Calculating Ksp with Known Info & Predicting Precipitation

At times the equilibrium state requires a solution with a common ion in order to remain productive. This means that one of the ions made by the salt will also be present in the solution. This slightly complicates our situation since one of the concentrations of the ions will be affected. This will apparently break our rule for establishing an equilibrium system, but is an important type of reaction to maintain equilibrium, not establish one. This is more apparent with an example:

The molar solubility of PbCl2 in a 0.10 M solution of NaCl is 1.7 x 10-3 M. What is the Ksp for PbCl2?

PbCl2(s) = Pb+2 (aq) + 2 Cl- (aq)

I-0 0.10

-1.7 x 10 -32(1.7 x 10-3)

E-1.7 x 10-30.10 + 0.0034 = 0.10 (2 sig figs)

Ksp = [Pb+2 (aq) ] [Cl- (aq)] 2

= (1.7 x 10-3 ) (0.10)2 = 1.7 x 10-5

Example II :What is the molar solubility of PbI2 in a 0.10 M solution of NaI? Ksp = 7.9 x 10-9.

PbI2 (s)=Pb+2 (aq) +2I- (aq)

I-00.10

-+x+2x

E -x0.10 + 2x `= 0.10

Ksp = 7.9 x 10-7 = (x) (0.10)2x = 7.9 x 10-7 (assumption Ok)

Molar solubility of PbI2 is 7.9 x 10-7 M.

Predicting a precipitate

A precipitate will form in solution if the calculated Ksp is less than the concentration of the actual ions in solution. The product of the ion concentrations is called the ion product. This will create a supersaturated solution and solid will fall out of solution. If Ksp = the ion product, the no precipitate will form since the solution is at he saturation point. If Ksp > the ion product, then the solution is unsaturated and no preciptitate will form. This can be shown with:

A student wishes to form a 1.00 L solution with 0.015 mol of NaCl and 0.15 mol of Pb(NO3)2. The concern is that PbCl2 may form. If the Ksp of PbCl2 is 1.7 x 10-5 will a precipitate form?

Ksp = 1.7 x 10-5 = [Pb+2 (aq)] [ Cl- (aq) ] 2

IP = (0.15 mol / 1L) (0.015 mol / 1 L)2 = 3.4 x 10-5

IP > Ksp  a ppte forms

Work:Q17 p 578Q20, 21 p 582Q22-25 p. 583