Equilibrium - Result Question 86
####90. $H_2 S$ gas when passed through a solution of cations containing $HCl$, precipitates the cations of second group of qualitative analysis but not those belonging to the fourth group. It is because
[2005]
(a) presence of $HCl$ 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 $HCl$.
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Solution:
- (a) $IV^{\text{th }}$ group needs higher $S^{2-}$ ion concentration.
In presence of $HCl$, the dissociation of $H_2 S$ decreases hence produces less amount of sulphide ions due to common ion effect, thus $HCl$ decreases the solubility of $H_2 S$ which is sufficient to precipitate $II^{\text{nd }}$ group radicals.
