A `1.100 g` sample of copper ore is dissolved and the `Cu_((aq.))^(2+)` is treated with excess `KI`. The liberated `I_(2)` requires `12.12 mL` of `0.10M Na_(2)S_(2)O_(3)` solution for titration. What is `%` copper by mass in the ore?

Explanation

The complete balanced equation for the redox reaction is
`2cu^(2+)+4I^(-)+2S_(2)O_(3)_^(2)rarrCu_(2)I_(2)+S_(2)O_(6)^(2-)+2I^(-)`
No of moles of `S_(2)O_(3)^(2-)` used =`(12.12)/(1000)xx0.1=1.212xx10^(-3)` moles
From the balance equation
2 moles of `S_(2)O_(3)^(2-)` reduce `Cu^(2+)` =2 moles
`therefore 1.212 xx10^(-3)` moles of `S_(2)O_(3)^(2-)` will reduce `Cu^(2+)=1.212 xx10^(-3)` moles
Thus % age of Cu in the are =`(0.77)/(1.1)xx100=7%`