Un diagrama de distribución consiste en representar la fracción de una especie respecto a la total de soluto (concentración analítica) en función del $\ce{pH}$.
$ \def\upalpha{\smash{\unicode{x03B1}}} \upalpha = \dfrac{\text{concentración de la especie}}{\text{conc. analítica}} $
Se representa $\upalpha$, cuyo valor como máximo puede ser $1$, frente al $\ce{pH}$.
A continuación, ejemplo de un ácido diprótico $\ce{H2A}$ con $\ce{p$K$1} \! = 4$ y $\ce{p$K$2} = 10$.
$ \begin{array}{l} \upalpha_{\ce{H2A}} = \dfrac{[\ce{H2A}]}{c_{\rm a}} \\[1ex] \upalpha_{\ce{HA-}} = \dfrac{[\ce{HA-}]}{c_{\rm a}} \\[1ex] \upalpha_{\ce{A^{2-}}} = \dfrac{[\ce{A^{2-}}]}{c_{\rm a}} \end{array} $
Así pues:
$ \begin{array}{l} \begin{split} \upalpha_{\ce{H2A}} &= \dfrac{[\ce{H2A}]}{c_{\rm a}} = \dfrac{ \cancel{c_{\rm a}} \mspace{-1mu} [\ce{H3O+}]^2 }{ \cancel{c_{\rm a}} \mspace{-5mu} \left( [\ce{H3O+}]^2 + K_1 [\ce{H3O+}] + K_1 K_2 \right) } = \\[1ex] &= \dfrac{[\ce{H3O+}]^2}{[\ce{H3O+}]^2 + K_1 [\ce{H3O+}] + K_1 K_2} \end{split} \\[1em] \begin{split} \upalpha_{\ce{HA-}} &= \dfrac{[\ce{HA-}]}{c_{\rm a}} = \dfrac{ \cancel{c_{\rm a}} \mspace{-1mu} K_1 [\ce{H3O+}] }{ \cancel{c_{\rm a}} \mspace{-4mu} \left( [\ce{H3O+}]^2 + K_1 [\ce{H3O+}] + K_1 K_2 \right) } = \\[1ex] &= \dfrac{K_1 [\ce{H3O+}]}{[\ce{H3O+}]^2 + K_1 [\ce{H3O+}] + K_1 K_2} \end{split} \\[1em] \begin{split} \upalpha_{\ce{A^{2-}}} &= \dfrac{[\ce{A^{2-}}]}{c_{\rm a}} = \dfrac{ \cancel{c_{\rm a}} \mspace{-1mu} K_1 K_2 }{ \cancel{c_{\rm a}} \mspace{-5mu} \left( [\ce{H3O+}]^2 + K_1 [\ce{H3O+}] + K_1 K_2 \right) } = \\[1ex] &= \dfrac{K_1 K_2}{[\ce{H3O+}]^2 + K_1 [\ce{H3O+}] + K_1 K_2} \end{split} \end{array} $
Con estas expresiones pueden calcularse los valores de las $\upalpha$ a distintos $\ce{pH}$.
a) $\ce{pH} = \ce{p$K$1} - 2\,$ ; $\,[\ce{H3O+}] = 100 K_1$
$ \begin{array}{l} \upalpha_{\ce{H2A}} = \dfrac{ 100^2 \! \cancel{(K_1)^2} }{ 100^2 \! \cancel{(K_1)^2} \! + 100 \! \cancel{(K_1)^2} \! + \! \cancel{K_1 K_2} } = \dfrac{10000}{10100} = 0{,}99 \approx 1 \\[1em] \upalpha_{\ce{HA-}} = \dfrac{ 100 \! \cancel{(K_1)^2} }{ 100^2 \! \cancel{(K_1)^2} \! + 100 \! \cancel{(K_1)^2} \! + \! \cancel{K_1 K_2} } = \dfrac{100}{10100} = 0{,}01 \approx 0 \\[1em] \upalpha_{\ce{A^{2-}}} = 0 \end{array} $
b) $\ce{pH} = \ce{p$K$1} - 1$ ; $\,[\ce{H3O+}] = 10 K_1$
$ \begin{array}{l} \upalpha_{\ce{H2A}} = \dfrac{ 10^2 \! \cancel{(K_1)^2} }{ 10^2 \! \cancel{(K_1)^2} \! + 10 \! \cancel{(K_1)^2} \! + \! \cancel{K_1 K_2} } = \dfrac{100}{110} = 0{,}9 \\[1em] \upalpha_{\ce{HA-}} = \dfrac{ 10 \! \cancel{(K_1)^2} }{ 10^2 \! \cancel{(K_1)^2} \! + 10 \! \cancel{(K_1)^2} \! + \! \cancel{K_1 K_2} } = \dfrac{10}{110} = 0{,}1 \\[1em] \upalpha_{\ce{A^{2-}}} = 0 \end{array} $
Por tanto $\ce{H2A}$ está presente con un valor del $\pu{90\%}$ de la $c_{\rm a}$, y $\ce{HA-}\!$ del $\pu{10\%}$, cuando el $\ce{pH}$ está una unidad por debajo de $\ce{p$K$1}\!$. Cuando el $\ce{pH}$ está dos unidades por debajo del valor del equilibrio entonces las cantidades son del $\pu{99\%}$ y $\pu{1\%}$ respectivamente. Esto es así porque el equilibrio se desplaza a la izquierda, hacia especie más protonada, a $\ce{pH}$ más bajo.
