12 gm of an impure sample oxide (`As_(2)O_(3)` which is acidic in nature ) was dissolved in water containing sodium bicarbonate ( which is basic in nature ) and the resulting solution was diluted to 250 ml. 25 ml of this solution was completely oxidized by `22.4` ml of a solution `24.8` gm of hydrated sodium thiosulphate `(Na_(2)S_(2)O_(3).5H_(2)O)` in one litre .
Calculate the percentage of arsenious oxide in the sample.

Explanation

In this reaction , `As_(2)O_(3)` acts as acidic oxide and `NaHCO_(3)` as a base, giving acid base neutralization reaction, which is non-redox process. Here , `As_(2)O_(3)` does not act as basic oxide. Hence , it will form `As_(2)(CO_(3))_(3)`, which does not exist as non-metals do not form carbonates.
n-factor of `As_(2)O_(3)` is 6 and that of `NaHCO_(3)` is 1. After the reaction, `As^(3+)` is oxidized by `l_(2)" to " As^(+5)` while `l_(2)` is reduced to `l^(-)` .
Normality of `Na_(2)S_(2)O_(3).5H_(2)O = (24.8)/(248) = 0.1`
Normality of `l_(2)` = Normality of `Na_(2)S_(2)O_(3). 5H_(2)O = (24.8).248`
` :. ` Equivalents of `l_(2) = 0.1 xx 22.4 xx 10^(-3)` = Equivalent of `As^(3+)` reacted in 25 ml ` = 2.24 xx 10^(-3)`
`:. ` Equivalents of `As^(3+) ` reacted in 250 ml = `2.24 xx 10^(-2)`
Moles of `As^(3+)` in 250 ml ` = (2.24 xx 10^(-2))/2 = 1.12 xx 10^(-2)`
Moles of `As_(2)O_(3)` reacted ` = (1.12 xx 10^(-2))/2 = 5.6 xx 10^(-3) `
Percentage of `As_(2)O_(3) = (5.6 xx 10^(-3) xx 198 xx 100)/12 = 9.24 %`