What minimum concentration of clâ is required to begin to precipitate pbcl2? for pbcl2, ksp=1.17ã10â5?

Missing in your question : [Pb2+] = 0.085 M

So from this balanced equation:
PbCl2  ↔ Pb+2   +   2Cl-

when Ksp = [Pb+2][Cl]^2
when we have Ksp = 1.17x10^-5 & [Pb+2] = 0.085 M so by substitution:
1.17x10^-5 = 0.085 [Cl-]^2

∴[Cl]^2 = (1.17x10^-5) / 0.085 = 1.38x10^-4
∴[Cl-] = √0.0002
          = 0.0117 M
So 0.0117 M is the minimum concentration requires precipitating PbCl2

2.00 is the !

257 grams h2o i think