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FINDING THE VALUE OF AN EQUILIBRIUM
CONSTANT I: Determining the Kc of the vivid
red ferric thiocyanate reaction.
Objectives:
 Use spectrophotometry to determine the equilibrium constant for the reaction
Fe3+ + SCN- ⇄ FeSCN2+ ;
One of the most useful and important notions in chemistry is the concept of systems at
chemical equilibrium. We are used to thinking about chemical reactions going forward, from
reactant to product:
Fe3+
Ferric (iron (III) ion
+
SCNthiocyanate ion
 FeSCN2+ (forward reaction)
ferric thiocyanate complex
In fact, all chemical reactions can also proceed in the reverse direction, from products to
reactants:
FeSCN2+  Fe3+ + SCN- (reverse reaction)
If you start out with ONLY reactants (in our example, ferric ion (iron (III) ion) and thiocyanate
ion) and NO products (ferric thiocyanate complex), the reaction will proceed “as written” in the
forward direction. What else can it do? NoFeSCN2+ , the reactant for the reverse reaction,
hasn’t been made yet. However, as the forward reaction keeps going, bright red FeSCN2+ gets
made, and the reverse reaction can get started.
After a while, both reactions run at the same rate. As fast as the forward reaction makes
FeSCN2+, the reverse reaction shoots it back to Fe3+and SCN-. Eventually, even though both
reactions are going on, concentrations of Fe3+, SCN- , and FeSCN2+ remain constant. At this
point, the reaction is said to be at equilibrium. The reactions haven’t stopped, but nothing is
changing (concentrations, temperature, whatever). Unless something comes along to mess with
the system, like you heating it up with a Bunsen burner, or adding more iron (III) ion, the system
won’t change. That’s equilibrium. At equilibrium, you can write the balanced chemical equation
like this:
Fe3+ + SCN- ⇄ FeSCN2+ The double arrow (⇄) indicates that the reaction is going in both the
forward and reverse directions, and is at equilibrium. (⇌, ↔, and ⇆ are also used as equilibrium
symbols)
One of the nice things about systems at equilibrium is that they let you write equations (called
“equilibrium expressions” or “laws of mass action” describing them, using an equal sign. Equal
signs make your algebraic life much much easier.
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For any common or garden variety reaction:
aA + bB
⇄ cC + dD
you can write an equilibrium expression (Law of Mass Action):
K = [C]c [D]d
[A]a [B]b
where K is the equilibrium constant. [A], [B], are the concentrations of reactants at equilibrium.
[C] and [D] are concentrations of products at equilibrium. a, b, c and d are just the
stoichiometric coefficients.
The equilibrium expression describing the reaction is:
A Bestiary of Equilibrium Constants
Equilibrium constants are just what they say they
are, a numerical constant which is equal to the
concentration of products divided by the
concentration of reactants of a reaction at
equilibrium.
That said, they are given a variety of different
subscripts to tell the reader what kind of reaction
they are describing. A partial list of equilibrium
constants:
Kc = [FeSCN2+]
[Fe3+][SCN-]
.
The subscript “c” indicates that the K is an equilibrium
constant.
Tables of Kc’s can be found in the Handbook of Chemistry
and Physics.
Ka given to K’s for reactions of acids with water:
What do equilibrium constants tell you?
Consider:
(note: water is not
HA + H2O ⇄ A-+ H3O+ Ka = [A-][ H3O+ ]/[HA]
Kb given to K’s for reactions of bases with water:
CH4 (g) + 2O2 (g) ⇄ CO2(g) + 2H2O (l) expression. The
A-+ H2O ⇄ HA + OH- Kb = [HA][ OH- ]/[A-]
Kw given to reaction of water with water:
H2O + H2O ⇄ H3O+ + OH- Kw = [H3O+][OH-]
Ksp given to K’s for reactions of slightly soluble
salts with water:
MA + H2O ⇄
+
Ksp =
Keq and Kc given to K’s for reactions that aren’t
any of the above.
M+
A-
[M+][A-]
included in the equilibrium
Kc = [CO2]
[CH4][O2]2
= 2 x 10143
concentration of pure
water remains constant
and is incorporated into
Kc.)
