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JEE Main
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Chemistry
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Practical Chemistry

JEE Main · Chemistry · Practical Chemistry

In the qualitative analysis of Group II cations, hydrogen sulfide (...

In the qualitative analysis of Group II cations, hydrogen sulfide (H2S\mathrm{H_2S}) gas is passed through the original solution in the presence of dilute hydrochloric acid (HCl\mathrm{HCl}). The primary chemical reason for adding dilute HCl\mathrm{HCl} before passing H2S\mathrm{H_2S} gas is to :

Options

  1. Aincrease the concentration of S2−\mathrm{S}^{2-} ions for faster precipitation
  2. Bincrease the concentration of Cl−\mathrm{Cl}^{-} ions to form soluble chloride complexes
  3. Cdissolve any interfering phosphate ions present in the mixture
  4. Ddecrease the concentration of S2−\mathrm{S}^{2-} ions by common ion effect

Correct answer

D. decrease the concentration of S2−\mathrm{S}^{2-} ions by common ion effect

Step-by-step solution

HCl\mathrm{HCl} is a strong acid that completely dissociates to give H+\mathrm{H}^{+} and Cl−\mathrm{Cl}^{-} ions.

HCl→H++Cl−\mathrm{HCl} \rightarrow \mathrm{H}^{+} + \mathrm{Cl}^{-}

Hydrogen sulfide (H2S\mathrm{H_2S}) is a weak diprotic acid that ionizes as:

H2S⇌2H++S2−\mathrm{H_2S} \rightleftharpoons 2\mathrm{H}^{+} + \mathrm{S}^{2-}

The presence of H+\mathrm{H}^{+} ions from HCl\mathrm{HCl} suppresses the ionization of H2S\mathrm{H_2S} due to the common ion effect. This significantly decreases the concentration of S2−\mathrm{S}^{2-} ions in the solution. The reduced [S2−][\mathrm{S}^{2-}] is sufficient to exceed the extremely low solubility product (KspK_{\mathrm{sp}}) of Group II sulfides, but it is too low to precipitate Group IV sulfides, thereby ensuring selective precipitation.

Answer: decrease the concentration of S2−\mathrm{S}^{2-} ions by common ion effect
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