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.)
Determine the percentage of tin in the alloy
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In the previous parts related to this question the balance equation and the amount of` KMnO_4` used for the reaction is obtained.
` 2MnO_4^(-)+5Sn^(2+) +16H^+ rarr 2Mn^(2+)+5Sn^(4+)+8H_2O`
`MnO_4^-:Sn^(2+) = 2:5`
Amount of `MnO_4^-` used `= 0.0107/1000xx15.61 = 1.67xx10^(-4)`
Amount of `Sn^(2+)` in the mix `= (1.67xx10^(-4))/2xx5 = 4.175xx10^(-4)`
So there were 4.175xx10^(-4) moles in 25ml of the solution. But the total solution was 100ml.
Total amount of `Sn^(2+) ` `= 4.175xx10^(-4)xx4 = 1.67xx10^(-3)`
Molar mass of sn = 118.7g/Mol
Amount of Sn in the alloy `= 118.7xx1.67xx10^(-3) = 0.198g`
% of Sn in alloy `= 0.198/0.472xx100% = 41.998%.`
So there is 41.99% Sn in the alloy.
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