(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A 3.67 gram sample of bug spray was decomposed in acid. Any [tex]As^5^+[/tex] was reduced to [tex]As^3^+[/tex] and diluted to 250.0 mL in a volumetric flask. A 5.00 mL sample of this was added to 125.0 mL of 0.0500 M KI buffered to pH 7. A coulmetric titration was carried out with electrically generated [tex]I_3^-[/tex], which oxidized [tex]As^3^+[/tex] to [tex]As^5^+[/tex] according to the reaction:

[tex]As^3^+ + I_3^- ----> 3I^- + As^5^+[/tex]

The titration required 287 seconds at a constant current of 24.28 mA to reach the endpoint. Calculate the percentage of [tex]As_2O_3[/tex] (197.84 g/mol) in the bug spray.

2. Relevant equations

[tex]Moles Reacted = (I*t)/(nF)[/tex]

3. The attempt at a solution

[tex]moles reacted = (.02425 amps * 287 seconds)/(2 (# electrons) * 9.6482x10^4 (Faraday constant)[/tex]

[tex]moles reacted = 3.607x10^-^5[/tex]

Question 1) Am I right to assume it is a one to one ratio of moles? or is it a one to 3 ratio? Assuming its one to one then [tex](3.607x10^-^5) * (197.84 g/mol) = 7.136x10^-^3 grams of As_2O_3[/tex]

If you divide this by the original mass [tex](7.136x10^-^3)/(3.67g)[/tex] oh and multiply by 100 to calculate percent, I come up with 0.1944%. NOW: I was able to do this with about half of the information given in the original equation. I looked at sample problems in my book and found similar calculations can be done if you are given standard potentials for half reactions. We were given none and I cannot find a standard potential for [tex]As^3^+ ----> As^5^+ + 2e^-[/tex]

Question 2) where am I going wrong here? Thank you so much whoever is able to help me tackle this homework problem

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# Homework Help: Finding out % of As2O3 in a sample

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