Question

Re-write the chemical reaction for the equilibrium established between the chromate and dichromate ion. Label the colors expected for the chromate and dichromate ions. You may refer to the experimental procedure.

2CrO₄^2- (aq) + 2H₃O⁺ (aq) Û Cr₂O₇^2- (aq) + 3H₂O (l)

(chromate-yellow)                   (dichromate-yellow)

A completed data table #1.

Change to the Reaction	Drops Added	Visual Observations	Shift in the reaction (to products, reactants, or no change)
K2CrO4 solution only		The color after the change to the reaction is yellow. There are no solids, or bubbles that are produced. A transparent yellow color occurs.	Reactant
First addition of H2SO4	5	No solids or bubbles occurred after the reaction. The color simply changed from a transparent yellow, to a transparent reddish/orange color.	Product
Addition of NaOH	5	The color stayed almost the same, while the color was still orange I could observe that it was slightly more yellow. No precipitates were formed, the solution was still transparent.	Reactant
Second addition of H2SO4	6	The solution became a deeper red-orange color. The test tube holding the solution became warmer, meaning it was an exothermic reaction.. The solution only subtly changed in color.	Product

3. Using equation 2, interpret your observations in terms of Le Châtelier’s principle when you added sulfuric acid (the first time) to the test tube containing potassium chromate.

Addition of H₂SO₄ for the first time creates a higher concentration of CrCo₄^2- ions. In order to reestablish the equilibrium the reaction shifts to the products, increasing the concentration of Cr₂O₇^2- ions giving a orange color.

Please let me know if this is correct, also let me know if there is an equation I can include for

Answer 1

The equlibrium between the ions is -

2CrO₄^2- + H₃O⁺ -> Cr₂O₇^2- + 3H₂O

CrO₄^2- is yellow colored and Cr₂O₇^2- has orange-red color

According to Le Chatelier's principle the equilibrium shifts in that direction which balances out the disturbance caused. So, addition of H2SO4 will increase the H3O+ ions present in the left side and thus disturb the equilibrium so, to balance out this the equilibrium will shift in the forward direction so as to remove this extra H3O+ ions. So, more Cr₂O₇^2- ions will be formed and thus we see the change in color from yellow to orange-red as Cr₂O₇^2- is orange-red in color.