Experiment 14a: Chemical Equilibrium
Many chemical reactions do not go 100% to completion. Reactions in a closed system may go to some
intermediate state in which both the reactants and products have concentrations that do not change with time. The
system is in equilibrium and there is a constant relationship between the concentration of the reactants and
products as described by the Law of Chemical Equilibrium. For example, when solutions of iron(III) ion and
thiocyanate ion are mixed, they react spontaneously to form a distinct red colored complex ion known as the
thiocyanatoiron (III) ion. These ions form an equilibrium mixture whose concentrations can be related through an
equilibrium constant, Kc.
In this experiment you will study in a quantitative manner the equilibrium properties of the reaction described by the
following reaction:
Fe+3 (aq) + SCN-1 (aq)

FeNCS+2 (aq)
You will also determine the numerical value for Kc for the reaction at room temperature. Different amounts of the
reactants will be combined to produce equilibrium mixtures that can be analyzed for the concentration of the
colored complex ion. These colored samples will be analyzed using a spectrometer since the eye can only discern
general differences in color intensity. The instrument will provide quantitative information (absorbance) that can be
used to determine the equilibrium constant. When white light is passed through a colored solution certain
wavelengths of light are absorbed. The FeNCS+2 (aq) ion appears red-orange because it absorbs blue-green light
that is about 450 nm. At low concentrations the amount of light absorbed is proportional to the concentration of the
absorbing species, Absorbance = k [FeNCS+2]. This is known as the Beer-Lambert Law. Since a colorimeter or
spectrometer can be set to pass a selected wavelength of light through a sample, the amount of light that is not
absorbed is detected by a photocell and the intensity of this light is indicated as % transmittance, %T, or as
absorbance. Absorbance can be easily calculated by using the relationship A = 2-log %T.
When using an instrument, a blank that contains some of the reactants is used to adjust the instrument to measure
100 %T and 0.0 A.
Materials
6 medium test tubes numbered 1-6
1 test tube rack
4 beakers labeled: Fe+3, SCN-1, HNO3, & mixing
1 0.1-1.0mL micropipette per lab table
1 1.0-10.0mL micropipette per lab table
0.100M Fe(NO3)3
0.00100M KSCN
0.600M HNO3
Procedure
1. Use the variable pipettor to place 2.50 mL of 0.00100 M KSCN into each of the six of the test tubes.
2. Add 2.50 mL of 0.100 M Fe(NO3) to test tube #1. (You may use one of the squeeze-bulb pipets to carefully
but completely mix this solution by drawing into the bulb and then squeezing the solution back into test tube
#1 or by gently shaking the test tube).
**
This next step is VERY IMPORTANT and a bit ‘tricky’.
3. Be careful. Use the variable pipettor to measure 5.00mL of the 0.100M Fe(NO3) into the beaker that is marked
mixing. Add 7.50 mL of 0.600 M HNO3 into the mixing beaker and mix. Here is the tricky part…remove 5.0 mL of
this solution from the mixing beaker using a variable pipettor, empty the mixing beaker, and place the 5.0 mL of
solution back into the mixing beaker. The mixing cup will have 5.00 mL of a solution that is 0.0400 M Fe(NO3).
You can verify this by using C1V1 = C2V2.
4. Continue with the serial dilution by adding 7.50 mL of 0.600 M HNO 3 into the mixing beaker that has the 5.00mL
of 0.0400M Fe+3. Mix well. Use the variable pipettor to place 2.50 mL of this mixture into test tube #2. Remove
5.00 mL of this solution, empty the mixing beaker, put the 5.00 mL back into the mixing beaker and add 7.50 mL of
0.600 M HNO3 into the mixing cup.
5. Repeat this dilution sequence until you have added 2.50 mL of serially diluted Fe3+ (aq) into all six test tubes.
A spectrometer (Spec20) should be turned on and allowed to warm up for 15-20 minutes before it is used to
analyze solutions. The on-off knob (front left) is used to adjust the 0 % before the blank is inserted in the sample
compartment (in a cuvette) and is adjusted to 100%T with the right-hand knob when the blank is in the sample
compartment. The cuvettes must be wiped and properly aligned. Carefully transfer the 5.0 mL of sample in test
tube #1 into a cuvette and measure the Absorbance. Repeat with each solution that was prepared.
Data:
Test
Tube
Initial
(M)
Fe+3
Results:
Initial
(M)
SCN-1
Absorbance
Equilibrium
(M)
Fe(SCN)+2
Equilibrium
(M)
Fe+3
Equilibrium
(M)
SCN-
Kc
1
2
3
4
5
6
Calculations
1. Use the dilution equation, C1V1 = C2V2, to determine the initial concentration of each reactant (ignore the nitric
acid) at the moment the solutions in each test tube were mixed but in the split second before any reaction began.
2. Assume the reaction in test tube #1 was essentially 100% (research has shown it to actually be about 98%) and
use the mole ratio between the reactants to estimate the equilibrium concentration of [FeNCS+2] in tube #1.
3. Use the estimated value for [FeNCS+2] and calculate the proportionality constant, k, in the equation:
Abs1 = k[FeNCS+2]1.
4. Since k is a constant, it will be used with the absorbance values measured for the solutions in the other test
tubes to estimate the equilibrium concentration of [FeNCS+2]n for tubes 2, 3, 4, 5, and 6.
5. Calculate the equilibrium concentrations for the reactants, [Fe+3] and [SCN-1].
6. Use the Law of Chemical Equilibrium to calculate the equilibrium constant, Kc, for solutions 2, 3, 4, 5, and 6.
Why is solutions 1 not useful?
