The Chromate - Dichromate Equilibrium

What happens if we add some sodium hydroxide?

One of the simplest ways to think about the effect of adding sodium hydroxide is to realize that it removes H⁺ ions because of this reaction: H⁺  (aq) +  OH^-(aq)  H₂O (l)

So sodium hydroxide removes H⁺ ions.  If there was no chemical reaction then adding NaOH should decrease the amount of H⁺(aq).  The original  H⁺in the water would decrease to become H⁺

If we removed H⁺ from the chromate solution and there was no chemical reaction:

2 CrO₄^2- + 2 H⁺   Cr₂O₇^2- + H₂O (l)
would become
2 CrO₄^2- + 2H⁺  Cr₂O₇^2- + H₂O(l)

If we removed  H⁺ from the dichromate solution and there was no chemical reaction:

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

Since the H⁺ is colorless, we wouldn't see any color change.   But there is going to be a chemical reaction.  From Le Chatelier's principle, we know the reaction will try to restore some of the H⁺  we have removed.  The required H⁺ comes from turning some of the Cr₂O₇^2- back into the CrO₄^2-.
 
For the chromate solution:
 

	add some OH- (aq)  and it becomes		a slightly more yellow solution
Original condition 2 CrO42-(aq) + 2 H+ (aq)  Cr2O72-(aq) + H2O (l) Immediately after some OH- is added 2 CrO42-(aq)  + 2 H+(aq)  Cr2O72-(aq) + H2O(l) New equilibrium is established 2 CrO42-(aq) + 2 H+(aq)  Cr2O72-(aq) + H2O(l) Notice, that the H+ is greater than it would be if there was no reaction, but less than it was at the beginning.

  For the dichromate solution:
 

	add some OH- (aq)  and it becomes		a much more yellow solution
Original condition CrO42-(aq) + 2 H+ (aq)  Cr2O72-(aq) + H2O(l) Immediately after some OH- is added CrO42-(aq) +2 H+(aq)  Cr2O72-(aq) + H2O(l) New equilibrium is established CrO42-(aq) + 2 H+ (aq)  Cr2O72-(aq) + H2O(l) Notice, that the H+ is greater than it would be if there was no reaction, but less than it was at the beginning.