# Question da59d

Jan 7, 2015

The answer is $1.8 \cdot {10}^{- 7}$ $\text{M}$.

The general reaction looks like this

$M {g}_{\left(a q\right)}^{2 +} + 2 N a O {H}_{\left(a q\right)} \to M g {\left(O H\right)}_{2 \left(s\right)} + 2 N {a}^{+} \left(a q\right)$

The reaction that is of interest is actually this

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

In order to determine the maximum concentration of $M {g}^{2 +}$ ions permissible in the $N a O H$ solution before a precipitate will be formed, you'd need the value of the solubility product constant, ${K}_{s p}$.

The expression for ${K}_{s p}$ for this reaction is

${K}_{s p} = \left[M {g}^{2 +}\right] \cdot {\left[O {H}^{-}\right]}^{2}$

A precipitate will form when the solution is saturated with ions; if the concentrations of the ions are low enough, a precipitate will not form. The threshold for the formation of a precipitate implies that the above equation be satisfied.

For a saturated solution, on the verge of forming a precipitate, the maximum value for the product of the concentrations of the two ions is equal to ${K}_{s p}$.

From this point on, any increase in the concentrations of the ions will result in the formation of the precipitate.

Since your reaction takes place in 1 L, 0.01 M $N a O H$ solution, and knowing that $N a O H$ dissociates completely into $N {a}^{+}$ and $O {H}^{-}$ ions, the concentration of hydroxide ions will be

$\left[O {H}^{-}\right] = \left[N a O H\right] = 0.01$ $\text{M}$

This means that the maximum concentration of $M {g}^{2 +}$ ions the solution can hold before a precipitate is formed is

[Mg^(2+)] = K_(sp)/([OH^(-)]^2) = K_(sp)/(0.01)^2 = K_(sp)/(10^(-4)#

As a numerical value, I've found ${K}_{s p}$ to be equal to $1.8 \cdot {10}^{- 11}$ (you can use any value given to you, of course), which means that

$\left[M {g}^{2 +}\right] = \frac{1.8 \cdot {10}^{- 11}}{10} ^ \left(- 4\right) = 1.8 \cdot {10}^{- 7}$ $\text{M}$