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Stoichiometry

FeSO4 solutio...

Question

$F e S O_{4}$ solution mixed with $(N H_{4})_{2} S O_{4}$ solution in $1 : 1$ molar ratio gives the test of $F e^{2 +}$ ion but $C u S O_{4}$ solution mixed with aqueous ammonia in $1 : 4$ molar ratio does not give the test of $C u^{2 +}$ ion. Explain why?

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Solution

Ferrous sulphate and ammonium sulphate forms Mohr's salt (a double salt) which gives free ferrous ions in solution.
Copper forms complex with ammonia. Free cupric ions are not present in solution.
$F e S O_{4} + (N H_{4})_{2} S O_{4} + 6 H_{2} O \to F e S O_{4} (N H_{4})_{2} S O_{4} .6 H_{2} O (M o h r^{'} s s a l t) C u S O_{4} + 4 N H_{3} + 5 H_{2} O \to [C u (N H_{3})_{4} S O_{4}] 5 H_{2} O$

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