Consiste en representar la fracción, respecto a la total, de cada una de las especies del metal frente al logaritmo de la concentración de ligando.
Así pues, por ejemplo, para un complejo $\ce{ML2}$:
$ \def\upalpha{\smash{\unicode{x03B1}}} \begin{array}{l} \upalpha_0 = \upalpha_{\ce{M}} = \dfrac{[\ce{M}]}{c_{\ce{M}}} = \dfrac{1}{1 + \beta_1 [\ce{L}] + \beta_2 [\ce{L}]^2} \\[1ex] \upalpha_1 = \upalpha_{\ce{ML}} = \dfrac{[\ce{ML}]}{c_{\ce{M}}} = \dfrac{\beta_1 [\ce{L}]}{1 + \beta_1 [\ce{L}] + \beta_2 [\ce{L}]^2} \\[1ex] \upalpha_2 = \upalpha_{\ce{ML2}} = \dfrac{[\ce{ML2}]}{c_{\ce{M}}} = \dfrac{\beta_2 [\ce{L}]^2}{1 + \beta_1 [\ce{L}] + \beta_2 [\ce{L}]^2} \end{array} $
Ejemplo de cálculo del diagrama de distribución para un complejo $\ce{ML}$ tal que:
$ \ce{M + L <=> ML} \quad k_1 = 10^8 $
Las expresiones de las $\upalpha$ para este sistema:
$ \begin{array}{l} c_{\ce{M}} = [\ce{M}] + [\ce{ML}] \\[1ex] \beta_1 = \dfrac{[\ce{ML}]}{[\ce{M}][\ce{L}]} \, \Rightarrow \, [\ce{ML}] = \beta_1 [\ce{M}] [\ce{L}] \\[1ex] \Rightarrow \, c_{\ce{M}} = [\ce{M}] + \beta_1 [\ce{M}] [\ce{L}] \, \Rightarrow \, [\ce{M}] = \dfrac{c_{\ce{M}}}{1 + \beta_1 [\ce{L}]} \\[1ex] \Rightarrow \, \upalpha_{\ce{M}} = \dfrac{[\ce{M}]}{c_{\ce{M}}} = \dfrac{1}{1 + \beta_1 [\ce{L}]} \\[1em] [\ce{ML}] = c_{\ce{M}} - [\ce{M}] = c_{\ce{M}} - \dfrac{c_{\ce{M}}}{1 + \beta_1 [\ce{L}]} = \dfrac{c_{\ce{M}} \beta_1 [\ce{L}]}{1 + \beta_1 [L]} \\[1ex] \Rightarrow \, \upalpha_{\ce{ML}} = \dfrac{[\ce{ML}]}{c_{\ce{M}}} = \dfrac{\beta_1 [\ce{L}]}{1 + \beta_1 [\ce{L}]} \end{array} $
Teniendo en cuenta esto, a continuación pueden plantearse los siguientes casos:
a) $[\ce{L}] = 0{,}01 (k_1)^{-1}$ ; ${}\log[\ce{L}] = \ce{p$k$1} - 2$
$ \begin{array}{l} \Rightarrow \upalpha_{\ce{M}} = \dfrac{1}{1 + \beta_1 [\ce{L}]} = \dfrac{1}{1 + \! \cancel{k_1} 0{,}01 \dfrac{1}{\cancel{k_1}}} = \dfrac{1}{1{,}01} = 0{,}99 \approx 1 \\[1em] \Rightarrow \upalpha_{\ce{ML}} = \dfrac{\beta_1 [\ce{L}]}{1 + \beta_1 [L]} = \dfrac{ \cancel{k_1} 0{,}01 \dfrac{1}{\cancel{k_1}} }{ 1 + \! \cancel{k_1} 0{,}01 \dfrac{1}{\cancel{k_1}} } = \dfrac{0{,}01}{1{,}01} = 0{,}01 \approx 0 \end{array} $
