Question

Consider the following reaction: Cr2O72– + 6Br– + 14 H+ à 2 Cr3+ + 3Br2 + 7H2O
In a redox titration, 14.62 mL of 0.175 M Cr2O72– was needed to completely react with a 25.00 mL sample of Br–. Calculate the [Br–] in the original solution.

Answer 1

In the reaction

Cr₂O₇^2– + 6Br^– + 14 H⁺ $\rightarrow$ 2 Cr³⁺ + 3Br₂ + 7H₂O,

1 ion of Cr₂O₇^2– reacts with 6 ions of Br^-, therefore their number of the moles relate as:

$n(Cr_2O_7^{2–}) = \frac{n(Br^-)}{6}$ .

The number of the moles of the bichromate anion is:

$n(Cr_2O_7^{2–}) = cV = 14.62·10^{-3}·0.175 = 2.559·10^{-3}$ mol.

The number of the moles in 25.000 mL sample of Br^- is:

$n(Br^-) = 6·2.559·10^{-3} = 0.0154$ mol.

Finally, the concentration of Br^- in the original solution is the ratio of the number of the moles of Br^- that react in titration to the volume of the sample:

$c(Br^-) = \frac{n}{V} = \frac{0.0154}{25·10^{-3}} = 0.614$ M.

Answer: the Br^- concentration in the original solution is 0.614 M.