Calculating solubility of iron(III) hydroxide in water

Question

The following example in Skoog's Analytical Chemistry uses mass-balance and charge-balance equations to calculate the solubility of iron(III) hydroxide in aqueous solution.

$$ \begin{align} \ce{Fe(OH)3 (s) &<=> Fe^3+ (aq) + 3OH- (aq)} \\ \ce{2H2O (l) &<=> H3O+ (aq) + OH- (aq)} \end{align} $$

$$ \begin{align} K_\mathrm{sp} &= [\ce{Fe^3+}][\ce{OH-}]^3 = 2\times 10^{-39}\\ K_\mathrm{w} &= [\ce{H3O+}][\ce{OH-}] = 1\times 10^{-14} \end{align} $$

$$[\ce{OH-}] = 3[\ce{Fe^3+}] + [\ce{H3O+}]$$

After writing down the equations for the relevant equilibria and the corresponding equilibrium and mass balance equations (as seen above), the problem made an approximation that puzzled me:

Step 7a. Make approximations As in Example 11-5, assume that $[\ce{H3O+}]$ is very small so that $[\ce{H3O+}] \ll 3[\ce{Fe^3+}]$ and

$$3[\ce{Fe^3+}] \approx [\ce{OH-}]$$

How can we make an approximation that $[\ce{H3O+}] \ll 3[\ce{Fe^3+}]$ when $K_\mathrm{sp} \ll K_\mathrm{w}$? Shouldn't it be the other way around, actually?

Please explain the reasoning for this step in the process.

Answer 1

In my version of the book the original assumption (step 7a) is shown to lead to a contradiction when later checked (logically, this form of proof is called reductio ad absurdum or proof by contradiction). The point is that you should 1) learn not to be afraid to apply assumptions, but that you should then 2) verify that the original assumption was indeed acceptable. This is a strategy repeatedly encouraged in the book.

To show how this leads to a contradiction,

1. make the assumption, so that $ \ce{3[Fe^{3+}]} = \ce{[OH-]}$
2. solve for $S = \ce{[Fe^{3+}]}$ based on the $K_{sp}$ and the assumption
3. Compute $\ce{[OH-] = \ce{3[Fe^{3+}]} }$, following the original assumption
4. Compute $\ce{[H+]}$ from $K_w$ and $\ce{[OH-]}$
5. Check the accuracy of the assumption

It is by the way not difficult to solve the problem exactly, by computing the $\ce{[OH-]}$ as follows:

$$ \ce{[OH-]}=\mathrm{\sqrt\frac{K_w+\sqrt{K_w^2 +12K_{sp}}}{2}}$$

and then applying the expression for the solubility constant to compute the solubility. However the point of the problem is to illustrate how to follow this line of reasoning when solving problems (the use and verification of assumptions).

Answer 2

Yes, It should be the other way around.(i.e. $\ce{[Fe^3+] << [H3O+]}$)

You are absolutely right in the logic that $K_{sp} << K_w$. Thus due to the common ion effect, the dissociation of $\ce{Fe(OH)3}$ should be very less. So, the solubility of Ferric hydroxide in water should be extremely less.

By mathematical treatment also, this is justifiable. Consider the solubility of Ferric hydroxide in water to be $S$ $\mathrm{mol.L^{-1}}$. and consider the concentration of $\ce{H3O+}$ is $x \ \mathrm{mol.L^{-1}}$. Thus the total concentration of $\ce{OH-}$ will be $(3S +x) \ \mathrm{mol.L^{-1}} $ . So, for finding $S$ and $x$ we need to solve two equations which are as follows, $$x(3S +x)^3 = 2 \times10^{-39}$$ and, $$x(3S +x) = 10^{-14}$$ Exact solution of these two equations is difficult to find, but if we substitute $(3S+x)$ by $\frac{10^{-14}}{x}$ in first equation we will find that $S = 2\times10^3 \times x^3$. Now $x$ is between $0$ and $1$ ( roughly of the order of $10^{-7}$) and thus if it is cubed, it will become even smaller. So, it is seen that $S <<x$. So, thus it is justified that $\ce{[H3O+] >> [Fe^3+]}$.

In fact it will be found that $\ce{[H3O+] \approx 10 ^{-7} M}$ and $\ce{[Fe^3+] \approx 2 \times 10^{-18} M}$ . So, your justification is absolutely valid.