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Potassium Dichromate

When a chromi...

Question

When a chromite ore (A) is fused with sodium carbonate in free excess of air and the product is dissolved in water, a yellow solution of compound (B) is obtained. After treatment of this yellow solution with sulphuric acid, compound (C) can be crystallised from the solution. When compound (C) is treated with $K C l$ , orange crystals of compound (D) crystallise out. Identify A to D and also explain the reactions.

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Solution

$∙$ The formula of chromite ore (A) is $F e C r_{2} O_{4}$ . When chromite ore $(F e C r_{2} O_{4})$ is fused with sodium carbonate in free excess of air and the product is dissolved in water, a yellow solution of sodium chromate (B) is formed.

$∙$ The reaction is as follows:

$4 F e C r_{2} O_{4} + 8 N a_{2} C O_{3} + 7 O_{2}$
(A)
↓
$8 N a_{2} C r O_{4} + 2 F e_{2} O_{3} + 8 C O_{2}$
(B)
(Yellow Solution)

$∙$ When the yellow solution of sodium chromate (B) is treated with sulphuric acid then it gets converted to sodium dichromate (C) that is orange in colour and can be crystallized from the solution.

$2 N a_{2} C r O_{4} + H_{2} S O_{4}$
(B)
↓
$N a_{2} C r_{2} O_{7} + N a_{2} S O_{4} + H_{2} O$
(C)

$∙$ When sodium dichromate (C) is treated with $K C l$ , orange crystals of potassium dichromate (D) crystallise out due to their lower solubility.

$N a_{2} C r_{2} O_{7} + 2 K C I \to K_{2} C r_{2} O_{7} + 2 N a C I$
(C) (D)

Hence – $A : F e C r_{2} O_{4}, B : N a_{2} C r O_{4}, C : N a_{2} C r_{2} O_{7}, D : K_{2} C r_{2} O_{7}$

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