1.104g of impure sample of NaNO2 was acidified and analysed using excess iodine ion. The reaction is 2HNO2+ 3H+-----2NO + 2H2O+I3- . Then I3- is titrated with 0.300M Na2S2O3.
S2O3 2- + I3- ------ I- +S4O6 2-
if the titre volume is 29.25cm3, calculate the % of NaNO2 in the sample
I got 27.5% NaNO2 (w/w) ... post work if you get stuck.
NOTE => You do need 3I^- on reactant side of the initial equation and on the product side of the thiosulfate rxn in post.