Question

$H_{2}S$ 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:

• A. presence of $HCl$ decreases the sulphide ion concentration
• B. presence of $HCl$ increases the sulphide ion concentration
• C. solubility product of group II sulphides is more than that of group IV sulphides
• D. sulphides of group IV cations are unstable in $HCl$

Answer 1

Answer: (A)

presence of $HCl$ decreases the sulphide ion concentration

Answer 2

In qualitative analysis of cation of second group $H _{2} S$ gas is passed in presence of $HCI$, therefore due to common ion effect, lower concentration of sulphide ions is obtained which is sufficient for the precipitation of second group cations in form of their sulphides due to lower value of their solubility product $\left(K_{s p}\right) .$ Here, fourth group cations are not precipitated because they require for exceeding their ionic product to their solubility products and higher sulphide ions concentration due to their higher $K_{s p}$ which is not obtained here due to common ion effect.