Laboratory
15
Variation of Solubility with
Temperature and Solvent
Objectives
•
Utilize laboratory techniques to explore two factors that affect solubility:
temperature and solute–solvent interactions.
•
Quantify the solubility and use graphing techniques to study the solubility
trends as a function of temperature.
•
Compare the presentation of solubility data using different units.
Introduction
Solubility is an important topic in chemistry. It has been introduced previously,
here we come back to the idea, and later in this course we will talk about it again.
Solubility is fundamental to many life processes and a solid understanding is
necessary for scientists, health professionals, and engineers alike. Solubility is a
SK\VLFDOSURSHUW\WKDWPD\EHTXDQWL¿HG7KURXJKGDLO\H[SHULHQFHVZHNQRZWKDW
some things can become more or less soluble if we change the surrounding conditions. If you like to sweeten your tea you know that sugar will dissolve much more
easily in hot tea than it will in iced tea. In this lab we will consider some factors
that affect solubility.
17
Laboratory 15
Discussion
15.1 Connection to Lecture
We began the discussion of solubility in our previous chemistry course with the table of
solubility rules. This table summarizes the rules for solubility and allows us to classify
compounds as soluble or insoluble in water. In lecture now we are learning that we can
quantify the solubility of solutes in solvents; two compounds may be soluble in the same
solvent, but likely one compound is more soluble than the other. We are learning that differHQWIDFWRUVFDQLQÀXHQFHKRZVROXEOHVRPHWKLQJLV)DFWRUVWKDWDIIHFWWKHVROXELOLW\LQFOXGH
1. temperature effects
2. solute–solvent interactions
3. in gases, pressure effects
In this lab you will be looking at temperature effects (Parts A and B) and solute–solvent
interactions (Part C) on the solubility of a compound.
15.2 Temperature Effects
It is no surprise that solubility depends on temperature. The relationship for several substances is shown graphically in Figure 15.1. As illustrated, solubility of an ionic solute in
water generally increases with increasing temperature, although there are exceptions. Note
two other things about the data displayed. First, each substance has a unique curve, and
secondly, dependence of solubility on temperature is generally not a linear relationship.
Solubility is an experimentally determined property.
CaCl2
100
NaNO3
KNO3
80
g solute
100 g H2O
60
KCl
40
NaCl
20
Ce2(SO4)3
0
0
20
80
40
60
Temperature,°C
100
Figure 15.1 Solubility of ionic solutes in water.
18
Explanation of the effect of temperature on solubility is in terms of competing processes
and energies. Consider formation of an aqueous solution of potassium iodide. Before the
VROXWLRQFDQIRUPWKHVROLG.,PXVWEHSXOOHGDSDUW²IRUFHVRIDWWUDFWLRQEHWZHHQ.+ ions
and I– ions must be overcome. This process requires energy. The separated ions are then
Variation of Solubility with Temperature and Solvent
VXUURXQGHG E\ OD\HUV RI ZDWHU PROHFXOHV²WKH\ EHFRPH K\GUDWHG7KLV SURFHVV HYROYHV
energy because of the electrostatic attractions between the charged ions and the polar water
molecules.
Question 15.1: Is this process endothermic or exothermic? What is the sign of ΔH?
A much smaller amount of energy is involved in separating water molecules from each
other and it is included in the hydration energy. The process of solution can be summarized
using an energy diagram, Figure 15.2.
Separated K+ and I-
K+(aq) and I-(aq)
KI(s)
Hsoln
Figure 15.2 Energy level diagram for the process of solution.
7KHRYHUDOOHQHUJ\FKDQJHIRUWKHSURFHVVRIVROXWLRQǻ+soln, is the difference between the
energy of the solution K+(aq) and I–(aq) and the energy of the solute KI(s) to be dissolved.
,QWKLVFDVHǻ+soln is positive and energy is required when KI dissolves in water. This can
be indicated by writing heat as a reactant in the equation for the solution process.
KI(sZDWHUKHDWĺVROXWLRQ
If we increase the temperature by adding heat, we are supplying a reactant so more solution
should form, provided both KI and water are available.
Question 15.2: Does the solubility of KI increase or decrease with increasing
temperature?
The strength of attraction between the ions in the solid and the strength of hydration both
depend on the size and charge of the ions. If the energy of hydration is greater than the
energy required to pull the solid apart, the overall heat of the solution will be negative
ǻ+soln < 0) and heat will be evolved when the solution forms.
6ROLGZDWHUļVROXWLRQKHDW
In this case, increasing the temperature supplies a product, which is a reactant for the reverse reaction. Increasing the temperature causes the reverse reaction; solid crystallizes out
of solution and solubility decreases. This is the case for Ce2(SO4)3 in Figure 15.1.
