Liquid Solutions
Readings for this section
Petrucci: Chapter 13 (sections 1 - 6)
Introduction
Solutions are homogeneous mixtures of more than one substance.
- They occur in a single phase,
- They have uniform properties,
- They can be solid, liquid, or gas (liquid solutions are most interesting to chemists).
Gas phase
Gas phase solutions are easily formed from any mixture of gases since the molecules of the gas so rarely interact with each other. If the mixture of gases doesn't actually react, then a gas phase solution will almost certainly form (at least at room temperature and pressure)
Liquid phase
In the liquid phase, the molecules are close enough that intermolecular forces become important. In this phase, a solution will only form between (say) two species A and B if the A---A, B---B and A--B intermolecular forces are approximately the same.
For example, hexane and heptane are two non-polar liquids. The intermolecular forces in each of these pure liquids are primarily dispersion forces, due to temporary dipoles. These are quite weak forces. However, the intermolecular forces that would exist between hexane and heptane would also be primarily dispersion in nature. Hence, a liquid solution will form. The two liquids are said to be completely miscible in each other.
If the forces of one of the molecules for its own kind is much greater than for the other a solution may not form. Take, for example, Water and hexane. Water is a polar molecule and in addition, it bonds to other water molecules with hydrogen bonds. Those are two stronger (and strongest) of the intermolecular forces (compared to dispersion forces). Hexane, on the other hand cannot get involved in either of these two types of interactions and so will not mix with the water. These two liquids are said to be immiscible in each other.
Solid Phase (Crystals)
In the solid phase, not only are the intermolecular forces very well defined, but the crystals of solid form rigid arrangements of atoms whose spacing is quite regular. In order for a second type of molecule to fit, it must be of similar size and shape to the host molecules (or atoms).
Common Solid solutions of this type can be found in gem stones and in metal alloys, among others.
Composition of solutions:
Molarity
There are several common methods for reporting the composition of solutions that we are dealing with. The particular method we use depends largely on the use to which we will put it. In most relatively dilute solutions where we need quick, easy calculations that relate the number of moles in solution to the volume, we use Molarity. Molarity can be defined as:
M = moles of solute/Litres of solution
This gives us units of moles · Litres–1 or mol · L–1. we use an uppercase italics M for the units of molarity and would report the units as 'molar'.
For example:
A sample of 0.243 moles of a compound is dissolved in 1.45 L of a solvent. What is the concentration of the solution in moles per litre?
We can use equations to solve this such as
ci = ni / V
i
molality
In some cases, it is not easy to measure volumes of solutions after mixing and in sometimes, for simplicity of calculations, it is easier to use an alternate concentration unit called molality. The units of molality are defined as
m = moles of solute / kg of solvent
Measurements done in molality are simpler to do if we can easily determine the amount of solute and solvent and then simply mix them as measured. The concentration of the solution in molal units will not depend on possible changes in volume that may occur on mixing. Hence, we can use measurements made before mixing of both the solute and solvent.
Example:
What is the concentration of a solution formed by adding 0.213 g of oxalic acid (COOH)2 to 1200 g of water?
First, we need the number of moles of the solute oxalic acid
n(oxalic acid) = 0.213 g / 90.035 g/mol =0.002366 mol
Now, we can calculate the concentration of the solution
c(oxalic acid) = 0.002366 mol oxalic acid / 1200 mL water
= 0.00197 mol/L (Note the conversion of 1200 mL = 1.200 L)
(assume density of water is 1 g/mL and assume volume of water is same as volume of the solution since very little solvent was added)
Note that the variable for concentration in molal is a lowercase italics m and the units of molal are also a lowercase italics m. Also be aware that the variable of mass in equations where mass is involved is also a lowercase italics m. Don't get them mixed up.
Scales like Molarity and molality are only useful in the case of relatively dilute solutions where one of the species is clearly the most abundant (termed the solvent) and the other is in relatively small proportions (the solute). We usually use measurements of Mole fraction x when we discuss solutions which form over a wide range of concentrations.
