Anonymous
Anonymous asked in Science & MathematicsChemistry · 1 month ago

Titration Calculations Chemistry A level?

In a titration , 10.00 cm3 of sodium chlorate(I) solution was pipetted into a conical flask before excess potassium iodide and sulfuric acid were added . A 0.500 mol dm - 3 solution of sodium thiosulfate was then run into the conical flask . The end point was reached when 11.2 cm3 of sodium thiosulfate had been added . Calculate the concentration of the sodium chlorate(I) solution to 3 s.f.

My textbook says the answer is 0.56mol/dm^3 but I keep getting 0.28 so could someone please tell me what I’m doing wrong?

Thank you!

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  • Bobby
    Lv 7
    1 month ago
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    Sulfuric  acid reacts with sodium hypochlorite to form hypochlorous acid:

    2 NaOCl + H2SO4 = 2 HOCl + Na2SO4

    Hypochlorous acid reacts with acidified iodide to give I3-    ...triiodide 

    2 HOCl + H2SO4 + 2 I- = 2 I3- + SO4 2- + 2 H2O + 2 Cl- .....eq1

    A dark blue complex is formed when triiodide is combined with starch. 

    I3- + 2S2O3 - = 3I- + S4O6 2-  ..... eq 2 

    the end point is when the blue colour disappears 

    moles of thiosulfate = 0.0112 dm ^3  * 0.500 mol dm ^-3

    =  5.60 * 10^-3 moles of S2O3 - 

    = 2.80 * 10 ^-3 moles of I3- according to equation 2

    = 2.80 * 10 ^-3 moles of HOCl according to equation 1

    = moles of NaOCl

    these are in  0.0100 dm ^3 so moles / dm^3 = = 2.80 * 10 ^-3 moles / 0.0100 dm^3

    = 0.28 M NaOCl 

    I think your book may have an error .. I agree with you 

    they may have forgot to halve to get the number of moles of I3- as we did

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