# Finding out % of As2O3 in a sample

1. Jul 28, 2008

### ssb

1. The problem statement, all variables and given/known data

A 3.67 gram sample of bug spray was decomposed in acid. Any $$As^5^+$$ was reduced to $$As^3^+$$ 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 $$I_3^-$$, which oxidized $$As^3^+$$ to $$As^5^+$$ according to the reaction:

$$As^3^+ + I_3^- ----> 3I^- + As^5^+$$

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

2. Relevant equations

$$Moles Reacted = (I*t)/(nF)$$

3. The attempt at a solution

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

$$moles reacted = 3.607x10^-^5$$

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 $$(3.607x10^-^5) * (197.84 g/mol) = 7.136x10^-^3 grams of As_2O_3$$
If you divide this by the original mass $$(7.136x10^-^3)/(3.67g)$$ 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 $$As^3^+ ----> As^5^+ + 2e^-$$

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

2. Jul 29, 2008

### Staff: Mentor

Iodine works only as an intermediate in transfering charge, so you may simply assume reaction going is As3+ -> As5+ + 2e- - you know how to calculate the charge, you should be eable to calculate amount of As.

As moles number seems correct, but it is not identical to the number of moles of As2O3.

You forgot that sample was diluted and only part was titrated.

3. Jul 29, 2008

### ssb

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

Moles reacted $$= 3.607 x 10^{-22}$$

Since its a 1 to 1 ratio of mol $$As^{3+}$$ to mol $$I_3$$ then there are also $$3.607 x 10^{-22}$$ mol of $$I_3$$

So I will now solve for grams of $$As_2O_3$$

Grams $$As_2O_3 = (3.607 x 10^{-22}/5.00 mL) * (250 mL) * (1 mol As_2O_3/2 mol As^{3+}) (197.84 grams / mol As_2O_3) = 0.178 grams$$

$$0.178 grams / 3.67 grams (original sample size)$$ = .0485

.0485 (100) **percentage calculation** = 4.85%

Does this look good?

4. Jul 29, 2008

### Staff: Mentor

Much better now.