Titration chemistry

Question

100 mL of 0.50 M potassium iodide are combined with 50. mL of 0.75 M lead (II) nitrate. Calculate the mass of lead (II) iodide that will precipitate.

Answer 1

Write the correctly balanced equation for the reaction:

2KI(aq) + Pb(NO₃)₂(aq) ==> PbI₂(s) + 2KNO₃(aq) ... balanced equation

Find the limiting reactant. An easy way to do this is to divide mols of each reactant by the corresponding coefficient in the balanced equation, and whichever value is less represents the limiting reactant.

For KI we have: 100 ml x 1L / 1000 ml x 0.50 mol / L = 0.05 mols KI (÷2->0.025)

For Pb(NO₃)₂ we have: 50 ml x 1 L / 1000 ml x 0.75 mol / L = 0.0375 mols Pb(NO₃)₂ (÷1->0.0375)

Since 0.025 is less than 0.0375, KI IS LIMITING. We now use the MOLES of KI to determine the mass of

PbI₂ that will be formed.

Mass of PbI₂ formed:

molar mass PbI₂ = 461 g / mole

0.05 mols KI x 1 mol PbI₂ / 2 mols KI x 461 g PbI₂ / mol = 11.5 g PbI₂ precipitate (3 sig, figs.)