Lec XVI

 Chem 1200

Angel C. de Dios

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Of course, we will not settle with qualitative descriptions of solubility: soluble, slightly soluble, and insoluble. Using the equilibrium expression, we can in fact quantitatively assess how much of solid will dissolve. 

Here is a table that shows solubility product constants, Ksp, for various mostly insoluble salts:

Achieve Hwk 6: 6I and 6J starts with a question that summarizes the relationship between the solubility product constants, Ksp, and molar solubility. Molar solubility is the number of moles of a given salt that will dissolve in 1 Liter of an aqueous solution. Here, we see the importance of the formula of the salt, for this provides us with the proper equilibrium expression.

The first example is AB. There is one A and one B per formula unit. Thus, in this case, a mole of AB is equal to one mole of A and one mole of B. We express the number of moles of AB that will dissolve in 1 Liter of solution (the molar solubility) as x. The concentrations of [A+] and [B-], therefore, in this solution are both equal to x.

In the case of AB2, we should note that each AB2 unit produces 2 B- ions. This 2 also needs to be incorporated in the equilibrium expression so we need [B-]^2. 

For AB3, we have a similar situation, [B-] is 3 times the amount of AB3 that dissolves and the equilibrium expression also takes into account this coefficient. We see 3 in front of x and 3 as an exponent.

The last example, A3B2, illustrates the case when both A and B are present more than once in a formula unit. 

I hope this also illustrates why it is helpful to see these problems before lecture. In this specific instance, I actually go through a problem in the homework. In most cases, I do not, but if you are aware of these problems, there is a greater chance that you would see the relevance of a specific topic I am discussing.

As in acids and bases, ions participating in an equilibrium can come from more than one source. The following is an example. It is an application that is useful in removing toxic metal ions from water.

This concentration of cadmium ions is still high and this can cause brain damage if one drinks water with this level of cadmium ions. To remove additional cadmium ions from solution, we take advantage of the fact that we can increase the amount of OH- ions in the solution by adding a strong base.

Of course, the above is a very basic solution. The next step is to filter so as to remove the precipitated cadmium hydroxide. After this filtration, we can then add acid to neutralize the solution. 

The pH can affect the solubility of an ionic compound, but the compound does not have to be a hydroxide. If the ionic compound contains a weak base, the pH can also affect its solubility. For this reason, it is important to keep in mind which anions are not neutral. This is where it is useful once more to remember the seven strong acids because these acids contain the anions that are not basic at all.

So, let us work with sample problems.