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Totally stuck

  1. Mar 11, 2005 #1

    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!
     
  2. jcsd
  3. Mar 11, 2005 #2

    Borek

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    Staff: Mentor

    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.


    Chemical calculators for labs and education
    BATE - pH calculations, titration curves, hydrolisis
     
  4. Mar 15, 2005 #3
    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
     
  5. Mar 16, 2005 #4

    egg

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    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?!?!
     
  6. Mar 16, 2005 #5

    Gokul43201

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    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
  7. Mar 17, 2005 #6

    Borek

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    Staff: Mentor

    Seems I should spend more time relaxing :(


    Chemical calculators for labs and education
    BATE - pH calculations, titration curves, hydrolisis
     
  8. Mar 17, 2005 #7

    egg

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    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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