# 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).