Q22.61CP - Chemistry: Molecular Nature Of Matter And Change

Question

In the production of magnesium, ${\mathbf{\text{Mg(OH)}}}_{\mathbf{\text{2}}}$  is precipitated by using  ${\mathbf{\text{Ca(OH)}}}_{\mathbf{\text{2}}}$, which itself is “insoluble.” 

(a) Use ${\mathbf{\text{K}}}_{\mathbf{\text{sp}}}$  values to show that  ${\mathbf{\text{Mg(OH)}}}_{\mathbf{\text{2}}}$ can be precipitated from seawater in which  ${\mathbf{\text{[Mg}}}^{\mathbf{\text{2 + }}}\mathbf{\text{]}}$ is initially 0.052M. 

(b) If the seawater is saturated with  ${\mathbf{\text{Ca(OH)}}}_{\mathbf{\text{2}}}$, what fraction of the ${\mathbf{\text{[Mg}}}^{\mathbf{\text{2 + }}}\mathbf{\text{]}}$  is precipitated?

Step-by-Step Solution

Answer

(a) It is shown that  ${\text{Mg(OH)}}_{\text{2}}$ can be precipitated from seawater in which  $\left[{\text{Mg}}^{\text{2 + }}\right]$ is initially  $\text{0.052 M}$ as  ${\text{[OH}}^{\text{ - }}\text{]>1.1}\times {\text{10}}^{\text{ - 4}}\text{ M}$.

(b) The fraction of  ${\text{Mg}}^{\text{2 + }}$ that is precipitated is  $\text{100\% }$.

Step 1: Concept Introduction

The solubility product is an equilibrium constant whose value is temperature dependent. Due to increased solubility, ${\mathit{K}}_{\mathbf{s}\mathbf{p}}$  normally increases as temperature rises.

Step 2: Magnesium Oxide Precipitation

(a)

The solubility-product constants for ${\text{Mg(OH)}}_{\text{2}}$  and  ${\text{Ca(OH)}}_{\text{2}}$ are –

 $$\begin{gathered} {\text{K}}_{\text{sp}}\left(\text{Mg}\left({\text{OH}}_{\text{2}}\right)\right)\text{ = 6.3}\times {\text{10}}^{\text{ - 10}} \\ {\text{K}}_{\text{sp}}\left(\text{Ca}\left({\text{OH}}_{\text{2}}\right)\right)\text{ = 6.5}\times {\text{10}}^{\text{ - 6}} \end{gathered}$$

Now, interpret the $K_{sp}$  values –

 $$\begin{gathered} {\text{Mg(OH)}}_{\text{2}}\text{(s)}\to {\text{Mg}}^{\text{2 + }}{\text{(aq) + 2OH}}^{\text{ - }}\text{(aq)} \\ {\text{K}}_{\text{sp}}\text{ = }\left[{\text{Mg}}^{\text{2 + }}\right]·{\left[{\text{OH}}^{\text{ - }}\right]}^{\text{2}}...\left(1\right) \\ {\text{Ca(OH)}}_{\text{2}}\text{(s)}\to {\text{Ca}}^{\text{2 + }}{\text{(aq) + 2OH}}^{\text{ - }}\text{(aq)}....\text{(2)} \\ {\text{K}}_{\text{sp}}\text{ = }\left[{\text{Ca}}^{\text{2 + }}\right]·{\left[{\text{OH}}^{\text{ - }}\right]}^{\text{2}} \end{gathered}$$

Therefore, the bigger the $K_{sp}$  value, the more soluble the compound.

