Pbcl2 = Pb 2+ + 2cl-
[cl-] = 0.05M
[Pb2+] = ksp/[cl-] ^2= 2.4*10-4 / 0.05^2 = 9.6 * 10-2 M
Concentration of Pb 2+ should be greater than 9.6 * 10-2 M to initiate the precipitate
Lead (II) nitrate is added slowly to a solution that is 0.0500 M in CI^- ions....
Lead(II) nitrate is added slowly to a solution that is 0.0100 M in Cl^- ions. Calculate the concentration of Pb^2+ ions (in mol/L) required to initiate the precipitation of PbCl2. (Ksp for PbCl2 is 2.40 X 10^-4.)
Enter your answer in the provided box. Lead(II) nitrate is added slowly to a solution that is 0.0600 M in Cl− ions. Calculate the concentration of Pb2+ ions (in mol / L) required to initiate the precipitation of PbCl2. (Ksp for PbCl2 is 2.40 × 10−4.)
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