91影视

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.

(a)

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

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

Now, interpret the $K_{sp}$ values 鈥�

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

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

(b)

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

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

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

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

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

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

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

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

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