Answer to Question #303437 in Physical Chemistry for Chemistry

Question

You have two test tubes T1 and T2.

T1 contains 2.0 mL of 0.1 N CuCl2 solution and T2 contains 2.0 mL of 0.1 N NiCl2 solution. You add dil. HCl in both the test tubes till the solutions turns acidic. Then you pass H2S gas in both the test tubes. In which of the two test tubes precipitation will occur readily? What is the principle behind the formation of precipitate?

Answer 1

In qualitative analysis of cations of second group, H2​S is passed in presence of HCl. Due to common ion effect, ionisation of H2​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 (Ksp​). 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 (Ksp​)  which is not obtained here due to common ion effect.