Respuesta :
Answer:
Check below
Explanation:
Ksp of AgCl = 1.8*10^-10
Ksp = [Ag+] [Cl-]
[Ag+] [Cl-] = 1.8*10^-10
Because [Ag+] = [Cl-] ,
[Ag+]² = 1.8*10^-10
[Ag+] = √((1.8*10^-10)_
[Ag+] = 1.35*10^-5M
That is [Ag+] in the original saturated solution.
NaCl is added to the saturated AgCl solution:
If Cl- = 0.72M , equation becomes
[Ag+] [ 0.72] = 1.8*10^-10
[Ag+] = (1.8*10^-10)/0.72
[Ag+] = 2.5*10^-10M
% Ag+ remaining in solution =(2.5*10^-10) / { (1.35*10^-5) * 100
% Ag+ remaining = 1.85*10^-3% or 0.00185%
Note that an approximation has been made here: [Cl-] is 0.72M + the unprecipitated Cl- from the AgCl. Because this latter is an infinitely small amount compared to 0.72M , it is quite safe to make this approximation without any effect on the answer.
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