look at the files first(11 Rate Law….. and use_det_eq…..) Two lab

AP Inquiry Lab 11

What is the Rate Law of the Fading Crystal Violet Reaction Using Beer’s Law?

In this experiment, you will observe the reaction between crystal violet and sodium hydroxide. One objective is to study the relationship between concentration of crystal violet and the time elapsed during the reaction. The equation for the reaction is shown here:

A simplified (and less intimidating!) version of the equation is:

CV+ + OH– CVOH
(crystal violet) (hydroxide)
The rate law for this reaction is in the form: rate = k[CV+]m[OH–]n, where k is the rate constant for the reaction, m is the order with respect to crystal violet (CV+), and n is the order with respect to the hydroxide ion. Since the hydroxide ion concentration is more than 1000 times as large as the concentration of crystal violet, [OH-] will not change appreciably during this experiment. Thus, you will find the order with respect to crystal violet (m), but not the order with respect to hydroxide (n).
You will be using a colorimeter for this lab. A colorimeter shines a light through the solution and checks how much light is absorbed by the solution. As the reaction proceeds, a violet-colored reactant will be slowly changing to a colorless product. Using the green (565 nm) light source of a computer-interfaced Colorimeter, you will monitor the absorbance of the crystal violet solution with time. Absorbance is proportional to the concentration of crystal violet (Beer’s law). Absorbance will be used in place of concentration in plotting the graphs.
Beer’s law is A = abc
where A = absorbance, a = molar absorptivity constant, b = path length, and c = concetration
Once the order with respect to crystal violet has been determined, you will also be finding the rate constant, k, and the half-life for this reaction.
PreLab
Add this lab to your table of contents
Write a purpose for this lab
Create a table of reagents
Sketch a graph of concentration vs time, ln concentration vs. time, and 1/concentration vs. time for a zero, first and second order reaction.
Sketch a graph of the [CV+] and the [CVOH] over time during this reaction, reaction and write what you should visually see due this change in concentrations.
MATERIALS

Power Macintosh or Windows PC

0.020 M NaOH

Vernier computer interface

2.0 X 10–5 M crystal violet

Logger Pro

distilled water

Vernier Colorimeter

stirring rod

one plastic cuvette

two 10-mL graduated cylinders

250-mL beaker

Write a PROCEDURE For Graphically determining the order of the reaction
The following bits of information will avoid excess waste, and help you with your procedure.
1. To set up the program, Click or tap Mode to open Data Collection Settings. Change Rate to 1 samples/s and End Collection to 200 s. Click or tap Done.
2. Use 10.0 mL of 0.020 M NaOH solution. Use 10.0 mL of 2.0 X 10–5 M crystal violet solution. CAUTION: Sodium hydroxide solution is caustic. Crystal violet is a biological stain. Avoid spilling either on your skin or clothing.
3. Remember to Calibrate the Colorimeter, set the wavelength on the Colorimeter to 565 nm (Green).
4. Do this quickly! To initiate the reaction, simultaneously pour the 10 mL portions of crystal violet and sodium hydroxide into a 250 mL beaker and stir the reaction mixture with a stirring rod. Empty the water from the cuvette. Rinse the cuvette twice with ~1 mL amounts of the reaction mixture, fill it 3/4 full, and place it in the device. Close the Colorimeter lid. Click or tap Collect to start data collection.
5. To keep the solution from warming inside the Colorimeter, the cuvette should be removed from the Colorimeter between readings. However, make sure your absorbance is stable before clicking keep. That is too say don’t rush too much putting it and clicking keep or you will have error.
6. Data collection will end after 200 s.
7. Create a calculated column, ln Absorbance, and add a linear curve fit to the graph ln Absorbance vs. time:
a. Click or tap View, , and select Graph and Table.
b. In the Absorbance column header in the table, click or tap More Options, , and choose Add Calculated Column.
c. Enter ln Absorbance as the Name and leave the Units field blank.
d. Click or tap Insert Expression and choose Aln(X) as the expression.
e. Enter 1 as Parameter A and select Absorbance as Column X.
f. Click or tap Apply. A graph of ln absorbance vs. time is displayed. Double-click the graph to autoscale the graph.
g. To see if the relationship is linear, click or tap Graph Tools, , and choose Apply Curve Fit.
h. Select Linear as the curve fit and Dismiss the Curve Fit box.
i. Record the slope as the rate constant, k, and dismiss the Linear curve fit box.
8. Create a calculated column, 1/Absorbance, and then plot a graph of 1/Absorbance vs. time:
a. In the data table, click or tap More Options, , in the Absorbance column header, and then choose Add Calculated Column.
b. Enter 1/Absorbance the Name and leave the Units field blank.
c. Click or tap Insert Expression and choose A/X as the expression.
d. Enter 1 as Parameter A and select Absorbance as Column X.
e. Click or tap Apply.
f. Click or tap the y-axis label and select only 1/Absorbance to display a graph of 1/Absorbance vs. time.
g. To see if the relationship is linear, click or tap Graph Tools, , and choose Apply Curve Fit.
h. Select Linear as the curve fit and Dismiss the Curve Fit box.
i. Record the slope as the rate constant, k, and dismiss the Linear curve fit box.
9. To see any of the three graphs again, click or tap the y-axis label and choose the column you want to display.
10. Write up and explain your procedure to your instructor before beginning. Be prepared to answer questions.

