The molar solubility of PbCl2 at 298K is 1.4422×10-2. Determine the Ksp of this salt.
PbCl2(s) <=> Pb+2(aq) +2Cl-(aq)
Given solubility of PbCl2 = 1.4422×10-2 mol/L
Therefore [Pb+2] = 1.4422×10-2
and [Cl-] = 1.4422×10-2
Solubility product Ksp = [Pb+2] x [2 [Cl-]]2
= 4 x (1.4422×10-2 )3
= 11.998 x 10-6
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