Inorganic and Applied Chemistry

Download free books at BookBooN.com

Inorganic and Applied Chemistry

As mentioned in the prior example the order of solubility is determined by looking at the sizes of the

solubility products only when these have the same units. In the following example we shall look at a

situation where the order of solubility for three species is to be determined.

Example 4- C:

Relative solubility; second example

We wish to determine the order of solubility for the following three solid species:

CuS(s) having a Ksp-value of Ksp = 8.5 · 10-45M^2

Ag 2 S(s) having a Ksp-value of Ksp = 1.6 · 10-49 M^3

Bi 2 S 3 (s) having a Ksp-value of Ksp = 1.1 · 10-73 M^5

These three solid species give upon dissolution a different number of ions (which also may be seen from

the different units for Ksp) following the reaction scheme:

CuS(s)  Cu2+(aq) + S2-(aq)

Ag 2 S(s)  2 Ag+(aq) + S2-(aq)

Bi 2 S 3 (s)  2 Bi3+(aq) + 3 S2-(aq)

In order to determine the order of solubility a comparison of the Ksp values may not be used. On the

contrary the solubility x for each of the species must be determined as sketched in a previous example:

Ksp x [moles/l]

CuS(s)  Cu2+(aq) + S2-(aq)

x x x·x x = Ksp1/3 = 4,2·10-23

Ag 2 S(s)  2 Ag+(aq) + S2-(aq)

2x x (2x)^2 ·x = 4x^3 x = (Ksp/4)1/3 = 3,4·10-17

Bi 2 S 3 (s)  2 Bi3+(aq) + 3 S2-(aq)

2x 3x (2x)^2 ·(3x)^3 = 108x^5 x = (Ksp/108)1/5 = 1,0·10-15

Now the order of solubility may be determined by comparing the sizes of the solubility x. Therefore Bi 2 S 3

is most soluble followed by Ag 2 S. Least soluble is CuS and this has thereby been sketched that when the

number of ions in the species are different the sizes of Ksp cannot be used to determine the order of

solubility.

Equilibrium