PostLAb

1. Was the reaction zero, first, or second order, with respect to the concentration of crystal violet? Explain.
2. Calculate the rate constant, k, using the slope of the linear regression line for your linear curve (k = –slope for zero and first order and k = slope for second order). Be sure to include correct units for the rate constant. Note: This constant is sometimes referred to as the pseudo rate constant, because it does not take into account the effect of the other reactant, OH-.
3. Write the correct rate law expression for the reaction, in terms of crystal violet (omit OH-).
4. Using the printed data table, estimate the half-life of the reaction; select two points, one with an absorbance value that is about half of the other absorbance value. The time it takes the absorbance (or concentration) to be halved is known the half-life for the reaction. (As an alternative, you may choose to calculate the half-life from the rate constant, k, using the appropriate concentration-time formula.)
5. Print a copy of the graphs “Absorbance”, “ln Absorbance”, or “1/Absorbance”, and attach them to the lab
2
C
C
OH
OH
–
+
N(CH )
N(CH )
N(CH )
+
N(CH )
N(CH )
N(CH )
3
3
3
3
3
3
2
2
2
2
2
2
¾
®
¾

2

>Sheet

1

2

1

2

4

4

5

6

6

7

8

9

13

14

8

0.28

0	0.	6
0.	5	0.60	9
0.59	7
1.5	0.576
0.552
2.5	0.5	4
3	0.539
3.5	0.541
0.526
4.5	0.5	14
0.506
5.5	0.495
0.4	8
6.5	0.478
0.465
7.5	0.457
0.446
8.5	0.435
0.425
9.5	0.4	13
10	0.405
10.5	0.393
11	0.382
11.5	0.372
12	0.361
12.5	0.353
0.34
13.5	0.334
0.326
14.5	0.313
15	0.307
15.5	0.297
16	0.291
16.5		0.28
17
17.5	0.271

T

vs

A

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5 17 17.5 0.62 0.60899999999999999 0.59699999999999998 0.57599999999999996 0.55200000000000005 0.54400000000000004 0.53900000000000003 0.54100000000000004 0.52600000000000002 0.51400000000000001 0.50600000000000001 0.495 0.48599999999999999 0.47799999999999998 0.46500000000000002 0.45700000000000002 0.44600000000000001 0.435 0.42499999999999999 0.41299999999999998 0.40500000000000003 0.39300000000000002 0.38200000000000001 0.372 0.36099999999999999 0.35299999999999998 0.34 0.33400000000000002 0.32600000000000001 0.313 0.307 0.29699999999999999 0.29099999999999998 0.28799999999999998 0.28000000000000003 0.27100000000000002

