10 g of a mixture of `Cu_2S` and Cus was titrated with 200 " mL of " 0.75 M `MnO_4^(ɵ)` in acidic medium producing `SO_2,Cu^(2+)`, and `Mn^(2+)`. The `SO_2` was boiled off and the excess of `MnO_4^(ɵ)` was titrated with 175 " mL of " `1 M Fe^(2+)` solution. Find the percentage of CuS the in original mixture.

Explanation

Total `MnO_4^(ɵ)=200xx0.75M MnO_4^(ɵ)`
`=150xx5 mEq` of `MnO_4^(ɵ)`
`=750 mEq` of `MnO_4^(ɵ)`
Excess of `MnO_4^(ɵ)=175 " mL of " M Fe^(2+)`
`=175 m mo`l of `Fe^(2+)`
`=175 mEq ` of `Fe^(2+)`
`=175 mEq` of ` MnO_4^(ɵ)`
`MnO_4^(ɵ)` used up `=750-175=575 mEq`
`Cu_2to2Cu^(2+)+2e^(-)`
`S^(2-)toSO_2+6e^(-)`
1 " mol of "`Cu_2S` required `8e^(-)`
`Ew(Cu_(2)S)=159//8`
`CuStoCu^(2+)+S^(2-)`
`S^(2-)toSO_(2)+6e^(-)`
`therefore` Equivalent weight of `CuS=95.5//6`
`therefore` x g of `Cu_2S` and `(10-x)g of CuS` react.
`therefore((x)/((159)/(8))+(10-x)/((95.5)/(6)))xx1000=575`
`thereforex=` Weight of `Cu_2S=4.25g`
`therefore` Weight of `CuS=5.75g`
% of `CuS=57.5%`