Question

A solution is prepared by mixing 50 mL of 1 M Pb(NO3)2 and 75 mL of .5 M NaF. Calculate the concentration of F- ions present at equilibrium. (Hint: First, write and balance the double displacement reaction taking place between Pb(NO3)2 and NaF to form PbF2(s). Perform the necessary stoichiometry (including finding which reactant is limiting) to calculate how much of each reactant remains after the reaction goes to completion. Remember that stoichiometry has to be done in moles. Once you have calculated how much of each reactant remains, then calculate the concentration of Pb2+ and F- remaining in solution, and use those as your initial concentrations for the equilibrium calculation using the reaction below.) (In essence, we are assuming that the reaction goes to completion first, and then we are letting it come back to equilibrium.) PbF2(s) <--> Pb2+ (aq) + 2F- (aq) Kc=3.7E-8

Answer 1

millimoles of Pb(NO3)2 = 50 x 1 = 50

millimoles of NaF = 75 x 0.5 = 37.5

Pb(NO3)2 + 2 NaF   -----------------> PbF2 + 2 NaNO3

     1                  2                                  1

0.05                 0.0375                             

here limiting reagent is NaF . so

Pb(NO3)2 reacted = 0.0375 x 1 / 2 = 0.01875

Pb(NO3)2 remained = 0.05 - 0.01875 = 0.03125

Molarity of Pb(NO3)2 = 0.03125 / (50 + 75) x 10^-3 = 0.25 M

[Pb+2] = 0.25 M

PbF2(s) <---------------> Pb2+ (aq) + 2F- (aq)

Ksp = [Pb+2][F-]^2

3.7x 10^-8 = 0.25 x [F-]^2

[F-] = 3.85 x 10^-4 M

equilibrium concentratiion of [F-] = 3.85 x 10^-4 M