Question on Selective precipitation

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In a solution containing both I− and Cl− at approximately 0.01 M, gravimetric analysis can be used to determine their amounts independently. To achieve this, Ag+ is added to precipitate I− first, as its solubility product constant (Ksp) is smaller than that of Cl−. The calculation shows that 8.5E-11 M of Ag+ is needed to effectively precipitate I−. After I− is removed, increasing Ag+ concentration will allow for the precipitation of Cl−. This method ensures that the concentrations of both ions can be accurately measured after selective precipitation.
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can anyone check if this is right?

question :
Explain if a solution contains both I− and Cl− at about 0.01 M, can we determine the amount of Cl− and I− independently by using gravimetric analysis?

yes
add [AG+] to precipitate out the I- first ( ksp is smaller)

0.01M * (0.01%/100%) = 1.0E-6 M I-
ksp = [Ag+][I-]
8.5E-17 = [Ag+](1.0E-6)
[Ag+] = 8.5E-11 M <------ add to precipitate AgI

increase [Ag+] to precipitate Cl- left
 
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Why are you multiplying the molarity by (0.01%/100%) ?
 
to lower the concentration:

i found using google:
"What do we mean by complete separation--> the concentration in solution of the analyte of interest must be less than or equal to 0.01% of its original value."
 
Okay, so you want to remove 99.99% of I-. That's fine. You've calculated that it takes 8.5E-11M of Ag+ to precipitate most of the I-.

So far, so good. Now what do you need to calculate next?
 
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