SATHEE: Qualitative Analysis - Result Question 5

Qualitative Analysis - Result Question 5

####5. Upon treatment with ammoniacal $H_{2} S$ , the metal ion that precipitates as a sulphide is

(2013 Adv.)

(a) $F e$ (III)

(b) $A l$ (III)

(d) $Z n$ (II)

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Answer:

Correct Answer: 5. (d)

Solution:

PLAN $K_{s p} (Z n S)$ is very high and $Z n^{2 +}$ is precipitated as $Z n S$ by high concentration of $S^{2 -}$ formed when $H_{2} S$ is passed in ammoniacal solution.

$\begin{aligned} H_{2} S & ⇌ Z n^{+} + S^{2 -} (I) \\ H^{+} + O H^{-} & ⇌ H_{2} O (I I) \end{aligned}$

Reaction (I) is favoured in forward side if $H^{+}$ is removed immediately by $O H^{-} (N H_{4} O H)$ .

$Z n^{2 +} + S^{2 -} ⟶ \underset{White ppt}{Z n S ↓}$

$F e^{3 +}$ and $A l^{3 +}$ are precipitated as hydroxide.

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Qualitative Analysis - Result Question 5

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