# How to convert soluble colorless [HgI₄]²⁻ back to insoluble red HgI₂? [closed]

When a solution contains $$\ce{Hg^2+}$$ and $$\ce{I-}$$, depend on the concentration ratio of $$\ce{Hg^2+}$$ and $$\ce{I-}$$, the following reactions can happen:

$$\ce{Hg^2+ + 2 I- ⟶ \underset{insoluble red color solid}{HgI2}}$$

If there is excess $$\ce{I-}$$, then this red solid will continue to react with excess $$\ce{I-}$$ to form an colorless soluble complex:

$$\ce{HgI2 + I- ⟶ [HgI3]^-}$$

$$\ce{[HgI3]^- + I- ⟶ [HgI4]^2-}$$

Is there a way to revert $$\ce{[HgI4]^2-}$$ back to $$\ce{HgI2}$$?

• Have you any ideas? There is rather trivial solution... – Mithoron Mar 17 '19 at 22:26
• What I can think of is adding the product [hgi4]2- in the solution so that the equilibrium will be pushed to the left and hgI2 will be formed again. – HGK Mar 17 '19 at 23:47
• No, you should add something else to eliminate the excess, think about proportion of reagents here. – Mithoron Mar 18 '19 at 0:17
• I can add more Hg2+ in solution to react with the excess I-. – HGK Mar 18 '19 at 0:38
• Indeed, it should work. Technically they react with these complex anions, polymerising until solid particles precipitate AFAIK. – Mithoron Mar 18 '19 at 1:14

As noted in the comments, to attain mercuric iodide, you need to maintain a balance between the concentration of $$\ce{Hg^2+}$$ and iodide ion. If there is slight excess of iodide ion, it will form a complex (much like $$\ce{KI3}$$ from $$\ce{KI}$$). The complex is popularly known as Nessler's reagent and is prepared pretty much in this way.
Another solution is to decompose the complex to revert back into original products. But I won't recommend this solution. Though, it is expected to give the original reactants, occasionally, decomposition reactions give rise to undesired side products and may release toxic fumes of mercury, iodine, iodide, potassium oxide($$\ce{K2O)}$$ etc. Do not try to heat the complex. Instead opt for the 1st solution.