The best way to separate the'metal ions from a mixture of Cu(II) chloride and Fe(II) sulphate
The best way to separate the'metal ions from a mixture of Cu(II) chloride and Fe(II) sulphate in qualitative analysis is by treating it with
(a) H2S in mild acidic medium, where only Cu(II) sulphide will be precipitated
(b) NH4OH buffer, where only Fe (II) hydroxide will be precipitated
(c) H2S in mild acidic medium, where only Fe
(II) sulphide will be precipitated
(d) NH4OH buffer, where only Cu (II) hydroxide will be precipitated
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Correct Option (a) H2S in mild acidic medium, where only Cu(II) sulphide will be precipitated
Explanation:
Cu2+ is a second group radical and has lower solubility product than Fe2+. Thus, Cu2+ gets precipitated first.
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