The only cations present in a slightiy acidic solution are $F e^{3 +}$ , $Z n^{2 +}$ and $C u^{2 +}$ The reagent that when added in excess to this solution would identify and separate $F e^{3 +}$ in one step is :

2 M H C l

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6 M N H_{3}

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6 M N a O H

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H_{2} S

gas

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Solution

The correct option is B $6 M N H_{3}$
With $N H_{3}$ , $Z n^{2 +}$ and $C u^{2 +}$ forms complexes which are soluble in excess reagent $N H_{3}$ , but
$F e^{3 +} (a q) + 3 N H_{3} (a q) + 3 H_{2} O \to F e (O H)_{3} (s) + 3 N H_{4}^{+} (a q)$

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Similar questions

The only cations present in a slightly acidic solution are $F e^{3 +}, Z n^{2 +}$ and $C u^{2 +}$ . The reagent that when added in excess of this solution would identify and separate $F e^{3 +}$ in one step is:

Q. The only cations present in a slightly acidic solutio are

F e^{3 +}, Z n^{2 +}

and

C u^{2 +}

. The reagent that when added in excess to this solution would identify and separate

F e^{3 +}

in one step is :

Q. The only cations present in a slightly acidic solution are

F e^{3 +}

Z n^{2 +}

and

C u^{2 +}

. The reagent which when added in excess to this solution would identify the

F e^{3 +}

ion in one step is :

The only cations present in a slightly acidic solution are $F e^{3 +}, Z n^{2 +} a n d C u^{2 +}$ . The reagent, added in excess to this solution, that would identify and separate $F e^{3 +}$ in one step is:

Q. On passing

H_{2} S

into a solution containing both

Z n^{2 -}

and

C u^{2 -}

in acidic medium, only CuS gets precipitated. This is because :

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The only cations present in a slightiy acidic solution are Fe3+, Zn2+ and Cu2+ The reagent that when added in excess to this solution would identify and separate Fe3+ in one step is :

The correct option is B 6MNH3With NH3, Zn2+ and Cu2+ forms complexes which are soluble in excess reagent NH3, butFe3+(aq)+3NH3(aq)+3H2O→Fe(OH)3(s)+3NH+4(aq)

The only cations present in a slightiy acidic solution are $F e^{3 +}$ , $Z n^{2 +}$ and $C u^{2 +}$ The reagent that when added in excess to this solution would identify and separate $F e^{3 +}$ in one step is :

The correct option is B $6 M N H_{3}$
With $N H_{3}$ , $Z n^{2 +}$ and $C u^{2 +}$ forms complexes which are soluble in excess reagent $N H_{3}$ , but
$F e^{3 +} (a q) + 3 N H_{3} (a q) + 3 H_{2} O \to F e (O H)_{3} (s) + 3 N H_{4}^{+} (a q)$