Chemistry - Surface Chemistry Question with Solution | TestHub

It is obtained by treating gold (III) chloride solution with$\mathrm{S}\mathrm{n}\mathrm{C}{\mathrm{l}}_2$.

$2\mathrm{A}\mathrm{u}\mathrm{C}{\mathrm{l}}_3+\mathrm{S}\mathrm{n}\mathrm{C}{\mathrm{l}}_2\to 2\mathrm{A}\mathrm{u}+3\mathrm{S}\mathrm{n}\mathrm{C}{\mathrm{l}}_4$

$\mathrm{S}\mathrm{n}\mathrm{C}{\mathrm{l}}_4$so formed hydrolyses as follows

$\mathrm{S}\mathrm{n}\mathrm{C}{\mathrm{l}}_4+4\mathrm{H}\mathrm{O}\mathrm{H}\to \underset{\text{Stannic acid}}{\mathrm{S}\mathrm{n}(\mathrm{O}\mathrm{H}{)}_4}+4\mathrm{H}\mathrm{C}\mathrm{l}$

The colloidal precipitate of stannic acid absorbs colloidal particles of gold. This has beautiful purple colour and is known as purple of cassius (because it is discovered by Cassius in 1865). It is used for colouring glass (ruby red and pottery).