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An aqueous solution of coloureless metal sulphate M gives a white precipitate with $NH_4OH$. This was soluble in excess of $NH_4OH$. On passing $H_2S$ through this solution a white ppt is formed. The metal M in the salt is

(a) Ca
(b) Ba
(c) Al
(d) Zn
Can you answer this question?
 
 

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Answer: Zn
$Zn^{2+} + 2NH_4OH \longrightarrow Zn(OH)_2$ (White ppt.)$ + 2NH_4^+$
$Zn(OH)_2 + 2NH_4OH \longrightarrow (NH_4)_2 ZnO_2$ (soluble) $+ 2H_2O$
$(NH_4)_2ZnO_2 + H_2S \longrightarrow ZnS$ {White ppt.) $ + 2NH_4OH$
answered Jan 28, 2014 by mosymeow_1
 

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