The Solubility of Ba (NO3)2
The dissolution of barium nitrate, Ba(NO3)2,
in water occurs according to the reaction
|
Ba(NO3)2(s) ® Ba2+(aq) + 2NO3-(aq) |
(1) |
the solubility product constant is given by Ksp = (Ba2+)(NO3-)2.
If S is the molar solubility of barium
nitrate in pure water, then (NO3-) = 2S. For every one barium nitrate formula unit
there are 2 nitrate ions. The Ksp
expression can be written as equation (1).
|
Ksp =
(S)(2S)2 = 4S3 |
(2) |
On
the other hand, if the salt is dissolved in a solution which contains an
additional source of nitrate ion, such as nitric acid, then the concentration
of nitrate ion arises from two sources:
from the dissolution of the Ba(NO3)2, and from the
nitric acid. For example, in 0.5 M HNO3
solution (NO3-) = 2S’ + 0.5, where S’ is the solubility
of Ba(NO3)2 in 0.5 M HNO3(aq). The concentration of Ba2+ will be S’, and Ksp is given by
equation (2).
|
Ksp = (S’)(2S’ + 0.5)2 |
(3) |
In
Part 1 of this experiment you will measure the solubility of barium nitrate at
room temperature in pure water and in 0.5 M HNO3, and use your
results to calculate Ksp.
The values of Ksp for the two solutions should be equal since
Ksp depends only on temperature.
Temperature dependence of
Ksp
The
temperature dependence of Ksp can be used to determine the enthalpy
and entropy changes accompanying the dissolution of 1 mole of barium nitrate in
water (reaction (1)).
The
solubility product constant is an equilibrium constant and is related to the
Gibbs free energy change, DG,
for the solution process by
|
ln Ksp = |
(4) |
where the units of DG are J/mol, T is in kelvins, and R is 8.3145
J/(K-mol).
Since DG = DH -TDS ( where DH and DS, are the enthalpy and entropy changes, respectively), equation (3) can be rewritten as
|
|
(5) |
The units of DH and DS, are J/mol and J/(mol-K), respectively.
Equation (4) shows that if Ksp
is measured as a function of temperature, then a graph of ln Ksp
versus 1/T should be straight line with a slope given by
|
Slope = -DH/R |
(6) |
and an intercept given by
|
Intercept = DS/R |
(7) |
In Part 2 of the experiment,
you will measure Ksp as a function of temperature, and use your
results to obtain values of DH
and DS. The values
of Ksp will be calculated in terms of molal rather than molar
concentration.
The molal concentration, m,
is given by
![]()
|
Ksp = (mBa2+)(mNO3-)2
= 4(S’’)3 |
(8) |
Where S’’ is the molal solubility of barium nitrate.
Weigh
a clean, dry 250 mL beaker to ±0.01 g. Place
about 5 g of solid Ba(NO3)2 in the beaker and weigh to ±0.01 g. Add
50.0 mL of deionized water, measured with a graduated cylinder. Stir for 10 min. Measure the temperature of the solution, and then decant as much
as possible of the saturated solution into a waste container. Heat the beaker and contents on a hot plate. Evaporate to dryness (about 5 min.) being
careful to avoid spattering. Let cool
to room temperature, and then weigh the beaker and contents to ±0.01 g. Save
the dry solid for use in Part 2.
Repeat the procedure in Part 1a, except substitute 50.0
mL of 0.5 M HNO3 for deionized water, and evaporate the contents in
the hood.
Part 2
Label and weigh to ±0.01 g, four dry, clean 100 mL beakers. To the solid
saved from Part 1a in the 250 mL beaker, Add additional Ba(NO3)2
to bring the total mass to 20 g. Add
100 mL of deionized water. On a hot
plate, heat the solution with stirring, until the temperature of the solution
is about 70oC. Stop
heating. As soon as the solution cools
to 35oC, quickly decant about 10 mL of the solution into one of the
pre-weighed 100 mL beakers, and record the temperature. It is not necessary to know the exact volume
of the solution decanted since the concentration will be expressed as molality
and not molarity. Be careful not to
transfer any undissolved solid. Continue
monitoring the temperature of the solution.
When the temperature cools to 30oC, decant another 10 mL into
a second pre-weighed beaker. Repeat
this when the solution cools to 25oC and then to 0oC. The last temperature is attained by placing
the solution in an ice bath. Allow each
of the solutions in the 100 mL beakers to come to room temperature, and then
weigh to ±0.01 g.
Heat each of the decanted solutions on a hot plate and
evaporate to dryness. To avoid
splattering, reduce the heating rate as evaporation nears completion. Allow the beakers to cool to room
temperature, and then weigh each beaker and the remaining residue to ±0.01 g.
The molar
solubility of Ba(NO3)2 is equal to the moles of Ba(NO3)2
dissolved in the saturated solution divided by the volume of the solution,
0.050 L.
![]()
Once you have calculated S,
use equation (2) to calculate Ksp.
Calculate the molar solubility of Ba(NO3)2
using the same method as in Part 1a, and calculate Ksp using
equation (3).
Part 2:
A sample calculation for this part is shown in the example below.
Example:
In the determination of the
solubility of Ba(NO3)2 at 69oC, a sample of
saturated solution was decanted into a pre-weighed beaker, and the following
data were obtained. Calculate the molal
solubility and molal solubility product of Ba(NO3)2 at 69oC.
mass of beaker + sat. soln = |
111.08 g |
|
mass of beaker + dry Ba(NO3)= |
102.12 g |
mass of water in sat. soln = |
8.96 g |
|
|
|
|
mass of beaker + dry Ba(NO3)2
= |
102.12 g |
|
mass of empty beaker = |
99.62 g |
|
mass of dry Ba(NO3)2
in sat. soln= |
2.50 g |
|
|
|
The molal solubility, S’’ is
= 1.07 mol/kg
and Ksp is given
by equation (8)
Ksp = 4(1.07)3 = 4.90
Use your results to calculate the molal solubility and Ksp
at four different temperatures. Use
Excel to calculate lnKsp and 1/T, and make a plot of lnKsp
versus 1/T. Use linear regression to
obtain the slope and intercept of your graph.
From the slope and intercept, calculate DH and DS, respectively, for the solution process.
Answer
the questions below in complete sentences.
1)
Report and compare the
values of Ksp obtained in Parts 1a and 1b. Are the two values the same?
Explain why or why not.
2) Report the solubilities and the values of
the Ksp’s obtained at the different temperatures in Part 2. How does the solubility vary with
temperature?
3) What is the enthalpy change for the solution
process? Is this an exothermic or an
endothermic processs? Does your value,
combined with your answer to question 2, agree with Le Chatelier’s Principle? Please explain.
4) What is the entropy change for the
solution process? Does its value agree
with what you know about the relationship between entropy and disorder? Please explain.