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?
- BobbyLv 71 month agoFavorite Answer
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