Colorimetric Analysis &
Determination of the Equilibrium for
a Chemical reaction
Help Notes
AP Chemistry
PART II
How to determine Concentration of FeSCN2+ at equilibrium.
•In this step you are determining the
concentration of the product at equilibrium
using the absorbance reading you measured in
the data table.
•Use the calibration curve you generated in
Part I to determine the concentration of your
samples. Plug in the y value for absorbance to
find the x value for concentration.
PART II
How to determine Concentration of FeSCN2+ at equilibrium.
•Here is one set of sample data generated in
class. Your absorbance values should be
DETERMINING y WHEN GIVEN x
Steps for TI Graphing Calculators
The directions below are form your Arrhenius made easy notes. You will use
the same steps to find x except instead of looking for Kelvin temperature your
x will represent the concentration of your sample at equilibrium.
How to determine the initial concentrations
of diluted stock standard.
In this step you are determining the INITIAL concentrations of
reagents added to the reaction. You will use the calibration curve
you generated in Part I to determine the equilibrium concentrations
AFTER the reaction has completed.
Tube
#
0.002 M
Fe(NO3)3
(mL)
0.002 M
SCN(mL)
1
3.00
0.00
7.00
2
3.00
2.00
5.00
3
3.00
3.00
4.00
4
3.00
4.00
3.00
5
3.00
5.00
2.00
H2 O
(mL)
Initial
[Fe3+]
Initial
[SCN-]
?
?
?
?
?
?
?
?
?
?
How to determine the initial concentrations concentration
of diluted stock standard.
FIRST: Determine the total volume of the new, diluted concentration. To do
this, add up all the amounts of standard added.
Tube
#
0.002 M
Fe(NO3)3
(mL)
0.002 M
SCN(mL)
1
3.00
0.00
7.00
2
3.00
2.00
5.00
3
3.00
3.00
4.00
4
3.00
4.00
3.00
5
3.00
5.00
2.00
H2 O
(mL)
Initial
[Fe3+]
Initial
[SCN-]
?
?
?
?
?
?
?
?
?
?
Note that for
every sample
the total
volume is 10
ml = 0.01 L.
How to determine the initial concentrations concentration
of diluted stock standard.
Second: Calculate the molar concentration of
Fe(NO3)3 in each tube.
Tube
#
0.002 M
Fe(NO3)3
(mL)
0.002 M
SCN(mL)
1
3.00
0.00
2
3.00
3
3.00
4
3.00
5
3.00
H2 O
(mL)
7.00
Initial
[Fe3+]
Initial
[SCN-]
?
?
2.00
5.00
?
?
3.00
4.00
? concentration
?
Note that this
will be the
4.00
? samples
?
same3.00for ALL
because you add
5.00
2.00
the same
amount
?
? with the same final
volume each time.
How to determine the initial concentrations concentration
of diluted stock standard.
Finally: Calculate the molar concentration of
SCN- in each tube at each concentration.
Tube
#
0.002 M
Fe(NO3)3
(mL)
0.002 M
SCN(mL)
1
3.00
0.00
2
3
4
5
H2 O
(mL)
7.00
Initial
[Fe3+]
6 x 10-4
Initial
[SCN-]
?
6 x 10
3.00
2.00
5.00
For sample #1 it is 0.00 ?
6add
x 10 any
3.00
3.00
4.00
because
you didn’t
?
6 x 10
SCN-.
3.00
4.00
3.00
?
6 x 10
3.00
5.00
2.00
?
-4
-4
-4
-4
How to determine the initial concentrations concentration
of diluted stock standard.
Finally: Calculate the molar concentration of
SCN- in each tube at each concentration.
Tube
#
0.002 M
Fe(NO3)3
(mL)
0.002 M
SCN(mL)
1
3.00
0.00
2
3.00
2.00
3
3.00
3.00
4
3.00
4.00
5
3.00
5.00
Initial
[Fe3+]
Initial
[SCN-]
7.00
6 x 10-4
0
5.00
6 x 10-4
H2 O
(mL)
For sample ?#2-5 find the
6 x 10 of stock solution and
4.00
amount
?
calculate
for? a new final
6 x 10
3.00
volume
6 x 10 of 10 mL (0.01 L)
2.00
?