T

A

2

>Sheet

1

0

2

0.5

1

0.549

2

3

4

0.477

5

7

8

4

0.37

9

5

0.34

6

0.32

14

0.222

5

0.205

20

	0
0.2		0.	5	8
0.55	4
0.8		0.54	9
0.54	3
1.2
1.5	0.551
1.8	0.559
0.552
2.2	0.535
2.5	0.52	7
2.8	0.508
0.503
3.2	0.504
3.5	0.501
3.8	0.497
0.489
4.2		0.477
4.5
4.8	0.471
0.459
5.2	0.458
5.5	0.443
5.8	0.441
6	0.429
6.2	0.426
6.5	0.421
6.8	0.41
0.404
7.2	0.394
7.5	0.391
7.8	0.392
0.381
8.2		0.37
8.5	0.366
8.8
0.362
9.2	0.357
9.5	0.351
9.8		0.34
10
10.2	0.335
10.5		0.32
10.8	0.324
11
11.2	0.307
11.5	0.3	14
11.8	0.301
12	0.295
12.2	0.292
12.5	0.293
12.8	0.28
13	0.282
13.2	0.283
13.5	0.276
13.8	0.264
0.259
14.2	0.258
14.5	0.257
14.8	0.248
15	0.244
15.2	0.241
15.5	0.243
15.8	0.237
16	0.233
16.2	0.231
16.5	0.225
16.8	0.234
17	0.221
17.2		0.222
17.5
17.8	0.215
18	0.213
18.2	0.211
18.5	0.223
18.8		0.	20
19
19.2	0.203
19.5	0.196
19.8	0.198
0.188

T

vs

A

0 0.2 0.5 0.8 1 1.2 1.5 1.8 2 2.2000000000000002 2.5 2.8 3 3.2 3.5 3.8 4 4.2 4.5 4.8 5 5.2 5.5 5.8 6 6.2 6.5 6.8 7 7.2 7.5 7.8 8 8.1999999999999993 8.5 8.8000000000000007 9 9.1999999999999993 9.5 9.8000000000000007 10 10.199999999999999 10.5 10.8 11 11.2 11.5 11.8 12 12.2 12.5 12.8 13 13.2 13.5 13.8 14 14.2 14.5 14.8 15 15.2 15.5 15.8 16 16.2 16.5 16.8 17 17.2 17.5 17.8 18 18.2 18.5 18.8 19 19.2 19.5 19.8 20 0 0.58199999999999996 0.55400000000000005 0.54900000000000004 0.54300000000000004 0.54900000000000004 0.55100000000000005 0.55900000000000005 0.55200000000000005 0.53500000000000003 0.52700000000000002 0.50800000000000001 0.503 0.504 0.501 0.497 0.48899999999999999 0.47699999999999998 0.47699999999999998 0.47099999999999997 0.45900000000000002 0.45800000000000002 0.443 0.441 0.42899999999999999 0.42599999999999999 0.42099999999999999 0.41 0.40400000000000003 0.39400000000000002 0.39100000000000001 0.39200000000000002 0.38100000000000001 0.374 0.36599999999999999 0.37 0.36199999999999999 0.35699999999999998 0.35099999999999998 0.34499999999999997 0.34 0.33500000000000002 0.32600000000000001 0.32400000000000001 0.32 0.307 0.314 0.30099999999999999 0.29499999999999998 0.29199999999999998 0.29299999999999998 0.28000000000000003 0.28199999999999997 0.28299999999999997 0.27600000000000002 0.26400000000000001 0.25900000000000001 0.25800000000000001 0.25700000000000001 0.248 0.24399999999999999 0.24099999999999999 0.24299999999999999 0.23699999999999999 0.23300000000000001 0.23100000000000001 0.22500000000000001 0.23400000000000001 0.221 0.222 0.222 0.215 0.21299999999999999 0.21099999999999999 0.223 0.20499999999999999 0.20499999999999999 0.20300000000000001 0.19600000000000001 0.19800000000000001 0.188

T

A

Experiment8: DETERMINATION OF AN

EQUILIBRIUM CONSTANT

77

Purpose: The equilibrium constant for the formation of iron(III) thiocyanate complex
ion is to be determined.

