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    Here's the question:

    A sample of copper (II) sulfate was dissolved in water to produce 250cm^3 of solution. An excess of aqueous potassium iodide, Kl(aq) was added to 25.0cm^3 of this solution and iodine was formed in the reaction:

    2Cu^2+ + 4I^- -> 2CuI + I2

    The iodine produced was titrated with 0.200 mol dm^-3 sodium thiosulfate, Na2S2O3:

    I2 + 2S2O3^2- -> 2I- + S4O6^2-
    The average titre obtained was 20.15 cm ^3

    Calculate the concentration of the copper (II) sulphate solution.

    This is what Ive done:

    Na2S2O3 moles = vol x conc. = 20.15/1000 x 0.2 = 4.03 x 10^-3 mol

    Molar ratio means theres 8.06 x 10^-3 mol of copper in the 25.0cm^3 solution
    therefore 0.0806 mol of Cu2+ in the 250cm^3 solution

    0.0806 = 250/1000 x conc.
    conc. = 0.3324 mol

    Answer in the book was 0.161, which is close to a half of what I've got. If it is the case that I have to half it (and the books just rounded it) I don't understand why, because 2Cu was used in the reaction... Please can someone explain.

    Thanks in advance

    Oh dear I just typed my explanation out and accidently deleted it. Brilliant.
    In your first step you correctly calculated moles of Na2S203. Looking at the second equation you've written we can see that 2 moles of this reacts with 1 mole of Iodine, so you halve this number.
    Then you look at the first equation and see 1 mol of Iodine is formed from 2 moles of Copper, so you double the moles, which leaves you with an answer of 4.03x10^-3 moles of Copper in the 25cm^3 solution, using that number for the rest of your calculations should get you the right answer. I hope I explained that well enough sorry if I didn't!

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