You visited us 2 times! Enjoying our articles? Unlock Full Access!

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?

Open in App

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.
$F e S O_{4} + (N H_{4})_{2} S O_{4} + 6 H_{2} O \to F e S O_{4} (N H_{4})_{2} S O_{4} .6 H_{2} O (M o h r^{'} s s a l t) C u S O_{4} + 4 N H_{3} + 5 H_{2} O \to [C u (N H_{3})_{4} S O_{4}] 5 H_{2} O$

Suggest Corrections

Similar questions

F e S O_{4}

F e^{2 +}

S O_{4}^{2 -}

F e^{2 +}

2L of 0.5N NaOH solution is mixed with 1L of 1M KOH solution. The molarity of hydroxide ions in the resulting solution will be?

N O_{3}^{-}

2

3

A g N O_{3}

3

1

B a C l_{2}

F e (C N)_{2}

F e^{2 +}

C N^{-}

K_{2} S O_{4}

A l_{2} (S O_{4})_{3}

K^{+}, A l^{3 +}

S O_{4}^{2 -}

F e^{2 +}

C N^{-}

Join BYJU'S Learning Program