Calcule a fração de cada espécie numa solução 0,050 M de nitrato de mercúrio contendo cloreto nas seguintes concentrações : 1,10^-2, 10^-4, 10^-6, 10^-8 e 10^-10 M.
Dados : K1=5,5 times 10^6 ; K2=3,02 times 10^6 ; K3=7,08 ; K4=10.

Passo 1

Resumindo o enunciado: temos uma solução de Hg^2+ ;0,050 M (vindos do nitrato de mercúrio) à qual adicionamos cloreto nas concentrações ,[Cl^-]=10^-2,10^-4,10^-6,10^-8,10^-10,M. O que o exercício pede é a fração de cada espécie de mercúrio (i.e.\ Hg^2+, HgCl^+, HgCl2, HgCl3^-, HgCl4^2-) em equilíbrio. O conteúdo envolvido é a distribuição de espécies em complexação passo a passo, usando as constantes de formação passo a passo
Hg^2+ + Cl^- <=ftharpoons HgCl^+,;K1, HgCl^+ + Cl^- <=ftharpoons HgCl2,;K2,
HgCl2 + Cl^- <=ftharpoons HgCl3^-,;K3, HgCl3^- + Cl^- <=ftharpoons HgCl4^2-,;K4

Passo 2

Definimos as constantes de formação cumulativas
beta0=1, beta1=K1, beta2=K1K2, beta3=K1K2K3, beta4=K1K2K3K4
Substituindo os valores numéricos K1=5,5times10^6, K2=3,02times10^6, K3=7,08, K4=10, obtemos
beta1=5,5times10^6, beta2=1,661times10^13, beta3=1,177times10^14, beta4=1,177times10^15

Passo 3

A fração alphai da espécie com i cloretos é dada por
alphai =betai,[Cl^-]^i ,sumj=0^4betaj,[Cl^-]^j,,
onde beta0=1. Denominamos
D=sumj=0^4betaj,[Cl^-]^j =1+beta1[Cl^-]+beta2[Cl^-]^2+beta3[Cl^-]^3+beta4[Cl^-]^4
e então alphai=(betai[Cl^-]^i)/D.

Passo 4

Para [Cl^-]=10^-2,M calculamos
beta1[Cl^-]&=5,5times10^4, beta2[Cl^-]^2&=1,661times10^9, beta3[Cl^-]^3&=1,177times10^8, beta4[Cl^-]^4&=1,177times10^7, D&approx1,79times10^9
Logo
alpha(Hg^2+)=5,6times10^-10, alpha(HgCl^+)=3,1times10^-5, alpha(HgCl2)=0,928, alpha(HgCl3^-)=0,0657, alpha(HgCl4^2-)=0,00658

Passo 5

Para [Cl^-]=10^-4,M temos
beta1[Cl^-]=550,; beta2[Cl^-]^2=1,661times10^5,; beta3[Cl^-]^3=117,7,; beta4[Cl^-]^4=0,01177,
Dapprox1,6677times10^5,
alpha(Hg^2+)=5,99times10^-6, alpha(HgCl^+)=3,30times10^-3, alpha(HgCl2)=0,9961, alpha(HgCl3^-)=7,05times10^-4, alpha(HgCl4^2-)=7,06times10^-8

Passo 6

Para [Cl^-]=10^-6,M
beta1[Cl^-]=5,5,; beta2[Cl^-]^2=16,61,; beta3[Cl^-]^3=1,177times10^-4,; beta4[Cl^-]^4=1,177times10^-9,
Dapprox23,1101,
alpha(Hg^2+)=0,0433, alpha(HgCl^+)=0,2379, alpha(HgCl2)=0,7187, alpha(HgCl3^-)=5,09times10^-6, alpha(HgCl4^2-)=5,09times10^-11

Passo 7

Para [Cl^-]=10^-8,M
beta1[Cl^-]=0,055,; beta2[Cl^-]^2=0,001661,; beta3[Cl^-]^3=1,177times10^-10,; beta4[Cl^-]^4=1,177times10^-17,
Dapprox1,056661,
alpha(Hg^2+)=0,9463, alpha(HgCl^+)=0,05205, alpha(HgCl2)=0,001573, alpha(HgCl3^-)=1,11times10^-10, alpha(HgCl4^2-)=1,11times10^-17

Passo 8

Para [Cl^-]=10^-10,M
beta1[Cl^-]=5,5times10^-4,; beta2[Cl^-]^2=1,661times10^-7,; beta3[Cl^-]^3=1,177times10^-16,; beta4[Cl^-]^4=1,177times10^-25,
Dapprox1,0005502,
alpha(Hg^2+)=0,99945, alpha(HgCl^+)=5,50times10^-4, alpha(HgCl2)=1,66times10^-7, alpha(HgCl3^-)=1,18times10^-16, alpha(HgCl4^2-)=1,18times10^-25

Resposta

As frações alpha de cada espécie nos diferentes [Cl^-] são:
[Cl^-]=10^-2,M: alpha(Hg^2+)=5,6times10^-10, alpha(HgCl^+)=3,1times10^-5, alpha(HgCl2)=0,928, alpha(HgCl3^-)=0,0657, alpha(HgCl4^2-)=0,00658 [Cl^-]=10^-4,M: alpha(Hg^2+)=5,99times10^-6, alpha(HgCl^+)=3,30times10^-3, alpha(HgCl2)=0,9961, alpha(HgCl3^-)=7,05times10^-4, alpha(HgCl4^2-)=7,06times10^-8 [Cl^-]=10^-6,M: alpha(Hg^2+)=0,0433, alpha(HgCl^+)=0,2379, alpha(HgCl2)=0,7187, alpha(HgCl3^-)=5,09times10^-6, alpha(HgCl4^2-)=5,09times10^-11 [Cl^-]=10^-8,M: alpha(Hg^2+)=0,9463, alpha(HgCl^+)=0,05205, alpha(HgCl2)=0,001573, alpha(HgCl3^-)=1,11times10^-10, alpha(HgCl4^2-)=1,11times10^-17 [Cl^-]=10^-10,M: alpha(Hg^2+)=0,99945, alpha(HgCl^+)=5,50times10^-4, alpha(HgCl2)=1,66times10^-7, alpha(HgCl3^-)=1,18times10^-16, alpha(HgCl4^2-)=1,18times10^-25

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