Question

The value of enthalpy of a solid does not indicate whether the solid is soluble in water or not,so how can we predict the solubility of a solid in water ?

Answer 1

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A solution is a homogeneous mixture of two or more substances and can either be in the gas phase, the liquid phase, the solid phase. The enthalpy change of solution refers to the amount of heat that is released or absorbed during the dissolving process (at constant pressure). This enthalpy of solution (ΔHsolutionΔHsolution) can either be positive (endothermic) or negative (exothermic). When understanding the enthalpy of solution, it is easiest to think of a hypothetical three-step process happening between two substances. One substance is the solute, let’s call that A. The other substance is the solvent, let’s call that B.

Step 1: Breaking up the Solute
The first process that happens deals only with the solute, A, which requires breaking all intramolecular forces holding it together. This means the solute molecules separate from each other. The enthalpy of this process is called ΔH1ΔH1. This since this is always an endothermic process (requiring energy to break interactions), then ΔH1>0ΔH1>0.

A(s)−→−−−−energy inA(g)
A(s)→energy inA(g)
Step 2: Breaking up the Solvent
The second process is very similar to the first step. Much like how the solute, A, needed to break apart from itself, the solvent, B, also needs to overcome the intermolecular forces holding it together. This causes the solvent molecules separate from each other. The enthalpy of this process is called ΔH2ΔH2. Like the first step, this reaction is always endothermic (ΔH2>0ΔH2>0) because energy is required to break the interaction between the B molecules.

B(l)−→−−−−energy inB(g)
B(l)→energy inB(g)
At this point, let us visualize what has happened so far. The solute, A, has broken from the intermolecular forces holding it together and the solvent, B, has broken from the intermolecular forces holding it together as well. It is at this time that the third process happens. We also have two values ΔH1ΔH1 and ΔH2ΔH2. that are both greater than zero (endothermic).

Step 3: Combining the Two Together
The third process is when substance A and substance B mix to for a solution. The separated solute molecules and the separated solvent molecules join together to form a solution. This solution will contain one mole of the solute A in an infinite amount of the solvent B.The enthalpy of combining these two substances to form the solution is ΔH3ΔH3 and is an exothermic reaction (releasing heat since interactions are formed) with ΔH3<0ΔH3<0.

A(g)+B(g)−→−−−−−energy outA(sol)
A(g)+B(g)→energy outA(sol)
The enthalpy of solution can expressed as the sum of enthalpy changes for each step:

ΔHsolution=ΔH1+ΔH2+ΔH3.(1)
(1)ΔHsolution=ΔH1+ΔH2+ΔH3.
So the enthalpy of solution can either be endothermic, exothermic or neither ΔHsolution=0ΔHsolution=0), depending on how much heat is required or release in each step. If ΔHsolution=0ΔHsolution=0, then the solution is called an ideal solution and if ΔHsolution>0ΔHsolution>0 or ΔHsolution<0ΔHsolution<0, then these solutions are called non-ideal solutions.



Figure 11: Energy Diagram for Endothermic Dissolving Process (where ΔHsolution>0ΔHsolution>0). The combined magnitude of Steps 1 and 2 is greater than the magnitude of Step 3. This is also a non-ideal solution.

The diagrams below can be used as visuals to help facilitate the understanding of this concept. Figure 11 is for an endothermic reaction, where ΔHsolution>0.ΔHsolution>0. Figure 22 is for an exothermic reaction, where ΔHsolution<0ΔHsolution<0. Figure 33 is for an ideal solution, where ΔHsolution=0ΔHsolution=0.



Figure 22: Energy Diagram for Exothermic Dissolving Process (where ΔHsolution<0ΔHsolution<0). The combined magnitude of Steps 1 and 2 is less than the magnitude of Step 3. This is also a non-ideal solution.

Ideal Solutions
The enthalpy of solution depends on the strengths of intermolecular forces of the solute and solvent and solvent (Equation 11). If the solution is ideal, and ΔHsolution=0ΔHsolution=0, then

ΔHsolution=ΔH1+ΔH2+ΔH3ΔH1+ΔH2=0.=−ΔH3
ΔHsolution=ΔH1+ΔH2+ΔH3=0.ΔH1

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