Suppose the solubility of PbCl2 (s) 5.6 x 10^-2 M. Calculate the ksp
........PbCl2 ==> Pb^2+ + 2Cl^-
I......solid......0........0
C......solid......x........2x
E......solid......x........2x
so x = 5.6E-2 and
2x = 2*5.6E-2
Plug those into Ksp expression and solve for Ksp.
Ksp = (5.6E-2)(2*5.6E-2)^2 = ?
To calculate the Ksp (solubility product constant), we need to know the concentrations of the ions formed when PbCl2 dissolves in water. PbCl2 dissociates into one lead ion (Pb2+) and two chloride ions (2Cl-).
Given that the solubility of PbCl2 is 5.6 x 10^-2 M, the concentration of Pb2+ will be equal to this value because one mole of PbCl2 produces one mole of Pb2+. Therefore, [Pb2+] = 5.6 x 10^-2 M.
Since two chloride ions are produced from one mole of PbCl2, the concentration of Cl- ions will be twice the concentration of Pb2+. So [Cl-] = 2 × [Pb2+] = 2 × 5.6 x 10^-2 M = 1.12 x 10^-1 M.
Now, multiply the concentrations of the ions raised to the power of their stoichiometric coefficients to get the Ksp:
Ksp = [Pb2+][Cl-]^2 = (5.6 x 10^-2 M) × (1.12 x 10^-1 M)^2
Evaluating this expression will give you the value of the Ksp for PbCl2.