Hi Brittanyl-ga,
The concept behind the problem you pose is known as Raoult's Law.
This law can be posed in essentially two different ways:
The lowering of the freezing point of a solvent is directly
proportional to the number of solute molecules (or moles) in a given
weight of solvent. The proportionality constant can be measured for
each solvent.
Another way to put it is as follows (from InfoPlease.com):
"Quantitatively, Raoult's law states that the solvent's vapor pressure
in solution is equal to its mole fraction times its vapor pressure as
a pure liquid, from which it follows that the freezing point
depression and boiling point elevation are directly proportional to
the molality of the solute, although the constants of proportion are
different in each case."
________________________
So, what does this all mean?
Basically, adding solute (such as NaOH) to a solvent (such as water)
causes the solute to "hold on" to solvent molecules that would
otherwise escape from the liquid (in the case of boiling point
raising) or bind to other like solvent molecules during the formation
of the solid phase (in the case of freezing point reduction).
The mathematical relation is relatively simple, but is only accurate
for dilute solutions. More concentrated solutions become more
complicated.
You can read more about Raoult's law here:
http://www.infoplease.com/ce6/sci/A0841143.html
A good description of how to do calculations using this law can be
found at Unit5.org:
http://www.unit5.org/christjs/Molality_Freezing_Lowering.htm
As stated at this site, it's important to recognize how many particles
a solute dissociates into on solution. For example, in the case of
ionic salts, such as NaCl, the solute will dissociate into Na+ and
Cl-, two particles. In the case of NaOH, we have Na+ and OH-. For
Ba(OH)2, we have Ba+ and 2 ions of OH-, three particles. We must
multiple the molality of our solute by the number of particles
liberated upon formation of the solution to get the right answer.
That is the key to your problem.
For NaOH and Ba(OH)2, because the solvet and molar concentrations of
the solvents are the same, we can see that Ba(OH)2 will lower the
freezing point (same as the melting point) more, because it liberates
3 particles on solution. If you like, we can actually calculate the
freezing point of each of the solutions, although it's useful to
develop the skill to see this answer based on the formula without
having to resort to the actual calculations.
________________________
Here we go...
For NaOH:
The basic formula for freezing point depression is
Freezing point = FP(without solute) - DT_f (depression of freezing temperature)
All of the values need to be in degrees Celsius or Kelvin.
FP = 0 deg Celsius for water, as I'm sure you know.
The formula for DT_f is
DT_f = k_f x m x N, where N is the number of particles we get on
solution, as I discussed above; m is the molality of the solution, k_f
is a constant that depends on the solvent.
k_f = 1.86 deg C/mol for water.
so, DT_f = 1.85 deg C/mol x 0.20 mol x 2
= 0.74 deg C
Similarly for Ba(OH)2,
DT_f = 1.85 deg C/mol x 0.20 mol x 3
= 1.11 deg C
So, for 0.2 molar NaOH, the freezing point will be -0.74 deg Celsius.
For Ba(OH)2, the freezing point will be -1.11 deg Celsius.
________________________
Also, as an aside, one commenter mentioned that salt used on roads in
the winter is typically KCl. In most states, the most common road
salt is actually CaCl2. As you can see, compared to NaCl, CaCl2 gives
us 3 particles in solution instead of 2 from NaCl. A problem,
however, is that CaCl2 is corrosive and somewhat toxic. Also, in this
application, the solute is at a very high concentration, which
violates the assumption I mentioned above (that the solution is
dilute).
If you're interested, you can read more about winter road salt here:
http://www.usroads.com/journals/p/rmj/9712/rm971202.htm
A nice comparison of road salts can be found here:
http://www.city.vancouver.bc.ca/ctyclerk/cclerk/980407/a5.htm
________________________
I hope this was helpful. Best of luck in your studies.
-welte-ga |
