# How to decide from the solubility product constants which hydroxide precipitates first?

Salts of $\ce{Fe^3+}$ and $\ce{Al^3+}$ are in a beaker of water. If NaOH is added slowly, what compound would precipitate first?

For me, I think it's $\ce{Fe^3+}$ because of its lower $K_\mathrm{sp}$ value when it reacts with $\ce{NaOH}$. The two equations I wrote were

1. $\displaystyle\ce{Fe^3+(aq) +3NaOH(aq) -> 3Na+(aq) +Fe(OH)3(s)}$
2. $\displaystyle\ce{Al^3+(aq) + 3NaOH(aq) -> 3Na+(aq) + Al(OH)3(s)}$

The lower the $K_\mathrm{sp}$ value, the more insoluble the substance is (the precipitate is insoluble). So, $\ce{Fe}$ would precipitate first. Is this correct?

• This is a homework question. We have a policy which states that ‎you should show your thoughts and/or efforts into solving the problem. It'll make us certain that ‎we aren't doing your homework for you. Otherwise, this question may get closed.‎ – M.A.R. Mar 26 '16 at 21:14
• I think this deserves reopening. That being said, neither of those reactions will happen as drawn, but your reasoning is sound. Just make sure you don't have redox reactions occuring. I think you just left out + charges, though, but I want to make sure you don't think $\ce{Na^0}$ is forming. – SendersReagent Mar 27 '16 at 0:49
• Additonally, bear in mind that precipitation of both of these hydroxides would be sensitive to pH. For instance, at higher pH (more basic environment), you aluminium hydroxide precipitate would disappear due to the formation of a hydroxocomplex; you need "moderately" basic conditions..not sure of the exact values. – getafix Mar 27 '16 at 4:26

Given equal stoichiometry for the precipitation of $\ce{Fe(OH)3}$ and $\ce{Al(OH)3}$ and solubility product constants of $6 \cdot 10^{38}$ and $4 \cdot 10^{15}$ respectively, $\ce{Fe(OH)3}$ will be the correct answer.
$$\ce{Fe^{3+}_{(aq)} + 3 OH^{-}_{(aq)} -> Fe(OH)3_{(s)}}$$
However, in practice the precipitate will probably contain less water than stoichiometric $\ce{Fe(OH)3}$, and also contain small amounts of $\ce{Al(OH)3}$