19
Laboratory 15
15.3 Solute–Solvent Interactions
Solubility also depends on the nature of the solute and the solvent. The fact that each
substance has a unique curve suggests that solubility depends on the nature of the solute.
Similarly, if you use the same solute and change the solvent, you would also expect unique
behavior. A generalization can be made about solubility and the nature of solute and solvent: Substances with similar intermolecular forces of attraction tend to be soluble in each
other. This is stated concisely as “like dissolves like.”
For a solution to form, intermolecular attractions between particles in the solute must be
overcome, attractions of solvent particles for each other must be overcome, and new attractions must exist between solute and solvent. In other words, solute–solvent attractions must
be strong enough to compete with solute–solute and solvent–solvent attractions.
Consider again the process of dissolving potassium iodide in water. Attractions between
particles in the solute are electrostatic attractions between positive ions and negative ions,
K+ and Ií. Attractions between solvent particles (H2O molecules) are hydrogen bonding,
dipole–dipole, and London dispersion forces. The ion–dipole forces of attraction between
K+ ions and the negative end of a water molecule, and between Ií and the positive end of
a water molecule are strong enough to compete with the solute–solute and solvent–solvent
DWWUDFWLRQV7KHVROXWLRQIRUPV7KHVROXELOLW\RI.,LVJJ+2O) at 0 °C.
If we try to dissolve KI in a nonpolar solvent such as CCl4, the weak London forces of
attraction between ions and CCl4 molecules cannot compete with the ion–ion forces in KI.
Potassium iodide is essentially insoluble in CCl4.
Now consider dissolving I2 in water and also in CCl4. London forces hold the solid I2
together. The only attraction between the solute and solvent is the London force. Water has
very weak London forces; carbon tetrachloride has relatively strong London forces. The
solubility of I2 is 100 times greater in CCl4 than in water. Again, “like dissolves like.”
15.4 Data Analysis and This Lab
Recall from class that a saturated solution is a solution in which both dissolved and undissolved solute are in equilibrium with each other. That is to say, a saturated solution contains
as much solute as it can hold while in equilibrium with excess solid solute. We will make
use of the saturation, the point at which no further solute can dissolve in solution, to quantify the solubility of our compound. Solutes will be dissolved in heated solutions and then
allowed to cool. When the temperature is such that solute begins to crystallize, the saturaWLRQHTXLOLEULXPLVHVWDEOLVKHG,Q\RXUPHDVXUHPHQWV\RXZLOOEH¿QGLQJWKHWHPSHUDWXUH
DWZKLFKWKH¿UVWFU\VWDOIRUPV.QRZQDPRXQWVRIVROXWHDQGVROYHQWDUHKHDWHGDQGVWLUUHG
until all of the solute dissolves. The solution is stirred while it cools, and watched closely
XQWLOWKH¿UVWFU\VWDOVIRUP7KHWHPSHUDWXUHLVUHFRUGHG$GGLWLRQDOVROYHQWLVDGGHGWRWKH
same test tube and the heating and cooling processes are repeated. The temperature is recorded at each concentration. This is the general procedure for each part of the experiment.
20
You will use the same solute for all parts of the experiment; it will be either potassium
dichromate, K2Cr2O7, an orange, inorganic salt or oxalic acid dihydrate, H2C2O4 • 2 H2O, a
ZKLWHRUJDQLFDFLG)RU3DUWV$DQG%WKHVROYHQWLVZDWHU7KHVROYHQWIRU3DUW&LVDZDWHU
dioxane mixture. Dioxane is a nonpolar molecule. It is soluble in water because the oxygen
atoms in the molecule have unshared pairs of electrons that are used to form hydrogen
bonds to water molecules.
Variation of Solubility with Temperature and Solvent
When studying solubility it helps to quantify how much solute is able to dissolve in a
given solvent. Solubility is the amount of substance that will dissolve in a given amount
RIVROYHQWRUVROXWLRQ&RPPRQXQLWVRIVROXELOLW\DUHJVROXWHJVROYHQW$QRWKHU
concentration unit that we will use for solubility in this experiment is mole fraction, X.
Xsolute =
mole solute
mole solute + mole solvent
For each solution, determine grams of solute in 100 g solvent using dimensional analysis.
)RUDTXHRXVVROXWLRQVDVVXPHDGHQVLW\RIZDWHURIJP/DWƒ&7KHGHQVLW\RI
WKHZDWHUGLR[DQHVROYHQWLVJP/
)RUH[DPSOH,IJRIVROXWHLVGLVVROYHGLQP/RIWKHZDWHU
dioxane solvent, what is the solubility?
