What is the concentration of available chlorine in the household bleach?

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Discussion Overview

The discussion revolves around the calculation of the concentration of available chlorine in household bleach, specifically focusing on the interpretation of the term 'available chlorine' and the methodology used in a titration experiment. The context includes a homework problem involving sodium hypochlorite (NaOCl) and its measurement through titration against potassium thiosulfate.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a detailed calculation of available chlorine based on titration results, concluding a concentration of 1.9g/100ml, which contradicts the label claim of 4%(w/v).
  • Another participant questions whether the focus should be on chlorine (Cl) or sodium hypochlorite (NaOCl) in the context of available chlorine.
  • A different participant clarifies that 'available chlorine' refers to NaOCl and recalculates the concentration, arriving at a value of 3.95g/100g, which aligns with the answer provided in the original problem.

Areas of Agreement / Disagreement

Participants express differing interpretations of 'available chlorine,' leading to conflicting calculations. While one participant supports the label's claim, another challenges it based on their own calculations. The discussion remains unresolved regarding which interpretation is correct.

Contextual Notes

The calculations depend on the definitions of available chlorine and the assumptions made regarding the chemical species involved in the titration. There is also a potential discrepancy in the interpretation of the concentration units (w/v vs. w/w).

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Homework Statement


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?




Homework Equations





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?
 
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Cl or NaOCl?
 
I took the question as asking the available chlorine ions so Cl. I think some of the compounds are in aqeous solution so are in ion forms like OCl-(aq) although I could be easily wrong here.
 
I had another look at the question and realized ‘available chlorine’ was not Cl but NaOCl. Assuming this, I redid the question again and got 3.95g/100g which is exactly what the answers suggeted. Thanks for pointing out this.
 

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