# Question 43e92

May 26, 2017

(a).

#### Explanation:

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

$\textsf{{K}_{s p} = \left[M {g}_{\left(a q\right)}^{2 +}\right] {\left[O {H}_{\left(a q\right)}^{-}\right]}^{2} = 7.1 \times {10}^{- 12} \textcolor{w h i t e}{x} {\text{mol"^3."l}}^{-} 3}$

If this is carried out in 0.50 M NaOH solution the position of equilibrium will be driven to the left thus supressing further solution of $\textsf{M {g}^{2 +}}$ ions.

To make things much easier for ourselves I am going to assume that the vast majority of the $\textsf{O {H}^{-}}$ ions come from the NaOH.

$\therefore$$\textsf{\left[M {g}_{\left(a q\right)}^{2 +}\right] = {K}_{s p} / {\left[O {H}_{\left(a q\right)}^{-}\right]}^{2}}$

$\textsf{\left[M {g}_{\left(a q\right)}^{2 +}\right] = \frac{7.1 \times {10}^{- 12}}{{0.50}^{2}} = 2.8 \times {10}^{- 11} \textcolor{w h i t e}{x} \text{mol/l}}$

This means the solubility of $\textsf{M {g}_{\left(a q\right)}^{2 +} = 2.8 \times {10}^{- 11} \textcolor{w h i t e}{x} \text{mol/l}}$.

This gives (a).