Chemical Equilibrium s1

Name: ______Per______

Chemical Equilibrium

Purpose: To determine the concentration of each of the ions at equilibrium, and find an expression which relates these quantities mathematically.

In this experiment you will study in a quantitative manner the reaction Fe3+(aq) + SCN-(aq)  FeSCN2+(aq), which was presented qualitatively as a class demo by your teacher. This time you will determine the concentration of each of the three ions at equilibrium and then determine the equilibrium expression, so you can calculate value of equilibrium constant (kc).

This reaction is a colorimetric reaction, because the complex formed (FeSCN2+) as a result of reaction of Fe3+ and SCN-, is a colored compound and the intensity of the color is directly proportional to the concentration of this complex compound. The concentration of this complex at equilibrium can be found by taking readings on Spec-20.

Different concentrations of Fe3+ are reacted with same concentration of SCN- and then the equilibrium concentrations of all three ions are calculated. In preparing the standard solution in step a of this experiment, you will use a low, known concentration of thiocyanate ion, SCN-(aq) and add a large excess of ferric ion, Fe3+(aq). Ferric solution will be diluted using HNO3 instead of water because water makes it unstable. You can assume that essentially all the thiocyanate ion will be used in forming the colored complex thiocyanoiron(III) ion, FeSCN2+(aq) and that the equilibrium concentration of the FeSCN2+(aq) ion will be essentially the same as the concentration of the SCN-(aq) ion with which you started. Thus we know C1, the equilibrium concentration of the color species FeSCN2+.

What is serial dilution? Serial dilutions are a way for us to make solutions of varying concentrations very easily in succession. This method is desirable because it eliminates a lot of the uncertainty and imprecision involved in making very small concentrations relative to the stock (the concentrated) solution.

The first step in making a serial dilution is to take a known volume (we will use 10 ml) of stock and place it into a known volume of something stable to dilute it will (we will use 15 ml of .6 M HNO3). This produces 25 ml of dilute solution. The dilute solution has 10 ml of extract /25 ml.

The technique used to make a single dilution is repeated sequentially using a more and more dilute solution as the "stock" solution. At each step 10 ml of the previous dilution is added to 15 ml of nitric acid. Each step results in a change in the concentration from the previous concentration

1 Wear safety glasses and an apron.

Record ALL DATA and do ALL CALCULATIONS to three (3) significant figures.

Chemical Equilibrium Prelab Questions

1. What reversible reaction will be studied?

2. What are the three objectives of this lab?

3. a. What is the limiting reactant?

b. What reactant will be in excess?

4. a. What is serial dilution?

b. Which of the reactants is serially diluted?

c. Why is HNO3 used for diluting?

d. How many steps will the dilution take place in?

e. How many different concentrations will be used?

PROCEDURE

A. Preparing F Tubes (Serial Dilution of Fe3+ ) a. Obtain five clean test tubes and six clean beakers. Label the test tubes as F 1, F3, F4,F 5,F 6 and the beakers #1-6 using masking tape. (Already done for you by our wonderful TAs). Make sure that the tubes are clean. If you have any dirty tubes or beakers, see your teacher. b. Add 20.0 mL 0.200 M Fe(NO3)3 to beaker #1. Transfer using F1 pipet , 5.00 mL of this solution to F1 tube. Now, transer 10.0 mL of this Fe3+ solution from beaker #1 to beaker #2 using same F1 pipet. c. To beaker #2, add 15.00 mL 0.600 M HNO3, nitric acid. Mix well. Nitric acid turns the skin yellow. Avoid contact. The yellow goes away when the skin does. d. Remove 10.0 mL of the solution in beaker #2 to beaker #3 using same F2 pipet. e. Add 15.0 mL 0.6 M HNO3 to the solution in the beaker #3. Mix well. Remove 5.0 mL of this solution and add it to F 3 tube and transfer 10.0 ml of this solution (in beaker #3) to beaker #4 using F3 pipet.

2 f. Now add 15.0 mL 0.6 M HNO3 to the solution in the beaker #4. Mix well. Remove 5.0 mL of this solution to F4 test tube. Transfer 10.0 ml of this solution in beaker #4 to beaker #5 using F4 pipet. g. Add 15.0 mL 0.6 M HNO3 to the solution in the beaker #5. Mix well. Transfer 5.0 mL of this solution to test tube F5 . Transfer 10.0 mL of this solution in beaker #5 to beaker #6 using F 5 pipet. h. Add 15.0 mL 0.6 M HNO3 to the solution in the beaker #6. Mix well. Transfer 5.0 mL of this solution to test tube F6 using F6 pipet. B. Preparing S Tubes:

a. Take 5 clean test tubes and label them S1, S3, S4, S5, and S6 using masking tape b. Add 5.0 mL 0.00200 M KSCN, potassium thiocyanate, to test tubes S1, S3, S4, S5, and S6. Mix well.

C. Preparing Reaction Tubes (RT)

a. Take 5 larger clean test tubes and label them RT1, RT3, RT4, RT5 and RT6. b. Mix F 1 and S1 tubes in RT 1 tube and so on and take the readings. c. Readings should be taken with in 30 minutes of mixing F and S tubes, otherwise the color of the complex starts to fade, giving a wrong reading.

