state why Zn(OH)2 is soluble in excess of NH4OH.
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Hi friend,
The Zincate is formed , as follows :-
Zn(OH)2 + 2OH-( from the NaOH ) =[ Zn(OH)4 ]2-
more about zinc hydroxide including that reaction.....i hope it helps
The Zincate is formed , as follows :-
Zn(OH)2 + 2OH-( from the NaOH ) =[ Zn(OH)4 ]2-
more about zinc hydroxide including that reaction.....i hope it helps
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