b) $[\ce{L}] = 0{,}1 (k_1)^{-1}$ ; ${}\log[\ce{L}] = \ce{p$k$1} - 1$
$ \begin{array}{l} \Rightarrow \, \upalpha_{\ce{M}} = \dfrac{1}{1 + \beta_1 [\ce{L}]} = \dfrac{1}{1 + \! \cancel{k_1} 0{,}1 \dfrac{1}{\cancel{k_1}}} = \dfrac{1}{1{,}1} = 0{,}9 \\[1ex] \Rightarrow \, \upalpha_{\ce{ML}} = \, 0{,}1 \end{array} $
c) $[\ce{L}] = (k_1)^{-1}$ ; ${}\log[\ce{L}] = \ce{p$k$1}$
$ \begin{array}{l} \Rightarrow \, \upalpha_{\ce{M}} = \dfrac{1}{1 + \beta_1 [\ce{L}]} = \dfrac{1}{1 + \! \cancel{k_1} \dfrac{1}{\cancel{k_1}}} = \dfrac{1}{2} = 0{,}5 \\[1ex] \Rightarrow \, \upalpha_{\ce{ML}} = \, 0{,}5 \end{array} $
d) $[\ce{L}] = 10 (k_1)^{-1}$ ; ${}\log[\ce{L}] = \ce{p$k$1} + 1$
$ \begin{array}{l} \Rightarrow \, \alpha_{\ce{M}} = \dfrac{1}{1 + \beta_1 [\ce{L}]} = \dfrac{1}{1 + \! \cancel{k_1} 10 \dfrac{1}{\cancel{k_1}}} = \dfrac{1}{11} = 0{,}1 \\[1ex] \Rightarrow \, \alpha_{\ce{ML}} = \, 0{,}9 \end{array} $
e) $[\ce{L}] = 100 (k_1)^{-1}$ ; ${}\log[\ce{L}] = \ce{p$k$1} + 2$
$ \begin{array}{l} \Rightarrow \upalpha_{\ce{M}} = \dfrac{1}{1 + \beta_1 [\ce{L}]} = \dfrac{1}{1 + \! \cancel{k_1} 100 \dfrac{1}{\cancel{k_1}}} = \dfrac{1}{101} = 0{,}01 \approx 0 \\[1ex] \Rightarrow \, \upalpha_{\ce{ML}} = \, 0{,}99 \approx 1 \end{array} $
Todos estos datos pueden resumirse en la siguiente tabla:
| $\boldsymbol{\log[\ce{L}]}$ | ${\bf \upalpha}_{\ce{M}}$ | ${\bf \upalpha}_{\ce{ML}}$ |
|---|---|---|
| $\ce{p$k$1} - 2$ | ${\sim} 1 \ (0{,}99)$ | ${\sim}\mspace{2mu} 0 \ (0{,}01)$ |
| $\ce{p$k$1} - 1$ | $0{,}9$ | $0{,}1$ |
| $\ce{p$k$1}$ | $0{,}5$ | $0{,}5$ |
| $\ce{p$k$1} + 1$ | $0{,}1$ | $0{,}9$ |
| $\ce{p$k$1} + 2$ | ${\sim}\mspace{2mu} 0 \ (0{,}01)$ | ${\sim} 1 \ (0{,}99)$ |
Siendo la representación del diagrama de distribución para este sistema:
Diagrama de distribución de un sistema $\ce{ML2}$ partiendo de:
$ \begin{array}{l} \ce{M + L <=> ML} \quad k_1 = 10^8 \\[1ex] \ce{ML + L <=> ML2} \quad k_2 = 10^3 \end{array} $
En este ejemplo, la primera constante sucesiva es mayor que la segunda. Por tanto, la primera reacción es más fácil, hay una mayor facilidad para adquirir el ligando. En cambio, el orden de las constantes globales es al revés, la $\beta_2$ es mayor que la $\beta_1$.