Mole fraction is defined as
xi = # moles of i / total number of moles in the solution.
The mole fraction xi can range in value from 0 to 1 where 0 means there is no compound i in the solution and 1 means that the solution is 100% composed of compound i. Obviously, when using mole fraction, there is no solvent/solute issue and changes in volume don't come into play. Mole fraction is used in many circumstances where other concentration units like Molarity and molality are not useful.
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Liquid-Vapour equilibrium
In an ideal solution, the intermolecular forces between the molecules A---A, B---B and A---B are all identical. In reality, we can never get this to occur but we can find solutions where the forces are very close to equal. One example of a mixture which forms nearly ideal solutions is hexane and heptane. These two 'straight-chain' hydrocarbons have similar molecular mass (they are six and seven carbon atoms in length respectively). They are both non-polar and therefore, can interact only using dispersion-type intermolecular forces.
Consider a mixture of hexane (A) and heptane (B). Since both of these liquids are volatile, we expect that the solution too will have a vapour pressure. The vapour will be made up of a mixture of the two gases. The total pressure of this mixture, according to Dalton's law is:
P*Soln = pA + pB {sum of the partial pressures}
For Ideal solutions, we can determine the partial pressure component in a vapour in equilibrium with a solution as a function of the mole fraction of the liquid in the solution. This is Raoult's law:
- pA = xAP*A
and
pB = xBP*B
Substituting into the first equation, we get,
P*Soln = xAP*A + xBP*B or
P*Soln = xAP*A + (1-xA)P*B
= P*B + xA(P*A - P*B )
From this relationship, we see that the vapour pressure of a solution of A and B is a linear function of the mole fraction of A (or of B) where P*A is the intercept and P*A - P*B is the slope.
The Vapour that collects over the solution will have a composition that is
not necessarily the same as that of the liquid. The more volatile component will
have a higher mole fraction in the vapour phase than it does in the liquid
phase.
We can write
-
- Mole Fraction A in the vapour phase = yA
- Mole Fraction B in the vapour phase = yB
We can calculate these values from the solution concentrations as follows.
The vapour composition curve can be plotted as shown in the figure below. It is really two plots, one (the straight line) is the Vapour pressure of solution versus Liquid composition xA and the other, (the curved line) is the same Vapour pressure of solution, but plotted as a function of Vapour composition yA. It could be thought of as pulling the Liquid line to the right (towards the more volatile liquid A). Horizontal tie lines join the two curves such that for any given Vapour pressure the liquid composition xA and the the corresponding vapour composition yA can be determined as indicated by the arrows in the figures.
Normally, we don't carry out experiments as constant temperature as seemed to be indicated in the previous two figures and the corresponding discussion. To do so would involve complicated pressure measuring devices, sealed rigid containers and constant temperature devices. We can much more easily do a measurement of temperature at fixed pressure (say one atm) as a function of mole fraction. We would thus get a plot of boiling point of solution as a function of mole fraction of the solution. To this, we can add a plot of the vapour composition. This curve can be calculated using concepts much like those discussed above for the constant temperature case. The resulting curve (seen below) is shifted towards the higher vapour pressure component just as it was in the diagram above.
In this case, since we already know that the vapour pressure is not a linear function of Temperature (c.f. the Clausius-Clapeyron equation), we do not expect a straight line graph of boiling point as a function of composition. However, for an ideal solution the curvature of the line is only slight.
Boiling of Solutions (Distillation)
If we were to collect all the vapour above the liquid at the boiling point and then condense it we would have a liquid that was higher in the more volatile component than the starting material. If we then re-boil this liquid, we again increase the more volatile component in the resulting distillate. With repetitive steps of boiling, condensing, boiling again, we can eventually completely separate the two components. This would require, however an infinite number of steps.
