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A 0.276 g impure sample of copper ore is...

A 0.276 g impure sample of copper ore is dissolved and `Cu^(2+)` is titrated with KI solution. `I_(2)` liberated required 40 mL of 0.1 M `Na_(2)` `S_(2)` `O_(3)` solution for titration. What is the % of impurities in the ore ?

Text Solution

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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%`
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