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A 0.472 g sample of an alloy of tin and bismuth is dissolved in sulfuric acid to...

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roshan-rox | Valedictorian

Posted July 20, 2013 at 5:41 PM via web

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A 0.472 g sample of an alloy of tin and bismuth is dissolved in sulfuric acid to produce tin(II) and bismuth(III) ions. This solution is diluted to the mark in a 100 mL volumetric flask and 25.00 mL aliquots are titrated with a 0.0107 M solution of KMnO4, forming tin(IV) and manganese(II) ions. (The bismuth ions are unaffected during this titration.) b. If an average titration requires 15.61 mL of the MnO4- solution, calculate the number of moles of MnO4- used in an average titration.

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted July 20, 2013 at 5:50 PM (Answer #1)

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The reaction between `MnO_4^-` and `Sn^(2+)` is as follows. This is also described http://www.enotes.com/homework-help/0-472-g-sample-an-alloy-tin-bismuth-dissolved-444120#answer-626430

`2MnO_4^(-)+16H^++5Sn^(2+) rarr 2Mn^(2+)+8H_2O+5Sn^(2+)`

The titration has takes place with 0.0107M `KMnO_4 ` and 15.61ml was consumed for the reaction.

Amount of `KMnO_4` used `= 0.0107/1000xx15.61 = 1.67xx10^-4 `

 

So `1.67xx10^-4` `MnO_4^- ` moles has used in the average titration.

 

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