Precipitate Formation

Question

Will a precipitate form when 1.00 mL of 0.10 M Pb(NO3)2 is added to 9.00 mL of NaOH solution (pH =

9)? (Pb(OH)2 Ksp = 1.42×10-20)

Answer 1

The balanced reaction is Pb(NO₃)₂ + 2NaOH -----> Pb(OH)₂ + 2Na(OH).

The # moles Pb(NO₃)₂ is 0.10 M/L / 1000 mL/L x 1.00 mL = 0.001 moles.

The # of moles of NaOH can be found as follows:

pOH = 5 = -log[OH]; [OH] = 1 x 10^-5 M. then 1 x 10^-5 M/L / 1000 mL/L x 9 mL = 9 x 10^-8 moles

NaOH is a limiting reactant, so the # moles of Pb²⁺ produced is 4.5 x 10^-8 and OH^1- is 9.0 x 10^-8.

the molar concentrations, therefore, are 4.5 x 10^-8/10 mL x 1000 mL/L = 4.5 x 10^-6 molar.

[OH] = 9 x 10^-8 /10 mL x 1000 mL/L = 9.0 x 10^-6 molar.

The product [PB²⁺] [OH^1-]² = 3.6 x 10^-16. This is larger than the K_sp, so there will be a precipitate.