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

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SUMMARY

The concentration of available chlorine in household bleach is determined through a titration process involving sodium hypochlorite (NaOCl) as the active ingredient. In this discussion, students measured a 20.00ml sample of bleach, diluted it to 250.00ml, and performed titrations using 1M sulfuric acid and 1M potassium iodide. The mean titre of 17.22ml with standardized potassium thiosulfate (0.09877M) indicated that the concentration of available chlorine was approximately 3.95g/100g, aligning with the label claim of 4%(w/v) available chlorine. The confusion arose from interpreting 'available chlorine' as either Cl or NaOCl, but the correct interpretation is NaOCl.

PREREQUISITES
  • Understanding of titration techniques and calculations
  • Knowledge of chemical reactions involving sodium hypochlorite and iodine
  • Familiarity with molarity and molar mass calculations
  • Basic grasp of aqueous solution chemistry
NEXT STEPS
  • Study the principles of titration and how to calculate concentrations from titration data
  • Learn about the chemical properties and reactions of sodium hypochlorite
  • Explore the concept of 'available chlorine' and its significance in bleach formulations
  • Investigate the preparation and standardization of potassium thiosulfate solutions
USEFUL FOR

Chemistry students, laboratory technicians, and anyone involved in the analysis of household cleaning products will benefit from this discussion.

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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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