Aquí también, para llevar a cabo la representación del diagrama de distribución, se pueden plantear distintos casos:
a) $[\ce{L}] = 0{,}01 (k_1)^{-1}$ ; ${}\log[\ce{L}] = \ce{p$k$1} - 2$
$ \begin{array}{l} \upalpha_{\ce{M}} = \dfrac{1}{1 + \beta_1 [\ce{L}] + \! \cancel{\beta_2 [\ce{L}]^2}} = \dfrac{1}{1 + \! \cancel{k_1} 0{,}01 \dfrac{1}{\cancel{k_1}}} = \dfrac{1}{1{,}01} = 0{,}99 \approx 1 \\[1em] \upalpha_{\ce{ML}} = \dfrac{\beta_1 [\ce{L}]}{1 + \beta_1 [\ce{L}] + \! \cancel{\beta_2 [\ce{L}]^2}} = \dfrac{ \cancel{k_1} 0{,}01 \dfrac{1}{\cancel{k_1}} }{ 1 + \! \cancel{k_1} 0{,}01 \dfrac{1}{\cancel{k_1}} } = \dfrac{0{,}01}{1{,}01} = 0{,}01 \approx 0 \\[1ex] \upalpha_{\ce{ML2}} \! = 0 \end{array} $
b) $[\ce{L}] = 0{,}1 (k_1)^{-1}$ ; ${}\log[\ce{L}] = \ce{p$k$1} - 1$
$ \begin{array}{l} \upalpha_{\ce{M}} = \dfrac{1}{1 + \beta_1 [\ce{L}] + \! \cancel{\beta_{2} [\ce{L}]^2}} = \dfrac{1}{1 \! + \cancel{k_1} 0{,}1 \dfrac{1}{\cancel{k_1}}} = \dfrac{1}{1{,}1} = 0{,}9 \\[1em] \upalpha_{\ce{ML}} = \dfrac{\beta_1 [\ce{L}]}{1 + \beta_1 [\ce{L}] + \! \cancel{\beta_2 [\ce{L}]^2}} = \dfrac{ \cancel{k_1} 0{,}1 \dfrac{1}{\cancel{k_1}} }{ 1 + \! \cancel{k_1} 0{,}1 \dfrac{1}{\cancel{k_1}} } = \dfrac{0{,}1}{1{,}1} = 0{,}1 \\[1ex] \upalpha_{\ce{ML2}} \! = 0 \end{array} $
c) $[\ce{L}] = (k_1)^{-1}$ ; ${}\log[\ce{L}] = \ce{p$k$1}$
$ \begin{array}{l} \upalpha_{\ce{M}} = \dfrac{1}{1 + \beta_1 [\ce{L}] + \! \cancel{\beta_2 [\ce{L}]^2}} = \dfrac{1}{1 + \! \cancel{k_1} \dfrac{1}{\cancel{k_1}}} = \dfrac{1}{2} = 0{,}5 \\[1em] \upalpha_{\ce{ML}} = \dfrac{\beta_1 [\ce{L}]}{1 + \beta_{1}[\ce{L}] + \! \cancel{\beta_2 [\ce{L}]^2}} = \dfrac{\cancel{k_1} \dfrac{1}{\cancel{k_1}}}{1 + \! \cancel{k_1} \dfrac{1}{\cancel{k_1}}} = \dfrac{1}{2} = 0{,}5 \\[1ex] \upalpha_{\ce{ML2}} \! = 0 \end{array} $
d) $[\ce{L}] = 10 (k_1)^{-1}$ ; ${}\log[\ce{L}] = \ce{p$k$1} + 1$
$ \begin{array}{l} \upalpha_{\ce{M}} = \dfrac{1}{1 + \beta_{1} [\ce{L}] + \! \cancel{\beta_2 [\ce{L}]^2}} = \dfrac{1}{1 + \! \cancel{k_1} 10 \dfrac{1}{\cancel{k_1}}} = \dfrac{1}{11} = 0{,}1 \\[1em] \upalpha_{\ce{ML}} = \dfrac{\beta_1 [\ce{L}]}{1 + \beta_1 [\ce{L}] + \! \cancel{\beta_2 [\ce{L}]^2}} = \dfrac{ \cancel{k_1} 10 \dfrac{1}{\cancel{k_1}} }{ 1 + \! \cancel{k_1} 10 \dfrac{1}{\cancel{k_1}} } = \dfrac{10}{11} = 0{,}9 \\[1ex] \upalpha_{\ce{ML2}} \! = 0 \end{array} $
