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Q: Lowering the freezing point of water ( Answered 5 out of 5 stars,   7 Comments )
Subject: Lowering the freezing point of water
Category: Science > Chemistry
Asked by: brittanyl-ga
List Price: $10.00
Posted: 25 Jul 2005 06:21 PDT
Expires: 24 Aug 2005 06:21 PDT
Question ID: 547572
Which lowers the freezing point of 2.0 Kg of water more, 0.20 mol or
.20 mol of Ba(OH)2? Why?

Clarification of Question by brittanyl-ga on 25 Jul 2005 13:30 PDT
That is suppose to be .20 moles of NaOh or .20 mol of Ba(OH)2? Sorry.
Subject: Re: Lowering the freezing point of water
Answered By: welte-ga on 28 Jul 2005 16:40 PDT
Rated:5 out of 5 stars
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
"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

You can read more about Raoult's law here:

A good description of how to do calculations using this law can be
found at

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

If you're interested, you can read more about winter road salt here:

A nice comparison of road salts can be found here:


I hope this was helpful.  Best of luck in your studies.

brittanyl-ga rated this answer:5 out of 5 stars
Very helpful and very descriptive answer!

Subject: Re: Lowering the freezing point of water
From: pafalafa-ga on 25 Jul 2005 06:58 PDT
.2 mol, definitely.
Subject: Re: Lowering the freezing point of water
From: brittanyl-ga on 25 Jul 2005 07:45 PDT
.2 mol of what?
Subject: Re: Lowering the freezing point of water
From: stephanbird-ga on 26 Jul 2005 01:17 PDT
Why should either of them lower the freezing point at all?
Subject: Re: Lowering the freezing point of water
From: hfshaw-ga on 26 Jul 2005 15:52 PDT
Sounds like a homework problem.  

The formula for freezing point depression is :

  dT = i*K*m,

where K is a constant specific to the solvent in question (equal to
-1.858 Kelvins/molal for water); m is the molality of the solution;
and i is the "van't Hoff factor", a unitless number that indicates the
degree of dissociation of the dissolved solute.  It is equal to 1 for
solvents that do not dissociate, and equal to the number of moles of
ions produced per mole of solute for solvents that do dissociate.

NaOH dissociates into 2 moles of ions per mole of solute, while
Ba(OH)2 dissociates into 3 moles of ions per mole of solute.

From this, you should be able to say which solute will depress the
freezing point of water more.
Subject: Re: Lowering the freezing point of water
From: greyw01f-ga on 27 Jul 2005 07:55 PDT
Answer:  BaOH2 lowers the freezing point of water more, and here's why:

When water freezes, what happens is that the individual water
molecules organize themselves into very orderly crystals.  When
there's a foreign material (ie. BaOH2 or NaOH) in the mix, it disrupts
the "orderliness" of the crystals.  It therefore requires colder
temperatures to form ice crystals, thereby lowering the freezing

When either compound (BaOH2 or NaOH) is added to water, the ions
dissociate completely (because they're both strong bases).  Not only
(as hfshaw-ga correctly stated) are there more ions in the Ba solution
(two OH-'s instead of one with NaOH), but the Ba++ ion is much larger
and more positively charged than the Na+ ion.  This causes more
disruption of the crystallization process, making it harder for ice
crystals to form.

This is exactly why the expensive "road salt" that you buy in stores
is much better at removing ice than "rock salt."  "Rock salt" is
essentially NaCl, and the more expensive white "road salt" is
primarily KCl.  Potassium is much larger than sodium and, even though
they have the same charge (+1), the larger potassium ion disrupts the
orderly ice crystal formation, thereby lowering the freezing
temperature of water.
Subject: Re: Lowering the freezing point of water
From: njhvactech-ga on 28 Jul 2005 20:28 PDT
I know for a fact that sugar lowers the freezing point of water. An
Icee/slurpee has a freezing point of 26 degrees, that's why you get
brain freeze when you drink it.
Subject: Re: Lowering the freezing point of water
From: nice360t-ga on 08 Aug 2005 19:40 PDT
It should be mentioned that the answer depends on the solubility of the salts
involved.  Barium hydroxide has limited solubility, .2 moles in 2 kg
water is just under the limit for cool water.  For slightly different
conditions (e.g., less water) the answer would be the sodium
hydroxide, since it is much more soluble.  A better example to show
the principle would be to compare Sodium sulfate (Na2SO4), which
produces three ions and is very soluble.

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