c) $\ce{pH} = \ce{p$K$1}$ ; $\,[\ce{H3O+}] = K_1$
$ \begin{array}{l} \upalpha_{\ce{H2A}} = \dfrac{ [\ce{H3O+}]^2 }{ [\ce{H3O+}]^2 + [\ce{H3O+}]^2 + \! \cancel{K_1 K_2} } = 0{,}5 \\[1em] \upalpha_{\ce{HA-}} = 0{,}5 \\[1em] \upalpha_{\ce{A^{2-}}} = 0 \end{array} $
Es lógico que siga siendo $\upalpha_{\ce{A^{2-}}} \! \approx 0$ ya que el segundo equilibrio es más difícil que se dé. Esto se ve en la expresión:
$ \upalpha_{\ce{A^{2-}}} = \dfrac{[\ce{H3O+}] K_2}{[\ce{H3O+}]^2 + [\ce{H3O+}]^2 + K_1 K_2} $
El númerador es $10^{-6}$ veces más pequeño que en los otros.
d) $\ce{pH} = \ce{p$K$1} + 1$ ; $\,K_1 = 10 [\ce{H3O+}]$
$ \begin{array}{l} \upalpha_{\ce{H2A}} = \dfrac{ \cancel{[\ce{H3O+}]^2} }{ \cancel{[\ce{H3O+}]^2} \! + 10 \! \cancel{[\ce{H3O+}]^2} \! + \! \cancel{K_1 K_2} } = \dfrac{1}{11} = 0{,}1 \\[1em] \upalpha_{\ce{HA-}} = \dfrac{ 10 \! \cancel{[\ce{H3O+}]^2} }{ \cancel{[\ce{H3O+}]^2} \! + 10 \! \cancel{[\ce{H3O+}]^2} \! + \! \cancel{K_1 K_2} } = \dfrac{10}{11} = 0{,}9 \\[1em] \upalpha_{\ce{A^{2-}}} = 0 \end{array} $
e) $\ce{pH} = \ce{p$K$1} + 2\,$ ; $\,K_1 = 100 [\ce{H3O+}]$
$ \begin{array}{l} \upalpha_{\ce{H2A}} = \dfrac{ \cancel{[\ce{H3O+}]^2} }{ \cancel{[\ce{H3O+}]^2} \! + 100 \! \cancel{[\ce{H3O+}]^2} \! + \! \cancel{K_1 K_2} } = \dfrac{1}{101} = 0{,}01 \approx 0 \\[1em] \upalpha_{\ce{HA-}} = \dfrac{ 100 \! \cancel{[\ce{H3O+}]^2} }{ \cancel{[\ce{H3O+}]^2} \! + 100 \! \cancel{[\ce{H3O+}]^2} \! + \! \cancel{K_1 K_2} } = \dfrac{100}{101} = 0{,}99 \approx 1 \\[1em] \upalpha_{\ce{A^{2-}}} = 0 \end{array} $
f) $\ce{pH} = \ce{p$K$1} + 3\,$ ; $\,K_1 = 1000 [\ce{H3O+}]$
$ \begin{array}{l} \upalpha_{\ce{H2A}} = \dfrac{ \cancel{[\ce{H3O+}]^2} }{ \cancel{[\ce{H3O+}]^2} \! + 1000 \! \cancel{[\ce{H3O+}]^2} \! + \! \cancel{K_1 K_2} } = \dfrac{1}{1001} = 0{,}001 \approx 0 \\[1em] \upalpha_{\ce{HA-}} = \dfrac{ 1000 \! \cancel{[\ce{H3O+}]^2} }{ \cancel{[\ce{H3O+}]^2} \! + 1000 \! \cancel{[\ce{H3O+}]^2} \! + \! \cancel{K_1 K_2} } = \dfrac{1000}{1001} = 0{,}999 \approx 1 \\[1em] \upalpha_{\ce{A^{2-}}} = \dfrac{ \cancel{1000} \! \cancel{[\ce{H3O+}]} \! K_2 }{ \cancel{[\ce{H3O+}]^2} \! + \! \cancel{1000} \! [\ce{H3O+}]^{\cancel{2}} \! + \! \cancel{K_1 K_2} } = \dfrac{K_2}{[\ce{H3O+}]} = 0{,}001 \approx 0 \end{array} $
En este punto se acaba el $\ce{H2A}$ y empieza a ser significativo $\ce{A^{2-}}\!$. Esto es así porque al estar ya a $3$ unidades del $\ce{p$K$2}\!$ empieza a influir su equilibrio, en el que las especies implicadas son como especie ácida el $\ce{HA-}\!$ y básica el $\ce{A^{2-}}\!$.
