Consider the extraction of $\mathbf{100}\mathbf{mL}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{M}}^{\mathbf{+}\mathbf{2}}\mathbf{}\mathbf{}\mathbf{by}\mathbf{}\mathbf{2}\mathbf{mL}\mathbf{}\mathbf{of}\mathbf{}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{}\mathbf{}\mathbf{M}\mathbf{}\mathbf{dithizone}\mathbf{}\mathbf{in}\mathbf{}{\mathbf{CHCl}}_{\mathbf{3}}$for which ${\mathbf{K}}_{\mathbf{L}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{,}{\mathbf{K}}_{\mathbf{M}}\mathbf{=}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{24}}\mathbf{,}\mathbf{Ka}\mathbf{=}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{,}\mathbf{\beta }\mathbf{=}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{18}}\mathbf{,}\mathbf{n}\mathbf{=}\mathbf{2}$(a) Derive an expression for the fraction of metal ion extracted into the organic phase, in terms of the distribution coefficient and volumes of the two phases. (b) Prepare a graph of the percentage of metal ion extracted over the pH range 0 to 5.

For the extraction of $100\mathrm{mL}\mathrm{of}{\mathrm{M}}^{+2}\mathrm{by}2\mathrm{mL}\mathrm{of}1\times {10}^{-5}\mathrm{M}\mathrm{dithizone}\mathrm{in}{\mathrm{CHCl}}_3$as

${\mathrm{K}}_{\mathrm{L}}=1.1\times {10}^4$.

$K_M=7\times {10}^4$,

$\mathrm{Ka}=3\times {10}^{-5}$,

$\mathrm{\beta }=5\times {10}^{18}$

n=2,

The expression for the fraction between the concentration of${\mathrm{ML}}_2$ in organic and aqueous phase

$\mathrm{D}=\frac{{\left[{\mathrm{ML}}_2\right]}_{\mathrm{org}}}{{\left[{\mathrm{ML}}_2\right]}_{\mathrm{aq}}}=\frac{{\mathrm{c}}_{\mathrm{org}}\times {\mathrm{v}}_{\mathrm{org}}}{{\mathrm{c}}_{\mathrm{aq}}\times {\mathrm{V}}_{\mathrm{aq}}}$

Expression for${\mathrm{c}}_{\mathrm{org}}$ by rearranging the equation is

${\mathrm{c}}_{\mathrm{org}}=\mathrm{D}\times \frac{{\mathrm{c}}_{\mathrm{aq}}\times {\mathrm{V}}_{\mathrm{aq}}}{{\mathrm{V}}_{\mathrm{org}}}$

Expression for the percentage of the extracted metal ion

role="math" localid="1664883296728" $$\begin{gathered} \mathrm{Extracted}\%=\frac{100\times {\mathrm{c}}_{\mathrm{org}}}{{\mathrm{c}}_{\mathrm{org}}+{\mathrm{c}}_{\mathrm{aq}}} \\ \mathrm{Extracted}\%=\frac{100\times \mathrm{D}\times {\displaystyle \frac{{\mathrm{c}}_{\mathrm{aq}}\times {\mathrm{v}}_{\mathrm{aq}}}{{\mathrm{v}}_{\mathrm{org}}}}}{\mathrm{D}\times {\displaystyle \frac{{\mathrm{c}}_{\mathrm{aq}}\times {\mathrm{v}}_{\mathrm{aq}}}{{\mathrm{v}}_{\mathrm{org}}}}+{\mathrm{c}}_{\mathrm{aq}}} \end{gathered}$$

To eliminate these values in the equation is

$\mathrm{Extracted}\%=\frac{100\times \mathrm{D}\times {\displaystyle \frac{{\mathrm{v}}_{\mathrm{aq}}}{{\mathrm{v}}_{\mathrm{org}}}}}{\mathrm{D}\times {\displaystyle \frac{{\mathrm{c}}_{\mathrm{aq}}}{{\mathrm{v}}_{\mathrm{org}}}}+1}$

Spreadsheet of percentage of metal ion extracted over the pH range

Column B calculation

$\left[{\mathrm{H}}^{+}\right]={10}^{-\mathrm{pH}}$

Column C calculation

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

Column D calculation

$\mathrm{Extracted}\%=\frac{100\times \mathrm{D}\times {\displaystyle \frac{{\mathrm{v}}_{\mathrm{aq}}}{{\mathrm{v}}_{\mathrm{org}}}}}{\mathrm{D}\times {\displaystyle \frac{{\mathrm{v}}_{\mathrm{aq}}}{{\mathrm{v}}_{\mathrm{org}}}}+1}$

By using these values for column, A and column D values in the graph as shown below.