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 by22.4 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.

12 gm of an impure sample oxide (As_(2)O_(3)which is acidic in nature

1.	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 by22.4 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.
Answer» SOLUTION :In this reaction , `As_(2)O_(3)` acts as acidic oxide and `NaHCO_(3)` as a base, givingacidbase neutralization reaction, which is non-redoxprocess. 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 %`