The Chromate - Dichromate Equilibrium

What happens if we add some sodium hydroxide?

For simplicity, the symbol H+ is being used for the acid proton, rather than the Brönsted-Lowry hydronium ion H3O+.   All ions in these equations are aqueous, but the (aq) symbol has been omitted for brevity.

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+   Since the H+ is colorless, we wouldn't see any color change.   But there is going to be a chemical reaction.  From le Châtelier'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 Cr2O72- back into the CrO42-.
 
For the chromate solution (remember that the original solution is an equilibrium, so it contains mostly yellow chromate, but a bit of orange dichromate):

add some OH- (aq) 
and it becomes		 a slightly more yellow solution, and probably not noticeable
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.  Because there already was a lot of yellow dichromate, the change is not very noticeable.

For the dichromate solution (remember that the original solution is also an equilibrium, so it contains mostly orange dichromate, but a bit of yellow chromate):

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.  Because there is now a lot more yellow chromate, and a lot less orange dichromate, the change in color is very noticeable.