The solubility product for lead chloride (PbCl2) is Ksp=1.2x10^-5 at T=25 degrees Celcious. If solid lead chloride is added to a 0.0200 M solution of sodium chloride (NaCl), a strong electrolyte, what will the concentration of Pb2+ ion solution at equilibrium?
2007-06-03 11:50:33 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
To maintain electro-negativity, the total chloride content must equal that of Na+ and 1/2Pb+2. So if x moles of PbCl2 dissolves,
[Pb+2][0.02+ 2*Pb+2]^2 = 1.2x10^-5
Pb+2 = appx 0.00045 M
2007-06-03 12:04:30 · answer #1 · answered by cattbarf 7 · 0⤊ 0⤋
Ksp = [Pb2+][Cl-]^2 = 1.2 x 10^-5 = 12 x 10^-6
Let [Pb2+] = x Then [Cl-] = (2x + 0.0200) ~ 0.0200 because x is so small.
(x)(0.0200)^2 = (x)(2x10^-2)^2 = (x)(4x10^-4) = 12 x 10^-6
x = 3x10^-2
2007-06-03 19:22:24 · answer #2 · answered by steve_geo1 7 · 0⤊ 0⤋