20.00 mL solvent #
1.023 g
1 mL
7.0510 g solute
20.46 g solvent
=
= 20.46 g solvent
x g solute
100 g solvent
Then rearrange to solve for x:
100 g solvent #
7.0510 g solute
20.46 g solvent
= 34.46 g solute in 100 g solvent
Construct a graph using software (e.g., Excel) with solubility on the y-axis and temperature
on the x-axis. Data for aqueous solutions (Parts A and B) should be plotted as one set of
data and fall along a smooth curve. Data for the mixed solvent (Part C) should form another
curve. Plot both sets of data on the same graph. Have the computer software (e.g., Excel)
draw smooth curves for the two sets of data. As long as you’ve plotted the data for Parts A
and B as one set of data (seven data points) the software should be able to draw a smooth
curve (an exponential or polynomial generally works well).
Different substances in the same system are often related in molar amounts, whether they
are reacting or simply present as a mixture. For each of the aqueous solutions, calculate the
mole fraction of the solute.
For water as the solvent (Parts A and B), construct a second graph of temperature on the
y-axis versus mole fraction on the x-axis. The data should fall along one smooth curve (a
ORJDULWKPLFOQ¿WLQ([FHOZRUNVUHDVRQDEO\ZHOO
21
Laboratory 15
For example: If 5.4942 g of H2C2O4 • 2 H2O dissolves in 20.00 mL of water,
what is the mole fraction?
Xsolute =
mole solute
mole solute + mole solvent
mole solute = 5.4942 g H 2 C 2 O 4 • H 2 O #
mole solvent = 19.95 g H 2 O #
Xsolute =
1 mol
= 0.043581 mole
126.07 g
1 mol
= 1.107 mole
18.02 g
0.043581 mol
= 0.03788
0.043581 mol + 1.107 mol
Materials Required
Equipment
10-mL and 5-mL pipets
KRWZDWHUEDWKP/EHDNHU
cold water bath (400-mL beaker)
Bunsen burner, wire gauze
WKLQZDOOHGUXEEHUWXELQJ
ring stand, ring
solubility apparatus
Chemicals
oxalic acid dihydrate, H2C2O4 • 2 H2O
JVWXGHQW
potassium dichromate, K2Cr2O7JVWXGHQW
70:30 water:dioxane solution, H2O:C4H8O2
P/VWXGHQW
ice
Common Equipment
analytical balance
Cautions
Dioxane is a suspect carcinogen. The vapor might be irritating to lungs and mucous membranes. The student hoods should be on. Potassium dichromate is corrosive. Oxalic acid is
SRLVRQRXV,IDQ\RIWKHVHVXEVWDQFHVFRPHLQWRFRQWDFWZLWK\RXUVNLQRUFORWKLQJÀRRG
the affected area with water. Be careful to avoid burns from the ring, beaker, and open
ÀDPH *RJJOHV PXVW EH ZRUQ DW DOO WLPHV 6WLU VDPSOHV JHQWO\ DV YLJRURXV VWLUULQJ ZLOO
result in the test tube breaking in the bottom and the loss of samples.
Procedure
A. Determine solubility in water as a function of temperature.
22
1. Obtain the solubility apparatus from your lab instructor. It consists of a thermometer in
a rubber stopper in a 75-mL test tube. The coiled end of the wire stirrer should encircle
the thermometer, as shown in Figure 15.3.
Variation of Solubility with Temperature and Solvent
Thermometer
©Hayden-McNeil, LLC
-80
-70
-60
-50
-40
-30
-20
-10
0
10
20
30
40
50
60
70
80
90
100
110
120
130
Wire
Stirrer
Note:
coil encircles
thermometer
Figure 15.3 Solubility apparatus.
2. Your lab instructor will assign oxalic acid or potassium dichromate for use in the
experiment.
3. Dry the inside of the test tube with a Kimwipe. Weigh your assigned solid, between 6.3
to 6.7 g of oxalic acid or 5.6 to 5.9 g of potassium dichromate, on the analytical balance
and add it to the test tube.
4. Use a pipet to deliver 10.00 mL of distilled water to the test tube.
5. Dissolve the sample completely by heating in hot water. Periodically take the test tube
out of the beaker and gently swirl the test tube, and stir the solution with the wire stirrer, to speed the process. When you stir the solution with the wire stirrer, make sure the
coil encircles the thermometer and use slow up-and-down movements. Rapid stirring
can cause the thermometer or the test tube to break.
6. After the solid is completely dissolved, remove the test tube from the water bath and
allow the solution to cool, while gently stirring, to maintain a uniform temperature and
FRQFHQWUDWLRQ5HFRUGWKHWHPSHUDWXUHDWZKLFKFU\VWDOVDUH¿UVWREVHUYHG
7. Use a pipet to deliver an additional 5.00 mL of distilled water to the test tube. Repeat
steps 5 and 6.
