How Do I Calculate the Molarity of Potassium Iodate in a Redox Titration?

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Discussion Overview

The discussion revolves around calculating the molarity of potassium iodate in a redox titration involving sodium thiosulphate, sulfuric acid, potassium iodide, and starch. Participants are seeking clarification on the stoichiometric relationships between the reactants and the equations governing the titration process.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about how to calculate the molarity of potassium iodate after titrating with sodium thiosulphate and provides the relevant chemical equations.
  • Another participant suggests balancing the equations by adding electrons to account for charge differences, indicating that this is necessary to understand the titration reaction.
  • A third participant proposes a stoichiometric relationship, stating that for every mole of IO3(-), three moles of I2 are produced, and thus the relationship between IO3(-) and S2O3(2-) is 1:5 based on the second equation.
  • A later reply questions the ratio proposed by a previous participant, suggesting that the ratio of IO3 to S2O3 should be 1:6 instead of 1:5, prompting a request for clarification from others.
  • Another participant asserts that the reactions provided are balanced and that one mole of iodate reacts with six moles of thiosulphate, indicating a different interpretation of the stoichiometry.
  • One participant acknowledges a misunderstanding in how the reactions were presented, suggesting that clarity in writing could help avoid confusion.

Areas of Agreement / Disagreement

Participants express differing views on the stoichiometric ratios between the reactants, particularly regarding the relationship between IO3(-) and S2O3(2-). There is no consensus on the correct ratio, and the discussion remains unresolved.

Contextual Notes

Participants highlight the need for careful balancing of the equations and the importance of understanding the stoichiometric relationships, but there are unresolved assumptions regarding the ratios and the interpretation of the equations.

egg
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totally stuck!

hey guys and girls, I'm having some serious problems with this redox reaction titration thing i just did!
Basically i titrated a sodium thiosulphate solution with potasium iodate(v), sulphuric acid, potassium iodide and some starch.
I've worked out the molarity of my sodium thiosulphate solution is 1.54 but i have no idea how to get the molarity of the potassium iodate(v)!

we've been given these equations:
IO3- + 6H+ + 5I- = 3I2 + 3H2O
I2 + 2S2O3 2- = 2I- + S4O6 2-
(sorry about the equals instead of arrows and the lack of super/subscript!)
but I have no idea what to do with them!
my teachers seem to have missed something vital when explaining all this to us and i really need some help!
thanks in advance!
 
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egg said:
we've been given these equations:
IO3- + 6H+ + 5I- = 3I2 + 3H2O
I2 + 2S2O3 2- = 2I- + S4O6 2-

Look for charges on both sides of equations - they are different, so thera electrons missing. Add them to balance charges on both sides of the equations, then balance both half reactions together and you will know what your reaction during titration was.


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IO3- + 6H+ + 5I- = 3I2 + 3H2O
I2 + 2S2O3 2- = 2I- + S4O6 2-

You have to figure out the number of IO3(-)/S2O3(2-)

Let me help you out here:

For every IO3(-) 3 I2 are made and the relationship between them are

1/3 Okay.

In the equation number 2, you have 1 I2 to 2 S2O3(2-). This relationship is

1/2. So the relationship between IO3(-)/S203(2-) is 1/5.

If you have the mol of IO3 then you can find the mol of S2O3
 
thanks for the help guys, i got some more help from my teacher the other day and kind of got it sussed! although i am now a little confused about the ratios.
I've worked out that the ratio of IO3:S2O3 is 1:6 not 1:5! are you certain of this nicholas as it will make a big difference to my result!
surely if the IO3:I2 is 1:3 and the I2:S2O3 is 1:2 then IO3:S2O3 is 1:6?
anyone else clarify this??
 
Borek said:
Look for charges on both sides of equations - they are different, so thera electrons missing. Add them to balance charges on both sides of the equations, then balance both half reactions together and you will know what your reaction during titration was.


Borek
--
http://www.chembuddy.com
Chemical calculators for labs and education
BATE - pH calculations, titration curves, hydrolisis
These are not half-reactions. They are both completely balanced reactions, and all the electrons have been taken care of.

egg, you are correct. 1 mole of iodate is consumed along with 6 moles of thio.
 
Gokul43201 said:
These are not half-reactions. They are both completely balanced reactions, and all the electrons have been taken care of.

Seems I should spend more time relaxing :(


Chemical calculators for labs and education
BATE - pH calculations, titration curves, hydrolisis
 
woo hoo i got something right!

borek i think it was the way i wrote the reactions out that made you think they were unbalanced! Next time i'll do a better job!

Thanks for all your help guys, now i can start worrying about all the essays I've just been set! (anyone fancy doing those for me! ha ha ha!)
 

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