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
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.