Question:
During the preparation of \(\mathrm{K_2Cr_2O_7}\) from chromite ore, in one of the steps, the yellow solution of sodium chromate is converted into orange sodium dichromate crystals. This is achieved by
During the preparation of \(\mathrm{K_2Cr_2O_7}\) from chromite ore, in one of the steps, the yellow solution of sodium chromate is converted into orange sodium dichromate crystals. This is achieved by
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Remember the chromate--dichromate equilibrium: \(\displaystyle 2\mathrm{CrO_4^{2-}}+2\mathrm{H^+}\rightleftharpoons\mathrm{Cr_2O_7^{2-}}+\mathrm{H_2O}\).
- Basic medium \(\rightarrow\) Yellow chromate (\(\mathrm{CrO_4^{2-}}\)).
- Acidic medium \(\rightarrow\) Orange dichromate (\(\mathrm{Cr_2O_7^{2-}}\)).
Updated On: Jul 9, 2026
- Increasing the pH
- Decreasing the pH
- Maintaining neutral pH
- Adding NaCl \bigskip
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The Correct Option is B
Solution and Explanation
Step 1: Identify the species.
Yellow sodium chromate contains the chromate ion,
\[
\mathrm{CrO_4^{2-}}
\]
whereas orange sodium dichromate contains the dichromate ion,
\[
\mathrm{Cr_2O_7^{2-}}.
\]
Step 2: Effect of pH. On acidification (decreasing pH), \[ 2\mathrm{CrO_4^{2-}}+2\mathrm{H^+} \rightleftharpoons \mathrm{Cr_2O_7^{2-}}+\mathrm{H_2O} \] Thus, lowering the pH shifts the equilibrium towards dichromate ions, changing the solution from yellow to orange.
Step 3: Final conclusion. Hence, sodium chromate is converted into sodium dichromate by \[ \boxed{\text{decreasing the pH}.} \] Therefore, the correct option is \(\boxed{(B)}\).
Step 2: Effect of pH. On acidification (decreasing pH), \[ 2\mathrm{CrO_4^{2-}}+2\mathrm{H^+} \rightleftharpoons \mathrm{Cr_2O_7^{2-}}+\mathrm{H_2O} \] Thus, lowering the pH shifts the equilibrium towards dichromate ions, changing the solution from yellow to orange.
Step 3: Final conclusion. Hence, sodium chromate is converted into sodium dichromate by \[ \boxed{\text{decreasing the pH}.} \] Therefore, the correct option is \(\boxed{(B)}\).
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