Ferric chloride can be prepared by dissolving hematite in hydrochloric acid:

$$\ce{Fe2O3 + 6HCl -> 2FeCl3 + 3H2O }\tag{1}$$

But the resulting ferric chloride will be exposed from its formation to water.

After investigating a little, I found that anhydrous ferric chloride, when dissolved in water, hydrolysed and forms ferric hydroxide:

$$\ce{FeCl3 + 3H2O -> Fe(OH)3 + 3HCl }\tag{2}$$

But isn't it supposed that, when dissolved in water, hexahydrate is formed? What really happens, hydrolyses or hydrates?


1 Answer 1


Iron(III) chloride indeed hydrolyses in water, or to be precise, $\ce{Fe^3+}$ does assuring acidic medium. Hydrolysis isn't completely suppressed even in strongly acidic solutions. In the absence of other factors, what happens exactly depends on the concentration of chloride and temperature.

In diluted cold solution a typical formation of metal aqua ions takes place:

$$ \begin{align} \ce{FeCl3 + 4 H2O &<=> [Fe(H2O)4Cl2]+ + Cl-}\tag{R1.1}\\ \ce{[Fe(H2O)4Cl2]+ + H2O &<=> [Fe(H2O)5Cl]^2+ + Cl-} &\quad \mathrm{p}K_\mathrm{a} &= 0.30\tag{R1.2}\\ \ce{[Fe(H2O)5Cl]^2+ + H2O &<=> [Fe(H2O)6]^3+ + Cl-} &\quad \mathrm{p}K_\mathrm{a} &= 1.48\tag{R1.3}\\ \ce{[Fe(H2O)5Cl]^2+ + H2O &<=> [Fe(H2O)4(OH)Cl]+ + H3O+} &\quad \mathrm{p}K_\mathrm{a} &= 3.05\tag{R1.4}\\ \end{align} $$

In concentrated cold solution aquacomplex is also accompanied by tetrachloroferrate(III):

$$\ce{2 FeCl3 + 4 H2O <=> [Fe(H2O)4Cl2)]+ + [FeCl4]-}\tag{R2}$$

In boiling-hot solution iron(III) chloride decomposes to hydroxochloride

$$\ce{FeCl3 + 2 H2O ->[Δ] Fe(OH)2Cl(s) + 2 HCl}\tag{R3}$$

which is partially hydrolyzed further to colloidal iron(III) hydroxide ($\ce{Fe(OH)2Cl}$ is still more soluble than $\ce{Fe(OH)3}$).

Finally treatment with steam results in iron(III) oxide:

$$\ce{2 FeCl3 + 3 H2O ->[\pu{350 - 500 °C}] Fe2O3 + 6 HCl}\tag{R4}$$

Reactions were adopted from [1, pp. 420–421].


  1. R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.