pH and Solubility
The solubilities of many substances depend on the pH of the solution. Consider the solubility equilibrium of calcium hydroxide, Ca(OH)2:
Ca(OH)2(s) <==> Ca2+(aq) + 2 OH-(aq)
Increasing the pH (adding OH-) shifts the equilibrium from right to left, decreasing the solubility of Ca(OH)2. This is another example of the common ion effect. On the other hand adding H+ ions or decreasing the pH shifts the equilibrium to the right, and the solubility of Ca(OH)2 increases. This is because the H+ ions remove OH- ions from the right hand side of the equilibrium expression to form water, and according to Le Chatelier's principle the equilibrium will shift toward the side where something is removed.
Example 1
Calculate solubility of Ca(OH)2 in a solution that is buffered at a pH of (a)12.00 and (b) 13.00.
(a) First find the OH- concentration at pH = 12.00
pOH = 14.00 - pH = 14.00 - 12.00 = 2.00
[OH-] = antilog(-2.00) = 1.0 x 10-2
let x equal the molar solubility of Ca(OH)2 which is [Ca2+]
Ksp = [Ca2+][OH-]2 = 8.0 x 10-6 = (x)(1.0 x 10-2)2
x = 0.080 M
(b) Find the OH- concentration at pH = 13.00
pOH = 14.00 - pH = 14.00 - 13.00 = 1.00
[OH-] = antilog(-1.00) = 1.0 x 10-1
let x equal the molar solubility of Ca(OH)2 which is [Ca2+]
Ksp = [Ca2+][OH-]2 = 8.0 x 10-6 = (x)(1.0 x 10-1) 2
x = 8.0 x 10-4 M
The pH also influences the solubility of salts that contain a basic anion, an anion from a weak acid for example. In general we would expect the salts of weak acids to be more soluble in acidic solutions. For example consider CaF2
CaF2(s) <==> Ca2+(aq) + 2 F-(aq)
In an acidic solution the F- ions will be removed by reacting with H3O+ to form HF, thus shifting the equilibrium to the right and causing more CaF2 to dissolve.
H3O+ + F- <==> HF + H2O
The solubilities salts containing anions that do not hydrolyze, such as Cl-, Br-, I-, NO3- in other words anions from strong acids are not affected by pH.