Question

The solubility product of $Fe{\left(OH\right)}_3$ is $1\times {10}^{-36}$. What is the minimum concentration of ${OH}^{-}$ ions required to precipitate $Fe{\left(OH\right)}_3$ from $0.001M$ solution of $Fe{Cl}_3$?

Answer 1

$Fe(OH{)}_3=F{e}^{3+}+3O{H}^{⊝}$

$\therefore \left[F{e}^{3+}\right][O{H}^{-}{]}^3=K_{sp}$ of $Fe(OH{)}_3=1\times {10}^{-36}$

$\Rightarrow [O{H}^{⊝}{]}^3=\frac{1\times {10}^{-36}}{0.001}$

$\Rightarrow \left[O{H}^{⊝}\right]=({10}^{-33}{)}^{Y_3}={10}^{-11}M$

$\therefore$ The minimum concentration of $O{H}^{⊝}$ will be ${10}^{-11}M$ .