To determine the concentration of the lead ions in the solution, we need to use the stoichiometry of the reaction between lead ions Pb2+ and potassium iodide KI . The balanced chemical equation for this reaction is:Pb2+ + 2KI PbI2 + 2K+From the balanced equation, we can see that 1 mole of Pb2+ reacts with 2 moles of KI. Now, we can use the information given about the KI solution to find the moles of KI used in the titration:Moles of KI = Molarity VolumeMoles of KI = 0.1 M 0.025 L converting 25 mL to liters Moles of KI = 0.0025 molesSince 1 mole of Pb2+ reacts with 2 moles of KI, we can find the moles of Pb2+ in the solution:Moles of Pb2+ = Moles of KI / 2Moles of Pb2+ = 0.0025 moles / 2Moles of Pb2+ = 0.00125 molesNow, to find the concentration of Pb2+ in the solution, we need to divide the moles of Pb2+ by the volume of the solution. However, the volume of the lead ion solution is not provided in the problem. If you can provide the volume of the lead ion solution, we can calculate the concentration of Pb2+ in the solution.