Introduction: In the previous week, we qualitatively investigated how an equilibrium
shifts in response to a stress to re-establish equilibrium. This week we will quantitatively
assess the equilibrium constant for the same reaction: the reaction of iron(III) cation
complexing with a thiocyanate anion (SCN–) to form the iron(III) thiocyanate complex,
Fe(SCN)2+ (Equation 1). Its equilibrium expression is as shown in Equation 2.

Fe3+ (aq) + SCN (aq) Fe(SCN)2+ (aq) Equation 1

2+

eq 3+
[Fe(SCN) ]

K = -[Fe ][SCN ]
Equation 2

If Keq is a large number (>1), then the chemical equilibrium favors the formation of product
(large numerator). If Keq is a small number (<1) then the chemical equilibrium favors the formation of reactants (large denominator). In this experiment, several solutions of varying initial concentrations of the reactants are to be prepared. Despite the different concentrations, the equilibrium constants calculated from their equilibrium concentrations should be the same, as long as the temperature is kept constant. Before we begin the study of the equilibrium concentrations, we must first prepare a standard curve to help us determine the concentration of Fe(SCN)2+ at equilibrium. Le Châtelier’s Principle states that if at equilibrium a change is applied to a system, the species will react to offset the change so as to maintain the equilibrium. We will use this principle to aid in the preparation of the standard curve. It will be made by plotting the absorbance versus concentration of the red iron(III) thiocyanate complex, (Fe(SCN)2+). If the concentration of the reactant, iron(III) nitrate, is increased (0.200 M), so as to become much larger than the thiocyanate anion concentration (0.00200M), then the reaction (Equation 1) will be forced almost completely to products. In this situation, the iron(III) concentration is 100 times that of the thiocyanate, therefore essentially all the SCN– anions will react to produce the red colored product, Fe(SCN)2+. Thus, within the limits of our detection apparatus, the final concentration of Fe(SCN)2+ is equal to the initial concentration of SCN–. The intensity of the red color will be measured spectrophotometrically and will be directly proportional to the equilibrium concentration of the Fe(SCN)2+ species. (Review Beer’s Law from Experiment 3.) After a standard curve is produced, the conditions will be altered so that the concentrations of each of the two reacting species (Fe3+ and SCN–) will be the same order of magnitude (~0.00200 M each). Because the concentrations will be so similar, the system will no longer be forced all the way to the right (towards the products) and you will be able to determine an equilibrium constant from the data. The concentration of Fe(SCN)2+ at equilibrium will be determined spectrophotometrically according to its absorbance in the standard curve. Since for every mole of the red complex, Fe(SCN)2+ produced, one mole of Fe3+ and one mole of

78 EXPERIMENT 8: DETERMINATION OF EQUILIBRIUM CONSTANT

SCN – will have reacted, the equilibrium concentrations (unreacted species) of Fe3+ and
SCN- can be determined by subtracting the concentration of Fe(SCN)2+ formed from the
initial concentrations before the reaction took place. We can set up an “ICE” table, find the
equilibrium concentrations for each of the three species, and solve for Keq.

Each of the initial solutions will be made up so as to contain 0.500 M H+. Therefore when
mixing the solution of 0.00200 M Fe3+ made up in 0.500 M H+ and the solution of
0.00200 M SCN– made up in 0.500 M H+, no matter what the proportions, the 0.500 M H+
concentration will be constant. The reason for this is that the iron(III) thiocyanate formation
reaction must be run around 0.5 M acid to prevent significant iron hydrolysis (Equation 3)
that affects the concentration of iron(III) ions.

Fe3+(aq) + 3H2O (l) Fe(OH)3 (s) + 3H+ (aq) Equation 3

Also, the reaction must be run at acid concentration below 0.7 M because otherwise the acid
reacts with the thiocyanate reducing the available SCN− as well (Equation 4).

