LAB 22 CHEMICAL EQUILIBRIUM (QUANTITATIVE)

This lab will be the same as lab 21 except you will collect data and explore the phenomenon of equilibrium quantitatively. Remember the equilibrium reaction is:

Fe+3(aq) + SCN-(aq) <====> FeSCN+2(aq)

Also remember the color is caused by the FeSCN+2(aq). The greater the concentration of the FeSCN+2(aq), the darker the color will be.

This time you will determine the actual concentration of the ions at equilibrium and then look for a mathematical relationship that relates these quantities in a simple, easy manner.

The initial concentrations of the reacting ions will be calculated by you from dilution data. The equilibrium concentrations of the FeSCN+2(aq) will be calculated colorimetrically, that is, by comparing the color intensity of a known equilibrium concentration to one that has been diluted. You have already seen the tea demo in the pre-lab lecture and know that color intensity, i.e. concentration, is inversely proportional to the depth of solution. Since you will be diluting the equilibrium concentrations the color intensity will be less. Therefore you will have to remove solution from the known concentration until the colors match. Once the color intensity between the two solutions is equal, you will measure the depth of each solution and determine the concentration of the diluted solution by using:

C1 x d1 = C2 x d2

The concentration of the standard equilibrium solution ions will be known because the thiocyanate ion,

SCN-(aq) will be so low that you can assume all of them will be consumed to produce the product, FeSCN+2(aq), since they will be mixed with a large excess of Fe+3(aq). Therefore the equilibrium concentration of the FeSCN+2(aq) will be equal to the initial concentration of the SCN-(aq). The SCN-(aq) is therefore the limiting reagent. This is only true for solution #1 however, because in subsequent solutions the concentration of the Fe+3(aq) will not be as great and you will reach equilibrium before you run out of the SCN-(aq)

PROCEDURE : prelab assignment

Before coming to class study the dilution procedure outline in your packet. This is a visual diagram of steps 1,2 and 3 in the procedures. Calculate the following concentrations:

1. [SCN-(aq)] in each vial after mixing but before any reaction occurs. Each vial has 5.0 ml of .00200 M KSCN diluted with 5.0 ml of Fe(NO3)3 solution.

2. [Fe+3(aq)] ion in each vial after mixing but before any reaction takes place. Remember the [Fe+3(aq)] in the beaker is not the same as the [Fe+3(aq)] in the vial.

You will need to construct a data table that will reflect the depth of the two solutions being compared for color intensity. Remember the depth of the darker control solution is the only vial that has any material removed.

LAB PROCEDURES:

1. Label 5 flat bottom vials 1,2,3,4, and 5. To each vial add 5.0 ml of .00200 M potassium thiocyanate. It will be easier to write the concentration in scientific notation, 2.00 x 10-3 M KSCN.

2. Add 5.0 ml of .200 M iron (III) nitrate, Fe(NO3)3, to vial #1. Again it will be easier in the calculations to deal with this concentration in scientific notation;

2.00 x 10-1 M Fe(NO3)3. This is now the control solution. You do not have an equilibrium established. You have run out of SCN-(aq) because you have overloaded the system with Fe+3(aq). You now know the [FeSCN+2(aq)] since the equation shows a 1:1 reacting ratio between the SCN-(aq) and FeSCN+2(aq).

3. Measure 10.0 ml of .200 M Fe(NO3)3 in a 25.0 ml graduated cylinder and add distilled water until 25.0 ml of solution has been made. Pour the solution into a clean beaker to completely mix the solution. Now place 5.0 ml of this solution into vial #2. Save the remaining solution for the next step.

4. Pour 10.0 ml of the solution from the beaker into a 25.0 ml graduated cylinder and again add distilled water until the cylinder is filled to 25.0 ml. Pour the solution into another clean beaker to mix and place 5.0 ml into vial #3.

5. Continue the dilution in this way until all 5 vials have successively more dilute solutions.

6. Since vial #1 is the standard solution, or control, you must now make this vials color intensity equal to the other vials color intensity by removing solution from vial #1 until its color equals the color of the vial being compared to it. The way to accomplish this is to wrap a piece of paper around each of the two vials being compared. This will make all light go through the solution. Now look vertically down through the vials, vial #1 should be darker. Using an eyedropper remove solution from vial #1 until its color intensity is less than vial #2. Place the solution that is removed into a clean beaker so it can be used again. Once the color is less in vial #1, re-add the solution until the color intensity is equal in both vials. You and your lab partner need to agree on this point.

Measure and record the depths of the two vials. This is the data you are after.

7. Repeat the procedure with vials 1 & 3, 1 & 4, and

1 & 5.

CALCULATIONS:

There are some assumptions that must be made in your calculations:

A. the Fe(NO3)3 and the KSCN are completely

ionized in solution.

B. In vial #1, all of the SCN-(aq) have reacted to

form the colored FeSCN+2(aq).

The symbol [ ] will represent the equilibrium concentration in molarity.

1. Calculate the following for vials 1-5:

A. depth ratios= depth of liquid in vial #1(when compared w/vial #2)

depth of liquid in vial #2

B. Calculate the equilibrium concentration of the[FeSCN+2] by using the relationship between depth of solution and concentration discovered in the tea demo.

C1 x d1 = C2 x d2 or C2 = C1 x d1

d2

where d1 = depth ratio from 1 (A) above.

d2

2. Calculate the [Fe+3] at equilibrium, by subtracting the [FeSCN+2] at equilibrium from [Fe+3] initially.

3. Calculate the [SCN-] at equilibrium, by subtracting the [FeSCN+2] at equilibrium from [SCN-] initially.

4. You will now try to find some constant numerical relationship between the equilibrium concentrations of the ions in each vial by multiplying and dividing the values obtained for each vial in various combinations. These different methods are shown on the data table supplied in your packet.

5. For each of the combinations A through D of calculation 4, find the ratio: largest value ( vials 2 through5)

smallest value (vials 2 through 5)

6. Which of the combinations of concentrations A,B,C, or D gives the most consistent numerical value? In other words in which combination did the largest and smallest value stay very much the same? (i.e. closest to the value 1)? This is called the equilibrium constant expression.

7. Restate this expression in words, using the names of the ions involved in the equilibrium.

8. Restate this expression in words, using the terms reactants and products.