If the Ksp of Pbl2 is 1.4 x 10 ^ - 8 M what is the molar solubility of Pb^ 2+ ?

Question

If the Ksp of Pbl2 is 1.4 x 10 ^ - 8 M what is the molar solubility of Pb^ 2+ ?

Answer 1

what is the Solubility in M of Pb^+2

in the compound PbI2 with a Ksp of 1.4x10^-8

The solubility of a compound indicates the maximum M or mol/L that can dissolve in solution when the compound ionizes into its respective ions at equilibrium.

Given the Ksp the problem solution proceeds as follows:

1) PbI2 <——> Pb^+2 + 2I^-

The rate of dissolution = rate of formation of solid compound at equilibrium and the concentrations are constant. The Ksp is a relatively small number indicating its maximum solubility at equilibrium will be small meaning the majority of the compound does not dissolve. At equilibrium the solution is saturated, the maximum amount of dissolved ions exist.

2) using an ICE table (I = initial) (C= Change)

(E=Equilibrium)

PbI2 <——>. Pb^+2 + 2I^-

I 0 0

C +X +2X

E X 2X

3) The equilibrium equation:

Ksp = {Pb^+2}{I^ -}^2

1.4 x 10^-8 = (X)(2X)^2

1.4 x 10^-8 = 4X^3

X^3 = 3.5 x 10^-9

Take the cube root of both sides

X = 1.518 x 10^-3 M

4) X = the solubility of PbI2 = 1.518 x 10^-3 M

as well as the solubility of Pb^2 = 1.518 x 10^-3 M

The solubility of 2I^- = 2X = 3.036 x 10^-3 M