Question

How to prepare stannous chloride in HCL for estimation of Ferric ion?

Answer 1

I cannot tell because I am in 8th class

Answer 2

Tin(II) chloride can dissolve in less than its own mass of water without apparent decomposition, but as the solution is diluted, hydrolysis occurs to form an insoluble basic salt:

SnCl2 (aq) + H2O (l) ⇌ Sn(OH)Cl (s) + HCl (aq)

Therefore, if clear solutions of tin(II) chloride are to be used, it must be dissolved in hydrochloric acid (typically of the same or greater molarity as the stannous chloride) to maintain the equilibrium towards the left-hand side (using Le Chatelier's principle). Solutions of SnCl2 are also unstable towards oxidation by the air:

6 SnCl2 (aq) + O2 (g) + 2 H2O (l) → 2 SnCl4 (aq) + 4 Sn(OH)Cl (s)

This can be prevented by storing the solution over lumps of tin metal.[3]

There are many such cases where tin(II) chloride acts as a reducing agent, reducing silver and gold salts to the metal, and iron(III) salts to iron(II), for example:

SnCl2 (aq) + 2 FeCl3 (aq) → SnCl4 (aq) + 2 FeCl2 (aq)

It also reduces copper(II) to copper(I).

Solutions of tin(II) chloride can also serve simply as a source of Sn2+ ions, which can form other tin(II) compounds via precipitation reactions. For example, reaction with sodium sulfide produces the brown/black tin(II) sulfide:

SnCl2 (aq) + Na2S (aq) → SnS (s) + 2 NaCl (aq)