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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 |
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Answer» `0.2496g` 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` |
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