Question

Including activity coefficients, calculate the concentration of Pb²⁺ in a 0.100 M aqueous solution of NaIO₃ saturated with Pb(IO₃)₂. (Hint: look up the Ksp value for Pb(IO₃)₂.

Answer 1

Ksp of Pb(IO3)2 = [Pb2+][IO3-]^2 = 2.5 x 10^-13

Using debye-huckel equation for activity coefficient

Y = inv.log.[(-0.51zi^2(sq.rt.(u))/(1+3.3r(sq.rt.(u))]

with,

zi = charge ; u = ionic strength ; r = radius

we get,

u = 1/2cizi^2

   = 1/2(0.1 x 2^2 + 2 x 0.1 x 1^2)

   = 0.3

So,

Y[Pb2+] = inv/log[(-0.51 x 2^2 x sq.rt.(0.3))/(1+3.3 x 0.45 x sq.rt.(0.3))]

              = 0.242

Y[IO3-] = inv/log[(-0.51 x 1^2 x sq.rt.(0.3))/(1+3.3 x 0.45 x sq.rt.(0.3))]

              = 0.701

so,

2.5 x 10^-13 = [Pb2+](0.242)(0.1 x 0.701)^2

[Pb2+] concentration = 2.10 x 10^-10 M