| In this experiment you will study a reaction in which there is considerable reversibility. This is the reaction beween chromate ions, CrO42- (aq) which are yellow, and dichromate ions Cr2O72- (aq) which are orange. |
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| A solution of chromate ions | A solution of dichromate ions |
Caution: chromium(VI) compounds are known carcinogens. Long term exposure is known to cause cancer in humans. While the quantities that you are handling in this lab are very small, you must nevertheless use extreme caution not to ingest them in any way. Make sure that you wear the appropriate safety equipment (goggles, gloves, and an apron) You are required to read the MSDS for potassium chromate before carrying out this lab. Wash up the equipment thoroughly at the end of the experiment, including your work space. Wash your hands with soap and water at the conclusion, and clean under your finger nails.
The reaction that you will investigate is: 2 CrO42- (aq) +
2 H+(aq) 
Cr2O72-
(aq) + H2O (l)
The procedure involves varying the concentration of the H+ ion in order to see how the concentrations of the yellow and orange species change. In part I of the procedure you are looking only for a change in color:
In part II you will be looking for the formation of a precipitate of BaCrO4 (s). The anount of precipitate formed tells you how many CrO42- (aq) ions are present.
In Part I, you will add HCl which is a direct source of H+ ions. Therefore, adding HCl is equivalent to increasing the [H+ (aq)] in the reaction. You will also be adding NaOH solution. NaOH removes H+ ions, because of acid-base neutralization. Adding NaOH is equivalent to reducing the [H+ (aq)] in the reaction.
Make sure you distinguinsh between color changes caused by dilution (the solution becomes paler because you are adding water) and a color shift from yellow to orange, or vice-versa.
Procedure:
Part I
1. Put approximately 1 mL (10 drops) of 0.1 M CrO42-(aq) solution into one clean 13 x 100 mm test tube. Put about the same amount (10 drops) of 0.1 M Cr2O72- (aq) into a second test tube.
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 CrO42-(aq) + HCl |
 Cr2O72-(aq) + HCl |
2. Repeat the first part of step 1 with fresh solutions (you will now have four test tubes with colored solutions).
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 CrO42-(aq) + NaOH |
 Cr2O72-(aq) + NaOH |
3. To one of the tubes from step 1, add 1 M NaOH drop by drop (maximum of 10 drops), noting any change in color.
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 CrO42-(aq) +HCl+NaOH |
 Cr2O72-(aq) +HCl+NaOH |
4. To one of the tubes from step 2, add 1 M HCl drop by drop (maximum of 10 drops), noting any change in color.
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 CrO42-(aq) +NaOH+HCl |
 Cr2O72-(aq) +NaOH+HCl |
Part II.
5. Put approximately 1 mL (10 drops) of 0.1 M CrO42-(aq) solution into one clean 13 x 100 mm test tube. Add 2 drops of 1 M NaOH (note: this will reduce the amount of H+ ions – what does this due to the amount of Cr2O72- (aq) ions present?). Add 0.1 M Ba+2 (aq) ions drop by drop, until a change is noted.
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| CrO42-(aq) | CrO42-(aq) + 2 drops NaOH |
CrO42-(aq) + 2 drops NaOH + 2 drops Ba(NO3)2 |
6. Put approximately 1 mL (10 drops) of 0.1 M Cr2O72-(aq) solution into one clean 13 x 100 mm test tube. Add 2 drops of 1 M HCl (note: this will increase the amount of H+ ions -- what does this due to the amount of CrO42- (aq) ions present?). Add 10 drops 0.1 M Ba+2 (aq) ions drop by drop noting any changes.
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| Cr2O72-(aq) | Cr2O72-(aq) + 2 drops HCl |
Cr2O72-(aq) + 2 drops HCL + 2 drops Ba(NO3)2 |
7. To the test tube from step 5, add 1 M HCl drop by drop until a change is noted.
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| (Original) CrO42-(aq) + 2 drops NaOH + 2 drops Ba(NO3)2 |
CrO42-(aq) + 2 drops NaOH + 2 drops Ba(NO3)2 +5 drops HCL |
8. To the test tube from step 6, add 1 M NaOH drop by drop until a change is noted.
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| (Original) Cr2O72-(aq) + 2 drops HCL + 2 drops Ba(NO3)2 |
Cr2O72-(aq) + 2 drops HCL + 2 drops Ba(NO3)2 + 5 drops NaOH |
9. Put approximately 1 mL (10 drops) of 0.1 M CrO42-(aq) solution into one clean 13 x 100 mm test tube, and the same amount of 0.1 M Cr2O72-(aq) into another. Add no acid or base to either test tube, but add about 5 drops of Ba+2 (aq) solution to each. Record your observations, and compare them to the above results when you added HCl or NaOH before adding the Ba+2 (aq) ions.
Previous test |
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| CrO42-(aq) solution alone | Cr2O72-(aq) solution alone | CrO42-(aq) + 2 drops Ba(NO3)2 |
Cr2O72-(aq) + 2 drops Ba(NO3)2 |
CrO42-(aq) + 2 drops NaOH + 2 drops Ba(NO3)2 |
Cr2O72-(aq) + 2 drops HCL + 2 drops Ba(NO3)2 |
Conclusion:
1. Create a table in your notebook, similar to the following. Use a blank row for each test that you carried out in part I.
| Step | Change Introduced | Reaction Response |
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| 2 CrO42- (aq) | + 2 H+(aq) | |
Cr2O72- (aq) | + H2O (l) | ||
| 1 | ||||||
| 2 | ||||||
| etc. | ||||||
Use arrows (up or down) for each change that results. Remember that in a chemical equation, everything on the same side of the equilibrium symbol must respond in the same way. So if you can see a visible increase in the orange color, you will put an up arrow in the Cr2O72- (aq) column. Although you can't see it, since the amount is very small compared to the amount of water present, there must also be an increase in the amount of H2O (l), so it will also have an up arrow.
2. Explain each color change that occurs by referring to the above table, and explaining what is going on using Le Chatelier’s principle.
3. Using evidence from the addition of Ba+2 ions, and the table you have prepared in conclusion question 1, explain the results of the precipitation reaction. Explain why HCl or NaOH were added, and why they effect the precipitation reaction.
4. Using results from the final step in the procedure, explain how these results prove that this reaction is truly an equilibrium.
