When sodium chromate is acidified with sulfuric acid and then further treated with potassium chloride, we obtain....
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Potassium dichromate
Sodium chromate ($Na_{2}CrO_{4}$) contains the chromate ion ($CrO_{4}^{2-}$). When the solution is acidified with sulfuric acid, chromate ions convert into dichromate ions.
Reaction:
$2CrO_{4}^{2-} + 2H^{+} \rightleftharpoons Cr_{2}O_{7}^{2-} + H_{2}O$
Thus, sodium chromate forms sodium dichromate ($Na_{2}Cr_{2}O_{7}$) in acidic medium.
When potassium chloride ($KCl$) is added to this solution, an ion exchange occurs. Potassium dichromate ($K_{2}Cr_{2}O_{7}$) is less soluble than sodium dichromate, so it crystallizes out from the solution.
Reaction:
$Na_{2}Cr_{2}O_{7} + 2KCl \rightarrow K_{2}Cr_{2}O_{7} + 2NaCl$
Therefore, the final product obtained is potassium dichromate.
$Na_2Cr_2O_7 + 2KCl \longrightarrow K_2Cr_2O_7 + 2NaCl$
$K_2Cr_2O_7$ (F) is a well-known oxidizing agent.