Question

DQ 11: Given that the solubility product of PbI₂ is expressed as K_sp = [Pb²⁺][I^–]², calculate the K_sp of PbI₂ from the concentration of Pb²⁺ found in Step 6 of the Data Analysis. (Hint: Think about the stoichiometry involved when PbI₂ was initially formed… how much I^– was left unreacted?) Is PbI₂ a soluble or insoluble salt? Explain.

3mL of 0.05M KI was used and 1mL of 0.05 Pb(NO3)2 was used to make the PbI2 solution/yellow precipitate

All work must be shown below for credit.

(2 pts) Show a properly balanced chemical equation:

Pb(NO₃)_2(aq)+ 2KI_(aq)--> PbI_2(s)+2KNO^-_3(aq)

(4 pts) Determine the limiting reagent:

(2 pts) Calculate the K_sp of PbI₂:

(2 pts) Is PbI₂ a soluble or insoluble salt? Explain:

Answer 1

moles of KI = 3 x 0.05 / 1000 = 1.5 x 10^-4

moles of Pb(NO3)2 = 1 x 0.05 / 1000 = 5.0 x 10^-5

a)

balanced chemical equation :

Pb(NO3)2 (aq)   + 2 KI (aq)    ------------->   PbI2 (s) + 2 KNO3 (aq)

b)

Pb(NO3)2 (aq)   + 2 KI (aq)    ------------->   PbI2 (s) + 2 KNO3 (aq)

    1                           2

5 x 10^-5             1.5 x 10^-4

here limting reagent is Pb(NO3)2.

because the mole ratio of Pb(NO3)2 is snall.

c)

PbI2 (s)   ----------------->   Pb2+   +   2 I-

Ksp = [Pb2+][I-]^2

       = (0.0125) (0.0375)^2

Ksp of PbI2 = 1.76 x 10^-5

d)

PbI2 is a insoluble salt.