InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A mixture containing `As_(2)O_(3)` and `As_(2)O_(3)` requried 20 " mL of " 0.05 N `I_(2)` for titration. The resulting solution is then acidified and excess KI was added. The liberated `I_(2)` required 1.24 g hypo `(Na_(2)S_(2)O_(3).H_(2)O)` for complete reaction. Calculate the mass of the mixture. The reactions are `As_(2)O_(3)+2I_(2)+2H_(2)OtoAs_(2)O_(3)+4H^(o+)+4I^(ɵ)` `As_(2)O_(5)+4H^(o+)+4I^(ɵ)toAs_(2)O_(3)+2I_(2)+2H_(2)O` |
|
Answer» l" Eq of "`I_(2)` used `=20xx0.05=1.0` Let a and b are m" Eq of "`As_(2)O` and `As_(2)O_(5)` respectively on addition of `I_(2)` to mixture `As_(2)^(3+)` is converted `As_(2)^(5+)`. `thereforem" Eq of "As_(2)O_(3)-=m" Eq of "I_(2) used-=1.0-=m" Eq of "As_(2)^(5)` formed. `thereforea=1.0` After reaction with `I_(2)`, mixture contains all the arsenic in `+5` oxidation state which is then titrated using `KI+` hypo. Thus, `m" Eq of "As_(2)_(3) as As_(2)^(5+) m" Eq of "As(2)O_(5) as As^(5+)=m" Eq of "I_(2)` liberated `=m" Eq of "`hypo used `thereforea+b=(1.24)/(248)xx10^(3)` Ew of `Na_(2)S_(2)O_(5) as 5H_(2)O=(248)/(1)` `thereforea+b=5` ...(ii) weight of `As_(2)O_(3)=1.0xx10^(-3)xx(198)/(4)=0.0495g` `{:(As_(2)^(3+)toAs_(2)^(5+)+4e^(-)),(Ew of As_(2)O_(3)=(198)/(4)):}` Weight of `As_(2)O_(3)=4xx10^(-3)xx(230)/(4)` `(Ew of As_(2)O_(5)=(230)/(4))` `=0.23 g` Weight of mixture`=0.0495+0.23=0.2795g` |
|