Chemistry, 21.02.2020 03:38 juliannabartra
Given the two reactions PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.85×10−10, and AgCl(aq)⇌Ag+(aq)+Cl−(aq), K4 = 1.17×10−4, what is the equilibrium constant Kfinal for the following reaction? PbCl2(aq)+2Ag+(aq)⇌2AgCl(aq)+Pb2+(a q)
Answers: 1
![K_3=\frac{[Pb^{2+}][Cl^-]^2}{[PbCl_2]}](/tpl/images/0518/7310/bd2c4.png)
...[1]
![K_4=\frac{[Ag^+][Cl^-]}{[AgCl]}](/tpl/images/0518/7310/28023.png)
..[2]![[Cl^-]=\frac{K_4\times [AgCl]}{[Ag^+]}](/tpl/images/0518/7310/ddb5a.png)
![K=\frac{[AgCl]^2[Pb^{2}]}{[PbCl_2]][Ag^+]^2}](/tpl/images/0518/7310/7360a.png)
![\frac{K_3}{K_4}=\frac{\frac{[Pb^{2+}][Cl^-]^2}{[PbCl_2]}}{\frac{[Ag^+][Cl^-]}{[AgCl]}}](/tpl/images/0518/7310/44e82.png)
![\frac{K_3}{K_4}=\frac{[Pb^{2+}][Cl^-][AgCl]}{[PbCl_2][Ag^+]}](/tpl/images/0518/7310/78721.png)
![[Cl^-]](/tpl/images/0518/7310/0726e.png)
from [2] :![\frac{K_3}{K_4}=\frac{[Pb^{2+}][AgCl]}{[PbCl_2][Ag^+]}\times \frac{K_4\times [AgCl]}{[Ag^+]}](/tpl/images/0518/7310/d7960.png)
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Given the two reactions PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.85×10−10, and AgCl(aq)⇌Ag+(aq)+Cl−(aq),...
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