Question

In presence of HCl, $H_{2}S$ results the precipitation of Group – 2 elements but no Gp – 4 elements during quantitative analysis. It is due to

• A. Higher concentration of $S^{2-}$
• B. Higher concentration of $H^{+}$
• C. Lower concentration of $S^{2-}$
• D. Lower concentration of $H^{+}$

Answer 1

The correct option is C

Lower concentration of $S^{2-}$

In HCl, the dissociation $H_{2}S$ is suppressed. Hence it will have lower concentration of $S^{2-}$ ions, which results in the precipatation of those sulphides having very low solubility product.