A mixture containing `As_(2)O_(3)` and `As_(2)O_(5)` required `20.10mL` of `0.02N` iodine for titration. The resulting solution is then acidified and excess of KI was added. The liberated iodine required `1.113g `hypo `(Na_(2)S_(2)O_(3),5H_(2)O)` for complete reaction. Calculate madd of the mixture. The reactions are:
`As_(2)O_(3)+2I_(2)+2H_(2)O rarr As_(@)O_(5)+4H^(+)+4I^(-)`
`As_(2)O_(5)+4H^(+)+4I^(-) rarr As_(2)O_(3) +2I_(2) +2H_(2)O`

Options

A. `0.2496g`
B. `0.0497g`
C. `0.1997g`
D. `0.5g`

Answer

A

Explanation

Meq of `I_(2)` used `= 20.10 xx 0.05 = 1.005`
Let Meq of `As_(2)O_(3)` and Meq of `As_(2)O_(5)` in mixture be a and b respectively. On addition of `I_(2)` to mixture `As_(2)O_(3)` is converted to `As_(2)O_(5)`.
Meq of `As_(2)O_(3) =` Meq of `I_(2)` used `=1.005` Meq of `As^(5+)` formed or `a = 1.005` .....(1)
After the reaction with `I_(2)`, mixture contains all the arsenic in `+5` oxidation state which is then titrated using `KI+` hypo. Thus,
Meq. of `As_(2)O_(3)` as `As^(+5) +` Meq. of `As_(2)O_(5)` as `As^(+5) =`
Meq. of liberated `I_(2) =` Meq. of hypo used or `a +b = (1.113)/(248) xx 1000` or `a+b = 4.481`
By equations (1) and (2), `b = 4.481 - 1.005 =`
`3.476 :.` Wt. of `As_(2)O_(3) =`
`("Meq" xx Eq.Wt)/(1000) = (1.005 xx 198)/(4 xx 1000) = 0.0497g`
and Wt. of `As_(2)O_(5) = (3.476xx 230)/(4 xx 1000) = 0.1999g`