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


$\displaystyle\ce{Fe^3+(aq) +3NaOH(aq) -&gt; 3Na+(aq) +Fe(OH)3(s)}$
$\displaystyle\ce{Al^3+(aq) + 3NaOH(aq) -&gt; 3Na+(aq) + Al(OH)3(s)}$


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?
If the question is about what compound would precipitate first, then there can be at most one answer.

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.

$$\ce{Fe^{3+}_{(aq)} + 3 OH^{-}_{(aq)} -&gt; Fe(OH)3_{(s)}}$$

However, in practice the precipitate will probably contain less water than stoichiometric $\ce{Fe(OH)3}$, and also contain small amounts of $\ce{Al(OH)3}$