The metal salt A is blue in colour. When salt A is heated strongly over a burner, then a substance B is eliminated and a white powder C is left behind. When a few drops of a liquid D are added to powder C, it becomes blue again. What could be A, B, C and D?

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Solution

Copper sulphate pentahydrate $(C u S O_{4} .5 H_{2} O)$ crystals (salt A) are blue in colour; intense heating causes loss of all water (substance B) and leaves white anhydrous copper sulphate (salt C) behind.

$C u S O_{4} .5 H_{2} O A ⟶ C u S O_{4} B + 5 H_{2} O C$

Addition of a few drops of water (liquid D) turns the anhydrous copper sulphate into blue coloured copper sulphate pentahydrate.

$C u S O_{4} C + 5 H_{2} O D ⟶ C u S O_{4} .5 H_{2} O A$

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