Question

Estimate the solubility of lead (ll) chloride in grams per liter.
ksp= 1.7 x 10^-5

Answer 1

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