Kc is HUGE. This means that if you ignite methane (CH4)
in oxygen, when the system comes to equilibrium mostly
you will have CO2 and water, and not much methane left.
Maybe 1 molecule, if you’re lucky. Which is in fact what
happens, based on our experience.
On the other hand, consider ammonia breaking down to nitrogen gas and hydrogen gas:
2NH3(g) ⇄ 3H2(g) + N2(g)
Kc = [H2]3[N2] = 1.6 x 10-6
[NH3]2
Kc is pretty small. This means that if you wait for ammonia to come to equilibrium with hydrogen
gas and nitrogen gas, you’ll still have mostly ammonia, and only a very little H2 and N2.
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The magnitude (size) of Kc tells you how far the reaction will proceed as written before equilibrium
sets in. It doesn’t tell you how fast it will get to equilibrium, just what the ratio of reactants to
products will be when the system gets to equilibrium.
Which brings us to this week’s lab.
Your goal: Find the Kc for the reaction Fe3+ + SCN- ⇄ FeSCN2+ .
You will do this by mixing known volumes of known solutions of iron(III) nitrate and sodium
thiocyanate together, letting them come to equilibrium, and then measuring the concentration of
FeSCN2+ made spectrophotometrically. You will first need to construct a standard curve of known
[FeSCN2+ ] vs. absorbance at 442 nm.
To make 5 L of 0.25 M HNO3:
THE EXPERIMENT:
Conc HNO3 = 16 M
Add 80 mL of conc HNO3 to 5 L of DI water in a carboy. Stir
well.
You will need:
0.002 M NaSCN in 0.25 M
HNO3
0.25 M HNO3
50
mL
500
mL
125
mL
25
mL
0.25 M Fe(NO3)3 in 0.25 M
HNO3
0.002 M Fe(NO3)3 in 0.25 M
HNO3
To make 2 L of 0.002 M NaSCN in 0.25 M HNO3:
MW NaSCN = 81.08 g/mol. Note: KSCN can be substituted.
Weigh 0.16 g NaSCN on the analytical balance. RECORD
THE MASS to the nearest 0.1 mg! Dissolve it in 0.25 M
HNO3 in a 2.00 L volumetric flask, and dilute to the mark.
Mix thoroughly. Record the calculated concentration to 4
significant figures. RECORD THAT CONCENTRATION
ON THE BOTTLE!
To make 2 L of 0.25 M Fe(NO3)3 in 0.25 M HNO3:
100.00 mL volumetric
flasks
10.00 mL volumetric flasks
Pastuer pipets and bulbs
Pipet set: 1 mL, 2 mL, 3 mL,
4 mL, 5 mL, 10 mL, 15 mL,
25 mL volumetric pipets,
with blue bulb
100 mL beakers
UV-Vis spectrophotometer,
cuvettes, kimwipes, waste
beaker
5
5
1 set
MW Fe = 55.85 g/mol
Weigh 28 g of Fe filings and place in a 2 L flask. Dissolve
them in 32 mL of conc HNO3. Carefully dilute to 2 L with DI
water. Stir.
To make 2 L of 0.002 M Fe(NO3)3 in 0.25 M HNO3:
2
Weigh 0.22 g of Fe filings on the analytical balance.
RECORD THE MASS TO THE NEAREST 0.1 mg!
Place the filings in a 2.00 L volumetric flask, and dissolve
them in 32 mL of conc. HNO3. Carefully dilute to the mark
with DI water, stir. CALCULATE AND RECORD ON THE
FLASK the [Fe(NO3)3 to 4 significant figures!
Make a blank (25 mL of 0.25 M Fe(NO3)3 in 100 mL of 0.25
M HNO3) for the spectrophotometer.
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Caution: HNO3 is caustic and corrosive. Protect your eyes with safety glasses. NEVER pipet by
mouth!.
THE PROCEDURE:
Before you come to lab: Make an Excel Template of the data charts n page
7-8. PUT IT ON A FLOPPY DISK & BRING IT TO LAB.
BRING YOUR KNIGHTCARD TO LAB!!!!
1. WRITE DOWN the [Fe(NO3)3] concentrations listed on the stock bottles, in your notebook. NOT
THERE AT THE END OF LAB? 25 POINTS OFF.
