le Châtelier's Principle:
The Chromate - Dichromate Equilibrium

In this experiment you will study a reaction in which there is considerable reversibility.  This is the reaction between chromate and dichromate ions:

2 CrO₄^2- (aq) + 2 H⁺(aq) Cr₂O₇^2- (aq) + H₂O (l)

Chromate ions, CrO₄^2- (aq), are yellow.  Dichromate ions, Cr₂O₇^2- (aq), are orange.  Since this is an equilibrium, both ions are present in a solution at the same time.  The color of the solution indicates which of the ions is present in the greater amount.

Procedure:

There are two parts to the following procedure.  In part I you will investigate how the chromate-dichromate equilibrium is changed by the concentration of acid.   There are 4 steps to the part I procedure.  Each step has some questions that must be answered before you can go on to the next step.  In part II you will investigate how many chromate ions are present by seeing how much precipitate forms.   There are also 4 steps to the part II procedure.

Part I  The reaction: 2 CrO₄^2- (aq) + 2 H⁺(aq) Cr₂O₇^2- (aq) + H₂O (l) involves hydrogen ions.  Part I of the procedure changes the concentration of the H⁺ ion in order to see how the concentrations of the yellow and orange species respond.  You are looking only for a change in color:

Notes:

• Make sure you distinguish between color changes caused by dilution (the solution becomes paler because you are adding water, and this is not important) and a color shift from yellow to orange, or vice-versa (which is the change you are looking for).
• If a solution is already orange, and becomes a bit more orange, you won't notice it.  Similarly, a yellow that gets a bit more yellow will appear unchanged.  However, it is fairly obvious if the color changes from yellow to orange, or vice-versa.
• You will add HCl (aq) to the reaction.  HCl (aq) is a direct source of H⁺ ions. Therefore, adding HCl (aq) is equivalent to increasing the [H⁺ (aq)] in the reaction.
• You will also be adding NaOH (aq) to the reaction. NaOH (aq) removes H⁺ (aq) ions, because of acid-base neutralization. Therefore, adding NaOH is equivalent to reducing the [H⁺ (aq)] in the reaction.

Click for help understanding the effect of HCl 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. Note the color of each solution initially. What ion is most predominant in test tube 1?      CrO42-   Cr2O72- What ion is most predominant in test tube 2?      CrO42-   Cr2O72- Initial colors Test Tube 1 Test Tube 2 Add 1 M HCl drop by drop (maximum of 5 drops) to each test tube and record the color change. What ion is now most predominant in test tube 1?      CrO42-   Cr2O72- What ion is now most predominant in test tube 2?      CrO42-   Cr2O72- Color after adding 1M HCl Test Tube 1 Test Tube 2 You have added HCl to the test tubes.  Which of the following is correct about the concentration of H+ ions ([H+ (aq)]) after completing the above step I-1? a)  [H+ (aq)] has not changed b)  [H+ (aq)] is greater in test tube 1, but unchanged in test tube 2 c)  [H+ (aq)] is greater in test tube 2, but unchanged in test tube 1 d)  [H+ (aq)] is greater in both test tubes 1 and 2 Write a paragraph that uses le Châtelier's principle to predict how the reaction should respond to the above stress.  Does this prediction agree with the observed results? When you have completed the above questions, you can go on to step I-2.

Click for help understanding the effect of NaOH I-2. In two new test tubes, repeat the first part of step I-1 with fresh solutions (you will now have four test tubes with colored solutions). The color of these two test tubes will be the same as the initial colors of test tubes 1 and 2 (see part 1).  Initially, there is more CrO42- in test tube 3, and more Cr2O72- in test tube 4. Initial colors Test Tube 3 Test Tube 4 Add 1 M NaOH drop by drop (maximum of 5 drops) to each test tube and record the color change.  Remember that NaOH removes H+ ions from solution. What ion is now most predominant in test tube 3?      CrO42-   Cr2O72- What ion is now most predominant in test tube 4?      CrO42-   Cr2O72- Color after adding 1M NaOH Test Tube 3 Test Tube 4 Which of the following is correct about the concentration of H+ ions ([H+ (aq)]) after completing the above step I-2? a)  [H+ (aq)] has not changed b)  [H+ (aq)] is lower in both test tubes 3 and 4 c)  [H+ (aq)] is greater in both test tubes 3 and 4 d)  [H+ (aq)] is lower in  test tube 3 and higher in test tube 4 Write a paragraph that uses le Châtelier's principle to predict how the reaction should respond to the above stress.  Does this prediction agree with the observed results? When you have completed the above questions, you can go on to step I-3.