g) $\ce{pH} = \ce{p$K$2} - 2$ ; $\,[\ce{H3O+}] = 100 K_2$
$ \begin{array}{l} \upalpha_{\ce{HA-}} = \dfrac{ 100 \! \cancel{K_1 K_2} }{ \cancel{100^{\mspace{2mu} 2} (K_2)^2} \! + 100 \cancel{K_1 K_2} \! + \! \cancel{K_1 K_2} } = \dfrac{100}{101} = 0{,}99 \approx 1 \\[1em] \upalpha_{\ce{A^{2-}}} = \dfrac{ \cancel{K_1 K_2} }{ \cancel{100^{\mspace{2mu} 2} (K_2)^2} \! + 100 \! \cancel{K_1 K_2} + \cancel{K_1 K_2} } = \dfrac{1}{101} = 0{,}01 \approx 0 \\[1em] \upalpha_{\ce{H2A}} = 0 \end{array} $
h) $\ce{pH} = \ce{p$K$2} - 1$ ; $\,[\ce{H3O+}] = 10 K_2$
$ \begin{array}{l} \upalpha_{\ce{HA-}} = \dfrac{10}{11} = 0{,}9 \\[1em] \upalpha_{\ce{A^{2-}}} = \dfrac{1}{11} = 0{,}1 \\[1em] \upalpha_{\ce{H2A}} = 0 \end{array} $
i) $\ce{pH} = \ce{p$K$2}$ ; $\,[\ce{H3O+}] = K_2$
$ \begin{array}{l} \upalpha_{\ce{HA-}} = \dfrac{1}{2} = 0{,}5 \\[1em] \upalpha_{\ce{A^{2-}}} = \dfrac{1}{2} = 0{,}5 \\[1em] \upalpha_{\ce{H2A}} = 0 \end{array} $
j) $\ce{pH} = \ce{p$K$2} + 1$ ; $\,K_2 = 10 [\ce{H3O+}]$
$ \begin{array}{l} \upalpha_{\ce{HA-}} = \dfrac{ \cancel{K_1 [\ce{H3O+}]} }{ \cancel{[\ce{H3O+}]^2} \! + \! \cancel{K_1 [\ce{H3O+}]} \! + 10 \! \cancel{K_1 [\ce{H3O+}]} } = \dfrac{1}{11} = 0{,}1 \\[1em] \upalpha_{\ce{A^{2-}}} = \dfrac{ 10 \! \cancel{K_1 [\ce{H3O+}]} }{ \cancel{[\ce{H3O+}]^2} \! + \! \cancel{K_1 [\ce{H3O+}]} \! + 10 \! \cancel{K_1 [\ce{H3O+}]} } = \dfrac{10}{11} = 0{,}9 \\[1em] \upalpha_{\ce{H2A}} = 0 \end{array} $
k) $\ce{pH} = \ce{p$K$2} + 2\,$ ; $\,K_2 = 100 [\ce{H3O+}]$
$ \begin{array}{l} \upalpha_{\ce{HA-}} = \dfrac{1}{101} = 0{,}01 \approx 0 \\[1em] \upalpha_{\ce{A^{2-}}} = \dfrac{100}{101} = 0{,}99 \approx 1 \\[1em] \upalpha_{\ce{H2A}} = 0 \end{array} $
Teniendo en cuenta lo visto, puede construirse la siguiente tabla:
| $\vphantom{()} {\bf \upalpha}_{\ce{H2A}}$ | $\vphantom{() {\bf \upalpha}_{\ce{H2A}}} \smash{{\bf \upalpha}_{\ce{HA-}}}$ | $\vphantom{() {\bf \upalpha}_{\ce{H2A}}} \smash{{\bf \upalpha}_{\ce{A^{2-}}}}$ | |
|---|---|---|---|
