Answer:
The colour of the orange solution becomes yellow.
Explanation:
1. Before adding NaOH
Assume the picture showed a beaker of potassium chromate and one of potassium dichromate.
Both solutions are involved in the same equilibrium:
[tex]\rm\underbrace{\hbox{2CrO$_{4}^{2-}$(aq)}}_{\text{yellow}} +2H^{+}(aq) \rightleftharpoons \, \underbrace{\hbox{Cr$_{2}$O$_{7}^{2-}$}}_{\text{orange}} + H_{2}O[/tex]
The first beaker contains mostly chromate ions with a few dichromate ions.
The position of equilibrium lies to the left and the solution is yellow.
The second beaker contains mostly dichromate ions with a few chromate ions.
The position of equilibrium lies to the right and the solution is orange.
2. After adding NaOH
According to Le Châtelier's Principle, when we apply a stress to a system at equilibrium, the system will respond in a way that tends to relieve the stress.
Beaker 1
If you add OH⁻ to the equilibrium solution, it removes the H⁺ (by forming water).
The system responds by having the dichromate react with water to replace the H⁺.
At the same time, the system forms more of the yellow chromate ion.
The position of equilibrium shifts to the left.
However, the solution is already yellow, so you see no change in colour.
Beaker 2
The reaction is the same as in Beaker 1.
This time, however, as the dichromate ion disappears, do does its orange colour.
Also, the yellow chromate is being formed and its yellow colour appears .
The colour changes from orange to yellow.
