Question

What color will the solution be if we add H₂SO₄ to the K₂CrO₄ solution?

H₂SO_{4 (aq)} ----> H⁺ (aq) + HSO₄^- (aq)

HSO₄^- (aq) ---> H⁺(aq) + SO₄^2- (aq)

How does increasing the concentration of H⁺ affect the chromate/dichromate equilibrium? What color will the solution be?

What if we add NaOH to the solution? How does it affect the equillibrium? What does OH^- react with in the solution? What color will the solution be?

Answer 1

Simplified Reaction:

[CrO4]2- + 2H+ <---------> [Cr2O7]2- + H2O

[CrO4]2- => Chromate Ion (Yellow)
[Cr2O7]2- => DiChromate Ion (Orange)

The reason why this demonstrates Le Chatlier's principle is:

If you add more Hydrogen Ions (Sulphuric Acid, H2SO4) to Chromate (Yellow) you will produce more of the Di-Chromate Ion (Orange), If you add Sodium Hydroxide (OH-), you'll produce more of the Chromate Ion (Yellow).

Reason:

The Sulphuric Acid adds more H+ ions, therefore more Di-Chromate Ions will be produced in order to mop up the excess H+ ions and equalise the system (we can observe this via the colour change, Yellow - Orange). If Sodium Hydroxide (OH-) is added, these will mop up the H+ ions and therefore more Chromate Ions (Yellow) will be produced since the system is equalising, (we can observe this via the colour change: Orange to Yellow).


I've also found this link for all your Chromium Needs and Le Chatlier's Principle.

http://www.chemguide.co.uk/inorganic/tra...