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    I got 1.54x10^-6 moles for the answer for the first question, and 3.3x10^-6 mg for the second, but im not confident with these answers, so if anyone who knows what they are talking about can help, that would be great thanks





    IO3 + 5I + 6H+ → 3I2 + 3H2O   Equation 24.1


    1. i. The student dissolves some of the iodised salt in a small volume of dilute hydrochloric acid. The student then transfers this solution to a 250 cm3 volumetric flask and makes up to the mark with distilled water.

      25.0 cm3 of this solution is then added to a conical flask containing a solution of excess iodide ions. The iodine produced is titrated with a standard solution of 0.00100 mol dm−3 sodium thiosulfate, Na2S2O3, using starch indicator.

      The reaction equation is:

         2Na2S2O3(aq) + I2(aq) → 2NaI(aq) + Na2S4O6(aq)   Equation 24.2

      The student’s results give a mean titre of 15.40 cm3.

      Calculate the amount (in mol) of IO3 in the 25.0 cm3 of iodate solution.













    amount of IO3 = ................................ .......... mol [2]





    ii Calculate the amount of IO3 in the original solution and hence the mass, in mg, of potassium iodate(V) in the iodised salt dissolved.
    Give Your answer to an appropriate number of significant figures.







    mass of KIO3 = ................................ .......... mg
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    (Original post by tryer1998)
    I got 1.54x10^-6 moles for the answer for the first question, and 3.3x10^-6 mg for the second, but im not confident with these answers, so if anyone who knows what they are talking about can help, that would be great thanks





    IO3 + 5I + 6H+ → 3I2 + 3H2O   Equation 24.1


    1. i. The student dissolves some of the iodised salt in a small volume of dilute hydrochloric acid. The student then transfers this solution to a 250 cm3 volumetric flask and makes up to the mark with distilled water.

      25.0 cm3 of this solution is then added to a conical flask containing a solution of excess iodide ions. The iodine produced is titrated with a standard solution of 0.00100 mol dm−3 sodium thiosulfate, Na2S2O3, using starch indicator.

      The reaction equation is:

         2Na2S2O3(aq) + I2(aq) → 2NaI(aq) + Na2S4O6(aq)   Equation 24.2

      The student’s results give a mean titre of 15.40 cm3.

      Calculate the amount (in mol) of IO3 in the 25.0 cm3 of iodate solution.














    amount of IO3 = ................................ .......... mol [2]





    ii Calculate the amount of IO3 in the original solution and hence the mass, in mg, of potassium iodate(V) in the iodised salt dissolved.
    Give Your answer to an appropriate number of significant figures.

    mass of KIO3 = ................................ .......... mg
    Hi,

    I'm not sure if my answer is right either but this is the method that I used:

    Moles sodium thiosulphate = CV = 0.001 x (15.4/1000) = 1.54 x 10^-5 mol.

    From equation 24.2: 2 moles of thiosulphate react with 1 mole of iodine so the moles of iodine used are 1.54x10^-5/2 = 7.7 x 10^-6.

    From equation 24.1: 3 moles of iodine were produced from 1 mole of the salt so the moles of the salt are moles of iodine/3 = 7.7×10^-6/3 = 2.57 x 10^-6.

    For the 2nd part. If there are 2.57 x 10^-6 moles of the salt in 25cm^3 then there are 2.57 x 10^-6 x 10 moles in 250cm^3 = 2.57 x 10^-5.

    m (KIO3)= n × Mr = 2.57 x 10^-5 × 214 = 5.49 × 10^-3 g x 1000 = 5.49 mg.

    Hope this helps
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    thanks very much i think youre right, youve been a huge help
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    (Original post by tryer1998)
    thanks very much i think youre right, youve been a huge help
    No problem, I'm not completely sure if it's right but glad that I could help a bit.
 
 
 
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