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Potassium Bromate is a powerful oxidizing agent which is reduced smoothly to bromide in the presence of HCl and which is then oxidized to give free bromine by bromate (excess).
BrO3- + 6H+ + 6e- ---> Br- + 3H2O
BrO3- + 5Br- + 6H+ ---> 3Br2 + 3H2O
The equivalent is therefore 1/6 moles, KBrO 3 /6 or 167/6 which is equal to 27.88. At the endpoint of the titration, free bromine appears.
BrO3- + 5Br- + 6H+ ---> 3Br2 + 3H2O therefore,
KBrO3 + 5Br- +6HCl. ---> 3Br2. + 6KCl + 3H2O
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- Principles
- Preparation of 0.016M KBrO 3 Solutions
- Preparation
- Application
- Essay
- Procedure for Assay
- IP Factors
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