>>Dr. Mark Morvant: In this video we'll look at solvation. The IUPAC says that solvation
is an interaction of a solute with a solvent, which leads to a stabilization of the solute
species in the solution. A way to think about this is that if we have a solute, let's say
sodium chloride, and we put it into water, before the sodium chloride can actually be
dissolved into a homogeneous solution, there needs to be an interaction between the water
molecules and the ions in the sodium chloride. A similar thing occurs with sugar, when we
dissolve sugar in tea. The sugar is going to need to interact with the water to become
solvated, or stabilized in solution -- to make a homogeneous solution. One of the phrases
you'll often hear with solvation is "like dissolves like." Let's take a look at this
idea. Let's look at these three possible substances. First one, sodium chloride, the sodium chloride
we need to think about what type of interactions this compound has. We know that the sodium
likes to give away electrons and the chloride will accept, so this is going to be an ionic
compound. If we look at water, H20, here we know that we have a strong dipole moment.
We also have a hydrogen attached to an oxygen, so this is going to be a hydrogen-bonding
compound. We can also say that it is very polar because it has a dipole, molecular dipole
moment in one direction. If we look at hexane, here we just have carbon and hydrogen covalent
bonds. We know that because the electronegativity difference between carbon and hydrogen, these
bonds are not polar. So therefore this molecule does not have any bond dipoles. And therefore
the molecule does not have a molecular dipole. What type of intermolecular forces would hexane
have? Well, without ions, hydrogen bonding or dipoles, what is left is London dispersion
forces. If we tried to dissolve sodium chloride in hexane, there would be a substantial difference
between the molecular interactions of the ionic compound and the London dispersion interactions
with the solvent hexane. These are not like enough to dissolve, so therefore they do not
follow the "like dissolves like," and we'd imagine that if we put salt into hexane that
it would just sit at the bottom of the hexane as a precipitant.
If we think about sodium chloride into water, well, we already know this one. This is salt
in water. This is going to dissolve. These are similar enough, like enough, so that it
dissolves. So what is the process? Well, what's actually happening with the sodium chloride
is that the water is interacting with the two ions. We can think about this as the water
is actually breaking the sodium chloride into its ionic character. So if we look at the
sodium, the sodium would interact with the partial negative of the oxygen, and the oxygen
would stabilize through the interaction of an induced dipole with the ion to give us
a stabilized sodium ion. A similar thing happens with the chloride, but on this side it's going
to be the partial positives that are on the hydrogens that will stabilize the chloride
ion. This is the stabilization that's referred to in the definition of solvation. Here we
can see the solvation of sodium chloride in water. Now in this case it's ionic so it breaks
into its ions. That is not always the case. If it is not an ionic compound, it is a covalent
compound, but we'd still have the same type of stabilization interaction. This interaction
is an intermolecular forces in solution. Most of the intermolecular forces you might've
learned about at this point are between pure compounds, dipole-dipole, London dispersion,
hydrogen bonding. When we have a mixture, we have a few additional intermolecular forces.
Those are the ion-dipole, which would be the case of the sodium ion in water. Dipole-induced
dipole, like CO2 in water -- so carbonation. And ion-induced dipole. An example of this
is the iron ion reacting with oxygen and hemoglobin. These interactions are important to whether
their molecule will dissolve into the solvent. The other thing we want to think about with
solvation is the enthalpy of solvation. The enthalpy of the solution is going to be equal
to three enthalpy terms. The first one is the separation of solute molecules. There's
an intermolecular interaction between the solute molecules, and we have to break this
interaction before we can start to create a solution. This is going to be an endothermic
reaction. The second enthalpic term is going to be the separation of solvent molecules.
Again, there is an intermolecular interaction between the solvent molecules, and we have
to overcome this energy. This is also an endothermic reaction. The last process is going to be
the interaction of solute and solvent molecules to form the solution. This will be an exothermic
process. So if we look at it, we have two endothermic processes, delta-H-1 and delta-H-2,
the separation of solute and separation of solvent molecules. And then we have the interaction,
or the coming together, of the solute and solvent molecules to form new intermolecular
forces so that we have an exothermic process. So therefore we can think about the enthalpy
of solution as one of these three possible cases. If delta-H-3, solute and solvent interactions,
is less than that of the enthalpy for the solute-solute and the solvent-solvent, then
it'll be endothermic. If delta-H-3 is equal to delta-H-1 plus delta-H-2, then that will
be at an equilibrium. And then if delta-H-3 is greater than delta-H-1 plus delta-H-2,
then that'll be exothermic.
Often cold packs or hot packs is actually just a solvation process, either an endothermic
solvation process in the case of a cold pack, or an exothermic solvation process in the
case of a heat pack. Solvation is pretty straightforward. Like dissolves like will be an important concept
to remember. The scientific basis for like dissolves like is the intermolecular interactions.
So whenever you see a substance, think about its intermolecular interaction. Is it ionic-ionic,
hydrogen bonding, dipole-dipole, or London dispersion forces? This will help you understand
whether something should be soluble or not soluble in a solvent. Practice, and become
confident.