Nernst's law of distribution

Illustration of Nernst's law of distribution based on a two-phase solution ( dispersion ). The concentration ratio in phase and is always constant and substance-specific. ${\displaystyle k}$ ${\displaystyle A_{2}}$ ${\displaystyle A_{1}}$

The Nernst's distribution law (named after the physical chemist and Nobel laureate Walther Herrmann Nernst ) describes the solubility of a substance in two adjacent phases . If a substance has the possibility of diffusion between two immiscible, highly dilute phases (e.g. a gaseous and a liquid phase or two liquid phases) : ${\displaystyle A}$

${\displaystyle A_{\mathrm {Phase\ 1} }\leftrightharpoons A_{\mathrm {Phase\ 2} }}$,

a distribution equilibrium arises, which is created by the relationship

${\displaystyle K(A)={\frac {c(A\ {\text{in Phase 1}})}{c(A\ {\text{in Phase 2}})}}={\frac {c_{i}^{\beta }}{c_{i}^{\alpha }}}}$

can be described, where is the Nernst partition coefficient ( equilibrium constant ) and the concentration of the substance in the respective phases. For concentrated solutions, however, the activity should be used instead of the concentration . ${\displaystyle K}$ ${\displaystyle c(A)}$

The partition coefficient is usually given in tables for the octanol / water system . Nernst drew up the Distribution Act in 1891. It is used when shaking out or extracting dissolved substances from water with diethyl ether , dichloromethane or other organic solvents, but also in pharmacy and cosmetics to determine the optimal addition of preservatives, which in emulsions accumulate especially in the lipophilic phase are only effective in the hydrophilic phase.

Generalization to Nernst's distribution theorem

The knowledge of the remaining amount of substance is often important for the further procedure (shaking it out again). A chain of sensible assumptions can be used to derive a system of equations from Nernst's law of distribution based on the remaining amount of substance. The larger the better the substance dissolves in phase and the worse in phase . When shaking out , the extraction agent must be selected according to the size of the distribution coefficient: ${\displaystyle K(A)}$ ${\displaystyle A}$ ${\displaystyle \beta }$ ${\displaystyle \alpha }$

${\displaystyle K(A)={\frac {c_{i}^{\beta }}{c_{i}^{\alpha }}}<1\Rightarrow c_{i}^{\alpha }>c_{i}^{\beta }\Rightarrow \alpha \ {\text{ist Extraktionsmittel}}}$

${\displaystyle K(A)={\frac {c_{i}^{\beta }}{c_{i}^{\alpha }}}>1\Rightarrow c_{i}^{\alpha }<c_{i}^{\beta }\Rightarrow \beta \ {\text{ist Extraktionsmittel}}}$

Since the amount of substance and mass does not change when shaken out, the sum of the amounts of substance dissolved in phase and phase is the same: ${\displaystyle \alpha }$ ${\displaystyle \beta }$

${\displaystyle n_{i}^{\alpha }+n_{i}^{\beta }=n_{i+1}^{\alpha }+n_{i+1}^{\beta }}$

(Since only their distribution changes in the phases)

It is now agreed to use the indices 0 and 1:

${\displaystyle n_{0}^{\alpha }+n_{0}^{\beta }=n_{1}^{\alpha }+n_{1}^{\beta }}$

Solving the relationship between concentration, amount of substance and volume gives:

${\displaystyle c={\frac {n}{V}}\Leftrightarrow n=c\cdot V}$

It is assumed that the amount of substance in phase (before shaking) is 0 at the beginning: ${\displaystyle \beta }$

${\displaystyle \Rightarrow c_{0}^{\alpha }\cdot V_{0}^{\alpha }+\underbrace {c_{0}^{\beta }\cdot V_{0}^{\beta }} _{\text{=0}}=c_{1}^{\alpha }\cdot V_{1}^{\alpha }+c_{1}^{\beta }\cdot V_{1}^{\beta }}$

This term is therefore omitted:

${\displaystyle c_{0}^{\alpha }\cdot V_{0}^{\alpha }=c_{1}^{\alpha }\cdot V_{1}^{\alpha }+c_{1}^{\beta }\cdot V_{1}^{\beta }}$

