Question

Question: Water that has been saturated with magnesia (MgO) at 25⁰C has a pH of 10.16. Write a balanced chemical equation for the equilibrium betweenMgO and the ions it furnishes in aqueous solution, and calculate the equilibrium constant at. What is the solubility, in moles per liter, ofMgO in water?

Short answer

Answer

The solubility of MgO in water ${\text{= 7.0×10}}^{\text{- 5}}{\text{moL}}^{\text{- 1}}$.

Step-by-step solution

Finding the equilibrium constant.

pH value for .

$$\begin{gathered} \text{pOH = 14 - pH} \\ \text{= 14 - 10.16} \\ \text{= 3.84} \end{gathered}$$ (It’s known as $\text{pH + pOH = 14}$) and (pH = 10.16)

$$\begin{gathered} {\text{pOH = - log[OH}}^{\text{-}}\text{]} \\ {\text{[OH}}^{\text{-}}{\text{]=10}}^{\text{- pOH}} \\ {\text{= 1.0×10}}^{\text{- 3.84}} \\ {\text{= 1.4×10}}^{\text{- 4}}\text{M} \end{gathered}$$

Half of the mole of produced is

$$\begin{gathered} {\text{Mg}}^{\text{2 +}}\text{]=}\frac{{\text{1.4×10}}^{\text{- 4}}\text{M}}{\text{2}} \\ {\text{= 0.7×10}}^{\text{- 4}}\text{M} \end{gathered}$$

Solubility in moles per liter of magnesium oxide.

First of all finding the equation when magnesium and hydrogen combine;

${\text{MgO(s) +H}}_{2}O\left(l\right)\to M{g}^{2+}\left(aq\right)+2O{H}^{-}\left(aq\right)$

Now for K value K is constant.$K=\left[M{g}^{2+}\right][O{H}^{-}{]}^2$

So,

$$\begin{gathered} dK=(0.7\times {10}^{-4})\times (1.4\times {10}^{-4}{)}^2 \\ =1.4\times {10}^{-12} \end{gathered}$$

Now for solubility$\left[M{g}^{2+}\right][O{H}^{-}{]}^2=K_{SP}$

We will denote this $\left[M{g}^{2+}\right]$= S and $\left[O{H}^{-}\right]$= 2S, Whereas S = solubility in ${\text{moL}}^{\text{- 1}}$.

Now equating the formula;

$$\begin{gathered} {\text{S×(2S)}}^{\text{2}}{\text{= 1.4×10}}^{\text{- 12}} \\ {\text{4S}}^{\text{3}}{\text{= 1.4×10}}^{\text{- 12}} \\ {\text{S = 7.0×10}}^{\text{- 5}} \end{gathered}$$

The solubility of in water${\text{S = 7.0×10}}^{\text{- 5}}$ .