-4
-4
-4
How to determine the initial concentrations concentration
of diluted stock standard.
Finally: Calculate the molar concentration of
SCN- in each tube at each concentration.
Tube
#
0.002 M
Fe(NO3)3
(mL)
0.002 M
SCN(mL)
1
3.00
0.00
2
3.00
3
Initial
[Fe3+]
Initial
[SCN-]
7.00
6 x 10-4
0
2.00
5.00
6 x 10-4
4 x 10-4
3.00
3.00
4.00
6 x 10-4
4
3.00
4.00
3.00
6 x 10-4
5
3.00
5.00
2.00
6 x 10-4
?
?
?
H2 O
(mL)
You will have to
calculate the
concentration for each
sample since the
amounts vary. However,
final volume will remain
10 mL for each sample.
PART II: Pre-Lab Questions
(a) Write the balanced net ionic equation for
the reaction taking place: (see pg 6)
Fe3+ (aq) + SCN- (aq)
FeSCN2+ (aq)
PART II: Pre-Lab Questions
(b) Write the equilibrium expression for the
reaction taking place.
Fe3+ (aq) + SCN- (aq)
FeSCN2+ (aq)
Remember, [products]coeff./ [reactants]coeff.
PART II: Post Lab Questions
3. Use a RICE table along with the initial
concentrations of the reactants you
calculated in Part II and equilibrium
concentrations to determine the value of
Keq for each experimental trial of this
experiment.
PART II: Post Lab Questions
RICE tables are a way of organizing your data to
quickly solve equilibrium problems.
R: Write the balanced net ionic equation.
I: List the initial concentrations for each species
(product and reactants)
C: On this row you will figure out what the
change in concentration for the reaction is.
E: Here you will list the equilibrium
concentrations for each species.
PART II: Post Lab Questions
For our Lab:.
R: Write the balanced net ionic equation.
I:
C:
E:
PART II: Post Lab Questions
For our Lab:.
R:
Fe3+ (aq) +
I:
C:
E:
SCN-
(aq)
FeSCN2+ (aq)
PART II: Post Lab Questions
For our Lab:.
R:
Fe3+ (aq) +
SCN-
(aq)
FeSCN2+ (aq)
I: List the initial concentrations. These are the
values you calculated in table 2 on page 7.
C:
E:
PART II: Post Lab Questions
For our Lab Test tube 2:
R:
Fe3+ (aq) + SCNI:
C:
E:
6x10-4M
(aq)
4x10-4 M
FeSCN2+ (aq)
0 (initially there is no product!)
PART II: Post Lab Questions
For our Lab Test tube 2:
R:
Fe3+ (aq) + SCNI:
6x10-4M
(aq)
4x10-4 M
FeSCN2+ (aq)
0 (initially there is no product!)
C:
E: the equilibrium concentrations are the
concentrations you measured in table 1, page 7. You
calculated these concentrations from absorbance
and your calibration curve.
PART II: Post Lab Questions
For our Lab Test tube 2:
R:
Fe3+ (aq) + SCNI:
6x10-4M
(aq)
4x10-4 M
FeSCN2+ (aq)
0 (initially there is no product!)
C:
E:
4.8x10-3-4 M
PART II: Post Lab Questions
For our Lab Test tube 2:
R:
Fe3+ (aq) + SCNI:
6x10-4M
(aq)
4x10-4 M
FeSCN2+ (aq)
0 (initially there is no product!)
C:
E:
4.8x10-3-4 M
Do you remember why we are doing the RICE table?
Our goal is to find Keq, the equilibrium constant. To do this
we need to know the concentrations of ALL species at
equilibrium.
PART II: Post Lab Questions
For our Lab Test tube 2:
R:
Fe3+ (aq) + SCNI:
6x10-4M
(aq)
4x10-4 M
FeSCN2+ (aq)
0 (initially there is no product!)
