Calcule as frações das espécies \mathrm{Hg}^{2+}, \mathrm{HgCl}^{+} e \mathrm{HgCl}_{2} em uma mistura de íons \mathrm{Hg}^{2+} e \mathrm{Cl}^{-}para concentração de \left[\mathrm{Cl}^{-}\right]=10^{-6} \mathrm{~mol} / \mathrm{L}.
Dados: \mathrm{K}_{\mathrm{f} 1}=5,50 \times 10^{6} ; \mathrm{K}_{\mathrm{f} 2}=3,02 \times 10^{6} ; \mathrm{K}_{\mathrm{f} 3}=7,08 ; \mathrm{K}_{\mathrm{f} 4}=10
\begin{aligned} \mathrm{Hg}^{2+}+\mathrm{Cl}^{-} \rightleftharpoons \mathrm{HgCl}^{+} & \mathrm{K}_{\mathrm{fl}} \\\ \mathrm{HgCl}^{+}+\mathrm{Cl}^{-} \rightleftharpoons \mathrm{HgCl}_{2} & \mathrm{~K}_{\mathrm{f} 2} \\\ \mathrm{HgCl}_{2}+\mathrm{Cl}^{-} \rightleftharpoons \mathrm{HgCl}_{3}^{-} & \mathrm{K}_{\mathrm{f} 3} \\\ \mathrm{HgCl}_{3}^{-}+\mathrm{Cl}^{-} \rightleftharpoons \mathrm{HgCl}_{4}^{2-} & \mathrm{K}_{\mathrm{f} 4} \end{aligned}

Passo 1

O enunciado pede a especiação de mercúrio em \mathrm{Hg}^{2+}, \mathrm{HgCl}^{+} e \mathrm{HgCl}_{2} numa solução onde \left[\mathrm{Cl}^{-}\right]=10^{-6},mol/L. O conteúdo é equilíbrio de complexação, usando constantes de formação
\mathrm{K}_{\mathrm{f}1},\mathrm{K}_{\mathrm{f}2},\mathrm{K}_{\mathrm{f}3},\mathrm{K}_{\mathrm{f}4}.

Passo 2

Expressamos cada espécie em função de [\mathrm{Hg}^{2+}] e de [\mathrm{Cl}^{-}]=C:
[\mathrm{HgCl}^{+}] = K_{\mathrm{f}1}\,[\mathrm{Hg}^{2+}]\;C,
[\mathrm{HgCl}_{2}] = K_{\mathrm{f}1}\,K_{\mathrm{f}2}\,[\mathrm{Hg}^{2+}]\;C^{2},
[\mathrm{HgCl}_{3}^{-}] = K_{\mathrm{f}1}\,K_{\mathrm{f}2}\,K_{\mathrm{f}3}\,[\mathrm{Hg}^{2+}]\;C^{3},
[\mathrm{HgCl}_{4}^{2-}] = K_{\mathrm{f}1}\,K_{\mathrm{f}2}\,K_{\mathrm{f}3}\,K_{\mathrm{f}4}\,[\mathrm{Hg}^{2+}]\;C^{4}.
O total de mercúrio é
C_{\text{tot}} = [\mathrm{Hg}^{2+}]\Bigl(1 + \beta_{1}C + \beta_{2}C^{2} + \beta_{3}C^{3} + \beta_{4}C^{4}\Bigr),
onde \beta_{n} são as constantes cumulativas.

Passo 3

Definimos
\beta_{1}=K_{\mathrm{f}1},
\beta_{2}=K_{\mathrm{f}1}K_{\mathrm{f}2},
\beta_{3}=\beta_{2}K_{\mathrm{f}3},
\beta_{4}=\beta_{3}K_{\mathrm{f}4}.
As frações são então
\alpha_{0} = \frac{[\mathrm{Hg}^{2+}]}{C_{\text{tot}}} = \frac{1}{D},\quad \alpha_{1} = \frac{[\mathrm{HgCl}^{+}]}{C_{\text{tot}}} = \frac{\beta_{1}C}{D},\quad \alpha_{2} = \frac{[\mathrm{HgCl}_{2}]}{C_{\text{tot}}} = \frac{\beta_{2}C^{2}}{D},
com
D = 1 + \beta_{1}C + \beta_{2}C^{2} + \beta_{3}C^{3} + \beta_{4}C^{4}.

Passo 4

Substituímos \,C=10^{-6},mol/L e os valores dados:
\beta_{1}=5{,}50\times10^{6},\quad \beta_{2}=5{,}50\times10^{6}\times3{,}02\times10^{6}=1{,}662\times10^{13},
\beta_{3}=1{,}662\times10^{13}\times7{,}08=1{,}176\times10^{14},\quad \beta_{4}=1{,}176\times10^{14}\times10=1{,}176\times10^{15}.
Calculamos os termos de D:
\beta_{1}C =5{,}50,\quad \beta_{2}C^{2}=16{,}62,\quad \beta_{3}C^{3}=1{,}176\times10^{-4},\quad \beta_{4}C^{4}=1{,}176\times10^{-9},
D = 1 + 5{,}50 +16{,}62 +0{,}0001176 +0{,}0000000012 \approx23{,}1201.

Passo 5

Agora as frações:
\alpha_{0} = \frac{1}{23{,}1201}\approx0{,}0432, \quad \alpha_{1} = \frac{5{,}50}{23{,}1201}\approx0{,}2379, \quad \alpha_{2} = \frac{16{,}62}{23{,}1201}\approx0{,}7191.

Resposta

\displaystyle \alpha(\mathrm{Hg}^{2+})\approx0{,}0432,\quad \alpha(\mathrm{HgCl}^{+})\approx0{,}2379,\quad \alpha(\mathrm{HgCl}_{2})\approx0{,}7191.

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