It is also assumed that the volume of the phase does not change because the phase is an extractant: ${\displaystyle \alpha }$ ${\displaystyle \beta }$

${\displaystyle V_{0}^{\alpha }=V_{1}^{\alpha }}$

${\displaystyle \Rightarrow c_{0}^{\alpha }\cdot V_{0}^{\alpha }=c_{1}^{\alpha }\cdot V_{0}^{\alpha }+c_{1}^{\beta }\cdot V_{1}^{\beta }\vert :V_{0}^{\alpha }}$

${\displaystyle \Leftrightarrow c_{0}^{\alpha }=c_{1}^{\alpha }+c_{1}^{\beta }\cdot {\frac {V_{1}^{\beta }}{V_{0}^{\alpha }}}}$

From the change in Nernst's law of distribution follows:

${\displaystyle K(A)={\frac {c_{i}^{\beta }}{c_{i}^{\alpha }}}}$

${\displaystyle \Leftrightarrow c_{i}^{\beta }=K(A)\cdot c_{i}^{\alpha }}$

Substituting in one gets the expression:

${\displaystyle c_{0}^{\alpha }=c_{1}^{\alpha }+K(A)\cdot c_{1}^{\alpha }\cdot {\frac {V_{1}^{\beta }}{V_{0}^{\alpha }}}}$

${\displaystyle =c_{1}^{\alpha }\left(1+K(A)\cdot {\frac {V_{1}^{\beta }}{V_{0}^{\alpha }}}\right)}$

Thus, the Nernst distribution theorem for the first shake out results in:

${\displaystyle c_{1}^{\alpha }={\frac {c_{0}^{\alpha }}{1+K(A)\cdot {\frac {V_{1}^{\beta }}{V_{0}^{\alpha }}}}}}$

It is assumed that the volumes of the phases do not change when shaken out and that the same volume is always used for extraction:

${\displaystyle V_{0}^{\alpha }=V_{1}^{\alpha }}$

${\displaystyle V_{0}^{\beta }=V_{1}^{\beta }}$

${\displaystyle \Rightarrow c_{1}^{\alpha }={\frac {c_{0}^{\alpha }}{1+K(A)\cdot {\frac {V^{\beta }}{V^{\alpha }}}}}}$

When shaking out n times , the following formula results:

${\displaystyle \Rightarrow c_{n}^{\alpha }={\frac {c_{0}^{\alpha }}{\left(1+K(A)\cdot {\frac {V^{\beta }}{V^{\alpha }}}\right)^{n}}}}$

Calculation of the extracted mass

The concentration (of the individual phases) adjusted according to the amount of substance

${\displaystyle c={\frac {n}{V}}}$

${\displaystyle \Leftrightarrow n=cV}$

combined with the molar mass (of the substance to be extracted)

${\displaystyle M={\frac {m}{n}}}$

${\displaystyle \Leftrightarrow n={\frac {m}{M}}}$

result in an expression that links mass and concentration:

${\displaystyle n={\frac {m}{M}}=cV}$

It can be resolved according to the mass:

${\displaystyle \Rightarrow m=c\cdot V\cdot M}$

In combination with the equation derived above, this results in:

${\displaystyle \Rightarrow {\frac {m_{n}^{\alpha }}{M\cdot V}}={\frac {\frac {m_{0}^{\alpha }}{M\cdot V}}{\left(1+K(A)\cdot {\frac {V^{\beta }}{V^{\alpha }}}\right)^{n}}}}$

Since it has already been agreed that the amount of substance and volume will not change, the following applies to the total mass of the extracted substance after shaking it out n times:

${\displaystyle \Rightarrow m_{n}^{\alpha }={\frac {m_{0}^{\alpha }}{\left(1+K(A)\cdot {\frac {V^{\beta }}{V^{\alpha }}}\right)^{n}}}}$

literature

H. Elias, S. Lorenz, G. Winnen: The experiment: 100 years of Nernst's distribution theorem, chemistry in our time, 26th year 1992, No. 2, p. 70, ISSN  0009-2851