PROCESSING THE DATA:

1) Record the absorbance value of the products from the Spec 20 in the given table.

2) Compare each tube’s absorbance to the darkest one by calculating Tube n Tube 1

3 Table 1: Color Intensity

Tube Number (n) Spec 20 Reading Ratio: Tube n Tube 1 1 #1/#1 = 1.00 *

* These values should be put into Column C (Color Ratios) on Table 2 – Ion Concentrations.

3) Fill in the blanks in Table 2 – Ion Concentrations – with the appropriate values. Use the directions found on pages 5 thru 7 to do the calculations needed. Show ALL calculations on the appropriate pages. Table 2. Ion Concentrations

Initial Conc Color Ratios Equilibrium Concentrations Tube # [Fe3+] [SCN-] Tube n [FeSCN2+] [Fe3+] [SCN-] Tube 1 A B C D E F 1 1.00 “0”

4 Calculations for column A in Table 2 for Fe 3+ Ion Concentrations:

3+ A. Initial concentrations of Fe in the diluted Fe(NO3)3 solutions. Follow the dilution chart for ferric nitrate (left-hand side). Calculate the initial concentration of the ferric ion in each of the reaction tubes (RT) just before the reaction.

Stage #1 Reaction tube (tt): ______M x ______mL = ______M mL

Stage #2 beaker: ______M x ______mL = ______M mL

reaction tube: ______M x ______mL = ______M mL

*Tube 2 is not reacted because the intensity is not different enough from tube 1.

STEPS FOR CALCULATING CONCENTRATION OF SCN - IONS FOR COLUMN B IN TABLE 2

B. Initial concentration of SCN- in the diluted KSCN solutions.

5 Each tube has 5.0 mL of 0.00200 M KSCN diluted by an equal volume of ferric nitrate solution. All tubes have the same initial concentration of thiocyanate ion. What is it?

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STEPS FOR CALCULATING FeSCN 2+ ION CONCENTRATION AT EQUILIBRIUM FOR COLUMN D IN TABLE 2

D. Equilibrium concentrations of FeSCN2+ in each reaction tube. 2+ Since each successive tube is a fraction of the original concentration, determine the [FeSCN ] at equilibrium by multiplying the decimal ratio of color intensity (Column C) by the equilibrium concentration of thiocyanoferrate ion in tube #1.

D = column C x column B

Tube #1 ( ) x [ ] = ______Tube #3 ( ) x [ ] = ______Tube #4 ( ) x [ ] = ______Tube #5 ( ) x [ ] = ______Tube #6 ( ) x [ ] = ______

STEPS FOR CALCULATING EQUILIBRIUM CONCENTRATION OF Fe 3+ ION FOR COLUMN E IN TABLE 2 E. Equilibrium concentrations of Fe3+ in each reaction tube. The [Fe3+] at equilibrium is the concentration left over after some of it has been used to form the product. Since [FeSCN2+] made is equal to [Fe3+] used, subtract [FeSCN2+] from the initial [Fe3+].

E = column A – column D

Tube #1 ( ) - [ ] = ______Tube #3 ( ) - [ ] = ______Tube #4 ( ) - [ ] = ______Tube #5 ( ) - [ ] = ______Tube #6 ( ) - [ ] = ______

STEPS FOR CALCULATING EQUILIBRIUM CONCENTRATIONS OF SCN - ION FOR COLUMN F IN TABLE 2

F. Equilibrium concentrations of SCN- in each reaction tube.

6 The [SCN-] at equilibrium is the concentration left over after some of it has been used to form the product. Since [FeSCN2+] made is also equal to [SCN-] used, subtract [FeSCN2+] from the initial [SCN-].

F = column B – column D

Tube #1 ( ) - [ ] = ______Tube #3 ( ) - [ ] = ______Tube #4 ( ) - [ ] = ______Tube #5 ( ) - [ ] = ______Tube #6 ( ) - [ ] = ______

4) The equilibrium concentrations from columns D, E, and F in Table 2 have changed, and there is a reasonable suspicion that they have changed in a consistent way. Now do some more calculations with different types of reactant-product rations to find a mathematical relationship which will give us a consistent result (a constant). That will allow the prediction of the results of any future concentration changes.

Type [FeSCN 2+ ] [Fe 3+ ] [FeSCN 2+ ] [FeSCN 2+ ] [Fe3+] + [SCN-] [SCN-] [Fe3+] [SCN-]

No. X Y Z [ ] [ ] [ ] + [ ] [ ] [ ] [ ] [ ] 3 [ ] = = =

4 [ ] [ ] [ ] [ ] [ ] + [ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ] 5 [ ] + [ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ] 6 [ ] + [ ] [ ] [ ] [ ] = = =

7 5. To recognize the constant-producing ratio you need to find consistent results. Compare the largest and smallest results in each column to see which gives the smallest range of values (smallest resulting value for the ratio).

X Y Z (Largest) = ______= ______= (Smallest)

6. Write in words the general meaning of the reactant- product ratio (X, Y, or Z), which gave the smallest value of the ratio in 5 above.