XBPoB
PTotal = XAPoA + XBPoB
Raoult's equation is simply dropping off one term (such as XBPoB because the solute is nonvolatile).
One last point about Raoult's Law….
It is for the ideal solutions.
BUT, even deviations from the law gives us information on the IMF of the solute to the solvent and IMF of the solvent molecules to solvent molecules.
If the solute-solvent IMF is strong, vp of solution would be lowered more than expected.
If the solvent-solvent IMF is strong, vp of the solution would not be lowered as much as expected (adding solute disrupts the IMF of the solvent and allows more solvent to vaporize).