How would you calculate the molar solubility of Mn(OH)2 (Ksp = 1.6e-13) in pure water?

Nov 11, 2015

$M n {\left(O H\right)}_{2} \left(s\right) r i g h t \le f t h a r p \infty n s M {n}^{2 +} + 2 O {H}^{-}$

${K}_{s p} = 1.6 \times {10}^{- 13}$

Explanation:

${K}_{s p} = 1.6 \times {10}^{- 13} = \left[M {n}^{2 +}\right] {\left[O {H}^{-}\right]}^{2}$

If we call the solubility of manganese(II) hydroxide $S$, then by the stoichiometry of the reaction:

${K}_{s p} = S {\left(2 S\right)}^{2} = 4 {S}^{3}$.

So $S = \sqrt[3]{\frac{1.6 \times {10}^{- 13}}{4}}$