e) $[\ce{L}] = 100 (k_1)^{-1}$ ; ${}\log[\ce{L}] = \ce{p$k$1} + 2$
$ \begin{array}{l} \upalpha_{\ce{M}} = \dfrac{1}{1 + \beta_1 [\ce{L}] + \! \cancel{\beta_2 [\ce{L}]^2}} = \dfrac{1}{1 + \! \cancel{k_1} 100 \dfrac{1}{\cancel{k_1}}} = \dfrac{1}{101} = 0{,}01 \approx 0 \\[1em] \upalpha_{\ce{ML}} = \dfrac{\beta_1 [\ce{L}]}{1 + \beta_1 [\ce{L}] + \! \cancel{\beta_2 [\ce{L}]^2}} = \dfrac{ \cancel{k_1} 100 \dfrac{1}{\cancel{k_1}} }{ 1 + \! \cancel{k_1} 100 \dfrac{1}{\cancel{k_1}} } = \dfrac{100}{101} = 0{,}99 \approx 1 \\[1ex] \upalpha_{\ce{ML2}} \! = 0 \end{array} $
f) $[\ce{L}] = 0{,}01 (k_2)^{-1}$ ; ${}\log[\ce{L}] = \ce{p$k$2} - 2$
$ \begin{array}{l} \upalpha_{\ce{ML}} = \dfrac{ \beta_1 \! \cancel{[\ce{L}]} }{ \color{blue}{\cancel{\color{#222}{1}}} \! + \beta_{1} \! \cancel{[\ce{L}]} \! + \beta_2 [\ce{L}]^{\cancel{2}} } = \dfrac{ \color{red}{\cancel{\color{#222}{k_1}}} }{ \color{red}{\cancel{\color{#222}{k_1}}} \! + \! \color{red}{\cancel{\color{#222}{k_1}}} \! \cancel{k_2} 0{,}01 \! \dfrac{1}{\cancel{k_2}} } = \dfrac{1}{1,01} = 0{,}99 \approx 1 \\[1em] \upalpha_{\ce{ML2}} = \dfrac{ \beta_2 [\ce{L}]^{\cancel{2}} }{ \color{blue}{\cancel{\color{#222}{1}}} \! + \beta_1 \! \cancel{[\ce{L}]} \! + \beta_2[\ce{L}]^{\cancel{2}} } = \dfrac{ \color{red}{\cancel{\color{#222}{k_1}}} \! \cancel{k_2} 0{,}01 \! \dfrac{1}{\cancel{k_2}} }{ \color{red}{\cancel{\color{#222}{k_1}}} \! + \! \color{red}{\cancel{\color{#222}{k_1}}} \! \cancel{k_2} 0{,}01 \! \dfrac{1}{\cancel{k_2}} } = \dfrac{0{,}01}{1{,}01} = 0{,}01 \approx 0 \\[1ex] \upalpha_{\ce{M}} \! = 0 \end{array} $
g) $[\ce{L}] = 0{,}1 (k_2)^{-1}$ ; ${}\log[\ce{L}] = \ce{p$k$2} - 1$
$ \begin{array}{l} \upalpha_{\ce{ML}} = \dfrac{ \beta_1 \! \cancel{[\ce{L}]} }{ \color{blue}{\cancel{\color{#222}{1}}} \! + \beta_1 \! \cancel{[\ce{L}]} \! + \beta_2 [\ce{L}]^{\cancel{2}} } = \dfrac{ \color{red}{\cancel{\color{#222}{k_1}}} }{ \color{red}{\cancel{\color{#222}{k_1}}} \! + \! \color{red}{\cancel{\color{#222}{k_1}}} \! \cancel{k_2} 0{,}1 \! \dfrac{1}{\cancel{k_2}} } = \dfrac{1}{1{,}1} = 0{,}9 \\[1em] \upalpha_{\ce{ML2}} = \dfrac{ \beta_2 [\ce{L}]^{\cancel{2}} }{ \color{blue}{\cancel{\color{#222}{1}}} \! + \beta_1 \! \cancel{[\ce{L}]} \! + \beta_2 [\ce{L}]^{\cancel{2}} } = \dfrac{ \color{red}{\cancel{\color{#222}{k_1}}} \! \cancel{k_2} 0{,}1 \! \dfrac{1}{\cancel{k_2}} }{ \color{red}{\cancel{\color{#222}{k_1}}} \! + \! \color{red}{\cancel{\color{#222}{k_1}}} \! \cancel{k_2} 0{,}1 \! \dfrac{1}{\cancel{k_2}} } = \dfrac{0{,}1}{1{,}1} = 0{,}1 \\[1ex] \upalpha_{\ce{M}} \! = 0 \end{array} $
