..........PbCl2(s) ==> Pb^2+ + 2Cl^-
x = solubility..x.......x.......2x
Ksp = (Pb^2+)(Cl^-)^2
Ksp = (x)(2x)^2
Solve for x which will give you the moles/L of PbCl2.
Mols/L x 0.48 L = moles in 480 mL.
Then moles x molar mass PbCl2 = grams PbCl2.
What mass of PbCl2 would be found in 0.48 L of a saturated solution of PbCl2 if the Ksp of PbCl2 is 1.6 x 10-5?
1 answer
1 answer