Introduction

In the first semester of this general chemistry course you discussed the properties of compounds in aqueous solutions, most notably solubility. In case you forgot, the solubility of a compound is specifically defined as the amount of solute that can be dissolved in a particular solvent, normally just plain water. Additionally, you have most likely treated solubility as a ‘yes’ or ‘no’ property of a substance. In other words, after being required to memorize the solubility rules, you can of course spit out that salts of nitrates, chlorates, and acetates are soluble, while the salts of carbonates, phosphates, and chromates are insoluble. Well, if the truth be told, ALL ionic compounds are soluble in water to some varying degree.

Solubility

If we consider the solubility of an ionic compound in terms of a reaction where a solid reagent dissolves into its component ions, then the reaction is governed by the same rules of equilibrium that rule all other chemical reactions. And, since the solubility of a particular ionic compound is affected by equilibrium, it seems logical that this concept may also make use of some sort of equilibrium constant. Specifically, this is referred to as the solubility product constant (Ksp) which describes the equilibrium between the soluble and insoluble portion of the solute.

When discussing solubility, it is also important to understand the factors affecting the process. In particular, the degree of an ionic compound’s solubility is dependent upon a variety of factors including: (1) the intermolecular forces between the solute and solvent; (2) the change in entropy accompanying the process; (3) the concentration of the products and the reactants; (4) the pressure of the system; and the factor which you will be directly observing in this experiment, (5) the temperature.

Calcium Hydroxide

The ionic compound that you will be dealing with in this experiment is unique. For most salts, the solubility of the compound is generally greater at higher temperatures than lower temperatures (Figure above). However, the solubility of calcium hydroxide is greatest at lower temperatures, a behavior completely opposite of what is generally expected.

In order to observe the effect of temperature upon solubility, and equilibrium, various solutions of calcium hydroxide will be produced at different temperatures. Once equilibrium has been established at each temperature, any excess solid will be removed by vacuum filtration, and then titrated with HCl to establish the concentration of hydroxide ion.

To further investigate some of the principles just discussed, let’s look at some sample data obtained from a similar experiment involving Ca(OH)₂. Let’s assume that after titrating the solution prepared at 25°C, we found that to reach the endpoint we had to add 15.1-mL of 0.05 mol/L HCl. Recalling that the endpoint occurs when the number of moles of acid equals the number of moles of base, we can easily figure out the concentration of the hydroxide ion in this particular solution. Mathematically this is accomplished by multiplying the volume of acid added by its respective concentration:

Setting the number of moles of acid equal to the number of moles of base, and then dividing by the volume of Ca(OH)₂, we arrive at the concentration of the hydroxide ion:

Drifting off topic for a second, the equilibrium reaction for dissolving Ca(OH)₂ is shown below along with the associated expression for its solubility product constant:

Using the equilibrium reaction to see that there is a 2:1 molar ratio of OH^- to Ca²⁺, we can easily arrive at the concentration of the calcium ion by simply dividing the concentration of the hydroxide ion by 2. If our math is correct, we arrive at a [Ca²⁺] of approximately 0.038 mol/L. Taking these values and plugging them into the K_sp expression shown above, we arrive at a value of ~2.17 x 10^-4 for the solubility product constant of Ca(OH)₂ at 25°C. With the same process being carried out at various temperatures, you will slowly begin to notice the temperature dependence of solubility.

Thermodynamics of Solubility

Now, remembering that K_sp is essentially a special form of an equilibrium constant, it seems logical that other thermodynamic factors such as enthalpy (ΔH), entropy (ΔS), and Gibbs free energy (ΔG) will also be dependent upon temperature changes. In fact, the equilibrium constant K_sp is directly correlated with Gibbs free energy of an ionic compound’s dissociation as shown in the following equation where R is the ideal gas constant (8.3145 J/mol K) and T is the experimental temperature in Kelvin.

For our specific example run at 25°C, 298 K, the value of ΔG comes out to be ~30.1 kJ/mol.

After we have calculated ΔG at the various temperatures, we can then use the following relationship to calculate ΔH and ΔS for the dissociation process.

Lucky for us, this equation is conveniently organized in the y = mx + b format. One with which we have become all too familiar with. Realizing this, we can generate a plot of ΔG versus temperature, in Kelvin, to obtain a slope and a y-intercept, -ΔS and ΔH respectively. At this point, it will be your job to figure out the significance of these two parameters!