1. The problem statement, all variables and given/known data Some brands of household bleach contain sodium hypochlorite, NaOCl as the active ingredient. The concentration of NaOCl in the preparation is recorded as the concentration of ‘available chlorine’. One brand of this bleach stated on the label that it contained ‘4%(w/v) available chlorine’. In an analysis of this bleach, each pair of students in a class measured out 20.00ml from a new, unopened bottle of the bleach. They then made this up to 250.00ml in a standard flask with distilled water, and removed 20.00ml aliquots for a series of repeat titrations. To each aliquot they added 5ml of 1M sulfuric acid and 5ml of 1M potassium iodide. The equation for the reaction that occurs is: OCl-(aq) + 2I-(aq) + 2H+(aq) -> I2(aq) + Cl-(aq) + H2O(l) Each sample immediately turned brown due to the production of iodine. The students then titrated this against a previously standardised potassium thiosulfate which was 0.09877M. The mean titre was 17.22ml. What is the concentration of the available chlorine in the original mixture? 2. Relevant equations 3. The attempt at a solution The reaction of the titration is I2 + (2S2O3)2- -> 2I- + (S4O6)2- So number of moles of potassium thiosulfate is 0.0017. So there is 0.00085 moles of I2 in the product formed of the reaction OCl-(aq) + 2I-(aq) + 2H+(aq) -> I2(aq) + Cl-(aq) + H2O(l) Hence 0.00085 moles of OCl- molecules were formed. 0.00085mole * 35.45g/mole (for chlorine) = 0.03g in the 20ml aliquot. There are 25/2 number of these in the 250ml flask. So 0.3768g in the 250ml flask. This came from 20ml from the bottle so 0.3768g/20ml = 0.019g/ml or 1.9g/100ml of available chlorine hence not even 2%(w/v). But the answer claimed 3.957%(w/w). Who is right?