Question

A 24.34 mL aliquot of a Pb2 solution, containing excess Pb2+, was added to 11.50 mL of a 2,3-dimercapto-1-propanol (BAL) solution of unknown concentration, forming the 1:1 Pb2+-BAL complex. The excess Pb2+ was titrated with 0.0157 M EDTA, requiring 9.88 mL to reach the equivalence point. Separately, 39.89 mL of the EDTA solution was required to titrate 32.75 mL of the Pb2+ solution. Calculate the BAL concentration, in molarity, of the original 11.50 mL solution concentration:

Answer 1

Excess moles of Pb²⁺ in original Pb²⁺ solution after BAL complexation = moles of EDTA in 9.88ml of 0.0157M EDTA

= 9.88ml * 0.0157M = 0.1551 milliMoles

Now, Molarity of initial Pb²⁺ solution = (Molarity of EDTA * Volume of EDTA) / (Volume of Pb²⁺ required to titrate)

= (0.0157M * 39.89 ml) / (32.75 ml) = 0.01912 M Pb2+ solution

Now, Total no. of moles of Pb²⁺ in original solution = (Molarity of Pb²⁺) * (Volume of Pb²⁺ solution)

= 0.01912 M * 24.34 ml = 0.4654 milliMoles.

Since Pb²⁺- BAL is a 1:1 complex,

Moles of BAL = Moles of Pb²⁺ used in complexation = (Total moles of Pb²⁺) - (Excess Moles of Pb²⁺)

= 0.4654 milliMoles - 0.1551 milliMoles

= 0.3103 milliMoles of BAL.

Thus, Molarity of BAL = Moles of BAL / Volume of BAL solution = 0.3103 milliMoles / 11.50 ml = 0.207M BAL solution (Ans)