I-3.  To test tubes 1 and 2 from step I-1, add 1 M NaOH drop by drop (maximum of 10 drops), noting any change in color. At the end of step I-1, test tubes 1 and 2 are both orange.  Addition of HCl in step 1 has forced the equilibrium reaction to the right, producing mostly orange Cr2O72-. Initial colors Test Tube 1 Test Tube 2 After the addition of 10 drops of NaOH, the color has turned back to yellow. What ion is now most predominant in test tube 1?      CrO42-   Cr2O72- What ion is now most predominant in test tube 2?      CrO42-   Cr2O72- Color after adding 1M NaOH Test Tube 1 Test Tube 2 Which of the following correctly describes what the addition of NaOH has done to the equilibrium reaction: 2 CrO42- (aq) + 2 H+(aq) Cr2O72- (aq) + H2O (l) after completing the above step I-3? a)  The added NaOH neutralizes the [H+ (aq)], causing the equilibrium to shift to the right. b)  The added NaOH neutralizes the [H+ (aq)], causing the equilibrium to shift to the left. c)  The color changed because the solution is now more diluted. When you have completed the above questions, you can go on to step I-4.

I-4.  To test tubes 3 and 4 from step I-2, add 1 M HCl drop by drop (maximum of 10 drops), noting any change in color. At the end of step I-2, test tubes 3 and 4 are both yellow.  Addition of NaOH in step 2 has forced the equilibrium reaction to the left, producing mostly yellow CrO42-. Initial colors Test Tube 3 Test Tube 4 After the addition of 10 drops of HCl, the color has turned back to orange. Color after adding 1M HCl Test Tube 3 Test Tube 4 Which of the following correctly describes what the addition of HCl has done to the equilibrium reaction: 2 CrO42- (aq) + 2 H+(aq) Cr2O72- (aq) + H2O (l) after completing the above step I-4? a)  The added HCl neutralizes the NaOH.  The extra [H+ (aq)] causes the equilibrium to shift to the right. b)  The added HCl neutralizes the NaOH.  The extra [H+ (aq)] causes the equilibrium to shift to the left. c)  The color changed because the solution is now more diluted. When you have completed the above question, you can go on to Part II.

Part II. 

The basis of part II is the precipitation reaction forming insoluble BaCrO₄:

Ba²⁺ (aq) + CrO₄^2- (aq) BaCrO₄ (s)

Notes:

• The added solution is Ba(NO₃)₂.  In this system, the NO₃^- ions are spectator ions.
• The amount of precipitate formed tells you how many CrO₄^2- (aq) ions are present.  More cloudiness indicates more ions.

Click for help understanding the precipitation reaction 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.  Add 0.1 M Ba+2 (aq) ions drop by drop, until a change is noted. Test Tube 5 CrO42-(aq) CrO42-(aq) + 2 drops NaOH CrO42-(aq) + 2 drops NaOH + 2 drops Ba(NO3)2 Considering what you learned in part I-1, and part I-3 of the procedure, what does the addition of the two drops of NaOH do to the amount of Cr2O72- (aq) ions present in test tube 5? a)  it will increase b)  it will decrease c)  it will stay the same When you have completed the above question, you can go on to step II-6.

II-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.  Add 0.1 M Ba+2 (aq) ions drop by drop, until a change is noted. Test Tube 6 Cr2O72-(aq) Cr2O72-(aq) + 2 drops HCl Cr2O72-(aq) + 2 drops HCl + 2 drops Ba(NO3)2 There is virtually no cloudiness in test tube 6 after the above procedure.  This observation, together with the observed results in reaction steps II-5 confirms that the identity of the precipitate is BaCrO4 True    False When you have completed the above question, you can go on to step II-7.