Asked by Donna
Estimate the solubility of lead (ll) chloride in grams per liter.
ksp= 1.7 x 10^-5
ksp= 1.7 x 10^-5
Answers
Answered by
Dr Russ
Given that
PbCl2 = Pb2+ + 2Cl-
then the Ksp = [Pb2+][Cl-]^2
the solubility of the lead(II)chloride in moles per litre is numerically equal to the concentration of [Pb2+] in moles per litre, because 1 mole of PbCl2 gives 1 mole fo Pb2+
so x moles of PbCl2 gives x moles of Pb2+ and 2x moles of Cl-
thus Ksp = (x)(2x)^2 = 1.7 x 10^-5
(I am assuming that your Ksp is appropriate for mole l-1)
so x = cube root [(1.7 x 10^-5)/4]
and find x
PbCl2 = Pb2+ + 2Cl-
then the Ksp = [Pb2+][Cl-]^2
the solubility of the lead(II)chloride in moles per litre is numerically equal to the concentration of [Pb2+] in moles per litre, because 1 mole of PbCl2 gives 1 mole fo Pb2+
so x moles of PbCl2 gives x moles of Pb2+ and 2x moles of Cl-
thus Ksp = (x)(2x)^2 = 1.7 x 10^-5
(I am assuming that your Ksp is appropriate for mole l-1)
so x = cube root [(1.7 x 10^-5)/4]
and find x