Chapter 12, Worksheet 3 (ws12.3)
Factors that Affect Solubility
Review
1. A solid was added to water and the mixture was stirred until no more solid would dissolve and
the excess solid remained at the bottom of the container. At this point, the solution is
___saturated____. We say that the dissolving reaction has “reached equilibrium”. What does
this mean?
At equilibrium, the rate of dissolving is equal to the rate of precipitating (coming out of solution).
Therefore, there is no change in the amount of undissolved solid or in the concentration of the
solution.
2. Starting with the solid compound, describe how to make 250.0 mL of 0.200 M NaCl. You have
an analytical balance, a 250 mL volumetric flask, and some water.
Step 1: Calculate the mass of NaCl needed: (0.200 mol/L)(0.2500 L)( 58.5 g/mol) = 2.93 g
Step 2: Weigh out 2.93 g of NaCl
Step 3: Transfer quantitatively to a 250 mL volumetric flask. (This probably requires that you rinse
the weigh boat with water and add the rinse to the flask.)
Step 4: Add enough water to dissolve the NaCl.
Step 5: Swirl the flask until the NaCl dissolves.
Step 6: Add water to the mark on the flask (250.0 mL total volume).
Step 7: Stopper the flask and invert several times to mix.
Note: If you add water to the mark BEFORE the NaCl dissolves, the final volume will probably be
more than 250.0 mL.
3. What is the molality of NaCl in the solution made in question 2? Assume that the density of the
solution is 1.00 g/mL.
Moles of NaCl = (0.200 mol/L)(0.2500 L) = 0.0500 mol
Mass of the water = mass of the solution – mass of NaCl = 250.0 g – 2.93 g = 247.07 g = 0.24707 kg
Molality = 0.0500 mol/0.24707 kg = 0.202 m (a bit higher than the molarity as expected.)
4. Why don’t water and oil mix?
Water is polar and oil is non-polar (oil is a mixture of hydrocarbons).
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5. Consider the trend shown in the table below. Which of the alcohols would be most soluble in a
non-polar solvent?
1-hexanol (the one with the longest non-polar tail)
6. Recall the solubility rules for ionic compounds dissolved in water. Many of the compounds
listed as “insoluble” in the table below are actually slightly soluble in water (the solubility is less
than 0.01 M) What type of ionic compounds would you expect to be only slightly soluble in
water? Consider solute-solute and solute-solvent interactions.
This is complex. Small ions
with large charges will form a
more stable crystal that will be
hard to disrupt (disfavors
dissolving!) but these same
ions will interact strongly with
water (favors dissolving!). Ions
that bind tightly to water cause
the water to become more
ordered (disfavors dissolving!)
Despite this complexity, notice
that the ions in most of the
soluble compounds are 1+ or 1while most of the insoluble
compounds have 2- or 3- ions.
7. Why are network solids such as diamond (carbon) and SiO2 (sand, glass) totally insoluble in
water?
They are held in the solid state by covalent bonds which are very strong.
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Effect of temperature on the solubility of solids and gases in water
The solubility of most solids increases as the temperature increases.
The solubility of all gases decreases as the temperature increases.
To understand the effect of temperature on the solubility, we must consider the effect of temperature on
the entropy of a system.
1. Let’s first review how the second law of thermodynamics allows us to explain why ammonium
nitrate is very soluble in water. Ammonium nitrate dissolving in water is an endothermic
process. This means that the solute-solvent interactions are weaker than the sum of the solutesolute and solvent-solvent interactions. Therefore, you might think that very little ammonium
nitrate would dissolve in water. In reality, it is very soluble (s = 19 M!) so there must be some
“driving force” for the process that we have not yet considered. What is it?
The other driving force is “disorder” or “entropy”. (The symbol for entropy is S.) For two processes
with similar H’s, a process that leads to more disorder (more entropy) occurs more readily than a
process that creates order. Creation of order requires energy. (Think about how your room tends to
get disordered and how much energy it takes to clean it up!) Even though ammonium nitrate
interacts with itself more strongly than with water, this is more than offset by the fact that dissolving
leads to a lot more disorder! Thus, the two driving forces for any chemical reaction are enthalpy and
entropy. A reaction is favored by a negative H (exothermic) and by a positive S (increase in
disorder).
Why should a reaction be favorable if there is an increase in disorder? This is the second law of
thermodynamic which you will study futher in chapter 18 . Basically, a more disordered state is a
more probable state (as you well know!). That’s because there are many more ways to be disordered
than ordered. This is similar to the fact that there are more ways to roll a 7 with two dice than to roll
a 12. Thus, rolling a 7 is more probable.
The second law says: For any spontaneous process, the entropy of the universe increases. This
means that even if you clean up your room (the system), the universe as a whole becomes more
disordered! We’ll talk about this a lot more later.
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2. Remember this catch phrase: Increasing the temperature favors the more disordered state.
(We will talk a lot about this idea as the course progresses.)
a. Which is more disordered, a pure crystalline solid or the same solid dissolved in water?
The solid dissolved in water.
b. Which is more disordered, a pure gas or the same gas dissolved in water?
The pure gas.
c. Now explain the effect of temperature on the solubility of solids and gases.
Since a dissolved solid is more disordered than the pure solid, raising the temperature leads to more
solid dissolving.