2. NOTICE that there are 2 (two), more than one, Fe(NO3)3 solutions. Notice which one is which. Look
in your lab protocol outline and UNDERLINE at what steps you use which solution. This is
IMPORTANT!!!
3. WRITE DOWN the [NaSCN] concentration listed on the stock bottle, in your notebook. NOT
THERE AT THE END OF LAB? 25 POINTS OFF.
4. NOTICE that there is a carboy of 0.25 M HNO3 at the end of bench 2. This does not contain NaSCN or
Fe(NO3)3. Look in your lab protocol outline and UNDERLINE at what steps you use this solution.
This is IMPORTANT!!!
Now you can start labeling glassware and getting organized. You will use the 1.00, 2.00, 5.00, and 10.00
mL pipets first, together with the 100.00 mL volumetric flasks. Label the flasks S1,S2…S5.
STANDARDS:
In this part of the lab, you will make up standard solutions that contain known [FeSCN2+].
The reaction is:
Fe3+ + SCN- ⇄ FeSCN2+ Kc = [FeSCN2+ ]/ [Fe3+][SCN-] > 1
We want the reaction to go virtually 100% to completion, so that we know the [FeSCN2+] and can use
that to make our standard curve. As it happens the Kc for this reaction is greater than one. At equilibrium,
much of the Fe3+ and SCN- have reacted to make FeSCN2+. Le Chatelier’s Principle, and simple algebra,
tell us that we can maximize the amount of product made . The equilibrium will move right if the
concentration of Fe3+ or SCN- is high. Let’s give this some thought. If you make [Fe3+] high at equilibrium,
then the [FeSCN2+] at equilibrium must be high, and [SCN-] really low so that the equilibrium conditions
are met.
Suppose Kc = 10 (it doesn’t, but suppose)
Start with 0.01 M Fe3+ , 0.01 M SCN-, and 0 M FeSCN2+. At equilibrium, the [Fe3+] is 0.0092 M, [SCN-]
is 0.0092 M and [FeSCN2+] is 0.00084 M. (See appendix 1 for help on how to figure this out). Plug these
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values into the equilibrium expression to convince yourself that the ratio of reactants to products comes out
equal to Kc
Now, see what happens when you swamp the system with one reactant. Start with 1 M Fe3+ , 0.01 M
SCN-, and 0 M FeSCN2+. At equilibrium, the [Fe3+] is 0.991 M, [SCN-] is 0.00091 M and [FeSCN2+] is
0.0091 M. Plug these values into the equilibrium expression to convince yourself that the ratio of reactants
to products comes out equal to Kc. Swamp the system with Fe3+ and almost all the SCN- gets converted
into FeSCN2+ If you know how much SCN- you started with, you can get a pretty good (not perfect, but
pretty good) estimate of the [FeSCN2+] you make at equilibrium.
All that being said, now you are ready to make up standard solutions, in which the [FeSCN2+] made is
approximately equal to the [NaSCN] you start with.
5. Make up the 5 standard solutions, following the chart, and NOTING AND RECORDING YOUR
OBSERVATIONS. Notice the concentrations of reagents you are to use. Write down the actual
concentrations as listed on the stock bottles in your notebook!
The TA will make up solution 0.
Use a set of labeled 100.00 mL volumetric flasks for this series.
Be sure you pipet and dilute correctly!
 Keep pipets vertical at all times;
 Never pipet by mouth;
 Bring the liquid level up in the pipet past the scratch on the neck. Drain it until the bottom of the
meniscus sits on the scratch;
 Drain the pipet into the labeled volumetric flask. Just touch the tip of the pipet to the side of the flask.
 Keep that pipet vertical!
 Carefully add 0.25 M HNO3 to the 100.00 mL volumetric flasks until the bottom of the meniscus sits on
the scratch;
 Cap or parafilm the volumetrics and mix the solutions thoroughly! 30 inversions is the canonical
number.
NOTE AND RECORD YOUR OBSERVATIONS!
