Question

If the Ksp of PbI₂ is 7.1 X 10^-9, will precipitation occur when 10 mL of 1.0 X 10^-4 M Pb(NO₃)₂ is mixed with 10 mL of 1.0X10^-3 M KI?

Will precipitation occur when 10mL of 1.0X 10^-2 M Pb(NO₃)₂ added to 10mL of 2.0 X 10^-2 M KI?

*Please show step-by-step, as I am really lost. Thank you in advance.

Answer 1

(1)

Pb(NO₃)₂ molarity in the mixture = molarity of Pb(NO₃)₂/ total volume

                                                      = 1.0 x 10^-4 / (10 +10)

                                                      = 5 x 10^-6 M

[ Pb+2 ] molarity = 5 x 10^-6 M

KI molarity in the mixture = 1.0 x 10^-3 / (10 + 10)

                                          = 5 x 10^-5 M

[I-] molarity = 5 x 10^-5 M

Q = ionic product = [Pb+2][I-]^2

                            = 5 x 10^-6 x (5 x 10^-5)^2

                             = 1.25 x 10^-14

Ksp of PbI₂ = 7.1 x 10^-9

here Ksp > Q (ionic product). so precipitate will not form.

(2)

Pb(NO₃)₂ molarity in the mixture = molarity of Pb(NO₃)₂/ total volume

                                                      = 1.0 x 10^-2 / (10 +10)

                                                      = 5 x 10^-4 M

[ Pb+2 ] molarity = 5 x 10^-4 M

KI molarity in the mixture = 2.0 x 10^-2 / (10 + 10)

                                          = 1 x 10^-3 M

[I-] molarity = 1 x 10^-3 M

Q = ionic product = [Pb+2][I-]^2

                            = 5 x 10^-4 x (1 x 10^-3)^2

                             = 5 x 10^-10

Ksp of PbI₂ = 7.1 x 10^-9

here Ksp > Q (ionic product). so precipitate will not form.