# Dissolution of iron(III) hydroxide

### Problem

From the theoretical tour of the Mendeleev Olympiad in Chemistry:

a) Calculate the minimal amount of sulfuric acid that should be diluted with water to yield $$\pu{100 mL}$$ of solution that could completely dissolve $$\pu{0.4 g}$$ of rust.

a) After the dissolution of a $$\pu{0.4 g}$$ portion of rust, the total concentration of iron(III) in a $$\pu{100 mL}$$ portion of solution is $$\displaystyle\frac{\pu{0.4 g}}{\pu{107 g mol-1}\times\pu{0.1 L}} = \pu{0.037 M}.$$ As there are no competing reactions (sulfate ion does not form stable complexes with iron), it is easy to find the minimal solution acidity after the dissolution using the solubility product $$K_\mathrm{s} = [\ce{Fe^3+}][\ce{OH-}]^3$$: $$[\ce{H+}] = K_\mathrm{w}\left(\frac{[\ce{Fe^3+}]}{K_\mathrm{s}}\right)^{1/3} = \pu{0.0097 M},$$ that is $$\pu{0.49 mmol}$$ of sulfuric acid in a $$\pu{100 mL}$$ solution. In addition, $$\pu{0.037 M}\times 3\times\pu{0.1 L}\times 1000/2 = \pu{5.6 mmol}$$ of sulfuric acid has reacted with iron hydroxide, therefore, the total amount of sulfuric acid required for the dissolution is $$\pu{6.1 mol}$$ (2 points).

### My thoughts

This task seemed incredibly easy, but the solution overcomplicated some aspects, from my point of view.

The concept of "rust" was simplified in this problem since it was stated from the beginning that here "rust" means, in fact, $$\ce{Fe(OH)3}.$$ So, I wrote the chemical equation (which, because it is a neutralization reaction, should have an astonishingly large constant and is an equilibrium practically shifted fully towards the products):

$$\ce{2 Fe(OH)3 + 3 H2SO4 -> Fe2(SO4)3 + 3 H2O}$$

$$\ce{Fe(OH)3}$$ has a molar mass of $$\pu{106.87 g/mol}$$. Finally, the amount of sulfuric acid is

$$n(\ce{H2SO4}) = \frac{\pu{0.4 g}}{\pu{106.87 g mol-1}}\times\frac{3}{2} ≈ \pu{5.61 mmol}.$$

After checking the solutions, not only I realized I were wrong (even though my final answer was part of theirs), but I also were confused by the explanations given. For some reason, the solubility product was included, even though its constant is so insignificantly small in comparison to the constant associated with the reaction I wrote above.

What I thought was happening in this system was that $$\ce{H2SO4}$$ would neutralize the rust and that was basically it. While I am familiar with the concept of equilibrium and I understand that $$\ce{H2SO4}$$ doesn't nullify the concentration of $$\ce{Fe(OH)3},$$ in the light of the huge constant the neutralization is expected to have, the solubility product, $$K_\mathrm{sp},$$ should be negligible.

However, it seems that not only it is not negligible, but it also seems to be able to influence the final result significantly. But why is that? Why do we bother calculating the additional acidity from the hydroxide ions resulted in the solubility reaction? Why can't we solve the problem stoichiometrically, the way an inorganic chemist would approach it? Please correct my reasoning wherever you consider it wrong.

• Even this query has a down vote. Someone must be really sick and disturbed to read this question. – M. Farooq Jul 17 '20 at 20:58

The $$K_\mathrm{sp}$$ of $$\ce{Fe(OH)3}$$ is varied from source to source, but a reliable source gives the value of $$2.79 \times 10^{-39}$$ at $$\pu{25 ^\circ C}$$. We'll use this value throughout the calculations.

Suppose, $$s$$ amount of $$\ce{Fe(OH)3}$$ dissolves some in water according to its $$K_\mathrm{sp}$$, but assume water is not ionized:

$$\ce{Fe(OH)3(s) <=>[H2O] Fe^3+(aq) + 3OH-(aq)}$$

$$K_\mathrm{sp} = s \times (3s)^3 = 27s^4 \ \Rightarrow \ \therefore \ s = \left(\frac{2.79 \times 10^{-39}}{27}\right)^{\frac 14} = 1.008 \times 10^{-10}$$

This is $$[\ce{Fe^3+}]$$ in water. Thus, $$[\ce{OH-}]$$ would be $$3s = 3.024 \times 10^{-10}$$. This value is much smaller than autoionization values of water. Therefore we have to consider $$[\ce{OH-}] = 1.00 \times 10^{-7}$$ for this calculation. Hence, if revise the calculation accordingly:

$$K_\mathrm{sp} = s \times [\ce{OH-}]^3 = (1.00 \times 10^{-7})^3s \ \Rightarrow \ \therefore \ s = \left(\frac{2.79 \times 10^{-39}}{1.00 \times 10^{-21}}\right) = 2.79 \times 10^{-18}$$

Thus, when autoionization of water is considered, $$[\ce{Fe^3+}]$$ in water at $$\mathrm{pH} \ 7$$ is much smaller (the actual value). I did these calculations to show the effect of autoionization of water. Similarly, when real $$[\ce{Fe^3+}]$$ is high in the solution like the case (a) here, $$K_\mathrm{sp}$$ plays a role.

