Determine as concentrações das espécies Pbl^+, Pbl2(aq), Pbl3 ^-e Pbl4 ^2- em uma solução saturada de Pbl2(s), contendo I^-na concentração de a) 0,001M e b) 1,0 M.
ll Pbl2(s) stackrelK<=ftharpoons Pb^2++2 I^- & Kps=[Pb^2+][I^-]^2=7,9 times 10^-9 Pb^2++I^- stackrelK1<=ftharpoons PbI^+ & K1=[PbI^+] /[Pb^2+][I^-]=1,0 times 10^2 Pb^2++2 I^- stackrel<=ftharpoons<=ftharpoons Pbl2(a q) & beta2=[PbI2(a q)] /[Pb^2+][I^-]^2=1.4 times 10^3 Pb^2++3 I^- stackrelstackrel1<=ftharpoons<=ftharpoons PbI3^- & beta3=[PbI3^-] /[Pb^2+][I^-]^3=8,3 times 10^3 Pb^2++4 I^- stackrel <=ftharpoons Pbl4^2- & beta4=[Pbl4^2-] /[Pb^2+][I^-]^4=3,0 times 10^4
R. a) Pbl^+=7,4 times 10^-4 M ; Pbl2(aq)=1,1 times 10^-5 M ; PbI3 ^-=6,6 times 10^-8 M ; Pbl4 ^2-=2,4 times 10^-10 M
b) PbI^+=7,9 times 10^-7 M ; Pbl2(aq)=1,1 times 10^-5 M ; Pbl3 ^-=6,6 times 10^-5 M ; Pbl4 ^2-=2,4 times 10^-4 M

Passo 1

O enunciado pede as concentrações dos complexos
PbI^+, PbI2(aq), PbI3^-, PbI4^2-
numa solução saturada de PbI2(s) em duas situações de [I^-] fixas (efeito do íon comum). Vamos usar:
1. A solubilidade de PbI2 dada por
Ksp = [Pb^2+],[I^-]^2 = 7,9times 10^-9
2. As constantes de formação ( beta) dos complexos:
K1=beta1&=[PbI^+][Pb^2+],[I^-]=1,0times10^2, beta2&=[PbI2(aq)][Pb^2+],[I^-]^2=1,4times10^3, beta3&=[PbI3^-][Pb^2+],[I^-]^3=8,3times10^3, beta4&=[PbI4^2-][Pb^2+],[I^-]^4=3,0times10^4

Passo 2

Em cada caso calculamos primeiro [Pb^2+] pela expressão do Ksp:
[Pb^2+] =Ksp[I^-]^2

Passo 3

Depois, cada complexo é obtido por
[PbI^+] &=K1,[Pb^2+],[I^-], [PbI2(aq)] &=beta2,[Pb^2+],[I^-]^2, [PbI3^-] &=beta3,[Pb^2+],[I^-]^3, [PbI4^2-] &=beta4,[Pb^2+],[I^-]^4

Passo 4

Caso (a): [I^-]=1,0times10^-3,M
[Pb^2+] =7,9times10^-9(1,0times10^-3)^2 =7,9times10^-3,M
[PbI^+] &=1,0times10^2times7,9times10^-3times1,0times10^-3 =7,9times10^-4,M, [PbI2(aq)] &=1,4times10^3times7,9times10^-3times(1,0times10^-3)^2 =1,1times10^-5,M, [PbI3^-] &=8,3times10^3times7,9times10^-3times(1,0times10^-3)^3 =6,6times10^-8,M, [PbI4^2-] &=3,0times10^4times7,9times10^-3times(1,0times10^-3)^4 =2,4times10^-10,M

Passo 5

Caso (b): [I^-]=1,0,M
[Pb^2+] =7,9times10^-9(1,0)^2 =7,9times10^-9,M
[PbI^+] &=1,0times10^2times7,9times10^-9times1,0 =7,9times10^-7,M, [PbI2(aq)] &=1,4times10^3times7,9times10^-9times(1,0)^2 =1,1times10^-5,M, [PbI3^-] &=8,3times10^3times7,9times10^-9times(1,0)^3 =6,6times10^-5,M, [PbI4^2-] &=3,0times10^4times7,9times10^-9times(1,0)^4 =2,4times10^-4,M

Resposta

a) [PbI^+]=7,9times10^-4,M;;[PbI2(aq)]=1,1times10^-5,M;;[PbI3^-]=6,6times10^-8,M;;[PbI4^2-]=2,4times10^-10,M
b) [PbI^+]=7,9times10^-7,M;;[PbI2(aq)]=1,1times10^-5,M;;[PbI3^-]=6,6times10^-5,M;;[PbI4^2-]=2,4times10^-4,M

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