10 mL of `1.0` M aqueous solution of `Br_(2) ` is added to excess of NaOH in order to disproportionate quantitatively to `Br^(-) " and " BrO_(3)^(-)` . The resulting soluting is made free from `Br^(-) ` ion by extraction and excess of `OH^(-)` neutralized by acidifying the solution. This solution requires `1.5 ` g of an impure `CaC_(2)O_(4)` sample for complete redox change . Calculate % purity of `CaC_(2)O_(4)` sample.

Explanation

`2Br_(2) + 6OH^(-) to 5Br^(-) + BrO_(3)^(-) + 3H_(2)O` reacts to give redox change as
`Br^(5+) + 6e to Br^(-) ` (Valence factor = 6)
`C_(2)^(3+) to 2C^(4+) + 2e^(-)`
Millimole of `BrO_(3)^(-)` formed ` = 10/3`
Meq. of `CaC_(2)O_(4) = ` Meq. of `BrO_(3)^(-) = 10/3 xx 6 = 20`
`W_(CaC_(2)O_(4))/((128)/2) xx 1000 = 20 " " W_(CaC_(2)O_(4)) = 1.28 ` g
` :. " of " CaC_(2)O_(4) = (1.28)/(1.5) xx 100 = 85.33`