Why do we have to prevent the hydrolysis of iron(III) nitrate?

Question

I want to make particular concentration of ferric ions from $\ce{Fe(NO3)3.9H2O}$, and then I found this video

It says that we have to add nitric acid to prevent iron from hydrolysis, what does it mean? Why do we have to prevent it? If not adding it, does it affect our concentrate solution later?

Answer 1

If not adding it (nitric acid), does it affect our concentrate solution later?

Yes it does because slow hydrolysis with water, some $\ce{Fe^3+}$ would separate out from the solution as solid particles (Ref.1). This review focuses on studies on the hydrolysis and precipitation of $\ce{Fe^3+}$ from aqueous solutions of its inorganic salts, chiefly the nitrate, perchlorate, chloride, and sulfate. Note that the $\ce{Fe^3+}$ ion has been shown to exist as the octahedral $\ce{[Fe(OH2)6]^3+}$ complex in aqueous solutions of $\ce{Fe(NO3)3.9H2O}$ (Ref.2).

The hydrolysis reactions of these inorganic salt solutions, most of which occur slowly, consists of several steps: (1) formation of low-molecular-weight species; (2) formation of a red cationic polymer; (3) aging of the polymer, with eventual conversion to oxide phases; and (4) precipitation of oxide phases directly from low-molecular-weight precursors.

The characterized crystalline iron(III) oxides and hydrous oxides are $\ce{Fe2O3}$ and $\ce{FeO(OH)}$, each of which is polymorphic and very insoluable in water. For example, the $\ce{\alpha-Fe2O3}$ (hematite) has a $\mathrm{p}K_\mathrm{sp}$ of $41.7$ while that of $\alpha-, \beta-,$ and $\gamma-$ phases of $\ce{FeO(OH)}$ ((goethite, akaganeite, and lepidocrocite) are $41.7, 36,$ and $38.7$, respectively. Thus, one can expect after preparing in aqueous solutions, $\ce{Fe^3+}$ can lost significant amount to hydrolysis.

The first step of the hydrolysis of inorganic Fe(III) solutions is the formation of low-molecular-weight species. Two such reactions are given in Ref.1 as: $$\ce{Fe(OH2)6]^3+ + H2O <=> Fe(OH2)5(OH)]^2+ + H3O+} \qquad K_\mathrm{eq} = 2.2 \tag1$$ $$\ce{Fe(OH2)5(OH)]^2+ + H2O <=> Fe(OH2)4(OH)2]^+ + H3O+} \qquad K_\mathrm{eq} = 3.5 \tag2$$

These formation reactions of low-molecular-weight species can be revised by adding appropriate strong acid (here it is nitric acid since we used $\ce{Fe(NO3)3}$) before they react further to make polymeric iron salts. Note that Hydrolyzed Fe(III) solutions containing the red polymer are not at equilibrium (Ref.1).

The polymer separated from hydrolysis has been studied to find the rate of their degradation by $\ce{HNO3}$, which results in immediate precipitation of an undetermined fraction of the iron. Thus, it should be advised that addition of the nitric acid should be immediate after the preparation of iron solution.

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References:

1. Charles M. Flynn Jr., "Hydrolysis of inorganic iron (III) salts," Chem. Rev. 1984, 84(1), 31-41 (DOI: https://doi.org/10.1021/cr00059a003).
2. Neil J. Hair and James K. Beattie, "Structure of hexaaquairon(III) nitrate trihydrate. Comparison of iron(II) and iron(III) bond lengths in high-spin octahedral environments," Inorg. Chem. 1977, 16(2), 245-250 (DOI: https://doi.org/10.1021/ic50168a006).
3. Barbara A. Sommer, Dale W. Margerum, John Renner, Paul Saltman, and Thomas G. Spiro, "Reactivity and aging in hydroxy-iron(III) polymers, analogs of ferritin cores," Bioinorganic Chemistry 1973, 2(4), 295-309 (DOI: https://doi.org/10.1016/S0006-3061(00)80202-1).

Answer 2

The Gist

As mentioned in the comments, we have to prevent the formation of aqua complexes of $\ce{Fe^{3+}}$ as these tend to hydrolyze:

$$\begin{multline} \ce{[Fe(H2O)_{2n}(NO3)_{3-n}]^{n+}.(NO3^-)_n.(H2O)_{9-2n} + H2O <=>} \\ \ce{[Fe(H2O)_{2n-1}(OH)(NO3)_{3-n}]^{n+}.(NO3^-)_n.(H2O)_{9-2n} + H3O+} \end{multline}$$

Note: In the above equilibrium, it was assumed that the denticity of $\ce{NO3^-}$ is 2. This may or may not be the case. I have avoided this ambiguity in the similar equation in the Remarks section.

Now, to prevent formation of aqua complexes, there are (at least) two ways to do this:

1. Decrease the pH to shift the above equilibrium to the reactants side in accordance with Le Chatelier's Principle.
2. Increase the concentration of $\ce{NO3^-}$ in the solution to prevent formation of aqua complexes in accordance with Le Chatelier's Principle, as explained in the section Hydration Isomerism of $\ce{Fe(NO3)3}$ below.