6. Find a UV-visible spectrophotometer and measure absorbance of your standard solutions..
Remember when using the cuvettes to touch only the frosted sides, not the clear sides. Find the
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absorbance of light by all 5 of your standard solutions and the 0 solution at 448 nm. FeSCN2+ are redorange, and will thus absorb blue-green light (the 448 nm light). Beer’s Law describes the relationship
between light absorption and concentration: Beer’s Law is given by the equation A = εbc, where A =
absorbance of light, b = the path length of light through the sample, c = concentration of lightabsorbing compound in the sample and ε is a constant of proportionality called the extinction
coefficient. As you can see, absorption of light is directly proportional to concentration.
Record your absorbance data in your notebook!. Once you are finished with the spectrophotometer, you
can discard the standard solutions in the waste jug in the hood, rinse your glassware and
7. Graph your standard curve. Graph your data, with absorbance at 448 nm on the Y-axis and
[FeSCN2+] on the x-axis. (See below for sample calculations)
MOVE ON TO MAKE UP THE “EXPERIMENTAL” SOLUTIONS. The next set of solutions you
make up, the “experimental” solutions, will NOT have one reactant added in vast excess to push the
equilibrium far to the right. You will make up solutions in which approximately equal amounts of each
reactant are mixed. You will then use your standard curve data to determine the [FeSCN2+] made, and how
much Fe3+ and SCN- are left over at equilibrium. Then you can calculate Kc.
EXPERIMENTAL
8.
Make up the 5 “Experimental” solutions, following the chart, and NOTING AND RECORDING
YOUR OBSERVATIONS. Notice the concentrations of reagents you are to use. They are NOT
the same as last time! Write down the actual concentrations as listed on the stock bottles in your
notebook!
Use a labeled set of 10.00 mL volumetric flasks for this series.
Be sure you pipet and dilute correctly!
 Keep pipets vertical at all times;
 Never pipet by mouth;
 Bring the liquid level up in the pipet past the scratch on the neck. Drain it until the bottom of the
meniscus sits on the scratch;
 Drain the pipet into the labeled volumetric flask. Just touch the tip of the pipet to the side of the flask.
 Keep that pipet vertical!
 Carefully add 0.25 M HNO3 to the 10.00 mL volumetric flasks until the bottom of the meniscus sits on
the scratch;
 Cap or parafilm the volumetrics and mix the solutions thoroughly! 30 inversions is the canonical
number.
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NOTE AND RECORD YOUR OBSERVATIONS!
9. Find a UV-visible spectrophotometer and measure absorbance of your “experimental” solutions..
Remember when using the cuvettes to touch only the frosted sides, not the clear sides. Find the
absorbance of light by all 5 of your solutions at 448 nm. Record your absorbance data in your notebook!.
Keep in mind that absorbance is proportional to the amount of light absorbed, A = εbc, and εb is the slope
of your standard curve. Once you are finished with the spectrophotometer, you can discard the standard
solutions in the waste jug in the hood, rinse your glassware and calculate the Kc for the reaction: Fe3+ +
SCN- ⇄ FeSCN2+ Kc = [FeSCN2+ ]/ [Fe3+][SCN-].
CLEAN UP! Wash all glassware with DI water and return them. Wash all pipets
with DI water and leave them to drain. Wash cuvet with DI water and leave it at the
instrument.
ALL WASTE FROM THIS EXPERIMENT GOES IN THE JUG LABELLED “Ferric
Thiocyanate Waste” IN THE HOOD.
CALCULATIONS:
1. SAMPLE CALCULATIONS FOR STANDARD SOLUTIONS:
Data: Note that the concentrations of stock reagents in this example are NOT what you used in your
experiment. Use your own!.
2. mmoles NaSCN: moles/L is the same as mmoles/mL. Convert moles/L to mmoles/mL and convince
yourself!. 1.00 mL NaSCN(1.000 x 10-3 mmoles NaSCN/mL) =
1.000 x 10-3 mmoles NaSCN = 1.000 x 10-3 mmoles SCN- ;
4. [SCN-]: 1.000 x 10-3 mmoles SCN- lives in a total volume of 100.00 mL of solution:
1.000 x 10-3 mmoles SCN-/ 100.00 mL = 1.000 x 10-5 mmole SCN-/mL = 1.000 x 10-5 mole SCN-/L =
1.000 x 10-5 M SCN5. [FeSCN2+]: Since we have swamped the system with Fe3+ and driven the equilibrium far to the right,
the [FeSCN2+] ≈ [SCN-]
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USE THIS DATA FOR YOUR STANDARD CURVE:
SAMPLE CALCULATIONS FOR “EXPERIMENTAL” DATE TO DETERMINE Kc
Data: Note that the concentrations of stock reagents in this example are NOT what you used in your
experiment. Use your own!.
