CHEM 110-Fall 2001-Debye
Dissolving KNO3: Entropy or Energy Driven?
The solubility of potassium nitrate, defined by the relative amounts of KNO3 and water
required to make a saturated solution, is described by the chemical reaction:
KNO3(s) --> K+(aq) + NO3-(aq)
If s moles of KNO3 are required to make a saturated solution in 1 L of water, we can
describe the solubility of KNO3 as s moles per liter of water.
The volumes of solute and solvent are found not to be additive when making aqueous
solutions of ionic compounds. Instead, the volume of the solution is very close to just the
volume of water used to make the solution. The ions from the solute appear to fit into the
empty spaces between the water molecules. Making use of this good approximation, the
concentrations (in moles per liter of solution) of the K+ and NO3- ions in a saturated
solution are equal to the solubility (in moles per liter of water) of KNO3:
[K+]eq = s and [NO3-]eq = s
The value of the solubility product varies with temperature as predicted by LeChatelier’s
Principle. An endothermic (ΔHsoln>0) dissolution process would absorb heat from the
heat + KNO3(s) --> K+(aq) + NO3-(aq)
and the solubility of the salt would increase at higher temperatures. But an exothermic
(ΔHsoln<0) dissolution solution process releases heat to the surroundings
KNO3(s) --> K+(aq) + NO3-(aq) + heat
and cause the solubility to decrease at higher temperatures. In this way, the sign of the
heat of solution determines whether the solubility of KNO3 should increase or decrease
with increasing temperature. Stated the other way around, the observed temperature
variation of the solubility if of KNO3 indicates if the dissolution process of KNO3 in
water is an endothermic or an exothermic process.
In this experiment, you will determine the solubility of potassium nitrate at 5 different
temperatures in the range 25–80 oC by dissolving a known amount of KNO3 in 20.0 mL
of hot water and then letting the solution cool until the KNO3 just begins to crystallize
and precipitate from the solution. At this particular temperature, you have a saturated
solution of known composition, defining the solubility of the KNO3 at this temperature.
In order to explain (or predict) the solubility of a substance in water, one must consider
both energy effects and structure effects.
The energy effects are summarized in the value of the heat of solution (ΔHsoln) and are
related to the relative magnitudes of the forces between water molecules, between the two
ions in solution and their surrounding water molecules (hydration spheres), and the forces
between the ions in the crystal lattice of the undissolved salt.
The structure effects are summarized in the value of the entropy of solution (ΔSsoln) and
are related to the differences in the structural disorder between the separated solid KNO3
and water, and the solution resulting when the KNO3 is dissolved in the water. If the
solution causes an overall increase in disorder, then ΔSsoln is positive. Remember that
entropy expresses disorder and not order.
By observing and recording how the solubility of KNO3 changes with temperature, you
not only can calculate the solubility product and but also can separate out the relative
importance of the energy (ΔHsoln) and structure (ΔSsoln) effects.
The Gibbs free energy change is calculated at each temperature from the observed
ΔGsoln = -RT Ln s2
where T is in Kelvin, R is 8.314 J mol-1 K-1, and s is in moles/liter. Remember that
ΔGsoln<0 favors lots of product (high solubility) and ΔGsoln>0 favors lots of reactant (low
There is a simple relationship between the Gibbs free energy of solution, the heat of
solution, and the entropy of solution which balances off these energy effects and
structure effects in determining the value of ΔGsoln :
ΔGsoln = ΔHsoln – T ΔSsoln
The temperature (T in kelvin) in this basic relationship acts as a scaling factor in
balancing the entropy change and the enthalpy change.
Note that at very low temperatures (T->0), the term TΔSsoln becomes very small so that
ΔGsoln = ΔHsoln (very low temperature)
resulting in the general statement that reactions are energy driven at low temperatures.
But at very high temperatures, – T ΔSsoln eventually always becomes much larger than
ΔHsoln so that
ΔGsoln = – T ΔSsoln (very high temperature)
and we can make the generalization that chemical reactions are entropy driven at high
A careful look at the above equation shows that it is the equation of a straight line
y = a + bx
if you graph your values of (ΔGsoln) vs. T. The entropy of solution (ΔSsoln in J K-1 mol-1)
then equals the negative slope of the line and the heat of solution (ΔHsoln in J mol-1) is
equal to its y-intercept.
1. Weigh 11.5-12.5 g of KNO3 to the nearest milligram directly into a clean, dry
2. Using a buret, add 20.0 mL deionized water to the beaker. Gently place a
magnetic stir bar and a small boiling chip into the beaker.
3. Place the beaker on a stirrer/hot plate and position the temperature probe of a pH
meter into the water so that its tip touches neither the beaker nor the stir bar.
4. Carefully turn on the magnetic stirrer to gently stir the mixture. CAUTION: If the
stirring is too vigorous, some solution will splash out of the beaker and you will
have to start over.
5. Set the heater control to Setting 4 or 5 and gently heat the contents of the beaker
until all of the KNO3 is dissolved. Do not overheat. Below are guidelines for the
temperatures at which the salt should dissolve. (If you overheat the solution, you
will suffer through excessive cooling times.)
Total KNO3 Heat to:
12 g 40 oC
6. Turn off the heat, place an insulating pad composed of seven paper towels doubly
wrapped in aluminum foil between the hot plate and beaker, and allow the
solution to cool.
7. Carefully watch the solution and, using your electronic thermometer, determine
the temperature at which the solid KNO3 begins to crystallize out of the solution.
8. Repeat Steps 3 through 7 four times, each time adding an additional 3.5-4.5 g
KNO3 (weighed to the nearest milligram) to your solution from the previous trial.
1.For each temperature, calculate the solubility (s) of potassium nitrate as moles KNO3
per liter of water and the Gibbs free energy of solution (ΔGsoln).
2. Make a graph of the solubility vs. temperature (in oC) and fit the points with the best
curve. This probably will not be a straight line. Report the equation of your fitted curve
and interpret the general appearance of your graph to determine if the dissolving of
KNO3 in water is an endothermic or an exothermic process.
3. Make a graph of ΔGsoln vs. T (in kelvin) and fit your points but now with a straight
line. From its slope and y-intercept, obtain the entropy of solution (ΔSsoln) and the heat of
4. Please answer the following questions and justify your answers:
a. Does dissolving potassium nitrate in water cause a local increase or decrease in
b. Is the solubility of potassium nitrate energy driven or entropy driven?
Technical Notes: The graphical analysis used here to obtain ΔHsoln and ΔSsoln by using a
straight line to fit the data holds well over the small temperature ranges investigated and
gives reasonable results. The potassium nitrate is slow to dissolve unless finely powdered
so students tend to overheat the solutions resulting in longer than necessary cooling times
so it is advisable to grind the stock KNO3 before weighing. The crystallization
temperatures are roughly with 5 oC of the indicated target heating times. A careful
analysis of the experiment should consider the following: the values of ΔHsoln and ΔSsoln
slowly change with temperature so that they are not truly constant as demanded by the
linear fit; the volumes of these concentrated solutions are not strictly equal to the volume
of water; the volumes of these solutions vary with temperature; the mean activity
coefficient of KNO3 surely is not equal to one and varies in these concentrated solutions.