C:
E:
4.8x10-3-4 M
We are missing the equilibrium concentrations of the
reactants. BUT, we know the change in concentration and
the ratios of those changes so we can calculate the final
concentrations of our reactants.
PART II: Post Lab Questions
For our Lab Test tube 2:
R:
Fe3+ (aq) + SCNI:
6x10-4M
(aq)
4x10-4 M
FeSCN2+ (aq)
0 (initially there is no product!)
C: That’s where “C” comes in. This represents the
change in concentration as equilibrium is reached.
E:
4.8x10-3-4 M
PART II: Post Lab Questions
For our Lab Test tube 2:
R:
Fe3+ (aq) + SCNI:
6x10-4M
(aq)
4x10-4 M
FeSCN2+ (aq)
0 (initially there is no product!)
C: Use the coefficients to determine the mole ratio
changes during the reactions. Our reaction is 1:1:1
so this problem is relatively simple.
E:
4.8x10-3-4 M
PART II: Post Lab Questions
For our Lab Test tube 2:
R:
Fe3+ (aq) + SCNI:
6x10-4M
(aq)
4x10-4 M
FeSCN2+ (aq)
0 (initially there is no product!)
C: Notice that if we produce one mole of Product
then one mole of each reactant is consumed.
E:
4.8x10-3-4 M
PART II: Post Lab Questions
For our Lab Test tube 2:
R:
Fe3+ (aq) + SCNI:
6x10-4M
(aq)
4x10-4 M
FeSCN2+ (aq)
0 (initially there is no product!)
C: Let “x” represent this amount.
E:
4.8x10-3-4 M
PART II: Post Lab Questions
For our Lab Test tube 2:
R:
Fe3+ (aq) + SCNI:
6x10-4M
(aq)
4x10-4 M
FeSCN2+ (aq)
0 (initially there is no product!)
C:
-x
-x
+x
One mole of each reactant was consumed (-x) to
produce one mole of product (+x)
E:
4.8x10-3-4 M
PART II: Post Lab Questions
For our Lab Test tube 2:
R:
Fe3+ (aq) + SCNI:
C:
E:
6x10-4M
-x
(aq)
4x10-4 M
-x
FeSCN2+ (aq)
0 (initially there is no product!)
+x
4.8x10-3-4 M
Now determine the equilibrium concentrations after the “C” values are
subtracted or added. Evaluating the product first we see the x = 4.8x10-3
PART II: Post Lab Questions
For our Lab Test tube 2:
R:
Fe3+ (aq) + SCNI:
C:
E:
6x10-4M
-x
(aq)
4x10-4 M
-x
6x10-4M
- 3.8x10-4
4x10-4
- 3.8x10-4
2.2x10-4
2.5x10-5
FeSCN2+ (aq)
0 (initially there is no product!)
+x
3.8x10-4 M
Now we can subtract “X” from our starting concentrations of
reactants
PART II: Post Lab Questions
For our Lab Test tube 2:
R:
Fe3+ (aq) + SCNI:
C:
E:
6x10-4M
-x
(aq)
4x10-4 M
-x
6x10-4M
- 3.8x10-4
4x10-4
- 3.8x10-4
2.2x10-4
2.5x10-5
FeSCN2+ (aq)
0 (initially there is no product!)
+x
3.8x10-4 M
Do you remember our goal? Find Keq .
Now all we need to do is plug our equilb. Concs. Into the
equlibrium expression!
PART II: Post Lab Questions
For our Lab Test tube 2:
R:
Fe3+ (aq) + SCNI:
C:
E:
6x10-4M
-x
(aq)
4x10-4 M
-x
6x10-4M
- 3.8x10-4
4x10-4
- 3.8x10-4
2.2x10-4
2.5x10-5
[FeSCN2+]
Keq = [Fe3+][SCN-] = plug and chug!
FeSCN2+ (aq)
0 (initially there is no product!)
+x
3.8x10-4 M