Chapter 1: Chemical Equilibrium Part 6 SABIS Grade 11 (Level M) Chemistry

1.3 Experiment of the FeSCN²⁺ System

In this experiment you will take a quantitative look at the reaction

Fe3+(aq) + SCN–(aq) ⇌ FeSCN2+(aq)

this was presented qualitatively as a class experiment by your teacher. This time you will determine the concentration of each of the ions at equilibrium, and then seek an expression that relates these quantities mathematically in a simple, convenient manner.

The determination of the concentrations will be done colorimetrically. If you have ever looked critically at a glass full of a colored liquid, such as iced tea, you know that the color intensity as viewed through the sides of the glass is much less than the color intensity as viewed from the top down. This is because the color intensity depends upon the concentration of the colored substance and on the depth of the solution. Thus, a 1 cm depth of a 1 M colored solution will appear to have the same color intensity as a 2 cm depth of a 0.5 M solution of the same material. The concentration of two such solutions may be compared by altering their relative depths until the color intensity appears the same.

The ratio of the concentrations is found to be the inverse of the ratio of the depths. Note that this procedure gives only relative values for the concentrations. To get absolute values, a standard solution of known concentration must be used.

1.3.1 Procedure

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 iron (III) ion, Fe3+(aq). You can assume that essentially all of the thiocyanate ions will be used in forming the 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.

Arrange your data sheet so that you can record the depth of the solution in each test tube and the depth of the standard solution compared with each of the other test tubes.

1. Line up five clean test tubes, all of the same diameter, and label them 1, 2, 3, 4, 5. Add 5.0 ml of 0.002 M potassium thiocyanate, KSCN, to each of these five test tubes. To test tube 1 add 5.0 ml of 0.2 M iron (III) nitrate, Fe(NO3)3. This tube will be used as the standard.
2. Measure 10.0 ml of 0.2 M Fe(NO3)3 in your 25 ml or 50 ml graduated cylinder and fill to the 25.0 ml mark with distilled water. Pour the solution into a clean dry beaker to mix it. Measure 5.0 ml of this solution and pour it into test tube 2. (Save the remainder of the Fe(NO3)3 solution for Part c.) Calculate the concentration of this solution as part of your pre-lab preparation.
3. Pour 10.0 ml of the solution from the beaker into your graduated cylinder. Discard the remainder. Continue to fill the graduated cylinder to the 25.0 ml mark with distilled water, and pour the solution into a clean dry beaker to mix. Pour 5.0 ml of this solution into test tube 3. Continue dilution in this manner until you have 5.0 ml of successively more dilute solution in each test tube. (Figure 1.8). Calculate the concentration of each of the solutions as part of your pre-lab preparation.

The goal of this experiment is to compare the solutions in each of the test tubes with the standard tube (number 1) in order to determine the concentration of the thiocyanoiron(III) ion, FeSCN2+(aq).

Wrap a strip of paper around test tubes 1 and 2 to exclude light from the side. Look vertically down through the solutions toward a diffuse light source. If the color intensities appear the same, measure the depth of each solution to the nearest millimeter and record this. If the color intensities do not appear the same, remove some of the solution from the standard tube with a medicine dropper until the color intensities are the same. Put the portion you removed into a clean dry beaker, since you may have to use some of this solution later. In fact, the matching may be accomplished by removing more standard than seems necessary and then replacing part of it drop by drop. When the color intensities are the same in each test tube, measure the depth of both solutions to the nearest millimeter. Repeat the procedure with test tubes 1 and 3, 1 and 4, and finally 1 and 5.

1.3.2 Calculations

Assume in your calculations that:

a) the iron (III) nitrate and the potassium thiocyanate exist in their respective solutions entirely as ions.

b) in the standard tube (number 1), essentially all the thiocyanate ions have reacted to form thiocyanoiron(III) complex ions. Also remember that both solutions are diluted on mixing.

The symbol [ ] will be used to represent the equilibrium concentration in moles per liter. The formula within the brackets denotes the species. Thus the notation [Fe3+] stands for the equilibrium concentration of the iron (III) ion, Fe3+(aq), in moles per liter.

Do all of the calculations for each test tube 2 through 5 as follows.

1. a) Calculate the ratio of depths in the color comparison. Example:

Ratio=Depth of standard match with tube 2Depth of liquid in tube 2${\displaystyle \text{Ratio}=\frac{\mathrm{D}\mathrm{e}\mathrm{p}\mathrm{t}\mathrm{h}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{a}\mathrm{r}\mathrm{d}\mathrm{m}\mathrm{a}\mathrm{t}\mathrm{c}\mathrm{h}\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}\mathrm{t}\mathrm{u}\mathrm{b}\mathrm{e}2}{\mathrm{D}\mathrm{e}\mathrm{p}\mathrm{t}\mathrm{h}\mathrm{o}\mathrm{f}\mathrm{l}\mathrm{i}\mathrm{q}\mathrm{u}\mathrm{i}\mathrm{d}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{u}\mathrm{b}\mathrm{e}2}}$

 
    b) From these ratios calculate the equilibrium concentration of thiocyanoiron(III) ion, [FeSCN2+]:

        [FeSCN2+] = Ratio of depth × conc. of standard

2. From your dilution data calculate the initial concentration of Fe3+(aq) ion.

3. Calculate the equilibrium concentration of Fe3+(aq) ion, [Fe3+], by subtracting the equilibrium concentration of the FeSCN2+(aq) ion from the initial concentration of the Fe3+(aq) ion.

4. Calculate the equilibrium concentration of the SCN–(aq) ion, [SCN–], in the same manner as for the Fe3+(aq) ion. Subtract the equilibrium concentration of the FeSCN2+(aq) ion from the initial concentration of the SCN–(aq) ion.

5. Now try to find some constant numerical relationship between the equilibrium concentrations of the ions in each test tube by multiplying and dividing the values obtained in each test tube in various combinations. For example, for each of the test tubes 2 through 5 calculate:

a) [Fe3+][FeSCN2+][SCN–]

b) [Fe3+][FeSCN2+][SCN−]${\displaystyle \frac{[{\mathrm{F}\mathrm{e}}^{3+}][{\mathrm{F}\mathrm{e}\mathrm{S}\mathrm{C}\mathrm{N}}^{2+}]}{[{\mathrm{S}\mathrm{C}\mathrm{N}}^{-}]}}$ 

c) [FeSCN2+][Fe3+][SCN−]${\displaystyle \frac{[{\mathrm{F}\mathrm{e}\mathrm{S}\mathrm{C}\mathrm{N}}^{2+}]}{[{\mathrm{F}\mathrm{e}}^{3+}][{\mathrm{S}\mathrm{C}\mathrm{N}}^{-}]}}$ 

1.3.3 Questions

1. Which of the combinations of concentrations, (a), (b), or (c), gives the most constant numerical value? This form is known as the mass action expression, or the reaction Quotient, Q.
2. Restate this expression, in words, using the terms reactants and products.
3. Give a possible explanation as to why such a relationship might exist.

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Chapter 1: Chemical Equilibrium  Part 7 SABIS Grade 11 (Level M) Chemistry