Solution Stoichiometry: Precipitation Calculation

Question

Calculate the number of milliliters of 0.732 M KOH required to precipitate all of the Pb²⁺ ions in 125 mL of 0.905 M Pb(NO₃)₂ solution as Pb(OH)₂. The equation for the reaction is:

Pb(NO₃)₂(aq) + 2KOH(aq) Pb(OH)₂(s) + 2KNO₃(aq)

Answer 1

Write the balanced equation:

2KOH(aq) + Pb(NO₃)₂(aq) ==> Pb(OH)₂(s) +2 KNO₃(aq)

Next find the moles of Pb(NO₃)₂ present:

0.125 L x 0.905 moles/L = 0.113 moles

Now find moles of KOH needed to react with 0.113 moles Pb(NO₃)₂:

0.113 moles Pb(NO₃)₂ x 2 moles KOH/mole Pb(NO₃)₂ = 0.226 moles

Finally, find the volume of KOH that contains 0.226 moles:

(x L)(0.732 mol/L) = 0.226 moles and x = 0.309 L = 309 mls