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Volume strength of H2O2 using iodometric titration

  1. May 10, 2016 #1

    Titan97

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    Gold Member

    1. The problem statement, all variables and given/known data
    To a 25ml H2O2 solution, excess acidified solution of potassium iodide is added. The libereated iodine required 20ml of 0.3N sodium thiosulphate solution. Calculate the volume strength of H2O2.

    2. Relevant equations
    Normality = n-factor × Molarity
    Volume strength = Molarity × 11.2

    3. The attempt at a solution
    I will use moles and not equivalence. But to do that, I need to convert normality to molarity.

    For the reaction of sodium thiosulphate with iodine, n-factor is 2 for sodium thiosulphate.
    Hence M = N/n = 0.15.
    So number of milli moles of thiosulphate is 3.

    $$I_2 + 2Na_2S_2O_3 \rightarrow 2Na_2S_4O_6 + 2NaI$$

    So number of millimoles of iodine is ##\frac{3}{2}##.

    $$2H^+ + H_2O_2 + 2KI \rightarrow 2H_2O+I_2+2K^+$$

    Hence millimoles of hydrogen peroxide is ##\frac{3}{2}##

    So molarity of hydrogen peroxide is ##\frac{3}{50}##

    So volume strength = ##\text{M}\times 11.2 = 0.672##

    But the answer given is double of what I got.
     
  2. jcsd
  3. May 10, 2016 #2

    Titan97

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    Gold Member

    oops...
    The number of electrons exchanged by 1 mole of thiosulphate is 1.
    $$2e^-+2S_2O_3^{2-}\rightarrow S_4O_6^{2-}$$

    Hence n-factor is 1 and not 2. Now I got the given answer.
     
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