A `1.10 g` sample of copper ore is dissolved and the `Cu^(2+)` of 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. Find the `%` copper by mass in ore.

Answer

7

Explanation

`{:(Cu^(2+),+e,rarr,Cu^(+),),(,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+)))/(63.6//1) xx 1000 = 1.212`
`:. w_(Cu) = w_(Cu^(2+)) = 0.077 (because Cu overset(H_(2)SO_(4))rarr CuSO_(4))`
`:. %Cu = (0.077)/(1.10) xx 100 = 7%`