1 L of 0.002 mol/L Pb(NO3)2 has been prepared. Given that ksp for PbCl2 is 1.7 x10^-5, what is the largest number of moles of nacl that can be added to this solution without causing a precipitate to form.

Answer 1

Ksp PbCl2 = 1.7E-5 = (Pb^2+)(Cl^-)^2
You know Ksp and (Pb^2+), substitute and solve for (Cl^-). That gives the molarity of Cl^- which = M NaCl and since that is in 1 L solution that is mols NaCl.

Answer 2

I have a similar question and I got 1.7x10^-8 but would we have to multiply this value by 2?