h) $[\ce{L}] = (k_2)^{-1}$ ; ${}\log[\ce{L}] = \ce{p$k$2}$
$ \begin{array}{l} \upalpha_{\ce{ML}} = \dfrac{ \beta_1 \! \cancel{[\ce{L}]} }{ \color{blue}{\cancel{\color{#222}{1}}} \! + \beta_1 \! \cancel{[\ce{L}]} \! + \beta_2 [\ce{L}]^{\cancel{2}}} = \dfrac{ \color{red}{\cancel{\color{#222}{k_1}}} }{ \color{red}{\cancel{\color{#222}{k_1}}} \! + \! \color{red}{\cancel{\color{#222}{k_1}}} \! \cancel{k_2} \! \dfrac{1}{\cancel{k_2}} } = \dfrac{1}{2} = 0{,}5 \\[1em] \upalpha_{\ce{ML2}} = \dfrac{ \beta_2 [\ce{L}]^{\cancel{2}} }{ \color{blue}{\cancel{\color{#222}{1}}} \! + \beta_1 \! \cancel{[\ce{L}]} \! + \beta_2 [\ce{L}]^{\cancel{2}} } = \dfrac{ \color{red}{\cancel{\color{#222}{k_1}}} \! \cancel{k_2} \! \dfrac{1}{\cancel{k_2}} }{ \color{red}{\cancel{\color{#222}{k_1}}} \! + \! \color{red}{\cancel{\color{#222}{k_1}}} \! \cancel{k_2} \! \dfrac{1}{\cancel{k_2}} } = \dfrac{1}{2} = 0{,}5 \\[1ex] \upalpha_{\ce{M}} \! = 0 \end{array} $
i) $[\ce{L}] = 10 (k_2)^{-1}$ ; ${}\log[\ce{L}] = \ce{p$k$2} + 1$
$ \begin{array}{l} \upalpha_{\ce{ML}} = \dfrac{ \beta_1 \! \cancel{[\ce{L}]} }{ \color{blue}{\cancel{\color{#222}{1}}} \! + \beta_1 \! \cancel{[\ce{L}]} \! + \beta_2 [\ce{L}]^{\cancel{2}} } = \dfrac{ \color{red}{\cancel{\color{#222}{k_1}}} }{ \color{red}{\cancel{\color{#222}{k_1}}} \! + \! \color{red}{\cancel{\color{#222}{k_1}}} \! \cancel{k_2} 10 \dfrac{1}{\cancel{k_2}} } = \dfrac{1}{11} = 0{,}1 \\[1em] \upalpha_{\ce{ML2}} = \dfrac{ \beta_2 [\ce{L}]^{\cancel{2}} }{ \color{blue}{\cancel{\color{#222}{1}}} \! + \beta_1 \! \cancel{[\ce{L}]} \! + \beta_2 [\ce{L}]^{\cancel{2}} } = \dfrac{ \color{red}{\cancel{\color{#222}{k_1}}} \! \cancel{k_2} 10 \dfrac{1}{\cancel{k_2}} }{ \color{red}{\cancel{\color{#222}{k_1}}} \! + \! \color{red}{\cancel{\color{#222}{k_1}}} \! \cancel{k_2} 10 \dfrac{1}{\cancel{k_2}} } = \dfrac{10}{11} = 0{,}9 \\[1ex] \upalpha_{\ce{M}} \! = 0 \end{array} $
j) $[\ce{L}] = 100 (k_2)^{-1}$ ; ${}\log[\ce{L}] = \ce{p$k$2} + 2$
