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A solution is prepared by mixing 50 mL of 1 M Pb(NO3)2 and 75 mL of...

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Chemistry

A solution is prepared by mixing 50 mL of 1 M Pb(NO3)2 and 75 mLof .5 M NaF. Calculate the concentration of F- ions present atequilibrium. (Hint: First, write and balance the doubledisplacement reaction taking place between Pb(NO3)2 and NaF to formPbF2(s). Perform the necessary stoichiometry (including findingwhich reactant is limiting) to calculate how much of each reactantremains after the reaction goes to completion. Remember thatstoichiometry has to be done in moles. Once you have calculated howmuch of each reactant remains, then calculate the concentration ofPb2+ and F- remaining in solution, and use those as your initialconcentrations for the equilibrium calculation using the reactionbelow.) (In essence, we are assuming that the reaction goes tocompletion first, and then we are letting it come back toequilibrium.) PbF2(s) <--> Pb2+ (aq) + 2F- (aq) Kc=3.7E-8

Answer & Explanation Solved by verified expert
3.8 Ratings (473 Votes)

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
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