| $\ce{pH} = \ce{p$K$1} - 2$ | $\vphantom{\ce{pH} = \ce{p$K$1} - 2} \smash{\sim 1 \ (0{,}99)}$ | $\vphantom{\ce{pH} = \ce{p$K$1} - 2} \smash{\sim 0 \ (0{,}01)}$ | $\vphantom{\ce{pH} = \ce{p$K$1} - 2} \smash{0}$ |
| $\ce{pH} = \ce{p$K$1} - 1$ | $\vphantom{\ce{pH} = \ce{p$K$1} - 1} 0{,}9$ | $\vphantom{\ce{pH} = \ce{p$K$1} - 1} 0{,}1$ | $\vphantom{\ce{pH} = \ce{p$K$1} - 1} 0$ |
| $\ce{pH} = \ce{p$K$1}$ | $\vphantom{\ce{pH} = \ce{p$K$1}} 0{,}5$ | $\vphantom{\ce{pH} = \ce{p$K$1}} 0{,}5$ | $\vphantom{\ce{pH} = \ce{p$K$1}} 0$ |
| $\ce{pH} = \ce{p$K$1} + 1$ | $\vphantom{\ce{pH} = \ce{p$K$1} + 1} 0{,}1$ | $\vphantom{\ce{pH} = \ce{p$K$1} + 1} 0{,}9$ | $\vphantom{\ce{pH} = \ce{p$K$1} + 1} 0$ |
| $\ce{pH} = \ce{p$K$1} + 2$ | $\vphantom{\ce{pH} = \ce{p$K$1} + 2} \smash{\sim 0 \ (0{,}01)}$ | $\vphantom{\ce{pH} = \ce{p$K$1} + 2} \smash{\sim 1 \ (0{,}99)}$ | $\vphantom{\ce{pH} = \ce{p$K$1} + 2} \smash{0}$ |
| $\ce{pH} = \ce{p$K$1} + 3$ | $\vphantom{\ce{pH} = \ce{p$K$1} + 3} 0$ | $\vphantom{\ce{pH} = \ce{p$K$1} + 3} 1$ | $\vphantom{\ce{pH} = \ce{p$K$1} + 3} 0$ |
| $\ce{pH} = \ce{p$K$2} - 2$ | $\vphantom{\ce{pH} = \ce{p$K$2} - 2} \smash{0}$ | $\vphantom{\ce{pH} = \ce{p$K$2} - 2} \smash{\sim 1 \ (0{,}99)}$ | $\vphantom{\ce{pH} = \ce{p$K$2} - 2} \smash{\sim 0 \ (0{,}01)}$ |
| $\ce{pH} = \ce{p$K$2} - 1$ | $\vphantom{\ce{pH} = \ce{p$K$2} - 1} 0$ | $\vphantom{\ce{pH} = \ce{p$K$2} - 1} 0{,}9$ | $\vphantom{\ce{pH} = \ce{p$K$2} - 1} 0{,}1$ |
| $\ce{pH} = \ce{p$K$2}$ | $\vphantom{\ce{pH} = \ce{p$K$2} - 1} 0$ | $\vphantom{\ce{pH} = \ce{p$K$2} - 1} 0{,}5$ | $\vphantom{\ce{pH} = \ce{p$K$2} - 1} 0{,}5$ |
| $\ce{pH} = \ce{p$K$2} + 1$ | $\vphantom{\ce{pH} = \ce{p$K$2} - 1} 0$ | $\vphantom{\ce{pH} = \ce{p$K$2} - 1} 0{,}1$ | $\vphantom{\ce{pH} = \ce{p$K$2} - 1} 0{,}9$ |
| $\ce{pH} = \ce{p$K$2} + 2$ | $\vphantom{\ce{pH} = \ce{p$K$2} + 2} \smash{0}$ | $\vphantom{\ce{pH} = \ce{p$K$2} + 2} \smash{\sim 0 \ (0{,}01)}$ | $\vphantom{\ce{pH} = \ce{p$K$2} + 2} \smash{\sim 1 \ (0{,}99)}$ |
La representación del diagrama de distribución para este ejemplo: