# Na2CO3 + FeBr3 + H2O: why do we get Fe(OH)3?

$$\ce{3Na2CO3 + 2FeBr3 + 3H2O -> 6NaBr + 2Fe(OH)3 + 3CO2}$$

Why $\ce{Fe(OH)3}$? Where do these $\ce{-OH}$ groups come from?

This is from a series of reactions in a Russian sample exam question.

P.S.

Following Ivan't hints, I've gone through the process of reaction, and I get some water in the right-hand side of the equation. I wonder why the answer has no water in the right-hand side.

• From the reaction of the carbonate with water. – bon May 23 '16 at 8:50
• Carbonic acid is a weak acid, $\ce{Fe(OH)3}$ is a weak base, so their salt hydrolyzes completely. – Ivan Neretin May 23 '16 at 8:52
• @IvanNeretin - but isn't $\ce{Fe(OH)3}$ a salt? And I've just read that $\ce{Na2CO3}$ should only hydrolyze as far as $\ce{NaHCO3}$ without heating. – CowperKettle May 23 '16 at 9:01
• You might have to review your definition of salt. As for $\ce{Na2CO3}$, it is a salt of a weak acid and a strong base, therefore it hydrolyzes only partially. – Ivan Neretin May 23 '16 at 9:03
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You have certainly already recognized $\ce{Na2CO3}$ as the base in your equation.
When $\ce{FeBr3}$ is dissolved in water, the hexaaqua complex $\ce{[Fe(H2O)6]^3+}$ is formed. This complex is fairly acidic: $$\ce{[Fe(H2O)6]^3+ + H2O<=> [Fe(H2O)5OH]^2+ + H3O+}$$
Ignoring ligand exchange processes, you might state that the $\ce{OH}$ groups come from the deprotonation of the water ligands of your initial complex.
• I see. $\ce{Na2CO3}$ creates a basic environment, because CO3 binds protons tightly and leaves a lot of OH(-) groups floating about. These groups deprotonate three of the H2O molecules from the $\ce{[Fe(H2O)6]^3+}$ complex, and we get $\ce{Fe(OH)3}$. Now it's much more clear, thank you! – CowperKettle May 23 '16 at 17:10