pH and solubility | Equilibrium | AP Chemistry | Khan Academy

Changing the pH of a solution can affect the solubility of a slightly soluble salt. For example, if we took some solid lead(II) fluoride, which is a white solid, and we put it in some distilled water, the solid is going to reach an equilibrium with the ions in solution. Lead(II) fluoride forms lead(II) plus ions and fluoride anions in a 1:2 mole ratio. So, if we have two lead(II) plus ions in this diagram, we need four fluoride anions at equilibrium.

The rate of dissolution is equal to the rate of precipitation, and therefore the amount of solid and the concentration of ions in solution remains constant. This forms a saturated solution of lead(II) fluoride. To the system at equilibrium, we're going to add some H plus ions. By increasing the concentration of H plus ions in solution, we're decreasing the pH of the solution. When the H plus ions are added to the solution, most of them react with the fluoride anions that are present. So, H plus plus F minus forms HF.

Comparing the first diagram to the second diagram, I just happen to add three H plus ions, which will react with three of the fluoride anions that are present to produce three HF. Notice how the concentration of fluoride anions in solution has decreased from the first diagram to the second diagram because of the addition of the H plus ions.

So, the system was at equilibrium, and the concentration of fluoride anions was decreased. According to Le Chatelier's principle, the system will shift in the direction that decreases the stress. So, if the stress is decreased concentration of fluoride anions, the system will shift to the right to make more fluoride anions.

When the system shifts to the right, more lead(II) fluoride dissolves to increase the concentration of Pb2 plus and fluoride anion. We can see that by comparing the second diagram to the third diagram. The amount of solid has gotten smaller since some of that lead(II) fluoride dissolved, and we've increased the concentration of Pb2 plus and F minus in solution. The solid keeps dissolving, and the concentration of ions keeps increasing in solution until the system reaches equilibrium.

For a saturated solution of lead(II) fluoride at equilibrium, decreasing the pH or making the solution more acidic by increasing the concentration of H plus ions increases the solubility of lead(II) fluoride. This is why we saw more of the solid dissolve when the H plus ions were added.

This effect of decreasing the pH and increasing the solubility of a slightly soluble salt happens whenever the slightly soluble salt contains a basic anion. For this example, the basic anion is the fluoride anion, which reacts with the added H plus ions. When the basic anion reacts, that decreases the concentration of that basic anion, which caused the equilibrium to shift to the right.

There are many other examples of basic anions; two more would be the hydroxide anion and the carbonate anion. If a compound contains a basic anion such as the hydroxide anion, hydroxide functions as a base and reacts with H plus ions to form H2O. Therefore, the solubility of a compound containing a hydroxide ion would increase as H plus ions are added to the solution.

It's also important to note for this lead(II) fluoride problem, if the pH is decreased at a constant temperature, the Ksp value for PbF2 remains constant. So, the molar solubility does increase, but the Ksp value remains the same.

This time, instead of lead(II) fluoride, let's look at lead(II) chloride. Lead(II) chloride is also a white solid. If we dissolved some in solution, eventually we would reach an equilibrium between the solid and the ions in solution. So, this diagram here shows a saturated solution of lead(II) chloride, and the system is at equilibrium.

To the system at equilibrium, we decrease the pH by adding H plus ions to the solution. In this case, the chloride anions aren't basic enough to react with the H plus ions; therefore, we do not form HCl, and the concentration of chloride anions remains the same as it was in the original diagram.

So, if the concentration of chloride ions remains the same, then the concentration of lead(II) plus ions would remain the same. The system is still at equilibrium, and decreasing the pH had no effect on the solubility of the solids. Therefore, we have the same amount of lead(II) chloride solid on the bottom of the beaker in both diagrams.

Whenever an anion is an extremely weak base like the chloride anion, we say that this is an anion of negligible basicity. The solubility of salts with anions of negligible basicity is unaffected by changes in pH.