Question

For the reaction: PbCl2(s) ↔ Pb2+(aq)+2Cl1-(aq), what is Q* when 1.7 mL of 0.058 M lead nitrate is added to 16 mL of 0.025 M sodium chloride? Ksp of lead chloride is 1.6 x 10-5 M3.

Answer 1

concentration of Pb(NO3)2 = 1.7 x 0.058 / 1.7 + 16

                                            = 5.57 x 10^-3 M

concentration of NaCl = 16 x 0.025 / 16 + 1.7

                                    = 0.0226 M

PbCl2(s) <---------------> Pb2+(aq) + 2Cl- (aq)

Q = [Pb2+] [Cl-]^2

   = (5.57 x 10^-3) (0.0226)^2

Q = 2.84 x 10^-6

here Q < Ksp . so no precipitate forms.