Question

Consider the extraction of ${\mathbf{M}}^{\mathbf{nt}}$from aqueous solution into organic solution by reaction with protonated ligand HL : $\mathbf{D}\mathbf{=}{\mathbf{M}}^{\mathbf{n}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{nHL}\mathbf{\left(}\mathbf{org}\mathbf{\right)}\mathbf{\square }\mathbf{}{\mathbf{ML}}_{\mathbf{n}}\mathbf{\left(}\mathbf{org}\mathbf{\right)}\mathbf{+}{\mathbf{nH}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}{\mathbf{K}}_{\mathbf{extraction}}\mathbf{=}\frac{{\left[{\mathrm{ML}}_n\right]}_{\mathbf{org}}\mathbf{}{\left[H^{+}\right]}_{\mathbf{aq}}^{\mathbf{n}}}{{\left[M^{n+}\right]}_{\mathbf{aq}\mathbf{}}{\left[\mathrm{HL}\right]}_{\mathbf{org}}^{\mathbf{n}}}$.Rewrite Equation 23 - 13 in terms of role="math" localid="1654863844402" ${\mathbf{K}}_{\mathbf{e}\mathbf{x}\mathbf{t}\mathbf{r}\mathbf{a}\mathbf{c}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}}$and express ${\mathbf{K}}_{\mathbf{extraction}}$in terms of the constants in Equation 23 - 13 . Give a physical reason why each constant increases or decreases ${\mathbf{K}}_{\mathbf{extraction}}$

Short answer

• In this task let us rewrite the equation 23 - 13 in terms of ${\mathrm{K}}_{\mathrm{extraction}}$and express it in terms of the constants.
• Further, we shall give a physical reason why each constant increases/decreases the ${\mathrm{K}}_{\mathrm{extraction}}$

Step-by-step solution

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• Let us write the Equation 23 - 13 for distribution of metal-chelate complex

between phases as $\mathrm{D}=\frac{{\mathrm{K}}_{\mathrm{M}}{\mathrm{\beta K}}_{\mathrm{a}}^{\mathrm{n}}}{{\mathrm{K}}_{\mathrm{L}}^{\mathrm{n}}}\frac{[{\mathrm{HL}}_{\mathrm{org}}^{\mathrm{n}}}{{\left[{\mathrm{H}}^{+}\right]}_{\mathrm{aq}}^{\mathrm{n}}}$.

• Here, ${\mathrm{ML}}_{\mathrm{n}}$is the form extracted into organic solvent.

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• Here, let us further consider the following term for distribution coefficient (D) .

$\mathrm{D}=\frac{\left[{\mathrm{ML}}_{\mathrm{n}}\right]}{\left[{\mathrm{M}}^{\mathrm{n}+}\right]}$

• Now, we shall take the formula from the task and rearrange it in the way so that we make it equal to the term above

$$\begin{gathered} {\mathrm{K}}_{\mathrm{extraction}}=\frac{\left[{\mathrm{ML}}_{\mathrm{n}}\right]{\left[{\mathrm{H}}^{+}\right]}^{\mathrm{n}}}{\left[{\mathrm{M}}^{\mathrm{n}+}\right]{\left[\mathrm{HL}\right]}^{\mathrm{n}}} \\ \frac{{\left[\mathrm{HL}\right]}^{\mathrm{n}}}{\left[{\mathrm{H}}^{+}\right]}\times {\mathrm{K}}_{\mathrm{extraction}}=\frac{\left[{\mathrm{ML}}_{\mathrm{n}}\right]}{\left[{\mathrm{M}}^{\mathrm{n}+}\right]} \end{gathered}$$

• Then, let us substitute it in order to combine the resulting equation with the

one from 23 - 13

• $$\begin{gathered} \mathrm{D}=\frac{\left[{\mathrm{ML}}_{\mathrm{n}}\right]}{\left[{\mathrm{M}}^{\mathrm{n}+}\right]} \\ \mathrm{D}=\frac{{\left[\mathrm{HL}\right]}^{\mathrm{n}}}{{\left[{\mathrm{H}}^{+}\right]}^{\mathrm{n}}}\times {\mathrm{K}}_{\mathrm{extraction}} \end{gathered}$$

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• Let us combine the resulting equation with Equation :

$\mathrm{D}=\frac{{\mathrm{K}}_{\mathrm{M}}{\mathrm{\beta K}}_{\mathrm{a}}^{\mathrm{n}}}{{\mathrm{K}}_{\mathrm{L}}^{\mathrm{n}}}\frac{{\left[\mathrm{HL}\right]}_{\mathrm{org}}^{\mathrm{n}}}{{\left[{\mathrm{H}}^{+}\right]}_{\mathrm{aq}}^{\mathrm{n}}}$

$\frac{{\left[\mathrm{HL}\right]}^n}{{\left[{\mathrm{H}}^{+}\right]}^n}\times {\mathrm{K}}_{\mathrm{extraction}}=\frac{{\mathrm{K}}_{\mathrm{M}}{\mathrm{\beta K}}_{\mathrm{a}}^{\mathrm{n}}}{{\mathrm{K}}_{\mathrm{L}}^{\mathrm{n}}}\frac{{\left[\mathrm{HL}\right]}_{\mathrm{org}}^{\mathrm{n}}}{{\left[{\mathrm{H}}^{+}\right]}_{\mathrm{aq}}^{\mathrm{n}}}$

• Let us eliminate it further to make it as a simplified one. So we have

${\mathrm{K}}_{\mathrm{extraction}}=\frac{{\mathrm{K}}_{\mathrm{M}}{\mathrm{\beta K}}_{\mathrm{a}}^{\mathrm{n}}}{{\mathrm{K}}_{\mathrm{L}}^{\mathrm{n}}}$

Result:

• ${\mathrm{K}}_{\mathrm{M}}$increases due to ${\mathrm{ML}}_{\mathrm{n}}$being more soluble in organic phase
• $\mathrm{\beta }$- increases because ${\mathrm{ML}}_{\mathrm{n}}$is more soluble in organic form and ligand in this

case binds the metal more tightly

• ${\mathrm{K}}_{\mathrm{a}}$increases because ligand loses ${\mathrm{H}}^{+}$easier and therefore increases the

formation of${\mathrm{ML}}_{\mathrm{n}}$

• ${\mathrm{K}}_{\mathrm{L}}$decreases as HL is more soluble in organic phase and then ${\mathrm{L}}^{-}$cannot react

with ${\mathrm{MN}}^{+}$in order to form ${\mathrm{ML}}_{\mathrm{n}}$.