Chemistry, Third edition

SOLUBILITY OF SPARINGLY SOLUBLE IONIC COMPOUNDS

Relationship between solubility product and molar

solubility

The dissolution of silver chloride,

AgCl(s) Ag(aq) + Cl(aq)

shows that the molar solubility of AgCl(s), symbolised s(AgCl), equals either of the

ion concentrations:

[Cl(aq)][Ag(aq)]s(AgCl)

or,

Ks(AgCl)[Ag(aq)][Cl(aq)]s(AgCl)s(AgCl)s(AgCl)^2

Rearranging for s(AgCl) gives

s(AgCl)Ks(AgCl) mol dm^3

At 25 C,Ks(AgCl)1.6 10 ^10 mol^2 dm^6 , so that:

s(AgCl)1.6 10 ^10 mol dm^3

1.3 10 ^5 mol dm^3

(compare Example 11.1).

Use of solubility products in predicting whether or not

precipitation will take place: the common ion effect

For AgCl(s) at 25C,

[Ag(aq)][Cl(aq)]Ks(AgCl)1.6 10 ^10 mol^2 dm^6

This expression tells us that in a saturated solution of AgCl(s) the silver and chloride

ion concentrations can take on any values providedthat when these concentrations

are multiplied together they equal 1.6  10 ^10.

We can decide whether or not a salt precipitates in water by substituting the

actualion concentrations into the solubility product expression for that salt:

1.If the product of the actual ion concentrations is (temporarily) greaterthan the

solubility product, some of the salt precipitates out in order to restore the

product of ion concentrations to the value of Ks(here, 1.6  10 ^10 ).

2.If the product of the ion concentrations exactly equalsthe solubility product, the

solution is saturated with salt, but no precipitation takes place.

3.If the product of the ion concentrations is less thanthe solubility product, the ions

remain in solution, the solution is unsaturated and no precipitation takes place.

Suppose we add a minute grain of silver chloride to some water so that there is

1.0 10 ^5 mol of AgCl(s) in 1 dm^3 of water. Will all the silver chloride dissolve?

The easiest way to approach this question is to supposethat all the AgCl(s) did

completely dissolve and then to decide whether or not the resulting ion concentra-

tions would exceed the solubility product of AgCl(s).

One AgCl(s) dissociates into one Ag(aq) and one Cl(aq) ion, so that

[Ag(aq)][Cl(aq)]1.0 10 ^5 mol dm^3

The product of the actualion concentrations would be

[Ag(aq)][Cl(aq)]1.0 10 ^5 1.0 10 ^5 = 1.0  10 ^10 mol^2 dm^6

179

Solubility product

and molar

solubility

Use the data in Table 11.4

to calculate the molar

solubility of:

(i)barium sulfate and

(ii)magnesium fluoride at

25 C.

Exercise 11F