H+(aq) + SCN-(aq) HSCN (aq) Equation 4

Each reagent is labeled with its concentration. However, once you mix reagents together,
you will have diluted the concentration. The calculations that you use will need to account
for these dilutions. An example is below:

Example 1: If 5.08 mL of 0.00200 M Fe(NO3)3 is mixed with 3.10 mL of 0.00200M KSCN

and 2.00 mL of 0.500 M HNO3 , what is the final concentration of the Fe3+ ion?

1 1 2 2

2

3 31 1
2 3 3

2

M V = M V where V is the TOTAL volume in the final solution
(0.00200 M Fe(NO ) )(5.08 mL)M V

M = = 0.000998 M Fe(NO )
V 10.18 mL

= 9.98×10–4 M Fe3+
Check: Is the answer reasonable? M2 should be more dilute than M1.

ICE Table Construction:
ICE tables are useful tables that summarize what is occurring in an equilibrium reaction.
The use of ICE tables should have been covered in your lecture class. You need to know that
“I” stands for initial concentration of each species in the solution, before they are allowed to
react. “C” stands for the change in concentration of each species from the initial
concentrations to the equilibrium concentrations. And the “E” stands for equilibrium
concentration of each species (i.e. concentration after the reaction has reached equilibrium).
Below is an example of how the ICE table will be used.

Example 2: Assume an initial concentration of [Fe3+] = 0.00100 M and an initial
concentration of [SCN–] = 0.000600 M in a sample solution for which you are to determine
the concentration of Fe(SCN)2+ from the standard curve. You can set up an ICE table as
shown below:

EXPERIMENT 8: DETERMINATION OF EQUILIBRIUM CONSTANT 79

Species Fe3+ SCN – Fe(SCN)2+
I. (Initial) 0.00100 M 0.000600 M 0.00 M
C.(Change) – X – X + X
E.(Equilibrium) 0.00100 M – X 0.000600 M – X X

X = [Fe(SCN)2+] and is to be determined from the standard curve. You can then calculate
the equilibrium constant, Keq, using the equilibrium concentrations.

In your ICE tables on the Calculations & Results Page, do not write “X” but use the actual
concentration obtained from the standard curve. For example, if X = 0.000211 M, [Fe3+] at
equilibrium would be (0.00100 − 0.000211) M = 0.00079 M.
Species Fe3+ SCN – Fe(SCN)2+
I. (Initial) 0.00100 M 0.000600 M 0.00 M
C.(Change) – 0.000211 – 0.000211 + 0.000211
E.(Equilibrium) 0.00079 0.00039 0.000211

Use of the Standard Curve

The standard curve is a plot of Absorbance versus [Fe(SCN)2+] (Figure 8.1). It can be used
to give us the concentration of a solution when given the absorbance. We can either read it
off the graph visually or calculate the concentration from the trendline equation. Remember
that Beer’s Law indicates the relationship between the concentration and the absorbance is
linear. Thus A = mC +b where A is the absorbance, m is the slope, C is the concentration
and b is the y-intercept. Considering that A has not units and C has units of M, what is the
unit of the slope? What is the unit of the y-intercept? You should know the answers.

For example, the trendline equation from the curve in

Figure 8.1

is y = 4312x+0.0075. If an
unknown concentration of Fe(SCN)2+ has an absorbance reading of 0.250 then you can solve
for the concentration of the Fe(SCN)2+: y = 4312x + 0.0075 translates into
A = 4312M–1 [FeSCN2+] + 0.0075

2+
1

2+

5

1 1

5

A 0.0075
[FeSCN ] = and substituting A = 0.250

4312M
0.250 0.0075 0.242

[FeSCN ] = = = 5.62×10 M
4312M 4312M

Figure 8.1

80 EXPERIMENT 8: DETERMINATION OF EQUILIBRIUM CONSTANT

Procedure: Work with one partner.

