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21.83 mL of a Pb2 solution, containing excess Pb2 , was added to 10.75 mL of...

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Chemistry

21.83 mL of a Pb2 solution, containing excess Pb2 , was added to10.75 mL of a 2,3-dimercapto-1-propanol (BAL) solution of unknownconcentration, forming the 1:1 Pb2 –BAL complex. The excess Pb2 wastitrated with 0.0141 M EDTA, requiring 8.86 mL to reach theequivalence point. Separately, 40.70 mL of the EDTA solution wasrequired to titrate 31.00 mL of the Pb2 solution. Calculate the BALconcentration, in molarity, of the original 10.75-mL solution.

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4.2 Ratings (899 Votes)
23dimercapto1propanol or CH2SHCHSHCH2OH is a complexing agent which we will denote here as which we will write as RSH 2 This bidentate ligand reacts selectively to form a complex with Pb2 that is much more stable than PbY2 PbY2 2RSH2 PbRS2 2H Y4 2183 ml of Pb2 solution is added to 1075 ml of 23dimercapto1propanol And excess of Pb2 is titrated    See Answer
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