$ \begin{array}{l} \upalpha_{\ce{ML}} = \dfrac{ \beta_1 \! \cancel{[\ce{L}]} }{ \color{blue}{\cancel{\color{#222}{1}}} \! + \beta_1 \! \cancel{[\ce{L}]} \! + \beta_2 [\ce{L}]^{\cancel{2}} } = \dfrac{ \color{red}{\cancel{\color{#222}{k_1}}} }{ \color{red}{\cancel{\color{#222}{k_1}}} \! + \! \color{red}{\cancel{\color{#222}{k_1}}} \! \cancel{k_2} 100 \dfrac{1}{\cancel{k_2}} } = \dfrac{1}{101} = 0{,}01 \approx 0\\[1em] \upalpha_{\ce{ML2}} = \dfrac{ \beta_2 [\ce{L}]^{\cancel{2}} }{ \color{blue}{\cancel{\color{#222}{1}}} \! + \beta_1 \! \cancel{[\ce{L}]} \! + \beta_2 [\ce{L}]^{\cancel{2}} } = \dfrac{ \color{red}{\cancel{\color{#222}{k_1}}} \! \cancel{k_2} 100 \dfrac{1}{\cancel{k_2}} }{ \color{red}{\cancel{\color{#222}{k_1}}} \! + \! \color{red}{\cancel{\color{#222}{k_1}}} \! \cancel{k_2} 100 \dfrac{1}{\cancel{k_2}} } = \dfrac{100}{101} = 0{,}99 \approx 1 \\[1ex] \upalpha_{\ce{M}} \! = 0 \end{array} $
Tabla resumen de lo anterior:
| $\boldsymbol{\log[\ce{L}]}$ | ${\bf\upalpha}_{\ce{M}}$ | ${\bf\upalpha}_{\ce{ML}}$ | ${\bf\upalpha}_{\ce{ML2}}$ |
|---|---|---|---|
| $\ce{p$k$1} - 2$ | ${\sim} 1 \ (0{,}99)$ | ${\sim} \mspace{2mu} 0 \ (0{,}01)$ | $0$ |
| $\ce{p$k$1} - 1$ | $0{,}9$ | $0{,}1$ | $0$ |
| $\ce{p$k$1}$ | $0{,}5$ | $0{,}5$ | $0$ |
| $\ce{p$k$1} + 1$ | $0{,}1$ | $0{,}9$ | $0$ |
| $\ce{p$k$1} + 2$ | ${\sim} \mspace{2mu} 0 \ (0{,}01)$ | ${\sim} 1 \ (0{,}99)$ | $0$ |
| $\ce{p$k$2} - 2$ | $0$ | ${\sim} 1 \ (0{,}99)$ | ${\sim} \mspace{2mu} 0 \ (0{,}01)$ |
| $\ce{p$k$2} - 1$ | $0$ | $0{,}9$ | $0{,}1$ |
| $\ce{p$k$2}$ | $0$ | $0{,}5$ | $0{,}5$ |
| $\ce{p$k$2} + 1$ | $0$ | $0{,}1$ | $0{,}9$ |
| $\ce{p$k$2} + 2$ | $0$ | ${\sim} \mspace{2mu} 0 \ (0{,}01)$ | ${\sim} 1 \ (0{,}99)$ |
La representación del diagrama de distribución de este sistema:
Este gráfico es el típico de un sistema $\ce{ML2}$ en que $k_1 \mspace{2mu} {\gg} \, k_2$. El gráfico cambia cuando ambas constantes sucesivas se juntan. Esto es:
1) $k_1\mspace{2mu} {\gg} \, k_2$
2) $k_1 \mspace{2mu} {>} \, k_2$
En este caso $\ce{ML}$ no llega a predominar de forma total nunca.
3) $k_1 \! = k_2$
Aquí, en esta situación, $\ce{ML}$ nunca está en superioridad.
En general, esto todavía se puede complicar más según vayan aumentando el número de ligandos, por tanto de constantes. Este tipo de esquemas pueden tener una distribución de especies muy amplia, y sirven de ayuda para conocer que especie o especies se tienen en función de la concentración de ligando libre.