Question

If we mix 40.0 mL of 3.00 M Pb(NO3)2 (aq) with 20.0 mL of 2.00 x 10-3 M NaI (aq), does PbI2 (s) precipitate from solution? I yes, calculate how many moles of PbI2 (s) precipitate and the values of [Pb2+], [I-], [NO3-], and [Na+] at 25 C at equilibrium?

Answer 1

moles of Pb(NO3)2 = 40 x 3 / 1000 = 0.12

moles of NaI = 20 x 2 x 10^-3 / 1000 = 4 x 10^-5

Pb(NO3)2 + 2 NaI ----------------------> PbI2    + 2NaNO3

1                    2                                     1

0.12              4 x 10^-5                          2 x 10^-5        4 x 10^-5

NaI limiting reagent .

moles of PbI2 formed = 4 x 10^-5 / 2

moles of PbI2 formed    = 2.0 x 10^-5

[Na+] = 0.67 M

[NO3] = form reactant = 0.11998 / 60 x 10^-3 = 2 M

[NO3-] = 2.00 M

[Pb+2] =1.998 M

[I-] = 0.67 M

ionic product of PbI2 = 1.998 x 0.67 ^2

                          =   0.888

ionic product > solubility product (9.8 x 10^-9) . so precipitate wil form