Question

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?

Answer 1

`{:(Cu^(2+),+e,rarr,Cu^(1+),),(,2I^(-),rarr,I_(2)+2e,),(,2S_(2)O_(3)^(2-),rarr,S_(4)O_(6)^(2-)+2e,):}`
Meq. of `Cu^(2+) =` Meq. of liberated `I_(2)`
`=` Meq. of `Na_(2)S_(2)O_(3)`
`= 12.12 xx 0.1 xx 1 = 1.212`
`:. (w_(Cu^(2+)))/(M//1) xx 1000 = 1.212`
`:. w_(Cu^(2+)) = (1.212 xxM)/(1000)`
`= (1.212 xx 63.6)/(1000) = 0.077 g`
`:. w_(Cu) = w_(Cu^(2+)) = 0.077g`
`:. %Cu = (0.077)/(1.100 ) xx 100 = 7.00%`