Setting up the Four Burets:

Do not share buret stands and do not set up burets too close to each other. You do
not want to be bumping elbows with each other.

Students will work in pairs but each pair will need to dispense four different solutions by
buret. We don’t have enough burets to distribute four to each pair of students. Besides it
would be a waste of chemicals if an arrangement is not made to share burets. This is how it
will be done:

Students first pair up for the experiment. Each pair then selects another pair of students with
whom to share burets. For each group of four students, there should be a total of four burets.
Each student in the group is to be responsible for cleaning and setting up one buret that the
rest of the group will be using:
0.200 M Fe(NO3)3 0.00200 M KSCN 0.5 M HNO3 0.00200 M Fe(NO3)3

You must learn not to waste chemicals by taking too much from the stock bottle. As usual
you should not be returning extra chemicals to the stock bottle. The Total Volume shown
in the table below is for each team of 4:
Reagent Vol for

Standard
Curve

Vol for
Equilibrium
Data

Vol per
pair of
students

Vol per
team of 4
students

Vol for
Rinse

Total
Volume

0.200 M
Fe(NO3)3

5 x 2.50 mL + 0 mL 12.50 mL = 12.50 x 2
= 25 mL

+20 mL 45 mL

0.00200 M
KSCN

5.00 mL + 15.00 mL 20.00 mL 20.00×2
= 40 mL

+ 20 mL 60 mL

0.500 M
HNO3

32.50 mL +10.00 mL 42.5 mL 42.5×2
= 85 mL

+ 20 mL 105 mL

0.00200 M
Fe(NO3)3

none 5×5.00 mL 25 mL 25×2
= 50 mL

+ 20 mL 70 mL

You should know how to do this kind of estimation. STUDY the calculations shown above.

This is what EACH STUDENT has to do with the reagent assigned to him/her:
Obtain the Total Volume (see table above) of the reagent assigned to you in a clean and

dry beaker.
Obtain a buret and rinse it twice with about 10 mL each with the reagent.
Label a 400-mL beaker as “Waste.”
Fill the buret with only the amount needed by your group of four. Check with your

instructor as to how much is needed.
Make sure you get rid of the air bubble at the tip of the buret.
Label the buret with the concentration and the formula of the solute with the index card

provided.

EXPERIMENT 8: DETERMINATION OF EQUILIBRIUM CONSTANT 81

At each buret there should be a 50-mL “refill” beaker labeled the same way as the buret
to be used if the buret needs refilling.

You are not responsible for dispensing your reagent for everyone.

In the experiment, each pair of students will then work on the rest of the experiment
independent of the other pair, measuring out solutions and obtaining absorbance values.

At the end of the experiment, each student will then clean up the buret he/she had set up
initially. However, check to make sure the buret is no longer needed by the other students in
the group before cleaning up. Slow workers may end up having to clean up all 4 burets.
FOLLOW THE DIRECTIONS ON THE BLACKBOARD ON DISPOSAL OF CHEM-
ICALS.

Note that there are two different concentrations of the Fe(NO3)3!
As you begin to prepare the solutions, remember that you should not write on or put

stickers on the cuvets as this could interfere with the absorption readings.

Standard Curve
1. Obtain 5 clean and dry test tubes (NOT cuvets) labeled 1-5 and fill each with exactly

2.50 mL of 0.200 M Fe(NO3)3 using a buret. Record the exact volume to the nearest
0.02 mL.

2. Again using a buret, add to test tube #1, exactly 0.50 mL of 0.00200 M KSCN solution,
to test tube #2, 0.75 mL of 0.00200 M KSCN solution and so on in increments of 0.25
mL.

3. Finally, add enough 0.5 M HNO3 to each of the test tubes so that the final volume in
each tube totals 10.00 mL. (The volume of HNO3 should have been calculated
beforehand as part of the pre-lab assignment.) Mix thoroughly by covering with
Parafilm and inverting the tubes numerous times until the contents are well mixed.

