Page 53 - Vogel's TEXTBOOK OF QUANTITATIVE CHEMICAL ANALYSIS

SOLUBILITY PRODUCT 2.6

water, some of it passes into solution to form a saturated solution of the Salt
and the process appears to cease. The following equilibrium is actually present
(the silver chloride is completely ionised in solution):

The rate of the forward reaction depends only upon the temperature, and at
any given temperature:

where k, is a constant. The rate of the reverse reaction is proportional to the
activity of each of the reactants; hence at any given temperature:

where k2 is another constant. At equilibrium the two rates are equal, i.e.

In the very dilute solutions with which we are concerned, the activities may be
taken as practically equal to the concentrations so that [Ag'] x [Cl-] = const.
It is important to note that the solubility product relation applies with
sufficient accuracy for purposes of quantitative analysis only to saturated
solutions of slightly soluble electrolytes and with small additions of other salts.
In the presence of moderate concentrations of salts, the ionic concentration,
and therefore the ionic strength of the solution, will increase. This will, in general,
lower the activity coefficients of both ions, and consequently the ionic
concentrations (and therefore the solubility) must increase in order to maintain
the solubility product constant. This effect, which is most marked when the
added electrolyte does not possess an ion in common with the sparingly soluble
salt, is termed the salt effect.
It will be clear from the above short discussion that two factors may come
into play when a solution of a Salt containing a common ion is added to a
saturated solution of a slightly soluble salt. At moderate concentrations of the
added salt, the solubility will generally decrease, but with higher concentrations
of the soluble salt, when the ionic strength of the solution increases considerably
and the activity coefficients of the ions decrease, the solubility may actually
increase. This is one of the reasons why a very large excess of the precipitating
agent is avoided in quantitative analysis.
The following examples illustrate the method of calculating solubility
products from solubility data and also the reverse procedure.

Example 1. The solubility of silver chloride is 0.0015 g per L. Calculate the
solubility product.
The relative molecular mass of silver chloride is 143.3. The solubility is
therefore 0.0015/143.3 = 1.05 x IO-' mol per L. In a saturated solution, 1 mole
of AgCl will give 1 mole each of Ag + and Cl -. Hence [Ag + ] = 1.05 x 10 - '
and [Cl-] = 1.05 x IO-' mol L-'.
K,(,,,,, = [Ag'] x [Cl-] = (1.05 x IO-') x (1.05 x IO-')
= 1.1 x IO-'' mol2 L-2