Your suggestion to add $\ce{HNO3}$ does both simultaneously, decreasing the pH and increasing the concentration of $\ce{NO3^-}$.

Hydration Isomerism

For a hydrated complex $\ce{[ML_y].xH2O}$, generally, assuming maximum valency of $\ce{M}$ is $y$, $y+1$ hydration isomers are possible given by the formulae:

$$ \ce{[ML_{y-n}(H2O)_n].L_n.($x-n$)H2O} %nasty workaround, possibly mhchem bug $$

Note: This formula is most accurate for crystalline ($\ce{c}$) phase and the behavior is more complex in aqueous ($\ce{aq}$); however, for simplicity, we will consider this formula for our analysis.

Example

Consider the example of $\ce{[MX2].2H2O}$ where $\ce{M}$ has a maximum valency of 2:

$$ \ce{\underset{\text{isomer 1}}{[MX2].2H2O} <=> \underset{\text{isomer 2}}{[MX(H2O)].X^-.H2O} <=> \underset{\text{isomer 3}}{[M(H2O)2].(X^-)2}} $$

Hydration Isomerism of $\ce{Fe(NO3)3}$

$\ce{Fe(NO3)3}$ has several hydration isomers and is commonly represented as $\ce{[Fe(NO3)3].xH2O}$, where $x$ is usually 9, which we will consider.

The usual coordination number and geometry of $\ce{Fe}$ are $6$ and octahedral, respectively, commonly forming six-coordinated complexes.

Thus, in $\ce{aq}$ phase, you are likely to observe (assuming that the denticity of $\ce{NO3^-}$ maybe $\pu{1}$ or $\pu{2}$), among many other isomers, the following complexes.

1. $\ce{[Fe(H2O)_6].(NO3^-)3.3H2O}$,
2. $\ce{[Fe(H2O)_4(NO3)_2].NO3^-.5H_2O}$, and
3. (very) low amounts of $\ce{[Fe(NO3)3].9H2O}$.

Note: These formulae are not quite accurate, as this Wikipedia article states.

This happens because, in the spectrochemical series, $\ce{H2O}$ is a stronger field ligand than $\ce{NO3^-}$, forming more stable complexes with $\ce{Fe^3+}$. We could represent this as:

$$ \ce{\underset{\text{isomer 1}}{[Fe(NO3)3].9H2O} <=> \underset{\text{isomer 2}}{[Fe(H2O)_4(NO3)_2].NO3^-.5H_2O} <=> \underset{\text{isomer 3}}{[Fe(H2O)_6].(NO3^-)3}} $$

where, in aqueous phase, isomer 3 (hydrolyzed and otherwise) is present in abundance, and isomer 1 rarely observed. We don't want aqua complexes such as isomer 2 and isomer 3 as, reasoned in The Gist section, these tend to hydrolyze:

$$\begin{multline} \ce{[Fe(H2O)_x(NO3)_y]^{n+}.(NO3^-)_n.(H2O)_{9-x} + H2O <=> }\\ \ce{[Fe(H2O)_{x-1}(OH)(NO3)_{3-n}]^{n+}.(NO3^-)_n.(H2O)_{9-x} + H3O+} \end{multline} $$

Shifting the Equilibrium

As pointed out in other answer, increasing the amount of $\ce{NO3^-}$ in the solution will, in accordance with Le Chatelier principle, shift the equilibrium towards the right. You are essentially carrying out the following reaction:

$$ \ce{\underset{will hydrolyze}{[Fe(H2O)_6].(NO3^-)3.(H2O)3(aq)} ->[\ce{H+NO3^-}][high concentration] \underset{won't hydrolyze}{[Fe(NO3)3]\cdot 9H2O}(aq)} $$

Remarks

Denticity of Nitrate Anion

Denticity of the nitrate anion maybe 1 or 2. Therefore, in:

$$ \ce{[Fe(H2O)_x(NO3)_y]^{n+}.(NO3^-)_n.(H2O)_{9-x}} $$

$x+y \in [3,6]; x>0, y>0$.

Two-Forked Action of Addition of Nitric Acid

Addition of $\ce{HNO3}$ is a two-forked strategy, decreasing the pH and increasing the concentration of $\ce{NO3^-}$, both avoiding formation of aqua complexes of ferric cation.

Answer 3

The iron becomes bound to hydroxyl groups in water.

1. Iron(III) nitrate dissolves in water as ions: $\ce{Fe(NO3)3 <=> Fe^{3+} + 3NO3-}$
2. The ferric ions forms a complex with water: $\ce{Fe^{3+} + 6H2O <=> Fe(H2O)6^{3+}}$
3. The iron(III) pentaaquahydroxy complex ion forms: $\ce{Fe(H2O)6^{3+} + H2O <=> [Fe(H2O)5(OH)]^{2+} + H+}$

Adding $\ce{HNO3}$ pushes the equilibrium back towards $\ce{Fe^{3+} + 3NO3-}$.

BTW, heating $\ce{Fe(NO3)3}$ solution to dryness drives off the nitric acid, leaving iron oxide.