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Q4.14:
Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?

Solution :

Step 1: Conversion of Chromite Ore to Sodium Chromate

The preparation begins with the roasting of finely powdered chromite ore ($\text{FeCr}_2\text{O}_4$) with sodium carbonate ($\text{Na}_2\text{CO}_3$) in the presence of excess air. This process yields yellow-colored sodium chromate ($\text{Na}_2\text{CrO}_4$).

Chemical equation: $4\text{FeCr}_2\text{O}_4 + 8\text{Na}_2\text{CO}_3 + 7\text{O}_2 \rightarrow 8\text{Na}_2\text{CrO}_4 + 2\text{Fe}_2\text{O}_3 + 8\text{CO}_2$ [Per the principle of oxidative roasting].

Step 2: Acidification to Sodium Dichromate

The yellow solution of sodium chromate is filtered and acidified with concentrated sulfuric acid ($\text{H}_2\text{SO}_4$) to convert it into orange-colored sodium dichromate ($\text{Na}_2\text{Cr}_2\text{O}_7$).

Chemical equation: $2\text{Na}_2\text{CrO}_4 + 2\text{H}^+ \rightarrow \text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{Na}^+ + \text{H}_2\text{O}$ [By acid-base equilibrium shift].

Step 3: Conversion to Potassium Dichromate

Sodium dichromate is more soluble than potassium dichromate. Therefore, the sodium dichromate solution is treated with potassium chloride ($\text{KCl}$). Potassium dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$) is obtained through fractional crystallization as orange crystals.

Chemical equation: $\text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{KCl} \rightarrow \text{K}_2\text{Cr}_2\text{O}_7 + 2\text{NaCl}$ [Based on solubility product principles].

Step 4: Effect of pH on Dichromate Equilibrium

The dichromate ion ($\text{Cr}_2\text{O}_7^{2-}$) and chromate ion ($\text{CrO}_4^{2-}$) exist in a pH-dependent equilibrium. Increasing the pH (adding base/removing $\text{H}^+$) shifts the equilibrium toward the formation of the chromate ion, changing the color from orange to yellow.

Equilibrium: $\text{Cr}_2\text{O}_7^{2-} + 2\text{OH}^- \rightleftharpoons 2\text{CrO}_4^{2-} + \text{H}_2\text{O}$

Cr₂O₇²⁻ (Orange) CrO₄²⁻ (Yellow) Increasing pH (Add OH⁻)
Species Color pH Range
$\text{Cr}_2\text{O}_7^{2-}$ Orange Acidic (Low pH)
$\text{CrO}_4^{2-}$ Yellow Basic (High pH)

Final Solution: The preparation involves the oxidation of iron chromite to sodium chromate, acidification to sodium dichromate, and double decomposition with KCl. Increasing the pH shifts the orange dichromate solution to a yellow chromate solution due to the deprotonation of the dichromate ion.


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