# Why do we use a solubility product

Name: Andy 2017-11

### The solubility product

The solubility product is used to determine the maximum solubility of a salt in a solvent. This is a fixed value that can be calculated by multiplying the concentrations of the dissolved ions. Furthermore, the solubility, or the solubility product, depends on the equilibrium of the dissolved components of the salt and the undissolved sediment.

In the case of a salt A + B-, which when dissolved in water reacts to form the ions A + and B-, the MWG is:

K = ([A+] · [B+]): [AB]

The concentration of the solid can be regarded as constant, since the concentration is independent of the amount. Now we can simplify the equilibrium constant to a new constant, usually called KL.

K × [AB] = KL. = [A+] · [B+]

This is the definition of the solubility product. In general, the equation is, for a salt that is according to the reaction equation A.mB.n → Am + Bn divided into anions and cations, similar to the law of mass action, as follows:

KL.= [A+]m · [B-]n

If the product of the concentration of the dissolved ions is less than the solubility product, then there is no saturated solution. However, if it is higher, this means that a supersaturated solution is present and the sediment will crystallize out in the foreseeable future. With the help of the solubility product one can determine how much of a certain salt can dissolve in water. Furthermore, the solubility product is temperature-dependent. The smaller the solubility product, the less soluble the salt is. For example, silver chloride has a solubility product of 1.6 × 10-10 mol2/ l2, while that of calcium sulfate at about 6.1 × 10-5 mol2/ l2 lies. The solubility product can often be read from tables (see formula collection) and thus the concentration of the individual dissolved ions can be calculated.

### Examples

Let's take the example of silver chloride directly. The reaction equation is:

AgCl → Ag+ + Cl-

and thus the solubility product:

KL.= [Ag+] · [Cl-]

Since there are the same number of atoms of silver and chlorine in one molecule of silver chloride, the concentration of the two ions must be the same. So the following applies:

[Ag+] = [Cl-]

We can also write the solubility product as follows:

KL.= [Cl-]2 or [Cl-] = KL.½

with KL = 1.6 × 10-10 mol2 / l2 results for [Cl-]:

[Cl-] = (1,6·10-10 mol2/ l2)½ = 1.26 x 10-5 mol / l

The concentration of the chlorine and silver ions is therefore 1.26 · 10-5 minor.

Let us now turn to a more sophisticated example. The reaction equation for the process of dissolving iron (II) phosphate in water is:

Fe3(PO4)2 → 3Fe2+ + 2 (PO4)3-

So we now have three times the concentration x of iron ions in the solution, twice this concentration of phosphate ions. Substituting these variables in, one obtains according to the general formula for the solubility product:

KL.= [3x]3 · [2x]2 = x5 · 33 · 22

or

x5 = KL. / 33 · 22

we solve for x and get:

x = (KL. / 33 · 22)1/5

Now we set for KL. enter the value taken from the table and determine x. For x we ​​get 3.08 · 10-8 minor. The concentration of iron ions is therefore only 6.16 · 10-8 mol / l and that of the phosphate ions just 9.24 · 10-8 minor. From this it can be concluded that iron (II) phosphate is an extremely poorly soluble salt.