Question on Selective precipitation

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Discussion Overview

The discussion revolves around the feasibility of determining the concentrations of I− and Cl− ions independently in a solution containing both at approximately 0.01 M using gravimetric analysis. Participants explore the method of selective precipitation and the calculations involved in the process.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests adding Ag+ to precipitate I− first, noting that the solubility product constant (Ksp) for AgI is smaller than that for AgCl.
  • Another participant questions the multiplication of molarity by (0.01%/100%) and seeks clarification on its purpose.
  • A third participant explains that this calculation is intended to lower the concentration of I− to achieve complete separation, referencing a definition found online.
  • A later reply confirms the calculation of Ag+ needed to precipitate I− and prompts further calculations regarding the next steps in the process.

Areas of Agreement / Disagreement

Participants generally agree on the approach of using selective precipitation to separate I− from Cl−, but there is some uncertainty regarding the calculations and the rationale behind certain steps. The discussion remains unresolved as participants seek clarification and further calculations.

Contextual Notes

There are limitations in the discussion regarding the assumptions made in the calculations and the dependence on the definitions of complete separation. Some mathematical steps remain unresolved.

higherme
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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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