Q.When orange solution containing $Cr_2O_7^{2-}$ ion is treated with an alkali, a yellow solution is formed and when $H^+$ ions are added to yellow solution, an orange solution is obtained. Explain why does this happen?

The key idea is the dichromate–chromate equilibrium, which depends on pH.

In acidic solution, the dichromate ion ($\text{Cr}_2\text{O}_7^{2-}$, orange) is stable. Adding an alkali (OH⁻) shifts the equilibrium toward the chromate ion ($\text{CrO}_4^{2-}$, yellow) by consuming H⁺ ions. The reverse happens when H⁺ is added back.

Step 1: Write the equilibrium:

(orange) $\quad\quad\quad\quad\quad$ (yellow) …

The key here is that dichromate and chromate are not independent ions — they are linked by a chemical equilibrium that depends on the pH of the solution. This is a classic example of how the same element in the same oxidation state can exist in different forms depending on the environment.

In acidic solution, the dominant species is the orange dichromate ion, $\text{Cr}_2\text{O}_7^{2-}$. In basic (alkaline) solution, the dominant species is the yellow chromate ion, $\text{CrO}_4^{2-}$. The interconversion is a reversible reaction that involves water and hydrogen ions.

The equilibrium is:

Let’s walk through what happens step by step.

1. Starting point: orange dichromate solution.

   You have an acidic or neutral solution containing $\text{Cr}_2\text{O}_7^{2-}$ ions. The equilibrium above lies to the left because the concentration of $\text{H}^+$ is relatively high.

2. Adding alkali (OH⁻ ions).

   The added hydroxide ions react with the $\text{H}^+$ ions on the right-hand side of the equilibrium:

This removes $\text{H}^+$ from the system. According to Le Chatelier’s principle, the equilibrium shifts to the right to produce more $\text{H}^+$ ions. As it shifts right, dichromate ions are consumed and chromate ions are produced. The solution turns yellow because $\text{CrO}_4^{2-}$ is now the dominant species.

3. Adding acid (H⁺ ions) to the yellow solution. Now you increase the concentration of $\text{H}^+$ in the solution. The equilibrium shifts to the left to consume the added $\text{H}^+$ ions. Chromate ions combine with $\text{H}^+$ to reform dichromate and water. The solution turns orange again as $\text{Cr}_2\text{O}_7^{2-}$ becomes dominant. …

Method: Le Chatelier’s Principle & Acid–Base Equilibrium Analysis

This phenomenon is explained by the dichromate–chromate equilibrium in aqueous solution, which is pH-dependent.

Step 1 – Write the equilibrium reaction

The key equilibrium is:

• Orange species: dichromate ion ($\ce{Cr2O7^{2-}}$)
• Yellow species: chromate ion ($\ce{CrO4^{2-}}$)

Step 2 – Effect of adding alkali (OH⁻)

When an alkali is added, $\ce{OH-}$ ions neutralise the $\ce{H+}$ ions on the right side:

• Removal of $\ce{H+}$ shifts the equilibrium to the right (Le Chatelier’s principle).
• More $\ce{CrO4^{2-}}$ (yellow) is formed.
• Result: solution turns yellow.

Step 3 – Effect of adding acid (H⁺)

When $\ce{H+}$ ions are added, the concentration of $\ce{H+}$ on the right side increases.

• Equilibrium shifts to the left to reduce the excess $\ce{H+}$.
• More $\ce{Cr2O7^{2-}}$ (orange) is formed. …

Here are the common mistakes students make on this question about the dichromate-chromate equilibrium, along with how to avoid each.

Mistake 1: Confusing the colour of the ions

The error: Students often write that $Cr_2O_7^{2-}$ is yellow and $CrO_4^{2-}$ is orange — the exact opposite of the truth.

Why it happens: The names "dichromate" and "chromate" sound similar, and students memorise the colours without linking them to the chemical formula.

How to avoid: Use a simple memory trick:

• Dichromate = Di = two chromium atoms = orange.
• Chromate = one chromium atom = yellow.

Also, write the formulas clearly:

• Orange: $Cr_2O_7^{2-}$ (dichromate)
• Yellow: $CrO_4^{2-}$ (chromate)

Mistake 2: Writing the wrong chemical equation

The error: Students write an unbalanced equation or forget that the reaction is reversible.

Example of wrong equation:

This is unbalanced — it misses the $H^+$ on the left.

How to avoid: Always write the balanced reversible equation:

Check that atoms and charges balance:

• Left: 2 Cr, 7 O + 2 O = 9 O, 2 H, charge = -2 + (-2) = -4
• Right: 2 Cr, 8 O + 1 O = 9 O, 2 H, charge = -4

Mistake 3: Forgetting to mention the reversible nature of the reaction

The error: Students explain only the forward reaction (alkali → yellow) but don't explain why adding acid reverses it.

How to avoid: Always state that this is an equilibrium reaction. When you add alkali ($OH^-$), it removes $H^+$ (by forming water), shifting the equilibrium to the right (yellow). When you add $H^+$, it shifts the equilibrium to the left (orange).

Write the equilibrium clearly:

Then explain:

• Alkali removes $H^+$ → equilibrium shifts right → yellow $CrO_4^{2-}$ forms.
• Acid adds $H^+$ → equilibrium shifts left → orange $Cr_2O_7^{2-}$ reforms.

Mistake 4: Not mentioning the role of water

The error: Students write the equation without water, making it seem like $OH^-$ directly reacts with $Cr_2O_7^{2-}$.

How to avoid: Understand that $OH^-$ neutralises $H^+$ to form water. The actual equilibrium involves water:

Adding $OH^-$ removes $H^+$ (forming $H_2O$), pulling the equilibrium to the right.

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