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    Please help me with the following question:

    "1.20 g of potassium iodate(V) are dissolved and made up to 250cm^3 with distilled water in a standard flask. 25 cm^3 of this solution are taken and added to an excess of potassium iodide solution and a small amount of dilute sulphuric acid, then titrated with sodium thiosulphate in the burette. A few drops of starch solution are added when the solution turns a pale straw colour and the end point is taken as the appearance of a permanent blue-black colour in solution. Calculate the concentration of the sodium thiosulphate solution given that 27.6 cm^3 are required to reach the end point and given the following equations:

    Iodate(V) ions and iodide ions in acidic solution:
    IO3^- (aq) + 5I^- (aq) + 6H^+ (aq) --> 3I2 (aq) + 3H2O (l)

    Thiosulphate ions and iodine:
    2S2O3^2- (aq) + I2 (aq) --> 2I^- (aq) + S4O6^2- (aq)"

    Please include steps to the solution, or else I won't understand.

    Thanks
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    moles of KIO3 = 1.2/(39+127+48)
    molarity in 250cm3 = 3/535 x 1000/250 = 12/535
    moles in 25cm3 = 12/535 x 25/1000 = 3/5350

    1 iodate = 3 iodine = 6 thiosulphate
    moles of thiosulphate required = 3/5350 x 6 = 18/5350
    concentration of thiosulphate = 18/5350 x 1000/27.6 = 0.121901666

    final answer = 0.122mol (3.s.f.)
    • Thread Starter
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    What does molarity mean?
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    But anyways, I get it now, thanks a lot darkenergy.
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    (Original post by blah888)
    What does molarity mean?
    molarity describes the concentration of a solute per unit volume of a solvent. Units; moldm-3
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    (Original post by darkenergy)
    moles of KIO3 = 1.2/(39+127+48)
    molarity in 250cm3 = 3/535
    moles in 25cm3 = 3/535 x 25/1000 = 3/21400

    1 iodate = 3 iodine = 6 thiosulphate
    moles of thiosulphate required = 3/21400 x 6 = 18/21400
    concentration of thiosulphate = 18/21400 x 1000/27.6 = 0.030475416

    final answer = 0.0305mol (3.s.f.)
    Hmm, don't you mean, 25/250?
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    Something's wrong....
    • Thread Starter
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    Wouldn't the molarity or concentration in 250 cm^3 be 3/535 * 1000/250?
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    Widowmaker, would you mind giving this question a try and compare with darkenergy's answer. I think there's something wrong!
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    (Original post by blah888)
    Wouldn't the molarity or concentration in 250 cm^3 be 3/535 * 1000/250?
    yes, dark-energy's solution is wrong, no pun intended.

    The figure given is (3/535)moles per 250cm3
    so multiply by 4 to get the answer with units moles per dm3
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    (Original post by blah888)
    Widowmaker, would you mind giving this question a try and compare with darkenergy's answer. I think there's something wrong!
    yes I'll have a go.
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    I got 0.739 mol
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    No of moles of KIO3 used in 250cm3 = 1.2/(39+127+48) moles = 1.2/214 moles = 3/535 moles
    Molarity of 250cm3 KIO3 solution = (3/535) mol/250cm³
    Concentration of KIO3 solution = (3/535 x 1000/250) mol/dm³ = (12/535) mol/dm³
    No of moles in 25cm3 = 3/535 x 25/250 = 3/5350 moles

    Mole ratio
    1 iodate = 3 iodine = 6 thiosulphate
    No of moles of thiosulphate required = 3/5350 x 6 = 18/5350
    Concentration of thiosulphate solution = 18/5350 x 1000/27.6 = 0.122 mol/dm3
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    (Original post by blah888)
    I got 0.739 mol
    show me your working.
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    (Original post by Widowmaker)
    yes, dark-energy's solution is wrong, no pun intended.

    The figure given is (3/535)moles per 250cm3
    so multiply by 4 to get the answer with units moles per dm3
    Yes got it wrong.
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    (Original post by Widowmaker)
    No of moles of KIO3 used in 250cm3 = 1.2/(39+127+48) moles = 1.2/214 moles = 3/535 moles
    Molarity of 250cm3 KIO3 solution = (3/535) mol/250cm³
    Concentration of KIO3 solution = (3/535 x 1000/250) mol/dm³ = (12/535) mol/dm³
    No of moles in 25cm3 = 3/535 x 25/250 = 3/5350 moles

    Mole ratio
    1 iodate = 3 iodine = 6 thiosulphate
    No of moles of thiosulphate required = 3/5350 x 6 = 18/5350
    Concentration of thiosulphate solution = 18/5350 x 1000/27.6 = 0.122 mol/dm3
    yeah got that as well.
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    I trust you cause you got an A in chem.

    Today was the first day of my AS courses!!
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    (Original post by blah888)
    I trust you cause you got an A in chem.

    Today was the first day of my AS courses!!
    oh thanks.
    good luck with AS chemistry!
    One that requires the most work in my opinion.
    What board are you with? edexcel?
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    (Original post by Widowmaker)
    show me your working.
    Did something completely wrong:

    I calculated the number of moles of potassium iodate. I used this value and calculated the concentration using the formula:

    conc = mol/vol

    Obviously can't do that cause the moles I calculated is the amount of moles in 250 cm^3. Bare with me, it's my first day!
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    (Original post by blah888)
    Did something completely wrong:

    I calculated the number of moles of potassium iodate. I used this value and calculated the concentration using the formula:

    conc = mol/vol

    Obviously can't do that cause the moles I calculated is the amount of moles in 250 cm^3. Bare with me, it's my first day!
    oh don't worry. I only touched relatively complicated calculations like this after 2 months into the course. You're doing very well for the first day!
 
 
 
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