[tpowell35]

Question:
 A 2.769 g sample containing an unknown amount of AsCl₃ was dissolved into a NaHCO₃ and HCl aqueous solution. To this solution was added 1.550 g of KI and 50.00 mL of 0.00976 M KIO₃. The excess of I₃^- was titrated with 50.00mL of 0.02000 M Na₂S₂O₃ solution. What was the mass percent of AsCl₃ in the original sample?

What I've done:

From answering the question wrong and getting some feedback, I know that I have to a) determine the limiting reagent from the KI and KIO₃ reaction, b) determine the amount of I₃^- produced from said reaction and then c) determine its excess from titration with Na₂S₂O₃, which will be equal to the moles of As³⁺ in the original sample.

So...

a) The first reaction is IO₃^- + 8I^- + 6H⁺   3I₃^- + 3 H₂O.

Moles of KI = 1.550g * 1mol/166.00277g = 0.009337 mol
Moles of KIO₃ = 0.050L * 0.00976 M = 0.000488 mol

b)KIO₃ is the limiting reagent, right? So moles of I₃^- = 0.000488 * 3 = 0.001464 moles.

c) This reaction is I₃^- + 2S₂O₃^2- 3I^- + S₄O₆^2-.

The 0.001464 moles of I₃^- would react with S₂O₃^2-, leaving 0.0009640 moles of I₃^-.

(0.05000 L * 0.02000 M = 0.001000 moles S₂O₃^2-, 1:2 ratio requires 0.0005000 moles of  I₃^- to react).

So... if the moles of  I₃^- = the moles of As³⁺, then there were 0.0009640 moles of As³⁺ in the original sample, * the molecular weight of arsenic is 74.92g/mol = 0.072222g/2.769g (weight of sample) * 100 = 2.608%.

But this is obviously not the answer. I feel like I'm missing some key component due to lack of sleep and this being finals week. Can anyone clarify or tell me where I'm screwing up?

Thanks!

[Borek]

Perhaps I am missing something, but it looks to me like you have calculated mass percent of As, not of AsCl₃.

That is assuming everything else is OK, I have skimmed over the solution and it looks like it is OK, but then I am in a vacation mode 

[persianprincess]

Yes, IO3 is the limiting reagent. First you'd have to find the amount reacted by subtracting the moles of I3 by the excess. Sooo

(0.05)*(0.00976 M KIO3)*(3 mol I3 / 1 mol IO3) = 0.001464 moles I3

Now we find the excess.

((0.02)(0.05))/2 = 5 x 10^-4 moles I3

Now we find the amount reacted.

0.001464- 5 x10^-4 = 9.64 x 10^-4 moles I3 reacted

So..

9.64 x 10^-4 AsCl3 * (181.28 g / 1 mol) = 0.1748 g AsCl3

Finally, to find the mass percentage of AsCl3 in the original sample...

0.1748/2.769 *100 = 6.31%

That should give you your answer.

[tpowell35]

Thanks to both of you. I knew it was something ridiculously obvious, if it were a snaked I'd have been bitten to death 10 times over... 

[Borek]

9.64 x 10^-4 AsCl3 * (181.28 g / 1 mol) = 0.1748 g AsCl3

Finally, to find the mass percentage of AsCl3 in the original sample...

0.1748/2.769 *100 = 6.31%

That should give you your answer.

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