Question

Calculate the solubility of$\text{Mg}(\text{OH}{)}_2\left({\mathit{K}}_{\text{sp}}{\text{=8.9×10}}^{\text{-12}}\right)$in an aqueous solution buffered at$\text{pH = 9.42}$ .

Short answer

The solubility of $\text{Mg}(\text{OH}{)}_2\left({\mathit{K}}_{\text{sp}}{\text{=8.9×10}}^{\text{-12}}\right)$in an aqueous solution buffered at$\text{pH = 9.42}$ is $0.013mol/L.$.

Step-by-step solution

Definition of solubility

A solution's solubility is defined as the greatest amount of solute that may be dissolved in a specific amount of solvent. In 90 g of water, 10 g of glucose powder dissolves.

Dissociation equilibrium of Mg(OH)2

Molar solubility is the number of moles of the solute that can be dissolved per litre of solution before the solution becomes saturated, and it is directly connected to the solubility product.

We must first determine the equilibrium concentrations of ${\text{Mg}}^{\text{2 +}}$and ${\text{OH}}^{\text{-}}$in order to determine the molar solubility of ${\text{Mg(OH)}}_{\text{2}}$.

Below is a diagram of the dissociation equilibrium of${\text{Mg(OH)}}_{\text{2}}$.

${\text{Mg(OH)}}_{\text{2}}\text{(s)}\rightleftharpoons {\text{Mg}}^{\text{2 +}}{\text{(aq) + 2OH}}^{\text{-}}\text{(aq)}$

Write the value of Ksp

According to the law of mass action, we can formulate the following equilibrium expression for this process:

${\text{K}}_{\text{sp}}\text{=}\left[{\text{Mg}}^{\text{2 +}}\right]{\left[{\text{OH}}^{\text{-}}\right]}^{\text{2}}$

$\text{where}\left[{\text{Mg}}^{\text{2 +}}\right]\text{and}\left[{\text{OH}}^{\text{-}}\right]\text{are expressed in mol/L}$.

For the equilibrium expression, the constant ${\text{K}}_{\text{sp}}$is referred to as the solubility product constant, or simply the solubility product.

$K_{\text{sp}}{\text{=8.9×10}}^{\text{-12}}$

Calculate OH -  from the pH of solution

Now, $\left[{\text{OH}}^{\text{-}}\right]$can be calculated from the solution's.

$\text{pH + pOH = 14}$

$\text{pOH = 14 - pH = 14 - 9.42 = 4.58}$

$\text{pOH = - log}\left[{\text{OH}}^{\text{-}}\right]$

$\left[{\text{OH}}^{\text{-}}\right]{\text{= 10}}^{\text{- pOH}}$

localid="1663834016943" $$\begin{gathered} \left[{\text{OH}}^{-}\right]{\text{=10}}^{\text{-4.58}} \\ \left[{\text{OH}}^{-}\right]{\text{=2.63×10}}^{\text{-5}}M \end{gathered}$$

Calculate the solubility(s)

Finally, the solubility can be calculated.

The equilibrium concentration of${\text{Mg(OH)}}_{\text{2}}$ is equal to the solubility of data-custom-editor="chemistry" $\left[{\text{Mg}}^{\text{2 +}}\right]$ .

$$\begin{gathered} \left[{\text{Mg}}^{\text{2 +}}\right]\text{= s} \\ {\text{K}}_{\text{sp}}=\left[{\text{Mg}}^{\text{2+}}\right]{\left[{\text{OH}}^{-}\right]}^2 \end{gathered}$$

$\text{s =}\frac{{\text{K}}_{\text{sp}}}{{\left[{\text{OH}}^{\text{-}}\right]}^{\text{2}}}$

$\text{s=}\frac{{\text{8.9×10}}^{\text{-12}}}{{\text{2.63×10}}^{\text{-5}}}$

$\text{s=0.013mol/L}$

Thus, in an aqueous solution buffered at role="math" localid="1663833949258" $\mathrm{pH}=9.42$, the solubility of ${\text{Mg(OH)}}_{\text{2}}$is $\text{0.013mol/L}$.