SOLUBILITY AND SOLUBILITY PRODUCT 131SOLUBILITY AND SOLUBILITY PRODUCT
The solubility of a substance is defined as the concentration of
the solution when it is in equilibrium with some of the undissolved
substance. Such a solution in equilibrium is known as a saturated
solution. Supersaturated solutions of many substances are
possible when there is none of the undissolved substance in con-
tact with the solution. At the surface of contact two processes
are going on: (1) Molecules or ions are passing into solution at a
constant rate per unit of surface. (2) Molecules or ions are de-
positing on the surface and thus building up the crystal structure
of the solid, but the rate is dependent on the concentration of the
dissolved substance. When these opposing rates are equal the
solution is saturated. If the dissolved substance is a non-electro-
lyte there is a definite concentration which gives this equilibrium
condition. If the substance is a salt and ionized in the solution
there is also a definite concentration which gives the equilibrium
provided the ions of the salt are present in equivalent amounts.
We frequently have to consider cases in which we have two salts
in solution which have an ion in common, for example, silver ace-
tate and sodium acetate. The tendency to build up the silver ace-
tate crystals is proportional to the concentration of the Ag+ ions
and of the Ac" ions. The latter includes all the Ac" ions, which-
ever salt they originally came from. Equilibrium is reached
when the product of the concentration [Ag+] X [Ac"] has a
definite value which is called the solubility product.
The solubility of AgAc at room temperature is 0.06 F.W. per
liter. Therefore, in a saturated solution the concentration of
each ion is 0.06 and the solubility product = 0.06 X 0.06 =
0.0036. If we stir crystals of AgAc into a 0.1 formal NaAc solu-
tion the crystals will dissolve until equilibrium is reached and the
product of [Ag+] [Ac"] = 0.0036. Let the solubility of AgAc per
liter = x. Then x = [Ag+] and x + 0.1 = [Ac"] and
[Ag+] [Ac"] = x (x + 0.1) = 0.0036Solving, x = 0.028, and therefore the solubility of AgAc in 0.1 N
NaAc = 0.028 F.W. per liter.
When we consider the solubility of a salt in which the ions are
of different valence, as for example lead iodide Pbl2 ^ Pb++ +
I" + I", the product must include as many concentrations as
there are ions to make the neutral molecule. Thus the solubility