4. As usual, record the Instrument ID #. Examine the box of cuvets assigned to you. Be
sure they are clean and dry. If a cuvet is wet, rinse it a couple times with small
quantities of the solution you are about to use. Pour the contents of each test tube into a
cuvet, filling it about ¾ full. Set the spectrophotometer to 447 nm and zero the
instrument with a cuvet filled with 0.5 M HNO3. Remember to wipe the sides of each
cuvet with Kimwipes before placing it into the instrument. Record the absorbance
starting from the most weakly absorbing and working towards the most intensely
colored. Do not cleanup until you have produced an acceptable Standard Curve (see
below.)

CALCULATIONS FOR THE STANDARD CURVE (to be completed before leaving)

Summarize the data needed to produce the standard curve by completing the tables on
the Calculations & Results Page, remembering that the concentration of Fe(SCN)2+ is
equal to the initial concentration of SCN–. Prepare the graph using Excel. Include the
data for the blank in your graph. There should be 6 points in your graph. Display the
trendline and the R2 on your graph and record them also on the Calculations & Results
Page. Please review the Checklist in Experiment 1 (or Appendix 2) as to what else must
be on your graph. Your data points should all lie close to the trendline. If not, you may

82 EXPERIMENT 8: DETERMINATION OF EQUILIBRIUM CONSTANT

have to prepare fresh samples for one or more of your data points. Consult with your
instructor. This is why the graph should be completed in class before you cleanup!

Equilibrium Data (You must use the same spectrophotometer as in the calibration.)

5. Obtain 5 clean and dry test tubes and fill each with 5.00 mL of 0.00200 M Fe(NO3)3

using a buret. If a test tube is wet, rinse it several times with small portions of the
solution.

6. Add exactly 1.00, 2.00, 3.00, 4.00, and 5.00 mL of 0.00200 M KSCN, respectively, to
test tubes labeled 1, 2, 3, 4, and 5.

7. Finally, add enough 0.5 M HNO3 to each of the test tubes so that the final volume in
each tube totals 10.00 mL. (Again, the volume of HNO3 should have been calculated
beforehand.) Mix thoroughly by covering with Parafilm and inverting the tubes.

8. Record the temperature of one of your samples in your lab notebook.
9. Repeat Step 4 and record the absorbance for each of the 5 samples.
10. Dispose of all chemicals in the designated waste container in the hood.

CALCULATIONS FOR THE EQUILIBRIUM DATA (Complete in class if time permits.)

Use the trendline equation from the standard curve to calculate the concentration of
Fe(SCN)2+ in each tube and record on the Calculations and Results Page. Show your
calculations on a separate sheet of paper.

Complete the ICE Table for each of the 5 samples and enter the equilibrium constant
values on the Calculations & Results Page. (Reminder: Do not write “X” but put the
actual values in.) Calculate an average for the equilibrium constant, and the error and
percent error for your average.

Prepare your lab notebook by copying NEATLY, the Data Table from the next page into your notebook.

84 EXPERIMENT 8: DETERMINATION OF EQUILIBRIUM CONSTANT

Copy these Data Tables neatly in your lab notebook prior to arriving to class.
Spectrophotometer ID #: ____

STANDARD CURVE DATA
Table 8.1: Volume of reagents to be used

Tube
#

Vol. of 0.200M
Fe(NO3)3

Volume of
0.00200M KSCN

Volume of
0.500M HNO3

Total Vol.

1 2.50 mL 0.50 mL 7.00 mL 10.00 mL
2 2.50 mL 0.75 mL 6.75 mL 10.00 mL
3 2.50 mL 1.00 mL 6.50 mL 10.00 mL
4 2.50 mL 1.25 mL 6.25 mL 10.00 mL
5 2.50 mL 1.50 mL 6.00 mL 10.00 mL