Let's re do the calculation again considering $$[\ce{Fe^3+}] = 0.037 M$$. If all $$\ce{Fe(OH)3}$$ stay dissolved, them we can find the $$[\ce{OH-}]$$ in the solution using $$K_\mathrm{sp}$$ calculations:

$$K_\mathrm{sp} = [\ce{Fe^3+}][\ce{OH-}]^3 = 0.037 \times [\ce{OH-}]^3 \ \Rightarrow \ \therefore \ [\ce{OH-}] = \left(\frac{2.79 \times 10^{-39}}{0.037}\right)^{\frac 13} \\ = 4.22 \times 10^{-13}$$

Thus, you need to keep $$[\ce{OH-}] = 4.22 \times 10^{-13}$$ to keep the solution away from precipitating back to $$\ce{Fe(OH)3}$$. Thus, $$[\ce{H+}]$$ should be at:

$$[\ce{H+}] = \frac{K_\mathrm{w}}{[\ce{OH-}]} = \frac{1.00 \times 10^{-14}}{4.22 \times 10^{-13}} = 0.024$$

Thus, after completely dissolving rust, you need to add additional $$\ce{H2SO4}$$ to keep the $$[\ce{H+}]$$ of the $$\pu{100 mL}$$ solution at $$\pu{0.024 M}$$. I think you can calculate that amount in $$\pu{mmol}$$ easily now (Keep in mind that you already added $$\pu{5.6 mmol}$$ of $$\ce{H2SO4}$$ to dissolve $$\pu{0.4 g}$$ of rust).

Note: Any deviation from the given answer should be due to the $$K_\mathrm{sp}$$ value used here (The given answer didn't give that numerical value).

• This is a tricky question. You have answered it well. I am also somewhat confused. I am wondering that $K_{sp}$ is an equilibrium value, calculated from the equilibrium concentrations of iron (III), and hydroxide ion. I am imagining that the aqueous solution is saturated with rust. How "valid" it is put the total dissolved iron concentration, after acid dissolution there in this expression: $K_\mathrm{sp} = [\ce{Fe^3+}][\ce{OH-}]^3$. There is no "rust" anymore after dissolution. – M. Farooq Jul 17 '20 at 1:27
• @M. Farooq: After dissolving all rust, the $[\ce{Fe^3+}]$ is large enough so that $Q_\mathrm{sp} = [\ce{Fe^3+}] \times (1.00 \times 10^{-7})^3 \gt K_\mathrm{sp}$, the condition to precipitate. So you add extra acid to get $Q_\mathrm{sp} = [\ce{Fe^3+}][\ce{OH}]^3 = K_\mathrm{sp}$, – Mathew Mahindaratne Jul 17 '20 at 5:10
• Thanks, that makes more sense. – M. Farooq Jul 17 '20 at 6:42
• Excellent explanation! I now understand the point of further addition! So we only have to take into consideration the necessary condition for precipitation-dissolution, imposed by the constant. Now when I think about it, it has become obvious. Thank you very much! – TheRelentlessNucleophile Jul 17 '20 at 8:07

The solubility product of $$\ce{Fe(OH)3}$$ is about $$\ce{10^{-39}}$$. So its influence on the real solubility of $$\ce{Fe(OH)3}$$ is not significant. The following equilibrium constants are much more important : $$\ce{[Fe(H_2O)_6]^{3+} <=> [Fe(H_2O)5(OH)]^{2+} + H+ ; \ ...\ K = 10^{-3.05}}$$ $$\ce{[Fe(H_2O)_5(OH)]^{2+} <=> [Fe(H_2O)4(OH)_2]^{+} + H+ ; \ ...\ K = 10^{-3.26}}$$ Another difficulty is the fact that these ions has a tendency to dimerize, according to : $$\ce{2[Fe(H_2O)_6]^{3+}<=> [Fe(H2O)_4(OH))]_2^{4+} + 2 H+ + 2 H_2O \ ...\ K = 10^{-2.91}}$$ So it is not easy to calculate the pH of hydrolyzed Fe(III) solutions.

Ref. R.N. Greenwood, A. Earnshaw, Chemistry of the Elements, Pergamon Press, Oxford $$1984$$, § $$25.3.4$$, page $$1265$$.

• The Olympiad Tour provided only the values for Ksp and Kw, so I suggest we stick to these constants only. Yes, I appreciate correctness and rigour, but let's refrain ourselves from overcomplicating the solution. I would like to know why does the official solution (which I am by no means refuting) take into consideration the solubility product and why isn't the fully stoichiometric approach (which is widely used in inorganic chemistry) correct? – TheRelentlessNucleophile Jul 16 '20 at 20:02