2. mmoles Fe3+initial: moles/L is the same as mmoles/mL. Convert moles/L to mmoles/mL and convince
yourself!. 5.00 mL Fe(NO3)3(5.000 x 10-3 mmoles Fe(NO3)3/mL = 2.500 x 10-2 mmoles Fe(NO3)3 =
2.500 x 10-2 mmoles Fe3+ ;
4. mmoles SCN-initial: moles/L is the same as mmoles/mL. 1.00 mL NaSCN)(1.000 x 10-3 mmoles
NaSCN/mL) = 1.000 x 10-3 mmoles NaSCN = 1.000 x 10-3 mmoles SCN-
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2+
7. [FeSCN ]: In this set of experiments, we have NOT swamped the system with Fe3+. We must
calculate the [FeSCN2+] from the standard curve:
The equation of the best fit straight line is in the form y = mx + b, where y is the absorbance of any sample
of FeSCN2+, x is the [FeSCN2+], m is the slope and b is the y-intercept.
0.4701 = 4867.4[FeSCN2+] + 0.008958
[FeSCN2+] = (0.4701 - 0.008958)/ 4867.4 = 9.47 x 10-5 M
8. mmoles FeSCN2+: 10.00 mL solution (9.47 x 10-5 mmoles FeSCN2+/mL)
= 9.47 x 10-4 mmoles FeSCN2+ = mmoles Fe3+ used = mmoles SCN- used
9. mmoles Fe3+ unreacted: mmoles start –mmoles eq = 2.500 x 10-2 mmole - 9.47 x 10-4 mmole =
2.41 x 10-2 mmoles Fe3+ unreacted & present at equilibrium.
10. [Fe3+] at equilibrium: 2.41 x 10-2 mmoles Fe3+/10.00 mL = 2.41 x 10-3 M Fe3+
11. mmoles SCN- unreacted: mmoles start –mmoles eq = 1.000 x 10-3 mmole - 9.47 x 10-4 mmole =
5.26 x 10-5 mmoles SCN- unreacted & present at equilibrium.
12. [SCN3+] at equilibrium: 5.26 x 10-5 mmoles SCN- /10.00 mL = 5.26 x 10-6 M SCN13. Kc : [FESCN2+]eq/[Fe3+]eq[SCN-]eq = (9.47 x 10-5 M FeSCN2+)/(2.41 x 10-3 M Fe3+)(5.26 x 10-6 M SCN-) =
7490
LAB WRITE UP:
You should have all your data and all your results for your standard curve in Excel.
You should have a graph of your standard curve, with best fit straight line, equation for best fit straight line and
correlation coefficient.
You should have all your data and all your results for your “experimental” section in Excel.
You should show sample calculations.
You should have detailed observations.
Include a summary of your data and an analysis of your results.
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APPENDIX 1: EQUILIBRIUM CALCULATIONS:
To determine the concentration of reactants and products at equilibrium, given starting concentrations:
Using our example:
Start:
Equilibrium:
Fe3+
0.01 M
0.01 M – x
+ SCN⇄ FeSCN2+
0.01 M
0
0.01 M – x
x
Kc = [FeSCN2+ ]/ [Fe3+][SCN-] = 10
I don’t know how much will react,
but I know some will! Call that “x”.
Plugging equilibrium values into the equilibrium expression, I get:
x/(0.01-x)(0.01-x) = 10  0.001 –1.2x + 10x2 = 0 .
Now I use a quadratic equation solver (I’m lazy) such as the one at
http://www.edteach.com/algebra/quad_explorer/quadratic.html
x1 =
x2 =
0.11916079783099616
0.0008392021690038387
Notice that they ignore significant figures. You’ll have to fix that yourself.
Root 1 makes no sense; I’d be making more stuff than I started with.
Root 2 works. x is the [FeSCN2+] at equilibrium.