Azeotropes
In the case of two liquids that mix completely but which are not
that close in terms of the relative strength of the intermolecular forces, we
have a more complicated situation. There are two possibilities: a) The average
intermolecular forces in the solution is stronger than in the individual
liquids, or b) the average intermolecular forces in the solution is less than in
the individual liquids.
Since the intermolecular forces holding a liquid together determines the vapour
pressure (and hence the boiling point of a liquid) we can predict that in the
case of liquid a, the expected boiling point of the solution should be higher
than that of either pure liquid while in solution b the solution will boil at a
lower temperature than either of the two pure liquids.
Consider a solution of benzene and ethanol. In this case, the
intermolecular forces in the solution are less than that in the individual
liquids. We expect that there will be a minimum in the boiling point curve (see
the figure below). at this point (mole fraction of ethanol = 0.46) we also find
that the composition of the vapour is identical to that of the liquid. This is
called an azeotropic mixture and the particular point on the boiling point curve
is called the azeotrope.
A maximum boiling azeotrope happens when the intermolecular forces of the mixture are stronger than the individual liquids. This results in a mixture with a higher boiling point (lower vapour pressure) than the individual. In this case, vapour in equilibrium with the liquid have compositions away from the azeotropic mixture composition, towards pure liquid.
Freezing of solutions
If you cool a solution sufficiently, it will freeze. Allowing for the freezing to occur slowly enough and the solid which crystallizes out will be pure. The temperature at which the solution starts freezing depends on the composition of the solution. Take, for example, a mixture of acetic acid and water. Pure water freezes at 0ºC and pure acetic acid freezes at +16.6ºC.
The phase diagram above shows four different regions. The yellow and blue areas represent compositions/temperature ranges where only one phase exists...
The yellow area is single phase liquid solution
The blue area represents a temperature where ice and acetic acid crystals are mixed in a solid phase.
The red and purple areas correspond to two temperature/composition regions where two phases exist in equilibrium with each other.
The red area represents an equilibrium between pure ice and a solution where the composition of the solution for any given temperature is represented by the position of the line which separates the red from the yellow areas.
The purple area represents an equilibrium between pure solid acetic acid and solution whose composition (for any given temperature) is represented by the position of the line separating the purple and yellow areas.
The intersection between the red-yellow border and the purple-yellow border represents the eutectic point. This represents the lowest melting point composition for this solution. For Acetic acid, water, that point is at a temperature of -26.7ºC. Below this temperature, any mixture of ice and acetic acid is solid.
This diagram can be used to explain several kinds of phenomenon.
Suppose a liquid solution of with a mole fraction of acetic acid = 0.1 is cooled slowly, starting from room temperature. What phase transitions will occur as the cooling process progresses. Follow the vertical line (marked with an asterisk) at x = 0.1.
The first phase change occurs at the temperature corresponding to the point where the vertical line crosses into the red zone. At this temperature, ice starts to crystallize out of the solution. This removes water from the solution, making it more concentrated in acetic acid. Hence, the freezing point lowers. This, process continues as the temperature is lowered such that the composition of the solution follows the red-yellow border down to the Eutectic point.
Below the eutectic point both ice and solid acetic acid crystallize.
We use these properties in our every-day experiences, for instance, in the radiators of automobiles, we put a mixture of ethylene glycol and water. The correct proportions of these two compounds can give a solution that freezes at temperatures as low as -50ºC. Even if it freezes, it will do so slowly, lowering the freezing point as it does and creating a slushy mixture rather than a single solid phase. Thus, even at extremely cold temperatures such as those found in northern Canada, the radiator 'coolant' mixture will flow through the engine and not plug it up.
Look at the diagram again. If we do an experiment at 10ºC (below the melting point of pure Acetic acid) in which we start with pure water and slowly add crystals of acetic acid, we can trace the progress along the blue dashed line.
