Question

In a similar experiment, the conductimetric titration of 15.00 mL of an unknown lead(II) nitrate solution with a 0.105 M potassium iodide solution was carried out. The volume of potassium iodide solution required to reach the equivalence point was found to be 17.35 mL. (a) Write a balanced molecular equation for the reaction. (b) Calculate the molarity of the unknown lead(II) nitrate solution.

Answer 1

Part a)

Balanced molecular equation

Pb(NO₃)₂ (aq) + 2KI (aq)      PbI₂ (s) + 2KNO₃ (aq)

Part b)

Moles of iodide = Volume in Liter x Molarity = 0.01735 L x 0.105 M = 0.00182 moles

We have the net ionic equation

Pb²⁺(aq) + 2I^-(aq)    PbI₂(s)

Here, 0.00182 moles of iodide react with 0.00182/2 = 0.000911 moles of Pb

Therefore, molarity = 0.000911 moles / 0.01500 = 0.607 M