Hard chemistry question...determining solubility...?

Question

A precipitate of PBI2 is formed when 5 mL of 0.012 M Pb(NO3)2 is added to 5 mL of 0.030M KI. It was found experimentally that the final concentration of I^- in 10 mL solution after equilibrium was 8 x 10^-3

(1) How many moles of Pb^2+are originally in the solution?

(2) How many moles of I^- are originally present?

(3) How many moles of I^- remained in the solution after equilibrium?

(4) How many moles of I^- precipitated?

(5) How many moles of PbI2 precipitated?

(6) How many moles of PbI2 remain in soolution after equilibrium?

(7) What is [Pb^2+] at equilibrium?

(8) What is the Ksp for the formation of PbI2?

Thank you so very much for your help!!

Answer 1

(1) One mole of Pb(NO3)2 dissolves in water

to give one mole of Pb2+ and 2 moles of NO3-.

Pb(NO3)2(s)  ==>  Pb2+(aq) + 2NO3-(aq)

So moles of Pb(NO3)2 = moles of Pb2+

moles Pb(NO3)2+ = M Pb(NO3)2 x L Pb(NO3)2 = (0.012)(0.005) = 6.0 x 10^-5 moles Pb(NO3)2 = moles Pb2+

(2) One mole of KI dissolves in water to form one mole of K+ and one mole of I-.

KI(s)   ==>   K+(aq)  +  I-(aq)

So moles of KI = moles I-

moles KI = M KI x L KI = (0.030)(0.005) = 1.5 x 10^-4 moles KI = moles I-

(3) After equilibrium, the problem states that [I-] = 0.008 M in 10 mL of solution.

moles [I-] = M [I-] x L [I-] = (0.008)(0.010) = 8 x 10^-5 moles I-.

(4) The precipitation reaction is

......Pb2+(aq)  +  2I-(aq)  ==>  PbI2(s)

6.0 x 10^-5 moles Pb2+ x (2 moles I- / 1 mole Pb2+) = 1.2 x 10^-4 moles I- precipitated.

Since we have 1.5 x 10^-4 moles of I-, we have an excess of (1.5 x 10^-4) - (1.2 x 10^-4) = 3 x 10^5 moles I- unreacted.

(5) As shown in (3), the amount of Pb2+ that reacted is 6.0 x 10^-5 moles.

(6) The only Pb2+ in solution after equilibrium is what comes from dissociation of PbI2 (x). The I- in solution comes from the dissociation of PbI2 (2x) and the amount left unreacted from the precipitation (3 x 10^-5):

PbI(s)   <==>  Pb2+(aq) + 2I-(aq)

moles

at eq.. . . . . . . x . . . . . . .(3 x 10^-5) + 2x

But in (3) we calculated that at equilibrium there are 8 x 10^-5 moles of I-.

So 8 x 10^-5 = (3 x 10^-5) + 2x

5 x 10^-5 = 2x

2.5 x 10^-5 = x = moles Pb2+ at eq. = moles PbI2 since there's one Pb2+ in PbI2

(7) From (6), moles Pb2+ = 2.5 x 10^-5

Molarity Pb2+ = [Pb2+] = moles / L = 2.5 x 10^-5 / 0.010 = 2.5 x 10^-3 M

(8) Ksp under the conditions stated is

Ksp = [Pb2+][I-]^2 = (2.5 x 10^-3)(8 x 10^-3)^2 = 1.6 x 10^-7