Question

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

Answer 1

Moles of mixture = $\frac{50.00}{1000}×0.0500=2.5×10^{-3}$ Moles

Molarity = $\frac{Moles }{Volume}$

Molarity = $\frac{2.5×10^{-3}}{0.0105}=0.238M$

Molarity $=\frac{2.5×10^{-3}}{0.0255}=0.098M$