What change will occur to the equilibrium mixture $2CrO^{2-}_4(aq) + 2H^+(aq) ⇔ Cr_2O^{2-}_7(aq) + H_2O(l)$ when excess sodium hydrogen solution is added?

Analyze the question

The equilibrium is: 2CrO₄²⁻(aq) + 2H⁺(aq) ⇌ Cr₂O₇²⁻(aq) + H₂O(l)

• CrO₄²⁻ (chromate) is yellow.
• Cr₂O₇²⁻ (dichromate) is orange.

The question says "sodium hydrogen solution", which is almost certainly a typo for sodium hydroxide (NaOH) solution. I will solve it on that basis.

Concept: Le Chatelier's Principle

When a change is applied to a system at equilibrium, the system shifts to oppose that change.

Step-by-step solution

1. Adding excess NaOH supplies OH⁻ ions.
2. OH⁻ reacts with H⁺ to form water: H⁺ + OH⁻ → H₂O.
3. This removes H⁺ from the equilibrium mixture, so [H⁺] decreases.
4. To replace the lost H⁺, the equilibrium shifts to the left, the side with more H⁺.
5. Dichromate (orange) is converted to chromate (yellow): Cr₂O₇²⁻ + 2OH⁻ → 2CrO₄²⁻ + H₂O
6. The solution therefore changes from orange to yellow.

Why the other options are wrong

• A and B: Cr(OH)₃ contains chromium(III), but chromium here is in the +6 state. No reduction occurs, so no Cr(OH)₃ forms. Cr(OH)₃ is also green, not yellow or orange.
• D: This would happen if acid were added, which shifts the equilibrium right.
• E: There is a visible colour change.

**Answer: C. The colour changes from orange to yellow.

Exam tip: Remember "Acid → Orange, Base → Yellow" (Acid-Orange, Base-Yellow). Adding acid favours dichromate (orange), and adding alkali favours chromate (yellow).