At first, the acetic acid dissolves into the water. As the proportion of acetic acid increases, we reach the point where the dashed line crosses into the purple region. Beyond the purple-yellow border, we would see crystals of acetic acid sitting in the bottom of the beaker. The solution would be saturated (at equilibrium) and no matter how much more acetic acid solid we add to the beaker, there will be no further net increase in the amount that will dissolve. However, if we raise the temperature to room temp, we would see the rest of the acetic acid dissolve as we cross back into the yellow region on the phase diagram. Hence, we can use the diagram to determine the solubility (concentration at equilibrium) of acetic acid in water for any particular temperature.
All phase diagrams of this sort have the same features. The pure liquids have characteristic melting points and the eutectic point represents the lowest melting point composition of the solution. There are always the same four regions and always, we can explain the freezing/dissolving processes using these diagrams.
In organic chemistry, we often use the properties of solutions to tell if we have properly separated out a desired compound. For example, in the synthesis experiments, which you do in the lab, you test the purity of the crystals you make by measuring their melting point. If your crystals melt at the correct temperature at a well-defined temperature, then your crystals are probably close to pure. If, on the other hand, they melt over a large temperature range or well below the correct melting point, you can be sure that your crystals are not very pure.
Solubility
Sometimes, the components being mixed to form the solutions have melting points that are very different. Take for example, the mixing of water and a salt such as KCl. The salt melts at a very high temperature (770 ºC). The only part of the phase diagram which is of interest to us is the portion shown in the figure above. The same four regions are visible as we noticed in the water/acetic acid phase diagram. However, in this case, we are only looking at relatively low concentrations of KCl in water.
Let's follow from (left to right) the horizontal line representing room temperature. As we add salt to our water, the salt dissolves at first. Salt will continue to dissolve as long as the concentration is in the yellow zone. Eventually, the salt no longer dissolves, it merely settles to the bottom of the beaker. The solution concentration that exists in equilibrium with the solid salt is represented by the intersection of the horizontal line with the purple-yellow border line. This is the solubility of the salt in mole fraction. We normally measure solubility in moles solute per litre of solution. We can easily convert the mole fraction determined here more common units, such as molarity. We can easily see that as the temperature of the solution is raised, the solubility goes up too.
We can also see that as the salt is added to the water, just like in the previous case, the melting point of the water is lowered. Hence, adding salt to ice on the sidewalks and roads lowers the melting point and (hopefully) the ice melts. In many parts of Canada, such as Saskatchewan, the temperature in winter is often well below the point where salt will do any good and hence, it is rarely used there.
Henry's Law
Common experiences tell us that gases dissolve in liquids too. For example, fish can live under water by separating dissolved oxygen from the water using their gills. If the water becomes stagnant and the dissolved oxygen content is reduced because of lack of aeration (mixing with air), many species of fish cannot live in it. Other species have developed special coping mechanisms for dealing with the low oxygen levels... But that's another story.
We also see the effect of gas dissolved in liquid whenever we open a carbonated drink. The drink has carbon dioxide dissolved in it and while the can (or bottle) is closed, the pressure of the gas above the liquid is in equilibrium with the dissolved gas solution. This, of course is the vapour pressure of the CO2 in the solution. When the can is opened, the CO2, whose vapour pressure is higher than normal ambient pressure, is released to the atmosphere and the liquid starts bubbling as the dissolved CO2 starts evolving back into the gas phase. If we shake the can before opening it, the pressure of CO2 above the liquid is raised noticeably, why?
We can see from this set of observations that the amount of dissolved gas in a liquid is dependent on two things. The first is the partial pressure of the gas above the liquid. The second is the rate of dissolution/evolution of the gas.
We are going to concern ourselves with the first option only and assume that enough time has passed to achieve equilibrium.
Henry's Law expresses mathematically what we've seen experimentally,
pB = xBKH
Where pB is the partial pressure of the gas (B) and x is its mole fraction. KH is a constant that depends both on the solvent and the solute. It is called Henry's Law parameter.