Consider the titration of 50.00 mL of a mixture of NaI and NaBr with 0.0500 M silver nitrate. The first equivalence point was observed at 10.50 mL and the second at 25.50 mL. Calculate the molar concentration of the two salts in the sample
Moles of mixture = "\\frac{50.00}{1000}\u00d70.0500=2.5\u00d710^{-3}" Moles
Molarity = "\\frac{Moles }{Volume}"
Molarity = "\\frac{2.5\u00d710^{-3}}{0.0105}=0.238M"
Molarity "=\\frac{2.5\u00d710^{-3}}{0.0255}=0.098M"
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