Q9 / 125 · BITSAT 2011 · Chemistry

$\mathrm{H}_{2} \mathrm{~S}$ gas when passed through a solution of cations containing $\mathrm{HCl}$ precipitates the cations of second group of qualitative analysis but not those belonging to the fourth group. It is because

  • Correct answer
    A
    presence of HC1 decreases the sulphide ion concentration.
  • B
    solubility product of group II sulphides is more than that of group IV sulphides.
  • C
    presence of HCl increases the sulphide ion concentration.
  • D
    sulphides of group IV cations are unstable in $\mathrm{HC} 1$

Solution

IV $^{\text {th }}$ group needs higher $\mathrm{S}^{2-}$ ion concentration. In presence of $\mathrm{HCl}$, the dissociation of $\mathrm{H}_{2} \mathrm{~S}$ decreases hence produces less amount of sulphide ions due to common ion effect, thus $\mathrm{HCl}$ decreases the solubility of $\mathrm{H}_{2} \mathrm{~S}$ which is sufficient to precipitate II $^{\text {nd }}$ group radicals.

Hence, the correct answer is option: 'A'.

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