Chemistry, Third edition

11 · SOLUTIONS AND SOLUBILITY

At 25C,Ks(AgCl)1.6 10 ^10 mol^2 dm^6 , so that the product of the ion con-

centration is smallerthanKs. No precipitation takes place, and so we conclude that

all the AgCl(s)remains in solution. In other words, the grain of silver chloride

completely dissolves in water.

Now suppose that we add a few crystals of solid sodium chloride (NaCl) to the

same solution of AgCl. The chloride ions from the sodium chloride flood into

the solution. Suppose that the concentration of Clin the mixture from the NaCl is

1.0 10 –2mol dm^3. The product of the ion concentrations for silver chloride is

now

[Ag(aq)][Cl(aq)](1.0 10 ^5 ) (1.0  10 ^5 1.0 10 ^2 )

from AgCl only from AgCl plus NaCl

Compared with 1.0  10 ^2 , 1.0  10 ^5 is small, and makes a negligible contribution

to the chloride ion concentration. We therefore take 1.0  10 ^5 + 1.0  10 ^2 as

being equal to 1.0  10 ^2 , and

[Ag(aq)][Cl(aq)](1.0 10 ^5 )(1.0 10 ^2 )1.0 10 ^7

1.0 10 ^7 is greater than 1.6  10 ^10. Because the ion product exceeds Ks, virtually

all the silver ions precipitate out as silver chloride.

The effect of swamping a solution with one of the ions in the solubility product

expression is called the common ion effect. The addition of a common ion reduces the

molar solubilityof the sparingly soluble salt and so most of the salt precipitates out. In

our example, we can prove that the molar solubility of AgCl has fallen, as follows.

We start by remembering that, at 25C,

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

Upon the addition of NaCl in the above example,

[Cl(aq)]1.0 10 ^2 mol dm^3

We then write

[Ag(aq)](1.0 10 ^2 )1.6 10 ^10 mol^2 dm^6

or,

[Ag(aq)]

1.6 10 ^10

1.6 10 ^8 mol dm^3 

molar solubility

=s

1.0 10 ^2 of AgCl

so that the molar solubility of AgCl is now 1.6  10 ^8 mol dm^3. Before adding the

NaCl it was 1.3  10 ^5 mol dm^3 (page 179). In other words, AgCl is about 800

times less soluble in 0.01 mol dm^3 NaCl solution than in pure water.

The common ion effect is also observed with very soluble salts, although the

arithmetic is more complicated because the concentrations in the solubility product

expression need to be modified in order to apply in these cases. A dramatic example

of the common ion effect is seen by making a saturated solution of sodium chloride

and adding a few drops of concentrated hydrochloric acid. The chloride ions from

the hydrochloric acid flood into the salt solution, the solubility product of NaCl is

exceeded, and salt appears as a white precipitate (Fig. 11.4).

Example 11.2 considers a slightly different case, where the relevant ions are intro-

duced by adding two very soluble ionic substances (iron(II) sulfate and sodium

hydroxide) to water.

180

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