# Question #885cc

Apr 27, 2015

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

From $K s p$ of $C u {\left(O H\right)}_{2}$, it starts precipitating at:
$2.2 \cdot {10}^{-} 20$ = $\left[C {u}^{2 +}\right]$*${\left[O {H}^{-}\right]}^{2}$
$\left[O {H}^{-}\right]$=$\sqrt{0.22 \cdot {10}^{-} 20}$=$0.46 \cdot {10}^{- 10}$

and from $K s p$ of $M g {\left(O H\right)}_{2}$, it starts precipitating at
$6.3 \cdot {10}^{-} 10$ =$\left[M {g}^{2 +}\right]$*${\left[O {H}^{-}\right]}^{2}$
$\left[O {H}^{-}\right]$=$\sqrt{1.26 \cdot {10}^{-} 10}$ =$1.12 \cdot {10}^{- 5}$

therefore at $\left[O {H}^{-}\right]$=${10}^{- 5}$ all $C u {\left(O H\right)}_{2}$ is precipitated and all $M g {\left(O H\right)}_{2}$ is in solution