Question

$H_{2}S$ gas when passed through a solution of cations containing $HCl$ precipitaes 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

The correct option is A Presence of $HCl$ decreases the sulphide ion concentration

In qualitative analysis of cations of second group, $H_{2}S$ is passed in presence of $HCl$. Due to common ion effect, ionisation of $H_{2}S$ decreases and less sulphide ions are obtained. These sulphide ions are sufficient for the precipitation of second group cations in form of their sulphides due to lower value of their solubility product $\left(K_{sp}\right)$. Here, fourth group cations are not precipitated because they required for exceeding their ionic product to their solubility products and highers sulphide ions concentration due to their higher $\left(K_{sp}\right)$ which is not obtained here due to common ion effect.