8. Repeat step 7 two more times for a total of four determinations with different volumes.
Be careful to stir all of the solution, especially when a large volume is used.
9. Collect the solution in a waste beaker.
23
Laboratory 15
B. Determine solubility in water at lower temperatures.
10. Clean and dry the inside of the test tube. Weigh between 1.7 to 1.9 g of your assigned
solid on the analytical balance and add it to the test tube.
11. Use a pipet to deliver 10.00 mL of distilled water to the test tube.
12. Dissolve the sample completely by heating in hot water. Periodically take the test tube
out of the beaker and gently swirl the test tube, and stir the solution with the wire stirrer, to speed the process. When you stir the solution with the wire stirrer, make sure the
coil encircles the thermometer and use slow up-and-down movements. Rapid stirring
can cause the thermometer or the test tube to break.
13. After the solid is completely dissolved, allow the solution to cool, while gently stirring,
to maintain a uniform temperature and concentration. Record the temperature at which
FU\VWDOVDUH¿UVWREVHUYHG,WPD\EHQHFHVVDU\WRFRROWKHWXEHLQDQLFHEDWKDWORZ
concentrations.
14. Use a pipet to deliver an additional 5.00 mL of distilled water to the test tube. Repeat
steps 12 and 13.
15. Repeat step 14 one more time for a total of three determinations. Collect the solutions
in a waste beaker.
C. Determine the solubility in mixed solvent.
16. Clean and dry the inside of the test tube. Weigh your assigned solid, between 6.5 to
7.0 g of oxalic acid or 2.8 to 3.2 g of potassium dichromate, on the analytical balance
and add it to the test tube.
17. Use a pipet to deliver 10.00 mL of 70:30 water:dioxane mixture to the test tube.
18. Dissolve the sample completely by heating in hot water. Periodically take the test tube
out of the beaker and swirl or shake the test tube, and stir the solution with the wire
stirrer, to speed the process. When you stir the solution with the wire stirrer, make sure
the coil encircles the thermometer and use slow up-and-down movements. Rapid stirring can cause the thermometer or the test tube to break.
19. After the solid is completely dissolved, allow the solution to cool, while gently stirring,
to maintain a uniform temperature and concentration. Record the temperature at which
FU\VWDOVDUH¿UVWREVHUYHG
20. Use a pipet to deliver an additional 10.00 mL of 70:30 water:dioxane mixture to the
test tube. Repeat steps 18 and 19.
21. Repeat step 20 two more times for a total of four determinations. Collect the solutions
in a waste beaker and clean and dry the test tube.
24
Variation of Solubility with Temperature and Solvent
Waste Disposal
Collect the waste in a beaker at your desk. All solutions containing potassium dichromate
will be orange and must be disposed of in the inorganic salts beaker. All solutions containing oxalic acid will be colorless (may contain white solid) and must be disposed of in the
organic solvents beaker. Fill in the corresponding waste disposal sheet. Your lab instructor
will dispose of the total volumes in the appropriate containers.
Points to Consider
In this lab you are studying the solubility trends with temperature and solvent.
•
Include your report sheet with sample calculations and printouts of the two graphs.
•
What do your graphs show, with respect to the solubility trends as a function of
temperature?
•
Is the trend that you see what was expected based on the material you have read about
in the text and learned in lecture?
•
What do you notice about the solubility in water versus the water:dioxane solvent?
Does the solubility in the different solvents seem reasonable based on attractive forces? Include what the attractive forces are for both the solute and solvents.
•
Do all data points fit this trend? Are there any known errors to mention for this experiment? Make sure to discuss at least two inherent errors.
25
Laboratory 15
26
Laboratory 15 Report Sheet
Variation of Solubility with Temperature and Solvent
Name: _________________________________ Date: ______________ TA Name: ____________________
This sheet must be attached to your formal lab report.
Identity of solute: ________________
A. Water as a solvent.
Addition
Total Volume
(mL)
Mass of sample ________________ g
Temp at which
Crystals Form
Concentration
g Solute/100 g Solvent
Mole
Fraction
1
2
3
4
B. Water as solvent—dilute concentration.
Addition
Total Volume
(mL)
Temp at which
Crystals Form
Mass of sample ________________ g
Concentration
g Solute/100 g Solvent
Mole
Fraction
1
2
3
C. 70:30 water:dioxane as solvent.
Addition
Total Volume
(mL)
Temp at which
Crystals Form
Mass of sample ________________ g
Concentration
g Solute/100 g Solvent
1
2
3
4
NOTE: Attach a sample calculation of solubility for an aqueous solution and for a mixed
solvent solution, and a sample calculation of mole fraction for an aqueous solution. Attach a
printout of both graphs.
27
Laboratory 15
28
Report Sheet