What is the weight of arsenic trioxide (98.25% As2O3) that would be used as a sample so that 29.5 ml of 0.1 N iodine would be needed to titrate it?

Answer 1

So, to begin with, we must write a correctly balanced equation for the redox reaction taking place.

As₂O₃ + 2I₂ + 5H₂O ==> 2H₃AsO₄ + 4HI ... balanced equation

(if you want to know how to balance such a reaction, write a comment or a separate question).

Now, we will find the moles of I₂ used:

I₂ has 2 equivalents per mole, so 0.1N is the same as 0.05 M

29.5 ml x 1 L / 1000 mls x 0.05 mols / L = 0.001475 mols I₂ used

Next, find the moles of As₂O₃ needed to react with this many moles of I₂

0.001475 mols I₂ x 1 mol As₂O₃ / 2 mols I₂ = 0.0007375 moles As₂O₃

Finally, convert the moles of As₂O₃ to grams of As₂O₃

0.0007375 mols As₂O₃ x 197.8 g / mol = 0.1459 g As₂O₃

0.1459 g As₂O₃ / 0.9825 = 0.148 g As₂O₃ (corrected for the 98.25% purity)