Table 8.2: Concentrations for Standard Curve
Tube

#
Concentration

of Fe3+
Concentration

of SCN –
Conc. of

Fe(SCN)2+
Absorbance

0 0.0000 M 0.0000 M 0.0000 M 0.000

1 0.0500 M 1.0 x10 4M 1.0 x10 4M

2 0.0500 M 1.5 x10 4M 1.5 x10 4M

3 0.0500 M 2.0 x10 4M 2.0 x10 4M

4 0.0500 M 2.5 x10 4M 2.5 x10 4M

5 0.0500 M 3.0 x 10−4M 3.0 x 10−4M

EQUILIBRIUM DATA: Temperature of one of the samples: _______________

Table 8.3: Volume of reagents to be used
Tube

#
Vol. of 0.00200M

Fe(NO3)3
Volume of

0.00200M KSCN
Volume of

0.500M HNO3
Total Vol.

1 5.00 mL 1.00 mL 4.00 mL 10.00 mL
2 5.00 mL 2.00 mL 3.00 mL 10.00 mL
3 5.00 mL 3.00 mL 2.00 mL 10.00 mL
4 5.00 mL 4.00 mL 1.00 mL 10.00 mL
5 5.00 mL 5.00 mL 0.00 mL 10.00 mL

Table 8.4: Concentrations for Equilibrium Calculations

Tube #
Initial

Concentration of
Fe3+

Initial
Concentration

of SCN –
Absorbance Equilibrium* Conc. of Fe(SCN)2+

1 0.00100 M 2.00 x10 4M

2 0.00100 M 4.00 x10 4M

3 0.00100 M 6.00 x10 4M

4 0.00100 M 8.00 x10 4M

5 0.00100 M 10.0 x10 4M

EXPERIMENT 8: DETERMINATION OF EQUILIBRIUM CONSTANT 85

Calculations & Results: Name: ________________________
Sec: ____ Partner’s Name : ________________

STANDARD CURVE:
List in order from lowest concentration to highest.

[Fe(SCN)2+] Absorbance

0.000 M 0.000

Trendline Equation =
R2 =

EQUILIBRIUM RESULTS:

Temperature of samples =

Tube
# Initial [Fe

3+] Initial [SCN ] Absorbance
[Fe(SCN)2+] at
Equilibrium*

1

2

3

4

5
*Calculated from the standard curve

Show calculations for each test tube on a separate sheet of paper.

ICE Table Test tube # 1 Reminder: Do not write “X” but use the actual numbers. See p.79.

[Fe 3+] [SCN –] [Fe(SCN)2+]

Initial 0.00 M

Change

Equilibrium
Cont’d.
next page

86 EXPERIMENT 8: DETERMINATION OF EQUILIBRIUM CONSTANT

ICE Table Test tube # 2

[Fe 3+] [SCN–] [Fe(SCN)2+]

Initial 0.00 M
Change

Equilibrium

ICE Table Test tube # 3

[Fe 3+] [SCN–] [Fe(SCN)2+]
Initial 0.00 M
Change

Equilibrium

ICE Table Test tube # 4

[Fe 3+] [SCN–] [Fe(SCN)2+]
Initial 0.00 M
Change

Equilibrium

ICE Table Test tube # 5

[Fe 3+] [SCN–] [Fe(SCN)2+]
Initial 0.00 M
Change

Equilibrium

Equilibrium constant: Show calc. setups on your own paper.
Keq #1 Keq #2 Keq #3 Keq #4 Keq #5 Average Keq

The literature value for the equilibrium constant is 138 (Ref. 1). Calculate the error and
percent error for your average equilibrium constant. Watch your sign! Show set up here.

Reference 1: Day & Underwood, “Quantitative Analysis” 1958 p.181

Vernier Format 2
KdatA.txt 6/11/2020 20:14:44
Run 1
conc Abs @ 446.2 nm
c A
M
0 0.000
.0001 0.416
.00015 0.687
.0002 0.895
.00025 1.215
.0003 1.429

Vernier Format 2
kdta2.txt 6/11/2020 20:30:35
Run 1
conc Abs @ 446.2 nm
c A
M
0 0.000
.0002 0.104
.0004 0.239
.0006 0.462
.0008 0.585
0.001 0.718