# Question #17cc3

Apr 13, 2015

The answer is A) $2 \cdot {10}^{- 3} \text{M}$.

Because the solubility product constant, ${K}_{s p}$, of manganese (II) hydroxide is so small, the compound is considered insoluble in aqueous solution.

However, very, very little amounts will dissociate according to the balanced equilibrium equation

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

According to the definition of the solutbility constant product, you have

${K}_{s p} = \left[M {n}^{2 +}\right] \cdot {\left[O {H}^{-}\right]}^{\textcolor{red}{2}}$ $\textcolor{b l u e}{\left(1\right)}$

Every time the concentrations of the $M {n}^{2 +}$ and $O {H}^{-}$ ions satisfy the above equation, a precipitate will form.

You can determine the concentration of the hydroxide ions from the solution's pH

$p O H = 14 - p {H}_{\text{sol}} = 14 - 9 = 5$

$\left[O {H}^{-}\right] = {10}^{- p O H} = {10}^{- 5}$

Therefore, according to equation $\textcolor{b l u e}{\left(1\right)}$,

$\left[M {n}^{2 +}\right] = {K}_{s p} / {\left(\left[O {H}^{-}\right]\right)}^{2} = \frac{2 \cdot {10}^{- 13}}{{10}^{- 5}} ^ \left(2\right) = \frac{2 \cdot {10}^{- 13}}{10} ^ \left(- 10\right)$

$\left[M {n}^{2 +}\right] = \textcolor{g r e e n}{2 \cdot {10}^{- 3} \text{M}}$