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## A Discussion about Extraction Limit

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Abstract

Extraction separation is a common operation in chemical experiments. According to the distribution law, the Lagrangian multiplier method is used to derive and prove that the extraction efficiency can be optimized only in the case of using same aliquots of solvent. The limit extraction residual rate of the extraction is related to the physical properties and the dose of the solvent used, as well as other factors. This paper provides reference and guidance on designing extraction method for those engaged in chemical synthesis, analysis, separation, and chemical industry and can be extended to the fields of dilution and solid phase separation.

Keywords： Multiple aliquots solvent extraction ; Lagrangian multiplier method ; Extraction limit ; Extraction residual rate

WU Yadong. A Discussion about Extraction Limit. University Chemistry[J], 2019, 34(1): 89-91 doi:10.3866/PKU.DXHX201806005

${\mu _{\rm{B}}}(\alpha ) = {\mu _{\rm{B}}}(\beta );\;\;\mu _{\rm{B}}^{\rm{ \mathsf{ θ} }}(\alpha ) + RT{\rm{ln}}x_{\rm{B}}^\alpha = \mu _{\rm{B}}^{\rm{ \mathsf{ θ} }}(\beta ) + RT{\rm{ln}}x_{\rm{B}}^\beta$

${\rm{ln}}\frac{{x_{\rm{B}}^\alpha }}{{x_{\rm{B}}^\beta }} = \frac{{\mu _{\rm{B}}^{\rm{ \mathsf{ θ} }}(\beta ) - \mu _{\rm{B}}^{\rm{ \mathsf{ θ} }}(\alpha )}}{{RT}}$

## 1 萃取方式推导

\begin{align} &K=\frac{{{m}_{1}}/{{V}_{1}}}{({{m}_{0}}-{{m}_{1}})/{{V}_{2}}} \\ &{{m}_{1}}={{m}_{0}}\frac{K{{V}_{1}}}{K{{V}_{1}}+{{V}_{2}}} \\ \end{align}

\begin{align} &{{m}_{n}}={{m}_{0}}{{\left( \frac{K{{V}_{1}}}{K{{V}_{1}}+{{V}_{2}}} \right)}^{n}} \\ &{{m}_{0}}-{{m}_{n}}={{m}_{0}}\left( 1-{{\left( \frac{K{{V}_{1}}}{K{{V}_{1}}+{{V}_{2}}} \right)}^{n}} \right) \\ \end{align}

${{m}_{n}}={{m}_{0}}\left( \frac{K{{V}_{1}}}{K{{V}_{1}}+{{V}_{21}}} \right)\left( \frac{K{{V}_{1}}}{K{{V}_{1}}+{{V}_{22}}} \right)...\left( \frac{K{{V}_{1}}}{K{{V}_{1}}+{{V}_{2n}}} \right)={{m}_{0}}\prod\nolimits_{i=1}^{n}{\left( \frac{K{{V}_{1}}}{K{{V}_{1}}+{{V}_{2i}}} \right)}$

$F\left( {{b}_{1}}, {{b}_{2}}...{{b}_{n}}, \ \lambda \right)=\text{ln}\left[ \prod\limits_{i=1}^{n}{(\frac{a}{a+{{b}_{i}}})} \right]+\lambda \left( b-\sum\limits_{i=1}^{n}{{{b}_{i}}} \right)=n\text{ln}a+\lambda b-\sum\limits_{i=1}^{n}{\left[ \text{ln}(a+bi)-\lambda {{b}_{i}} \right]}$

\left\{ \begin{align} &\frac{\partial F}{\partial {{b}_{i}}}=-\frac{1}{a+{{b}_{i}}}-\lambda =0 \\ &\frac{\partial F}{\partial \lambda }=b-\sum\limits_{i=1}^{n}{{{b}_{i}}}=0 \\ \end{align} \right.

${{b}_{1}}={{b}_{2}}={{b}_{i}}={{b}_{n}}=\frac{b}{n}$

## 2 萃取极限

$\underset{n\to \infty }{\mathop{\text{lim}}}\, m(n)=\underset{n\to \infty }{\mathop{\text{lim}}}\, {{\left( \frac{a}{a+\frac{b}{n}} \right)}^{n}}=\underset{n\to \infty }{\mathop{\text{lim}}}\, \frac{1}{{{\left( 1+\frac{b}{na} \right)}^{n}}}=\underset{n\to \infty }{\mathop{\text{lim}}}\, \frac{1}{{{\left( 1+\frac{b}{na} \right)}^{\frac{na}{b}\frac{b}{a}}}}$

$\underset{n\to \infty }{\mathop{\text{lim}}}\, m(n)=\frac{1}{{{\text{e}}^{\frac{b}{a}}}}={{\text{e}}^{-\frac{b}{a}}}={{\text{e}}^{-\frac{1}{K}\frac{{{V}_{2}}}{{{V}_{1}}}}}$

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