Question

21.83 mL of a Pb2 solution, containing excess Pb2 , was added to 10.75 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 0.0141 M EDTA, requiring 8.86 mL to reach the equivalence point. Separately, 40.70 mL of the EDTA solution was required to titrate 31.00 mL of the Pb2 solution. Calculate the BAL concentration, in molarity, of the original 10.75-mL solution.

Answer 1

2-3-dimercapto-1-propanol or CH2SHCHSHCH2OH is a complexing agent which we will denote here as which we will write as R(SH) 2. This bidentate ligand reacts selectively to form a complex with Pb2+ that is much more stable than PbY^2-.

PbY^2- + 2R(SH)_{2 -->} Pb(RS)₂ + 2H⁺ + Y^4-

21.83 ml of Pb²⁺ solution is added to 10.75 ml of 2,3-dimercapto-1-propanol.

And excess of Pb2+ is titrated with 8.86ml of 0.0141M EDTA.

Therefor no. of moles of EDTA used = 0.0141 × 0.00886 L = 0.0001249 moles

Now we know that one mole of EDTA will chelate one mole of Pb2+ ion

So, moles of Pb2+ present in the solution after reacting with 2,3-dimercapto-1-propanol = 0.0001249 moles

Now as stated in the question above 40.70 mL of 0.0141M EDTA is use to reach equivalence point for 31ml of Pb2+ solution,

Moles of EDTA used = moles of pb2+ present in 31ml = 0.0407L × 0.0141M = 0.00057387 moles

Therefore moles of pb2+ present in 31ml = 0.00057387 moles

Molarity of 31ml of pb2+ solution = 0.00057387/ 0.031L = 0.01851M

Now, the total number of moles of Pb2+ present in 21.83ml of solution = 0.02183L × 0.01851M = 0.00040411moles

As we have calculated above that 0.0001249 mole of Pb2+ remains in the solution after reacting with 2-3-dimercapto-1-propanol and now we have total number of moles of Pb2+ present in 21.83 ml of solution. So we can calculate number of moles those actually reacted with 2-3-dimercapto-1-propanol as follows,

No. of moles reacted with 2-3-dimercapto-1-propanol= total no. of moles- no of moles remains in solution after reacting with 2-3-dimercapto-1-propanol

No. of moles of Pb2+ reacted with 2-3-dimercapto-1-propanol= 0.00040411moles - 0.0001249 = 0.000279 = 2.79 × 10^-4 moles

As we know from reaction stoichiometry

No. of moles of Pb2+ = no of moles of 2-3-dimercapto-1-propanol= 2.79 × 10^-4 moles

Therefore molarity of 10.75 mL of a 2,3-dimercapto-1-propanol = 2.79 × 10^-4 / 0.01075L = 0.0259 M