Question

When a solution of Pb(NO₃)₂ and NaF are mixed, a precipitate is formed. The net ionic equation is: Pb⁺²(aq) + 2F^-(aq) --> PbF₂(s) and the K_sp for this compound is 3.3 x 10^-8. What is the concentration of the ions remaining in solution?

To solve, first do the stoichiometry of the reaction to determine the limiting reactant. Then, any excess reactant is treated as a common ion when the net ionic reaction is reversed and treated as a K_sp problem.

Suppose that 7.4mL of 0.045M of Pb(NO3)2 is added to 20.0mL of 0.050M NaF. Determine the concentrations of Pb+2 and F- that remain after precipitation have occurred. Do not forget to account for the dilution effect when adding two solutions together!

Part A: What concentration, in moles per liter, of lead ion (Pb⁺²) would remain in solution?

Part B: What concentration, in moles per liter, of fluoride ion (F^-) would remain in solution?

Answer 1

The balanced reaction between Pb(NO3)2 and NaF is

Hence, 2 moles of NaF completely reacts with 1 mol of Pb(NO3)2.

Now, volume of Pb(NO3)2 added = 7.4 mL = 0.0074 L

Concentration of Pb(NO3)2 = 0.045 M = 0.045 mol/L

Hence, number of moles of Pb(NO3)2 added is

Hence, amount of Pb2+ in the solution is

Volume of NaF added = 20.0 mL = 0.0200 L

Concentration of NaF = 0.050 M = 0.050 mol/L

hence, the number of moles of NaF added is

Hence, amount of moles of F- produced is

Note that according to the balanced reaction, 1 mol of Pb(NO3)2 reacts with 2 moles of NaF.

Hence, Pb(NO3)2 will react with

Since we have more than , the limiting reactant is Pb(NO3)2.

Hence, the precipitation reaction table is

Initial, mol			0
Change, mol
Final, mol	0

Note that since Pb2+ is the limiting reactant, it is consumed in the precipitation reaction.

Now, the total volume of the solution = 7.4 mL + 20.0 mL = 27.4 mL = 0.0274 L

Hence, concentrations of the ions after precipitation reaction is

Now, we know that not all of the precipitate formed will remain as solid. Hence, some of it might dissolve based on the Ksp of the precipitate.

Where the Ksp is given as

With the known initial concentrations, we can create the following ICE table

Initial, M		0
Change, M	-x	+x	+2x
Equilibrium, M		x

hence, from the Ksp equation

Hence,

Part A

The equilibrium concentration of Pb2+ ion that remains in the solution is

Part B

The equilibrium concentration of F- ion that remains in the solution is