Of the general chemistry content covered on the MCAT, solubility is notable because it seems simple, but contains some surprises. On one hand, solubility is an accessible topic to anyone who has ever added sugar to their coffee or tea, but on the other hand, there are some formal aspects of how it’s analyzed that can cause uncertainty. In this short blog post, we’ll review how solubility constants are written and analyze how they can be used to make sense of the common ion effect, a common point of confusion.
Although we often discuss substances as being either soluble or insoluble within a given solvent, in reality, solubility is a spectrum. We can think of solubility as reflecting a reaction in which a solid dissolves, which can be expressed as follows for an ionic solid:
AB(s) → aA(aq) + bB(aq)
We can write an equilibrium constant for this reaction, using our usual formula of concentration of products over concentration of reactants; however, we should keep in mind that the solid-phase reactant shouldn’t be included. This equilibrium constant actually is the solubility product (Ksp):
Ksp = [A]a[B]b
Therefore, for example, the solubility constant for CaF2 would be expressed as Ksp = [Ca][F]2. Be sure to account for stoichiometric constants, which turn into the exponents of the concentration terms.
At this point, a natural question would be why we use solubility products. Why not just express solubility as mass per unit volume at a given temperature? In fact, we can do so. For instance, the solubility of CaF2 (which is relatively insoluble) in water at 20°C is 0.0016 g/100 mL. However, using the solubility product allows us to capture some other important dynamics associated with solubility. One of the most important of these phenomena is known as the common-ion effect.
Consider silver chloride, AgCl, which is only very slightly soluble in water (Ksp = 1.77×10−10). Let’s first consider what this value would imply for dissolving AgCl in pure water. We can write the solubility product as follows:
Ksp = 1.77×10−10 = [Ag+][Cl−]
Then, we can substitute x for the concentration of AgCl:
1.77×10−10 = [x][x]
1.77×10−10 = [x]2
[x] = 1.77×10−5 M
Now, let’s imagine that we dissolve AgCl in a solution of 0.5 M NaCl. In this case, the [Cl−] term will automatically be [0.5 M]. We can neglect the amount of chloride ion that comes from AgCl to simplify our calculations, and carry out the following math:
Ksp = 1.77×10−10 = [Ag+][Cl−]
1.77×10−10 = [x][0.5]
[x] = 3.54×10−10 M
This shows that dramatically less AgCl will dissolve now that we have flooded the reaction environment with [Cl−]. This is the common ion effect in action!
Hopefully this quick review has demystified solubility to some extent. If faced with a problem involving the solubility constant on Test Day, remember to take care to include stoichiometric constants as exponents and to account for the reaction conditions that you’re provided.
We wish you the best of luck!