Since a pure gas is more disordered than a dissolved gas, raising the temperature leads to more gas
leaving the solution.
d. Why do you think some solids buck the trend? (Look at cerium sulfate in the graph above.)
It must be that the system becomes more ordered when these solids dissolve! This is due to the
ordering of the water that surrounds the dissolved solute particles.
e. How do you think temperature affects the solubility of a liquid in water? Explain.
Temperature has very little effect on the solubility of a liquid in water because there is very little
change in entropy when a liquid dissolves in another liquid.
f. Why does an open soda taste better if stored in the frig rather than on the kitchen table?
At the lower temperature, more of the carbon dioxide stays dissolved in the soda and it doesn’t taste as
flat as the soda left out in the kitchen.
Actually, this isn’t really correct. Even at the lower temperature, the vast majority of the carbon
dioxide will EVENTUALLY leave (see below). The real reason the refrigerated soda tastes better is
that, at the colder temperature, the carbon dioxide leaves more SLOWLY! So the soda tastes fine the
next day, but wait a week and it will be just as flat as the soda left out on the kitchen table. It is true,
however, that once the two systems reach equilibrium, the cold soda will have a bit more carbon
dioxide in it than the warm soda. (But still not very much!)
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Henry’s law: Effect of partial pressure of a gas on its solubility in water
Henry’s Law: The solubility of a gas in liquid is directly proportional the partial pressure of the gas
above the solution.
c = kP
c is the molar solubility of the gas
P is the partial pressure of the gas over the solution (in atm)
k is the Henry’s law constant. The value of k depends only on temperature and the identity of the gas. (k
is nothing more than the equilibrium constant for the dissolving process!)
1. If, at equilibrium, the partial pressure of an undissolved gas triples, what has happened to its
solubility?
The solubility has also tripled.
2. Explain why the solubility of a gas increases as the partial pressure of the gas increases.
(Remember that in a saturated solution, the rate of dissolving is equal to the rate of leaving the
solution. What happens to these rates as P increases?)
When an aqueous solution of a gas is saturated, the dissolving process is at equilibrium so the rate of
dissolving is equal to the rate of coming out of solution. If the partial pressure of a gas doubles, the
rate of dissolving also doubles because the gas particles hit the liquid surface twice as frequently. The
rate of coming out of solution is not affected by the partial pressure of the gas. Thus more gas
dissolves. As the gas dissolves, the rate of dissolving decreases and the rate of coming out of solution
increases. The gas continues to dissolve until the two rates are equal and equilibrium has been reestablished.
3. Why does soda taste better if you store it on the kitchen table with the lid on than with the lid
off?
With the lid on, the partial pressure of the carbon dioxide above the soda is about 2 atm. With the lid
off, the partial pressure of carbon dioxide above the soda is the same as the partial pressure in the
atmosphere which is very small (about 0.04 atm). Thus, when the bottle is opened, most of the carbon
dioxide escapes because, suddenly, it is MUCH less soluble (fizz!).
When the PCO2 drops from 2 atm to 004 atm, the rate of dissolving decreases dramatically. The rate
of coming out of solution is not affected (initially) so there is a net loss of carbon dioxide. As the
carbon dioxide escapes, the rate of escape decreases. Eventually, the equilibrium will be reestablished and the rate of dissolving becomes equal to the rate of escape. It takes time to re-establish
equilibrium so it will take a while before the soda is flat. Since the partial pressure of CO2 will
remain at 0.04 atm, eventually, there will be 50 times less carbon dioxide in the soda.
In the previous question about the affect of storing the open soda in the refrigerator, at equilibrium,
there would be a bit more carbon dioxide in the cold soda because the Henry’s law constant (k) is
little bit larger at the lower temperature but most of the carbon dioxide will still escape due to the
large decrease in the partial pressure.
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4. The partial pressure of nitrogen in the atmosphere at sea level is 0.78 atm. At 25oC, the
solubility of nitrogen at sea level is 5.3 x 10-4 M.
a. What is the value of Henry’s law constant for nitrogen at 25oC. Include the correct unit.
k = c/P = 5.3 x 10-4 M/0.78 atm = 6.8 x 10-4 M-atm-1
b. What is the solubility of nitrogen in water when the partial pressure of nitrogen is 5.0 atm (in
a pressurized tank)?
c = kP = (6.8 x 10-4 M-atm-1)(5.0 atm) = 3.4 x 10-3 M (almost 10 times more soluble!)
c. What must happen to the value of k as the temperature increases?
It must decrease because the solubility of the gas decreases as the temperature increases.
5. Why do divers experience decompression sickness (“the bends”) if they surface too quickly?
At depth, the partial pressure of nitrogen in the lungs and in the capillaries is high so there is a lot of
gas (mostly nitrogen) dissolved in the blood. If the diver surfaces quickly, a lot of nitrogen comes out
of the blood very quickly and forms large bubbles. These bubbles circulate throughout the body and
wreak havoc with lots of systems. In particular the bubbles interfere with the nervous system causing
severe pain (the diver “bends” over in pain). The sickness is treated by putting the diver in a high
pressure chamber to force the nitrogen back into solution. Then the pressure is released gradually,
allowing the gas to escape slowly. The gas then diffuses away without forming large bubbles.
One way to help prevent decompression sickness, is to replace nitrogen with a gas that is much less
soluble in water. Look at the graph of gas solubility vs temperature above and guess what gas is used
for this purpose!
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