Let's do this bit by bit.
The IO3^-/I^- is an oxidation/reduction reaction. Do you know how to balance that type? The I in IO3^- has an oxidation state of +5 and the I^- is -1.
Hi, I'm really having trouble this this chemistry problem. I have the answers; however I don't have extensive knowledge about titrations and I'm struggling with working it out. If anyone could suggest how to work this out (or even where to start) that would be great. Thanks!
Triiodide ions are generated in a solution by the following reaction in acidic solution.
IO3- (aq) + I- ----> I3-
Triiodie ion is determined by titration with a thiosulfate solution (Na2S2O3). The products are iodide ion and tetrathionate ion (S4O6 2+).
a. Balance the equation for the reaction of IO3- with I- ions.
b. Write and balance the equation for the reaction of S2O3 -2 with I3- in acidic solution.
c. A 25.00 mL sample of a .01000 M solution of KIO3 is reacted with excess KI. It requires 32.04 mL of Na2S2O3 to titrate the I3- ions present. What is the molarity of the Na2S2O3 solution?
d. how would you prepare 500.0 mL of the KIO3 solution in part c, using pure dry solid KIo3.
5 answers
Yes, I know you must use the half reaction method to balance the equation. I put a 2 as the coefficient of I- but I'm not sure how to divide the equation into two reactions since there are only 3 parts.
Personally, I think using I3^- complicates a relatively easy equation to write. So let's use I2 and we can add I^- to it later if that's what the professor wants.
6H^+ + IO3^- + 5I^- ==> 3I2 + 3H2O Then the liberated I2 is titrated with thiosulfate in this reaction.
I2 + 2S2O3^-2 ==> S4O6^-2 + 2I^-
This gets part a and b EXCEPT if the prof wants I3^-, then just add I^- to make it work. For example, the iodate equation, replace 3I2 with 3I3^- and balance that by using 8I^- on the left instead of 5I^-. For the second one use 3I^- on the right and I3^- on the left.
For part c. calculate mols IO3^-, use the equation to convert to mols I2 and use the titration equation to convert to mols thiosulfate. Then L x M = mols. You have mols thiosulfate and L thioulfatre and you can calculate M thiosulfate.
For part d. remember M = mols/L. You want 0.01 mols in 1 L so you will need -.005 mols in 500 mL. grams = mols x molar mass.
6H^+ + IO3^- + 5I^- ==> 3I2 + 3H2O Then the liberated I2 is titrated with thiosulfate in this reaction.
I2 + 2S2O3^-2 ==> S4O6^-2 + 2I^-
This gets part a and b EXCEPT if the prof wants I3^-, then just add I^- to make it work. For example, the iodate equation, replace 3I2 with 3I3^- and balance that by using 8I^- on the left instead of 5I^-. For the second one use 3I^- on the right and I3^- on the left.
For part c. calculate mols IO3^-, use the equation to convert to mols I2 and use the titration equation to convert to mols thiosulfate. Then L x M = mols. You have mols thiosulfate and L thioulfatre and you can calculate M thiosulfate.
For part d. remember M = mols/L. You want 0.01 mols in 1 L so you will need -.005 mols in 500 mL. grams = mols x molar mass.
thank you so much!
wait, i'm sorry, do i get the moles of IO3 by divinding 174.9 (molar mass of IO3) by 1196.16g (the mass of the IO3, hydrogen and I3 combined?) (the whole left side of the equation?)
1 answer