Question

1.) Calculate the solubility of lead(II) iodate, Pb(IO₃)₂ in pure water and in a 0.50M solution of sodium iodate. The K_sp = 2.6 × 10^‑13.

Answer 1

Ksp = 2.6 x10^-13

a) In pure water

Pb(IO3)2(s) ---------------------------- Pb+2(aq) + 2 IO3-(aq)

                                                      S mol/L           2S mol/L                   where S= molar solubility

Ksp = [Pb+2] [IO3-]^2

Ksp = S x (2S)^2

Ksp = 4S^3

Ksp/4 = S^3

S^3 = Ksp / 4 = 2.6 x10^-13/4 = 0.65 x 10^-13

S = 4.02 x10^-5 mol/L

Solubility of Lead Iodate = S = 4.02 x10^-5 M

b) in 0.50M solution of Sodium Iodate ( NaIO3)

NaIO3 ------------------- Na+ + IO3-

                               0.50M     0.50M

Pb(IO3)2(s) -------------------------- Pb+2(aq) +   2 IO3-(aq)

                                                  S mol/L             ( 2S + 0.50)mol/L

Ksp = [Pb+2][IO3-]^2

Ksp = S x (2S + 0.50)^2

Ksp = S x [ 4S^2 + 0.25 + 2S]

Ksp = 4S^3 + 0.25S + 2S^2

Ksp = 0.25S                       where      4S^3 and 2S^2 is neglected

2.6 x10^-13 = 0.25 S

S= 2.6 x10^-13 / 0.25 = 10.4 x10^-13

S = 1.04 x10^-12 M

Solubility of Lead iodate = 1.04 x 10^-12M