Question

13. Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product: (a) KHC4H4O6 (b) PbI2 (C) Ag4[Fe(CN)6], a salt containing the Fe(CN)4- ion (d) Hg2I2

Answer 1

Answer 2

a) KHC4H4O6

This is potassium hydrogen titrate. It will dissociate as

                    KHC4H4O6   ---> K⁺ + HC4H4O6^-

At equilibrium                         s           s

where solubility is "s"

Ksp = 3.8 x 10-4 = [K⁺][HC4H4O6-] = s²

s = 1.95 X 10^-2 mole / L

b) PbI2

                      PbI₂ ---> Pb⁺² + 2I^-

at equilibrium                  s         2s

Where "s" is molar solubility of PbI2

Ksp = 1.4 x 10^-8 = [Pb⁺²][I-]² = [s][2s]²

1.4 x 10-8= 4s³

s³ = 3.5 X 10^-9

Therefore s= molar solubility of PbI2 = 1.518 X 10^-3 moles / L

c) Ag4[Fe(CN)6] ---> 4Ag⁺    + [Fe(CN)4]-

Ksp = [Ag+]⁴ [[Fe(CN)4]-] = [4s]⁴[s]

1.6 x 10^-41 = 256s^5

62.5 X 10^-45 = s^5

s = 2.286 X 10^-9 Moles / L

d) Hg2I2 ---> Hg₂⁺² + 2I-

Ksp = [Hg₂⁺² ] [I-]²

Ksp = 4.5 X 10^-29 = [s][2s]^2

4s³ = 4.5 X 10-29

s³ =11.25 X 10^-30

s = 2.25 x 10^-30 moles / L