7. Show the calculations that prove that Kc  [Fe+3] [SCN-1]
[FeNCS+2]
8. What is the best ‘average’ value of Kc shown by your data? What is the best ‘average’ value of Kc shown by the
class data?
Experiment 14b: Le Chatelier’s Principle
Since most reactions seem to go to completion (reactants are turned 100% into products), reactions seem to
only go in one direction. A classic example is the combustion of methane, the rusting of iron, and even the
formation of water from the gaseous elements hydrogen and oxygen.
Many chemical reactions like the one studied in Experiment 14a actually establish equilibrium when a system
with steady conditions is established; Fe+3 (aq) + SCN-1 (aq)  FeNCS+2 (aq).
Henri Louis Le Chatelier studied reactions and noticed an important relationship between reactants and
products in many reactions. He had a difficult time explaining the principle in 1884 when it was first published, and
it is generally stated today as: If a "stress" is applied to a system at equilibrium, the equilibrium
condition is upset; a net reaction occurs in that direction which tends to relieve the "stress," and
a new equilibrium is obtained.
In this experiment some reversible reactions will be studied so that products form reactants just as reactants
make products. Color changes and the formation or dissolving of precipitates can be used to observe shifts in
equilibrium. Assume that a reaction in solution is part of a ‘closed’ system since any change due to evaporation or
room temperature will be slow enough to not have an effect on the validity of the observations.
In an aqueous solution the yellow chromate ion, CrO4-2 (aq), can be changed into the orange dichromate ion,
Cr2O7-2 (aq) and the reverse reaction is just as easy to produce. The extent to which these reactions take place
depends upon the concentration of the hydrogen ion concentration in the solution. The [H +] will be increased by
adding HCl and decreased by adding NaOH since H+ and OH- react to produce H2O.
**
The solutions used in this experiment are known to be dangerous. Be safe!! Wash your hands when you are
through. Sodium or ammonium chromate solution will also work in place just as well as potassium compounds.
Other acids and bases will also adjust the [H+] of the solutions.
The reactions may be performed on an acetate sheet over a piece of white paper towel or paper. The solutions
should be rinsed into a hazardous waste container.
Procedure Part I
1. Place about 10 drops of 0.1 M potassium chromate and in a different location place about 10 drops of 0.1 M
potassium dichromate on a piece of acetate. Record the starting color of each solution. Ask for assistance if they
appear to be the same color.
2. Add 1 M NaOH a drop at a time to each of the solutions until an observed color change is noted for one of the
solutions. Record the color change. Did chromate or dichromate react with the NaOH?
3. Add 1 M HCl a drop at a time to the solutions produced in step 2 until an observed color change is noted for one
of the solutions. Record the color change. Did chromate or dichromate react with the HCl?
4. Again, place about 10 drops of 0.1 M potassium chromate and in a different location place about 10 drops of
potassium dichromate on a piece of acetate. Record the starting color of each solution. Add 1 M HCl a drop at a
time to the solutions until an observed color change is noted for one of the solutions. Record the color change. Did
chromate or dichromate react with the HCl?
5. Add 1 M NaOH a drop at a time to the solutions produced in step 4 until an observed color change is noted for
one of the solutions. Record the color change. Did chromate or dichromate react with the NaOH?
6. Write a balanced chemical equation that shows the change of CrO 4-2 into Cr2O7-2. Include H+, H2O, or OH- as
needed.
Part II
The equilibrium of solid barium chromate in a saturated solution of its ions can be shown by:
BaCrO4 (s)
 Ba+2 (aq) + CrO4-2 (aq).
1. Repeat step 1 in Part I (above). Add 2 drops of 1 M NaOH to the chromate solution and 2 drops of 1 M HCl to
the dichromate solution. Record the starting colors. Add 0.1 M Ba(NO 3)2 to each until a change is observed.
Record the relative solubility of the compound formed. Which is most soluble (lease insoluble); BaCrO4 or
BaCr2O7?
2. Slowly (observe) add 10 drops of HCl to the chromate solution in step 1. Slowly (observe) add 10 drops of
NaOH to the dichromate solution in step 1.
3. Repeat step 1 in Part I (above) and add 3-5 drops of 0.1 M Ba(NO3)2 to each. Describe the relative amounts of
precipitate (product) that was formed.
Questions
1. What can you conclude about the importance of H+ (aq) on CrO4-2 (aq)  Cr2O7-2 (aq)? Write a balanced
chemical equation using the proper number of H+ and H2O to the appropriate sides of the equation.
2. What can you conclude about the importance of OH- (aq) on Cr2O7-2 (aq)  CrO4-2 (aq)? Write a balanced
chemical equation using the proper number of OH- and H2O to the appropriate sides of the equation.
3. Using the color of the solutions and the reaction with H+, OH-, and Ba2+, what was the relative equilibrium
concentrations of CrO4-2 (aq) and Cr2O7-2 (aq) in the starting solutions?
4. Use one (or both) of the balanced equations you wrote in questions 1 and 2 above to make a summarizing
statement concerning the chromate-ion and dichromate-ion equilibrium including the principle of Le Chatelier.
5. Assume that the CrO4-2 (aq) and Cr2O7-2 (aq) ions were reacted with CH3COOH, HNO3, Ca(OH)2, H2SO4, KOH,
C2H5OH, NH3 (aq), Li+, and Fe3+ (aq). Predict what you would observe with each of the solutions and record the
hypothetical data table that included color changes and precipitates in your journal.
Chemical Reactant
CH3COOH
HNO3
Ca(OH)2
H2SO4
KOH
C2H5OH
NH3
Li+
Fe3+
CrO4-2 (aq)
Cr2O7-2 (aq)