The ksp of pbi2 is 1.4 x 10-8. what is the molar solubility of lead(ii iodide in a solution of 0.400 m sodium iodide?

The solubility product of a substance us calculated by the product of the concentration of the dissociated ions in the solution raise to the stoichiometric coefficient of the ions. Therefore, we need the dissociation reaction. For this, it will have the reaction:

PbI2 = Pb^2+ + 2I-

We solve as follows:

Ksp = [Pb2+][I-]^2 = 1.4 x 10-8
1.4 x 10-8 = x(2x)^2
1.4 x 10-8 = 4x^3
x = 1.5x10^-3 M

The molar solubility would be 1.5x10^-3 M.

Spontaneity tell us the direction of the reaction, but not how fast it goes. gibbs free energy brings together enthalpy, entropy and temperature in one easy formula. g = h - ts. and, finally the sign of delta g is always negative for a spontaneous reaction.

Answers are 1,2,4 its correct on edgenuity.