Answer to Question #133447 in General Chemistry for hi

Question

Show how to calculate the molar solubility of PbI2 in 0.1M KI (aq)? Ksp=7.1x10-9

Answer 1

The dissociation of Pbl, is as follows:

Pbl₂ gives Pb²⁺ + 2I^-

K_sp of Pbl₂ =7.1x10^-9

The expression for the K of the reaction is as follows:

Construct an ICE table for the reaction.

Pbl_{2 gives} Pb⁺²+ 2I^-

Initial: 0 0.1

Change +x +2x

Equilibrium: x (0.1+2x)

Substitute these values in the expression as follows:

K_sp= (Pb⁺²)(I^-)²

7.1x10 =(x){0.1+2x)²

Solve for x

x= 7.1x10^-7 M

Therefore, the molar solubility of Pbl₂ in KI is 7.1x10^-7 M.

The molar solubility of Pbl₂ is 1.22x10^-3 M