# Chemistry - Dissolution of iron(III) hydroxide

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).