 |
 |
Solubility Factors
Name: Michael G.
Status: educator
Age: 50s
Location: N/A
Country: N/A
Date: Wednesday, February 26, 2003
Question:
This test question has caused some debate among my
fellow faculty members. Please help us.
A factor that does not affect solubility is:
a. polarity
b. temperature
c. freezing
d. pressure
Replies:
Michael,
As the question and answer choices are posed, I can understand the
controversy because they make no reference as to the nature of the solute
and solvent. Given the opportunity to specify solutes and solvents of my
choice, I could make the case that any or all the answer choices would
have an effect on solubility.
Regards,
ProfHoff 599
It is of course important to know the solubility of "what solute" in "what
solvent". Ignoring that, I think that the targeted answer is (d) for two
reasons:
1. Elimination of the other three. Certainly solvent polarity has a
large effect on solubility and does temperature (in most cases). And
considering that most solutes that are soluble in a liquid solvent are
(usually) not soluble in the solid phase of the solvent.
2. The more sophisticated reason is thermodynamic. Solubility has a
certain change in free energy,dG, associated with the general reaction
A(solid) or A(liquid) ------> A(solution) in some liquid solvent call it
'B'. The free energy change with a change in pressure (at constant
temperature) is: dG / dP = V. Now almost all solids and liquids are
approximately "incompressible" at reasonable pressures changes say 1 to
10 atm (not 1000 atm) so the free energy change with pressure (and hence the
solubility change) is essentially independent of the change in pressure
because the volumes do not change.
Vince Calder
The factor that matters least when dissolving solids in liquids is
pressure, because neither solids nor liquids are affected very much by
pressure. But if you are interested in dissolving gases in liquids, or
anything in supercritical fluids, pressure matters very, very much. So my
vote is that the question is WRONG! Bad question! Bad question!
Richard E. Barrans Jr., Ph.D.
PG Research Foundation, Darien, Illinois
It is of course important to know the solubility of "what solute" in
"what solvent". Ignoring that, I think that the targeted answer is (d) for two
reasons:
1. Elimination of the other three. Certainly solvent polarity has a
large effect on solubility and does temperature (in most cases). And
considering that most solutes that are soluble in a liquid solvent are
(usually) not soluble in the solid phase of the solvent.
2. The more sophisticated reason is thermodynamic. Solubility has a
certain change in free energy,dG, associated with the general reaction
A(solid) or A(liquid) ------> A(solution) in some liquid solvent call it
'B'. The free energy change with a change in pressure (at constant
temperature) is: dG / dP = V. Now almost all solids and liquids are
approximately "incompressible" at reasonable pressures changes say 1 to 10
atm (not 1000 atm) so the free energy change with pressure (and hence the
solubility change) is essentially independent of the change in pressure
because the volumes don't change.
Vince Calder
There are some more considerations to what I provided above that
occurred to me after I responded.
If the solute is a gas and the solvent is a liquid then pressure DOES
affect solubility (Henry's Law). In addition, temperature and solvent
polarity also play a role. Freezing also plays a role (depending upon
whether the solute is soluble both in the liquid and solid phases of the
solvent. However, the "usual" case, for example water and ice, most
things -- dissolved minerals and air -- are not soluble in ice. That is
why ice cubes look cloudy. The reason ice cubes you by in the market,
motels, and bars are clear is that the freezing takes place in a flow of
water over a cold surface. Then the minerals and air are continuously
washed away and
pure ice forms clear ice cubes.
Second, neither the inquiry nor I mentioned the solubility of various
substances in super-critical fluids, i.e. fluids at temperatures and
pressures above the critical point temperature and pressure. Here
solvents like liquid CO2, and various other substances exhibit remarkable
solvent properties very different than their behavior at room temperature
and atmospheric pressure. This is a totally new phase of the solvent that
is used commercially to de-caffeinate coffee, and various other
interesting applications. Space here does not allow a discussion of
super-critical solvents, but a "Google" search on "supercritical
fractionation" or similar
search term will provide a wealth of web sites describing this phenomenon.
In the super-critical domain, solids do not form, so choice (C) above in
the inquiry is irrelavent.
Another "simple" question with a complicated answer!
Vince Calder (2)
Dear Michael and colleagues:
Any multiple-guess question and answer format necessarily has its
limitations. In this case one has to distinguish between the
equilibrium constant for solubility and the amount of material solvated.
In addition, what is meant by "solubility?" The usual
definition is that solubility is the concentration of a solvent when the
solution is saturated, i.e., when the dissolving of a solvent and
its recrystallization are in equilibrium. This implies the existence of an
equilibrium constant, and those are functions of temperature and not
pressure (which is something we learn how to prove in physical chemistry
and other thermodynamics courses).
Clearly polarity affects solubility because polar solvents are better at
dissolving ionic and polar substances than nonpolar solvents.
It is also clear that temperature affects solubility because equilibrium
constants are generally strong functions of temperature (given by the van
t'Hoff equation). If the solvation reaction is endothermic the solubility
will increase with increasing temperature (due to the increase in the
equilibrium constant). If exothermic, the solubility will decrease with
increasing temperature. NaCl is an example of the first case; NaOH is an
example of the second case. This is basically LeChatelier's principle in
action.
"Freezing" is a difficulty. I am not sure what is meant here. Clearly if
one has a saturated solution and then freezes the solvent, some solute is
going to have to precipitate out if the frozen solvent is pure (usually
the case, at least to a high degree of approximation. This is why
recrystallization is such a useful purification technique.). However, I
believe that the concentration would
not change and therefore neither would the solubility. On the other hand,
lowering the temperature to the freezing point would definitely affect the
solubility (see previous paragraph).
As far as pressure is concerned, well, equilibrium CONSTANTS are only
functions of temperature but equilibrium POSITIONS are functions of both.
If the solute in question is a dissolved gas, increasing the pressure
increases the amount of gas in the solution and therefore the
concentration of gas - and therefore the solubility. This is Henry's Law:
C = k P where C is the concentration, P is the pressure and k is the
Henry's law constant for the particular gas/solvent combination -
effectively an equilibrium constant.
This law describes the bubbles which rise out of solution when you open a
pressurized bottle of seltzer or other soda. If C is taken to be the
solubility then one must say that the solubility depends on pressure. On
the other hand, if the solute is a crystalline solid then pressure will
not make a bit of difference because liquids (and solids) are highly
incompressible and the solute is nonvolatile, so the concentration
depends only on the equilibrium constant's value, which is independent of
pressure in all cases.
So, if I had to choose one I would pick c. Clearly a and b would be wrong.
I think this is a vague question, however, and the question should instead
read "A factor that does not affect solubility OF GASES is..." and then
the answer would definitely be c.
I hope this is helpful to you all.
best, Prof. Topper
Click here to return to the Chemistry Archives
| |
Update: June 2012
|
|
