Question

A saturated solution is prepared by adding excess PbI2(s) to distilled water to form 1.0L of solution at 25oC. The concentration of Pb+2 in the saturated solution is found to be 1.3*10-3 M.

Calculate the Ksp for PbI2(s) in water at 25 °C

R/ Kps=[Pb^2+][I^-1]^2, where, [I^-1]=2[Pb^2+]= 2.6*10^-3,
Kps=[Pb^2+][I^-]^2= 8.8*10^-9

B) Calculate the solubility of PbI2(s) in distilled water at 25oC. Answer in mg/L. Answer: 599.57 mg/L

PbI2-------->Pb^2+ + 2I^-
[]0 0 0 0
[]rxn 0 x 2x
[]equilibrio 0 x 2x

Kps=[x][2x]^2, where Kps= 8.8*10^-9
8.8*10^-9=(s)(4s^2)
s^3=8.8*10^-9/4
s=1.30054*10^-3 mol/L
(1.30057*10^-3 mol/L)(461 g/1 mol)(1000 mg/1g)=599.57 mg/L

C) Calculate the solubility of PbI2(s) in 0.1M NaI. Answer in mg/L.
is similar and the answer is 0.40568 mg/L

Answer 1

Is your prof picky about significant figures. You have too many in some cases.