Chemistry - Ionic Equilibrium Question with Solution | TestHub

In qualitative analysis of cation of second group, ${\mathrm{H}}_2\mathrm{S}$ gas is passed in presence of $\mathrm{HCl}$, sulphide ions obtained is sufficient for the precipitation of second group cations in form of their sulphides due to lower value of their solubility product $\left({\mathrm{K}}_{\mathrm{sp}}\right)$. Hence, 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 ${\mathrm{K}}_{\mathrm{sp}}$ which is not obtained here due to common ion effect. common ion Caution $\mathrm{HCl}$ is added before ${\mathrm{H}}_2\mathrm{S}$ gas during qualitative analysis of second group radicals because the concentration of sulphide ions is suppressed by $\mathrm{HCl}$ due to common ion effect. The already present ${\mathrm{H}}^{+}$ions suppress the dissociation of ${\mathrm{H}}_2\mathrm{S}$ by moving the reaction in the backward reaction.