Step 3: Fraction of Magnesium Precipitating

(b)

The  ${\text{K}}_{\text{sp}}$ value is given as –

 $$\begin{gathered} {\text{K}}_{\text{sp}}\text{ = 6.3}\times {\text{10}}^{\text{ - 10}} \\ \left[{\text{Mg}}^{\text{2 + }}\right]\text{ = 0.052 M} \end{gathered}$$

To calculate the required concentration of  ${\text{OH}}^{\text{ - }}$ for ${\text{Mg(OH)}}_{\text{2}}$  to precipitate, use equation (1)  –

 $$\begin{gathered} {\text{K}}_{\text{sp}}\text{ = }\left[{\text{Mg}}^{\text{2 + }}\right]·{\left[{\text{OH}}^{\text{ - }}\right]}^{\text{2}} \\ {\left[{\text{OH}}^{\text{ - }}\right]}^{\text{2}}\text{ = }\frac{{\text{K}}_{\text{sp}}}{\left[{\text{Mg}}^{\text{2 + }}\right]} \\ {\left[{\text{OH}}^{\text{ - }}\right]}^{\text{2}}\text{ = 1.21}\times {\text{10}}^{\text{ - 8}} \\ \left[{\text{OH}}^{\text{ - }}\right]\text{ = 1.1}\times \text{1} \end{gathered}$$

For any concentration of  ${\text{OH}}^{\text{ - }}$ higher than $\text{1.1}\times {\text{10}}^{\text{ - 4}}\text{ M}$ ,  ${\text{Mg(OH)}}_{\text{2}}$ will precipitate.

If  $K_{sp}$ for  ${\text{Ca(OH)}}_{\text{2}}$ is known and it is known that the seawater is saturated with ${\text{Ca(OH)}}_{\text{2}}$ , the concentration of dissolved  ${\text{Ca(OH)}}_{\text{2}}$ can be calculated. Keep in mind that for every molecule of  ${\text{Ca(OH)}}_{\text{2}}$ dissolved, the solution gets one ${\text{Ca}}^{\text{2 + }}$  ion and two   ions. So, the concentration of  ${\text{OH}}^{\text{ - }}$ ions will be twice bigger than the concentration of  ${\text{Ca}}^{\text{2 + }}$ ions.

 

 $$\begin{gathered} {\text{K}}_{\text{sp}}\text{ = }\left[{\text{Ca}}^{\text{2 + }}\right]·{\left[{\text{OHH}}^{\text{ - }}\right]}^{\text{2}} \\ \left[{\text{Ca}}^{\text{2 + }}\right]\text{ = x} \\ \left[{\text{OH}}^{\text{ - }}\right]\text{ = 2x} \\ {\text{K}}_{\text{sp}}\text{ = x}·{\text{(2x)}}^{\text{2}} \\ {\text{K}}_{\text{sp}}{\text{ = 4x}}^{\text{3}} \\ \text{x = }\sqrt[\text{3}]{\frac{{\text{K}}_{\text{sp}}}{\text{4}}} \\ \text{x = 0.012 M} \\ \left[{\text{OH}}^{\text{ - }}\right]\text{ = 0.024 } \end{gathered}$$

 

 

Again, using equation  (2), calculate the fraction of  ${\text{Mg}}^{\text{2 + }}$ precipitated –

 $$\begin{gathered} {\text{K}}_{\text{sp}}\text{ = }\left[{\text{Mg}}^{\text{2 + }}\right]·{\left[{\text{OH}}^{\text{ - }}\right]}^{\text{2}} \\ \left[{\text{Mg}}^{\text{2 + }}\right]\text{ = }\frac{{\text{K}}_{\text{sp}}}{{{\text{[OH}}^{\text{ - }}\text{]}}^{\text{2}}} \\ \left[{\text{Mg}}^{\text{2 + }}\right]\text{ = 4.56}\times {\text{10}}^{\text{ - 6}}\text{ M} \end{gathered}$$

Therefore, the concentration of  ${\text{Mg}}^{\text{2 + }}$ ions in the seawater is  $\text{4.56}\times {\text{10}}^{\text{ - 6}}\text{ mol/L}$, so we can say that almost  $\text{100\% }$ of the magnesium has precipitated.