Observation 1: Lowering of the Vapor Pressure in Solution

To begin studying solutions, our rst task is to observe what impact, if any, the presence of a solute has on the properties of the solvent. We will begin with a simple two-component solution, with a solvent and a single solute. The type of solute will matter to us, as we will observe dierent behaviors for dierent solutes, particularly whether the solute is, in its pure form, a solid, a liquid, or a gas. To start, we will consider solutions formed by dissolving a solid solute into a liquid solvent. This choice is easiest to start with because the solid solute will be assumed to be non-volatile. That is, it does not readily evaporate and therefore has zero vapor pressure. Solids do have a vapor pressure, but for most solids, the vapor pressures are suciently small that we can ignore them. As a rst guess, then, we might assume that the solution formed from a volatile solvent and a non-volatile solute would have the same vapor pressure as the solvent alone, since the solute seems to contribute nothing to the equilibrium vapor pressure.

To study this question, we return to our familiar apparatus for studying vapor pressure. Consider trapping a quantity of liquid in a cylinder with a piston, which is then pulled back to create a volume above the liquid. We then measure the pressure in that volume, which is the pressure created by the vapor in equilibrium with the liquid. Recall that the pressure we observe is independent of the volume of liquid we trap, so long as there is some liquid remaining in the cylinder, or as long as the liquid does not all evaporate.

In this case, instead of trapping pure water in the cylinder, we will trap a solution of glucose in water. Just to be quantitative, we'll make this a 1.0 M solution of glucose. Recall that the vapor pressure of pure water at 25 ºC is 23.756 torr. When we measure the vapor pressure of the glucose solution, we nd a dierent value, 23.33 torr, lower than the vapor pressure of the pure water. To check this, we measure the vapor pressure of a 2.0 M glucose solution , and we observe 22.93 torr. Note that the higher concentration has an even lower vapor pressure. If we look at these numbers carefully, we notice that the amount by which the vapor pressure was lowered for the 2.0 M solution (0.82 torr) seems to be about double the lowering for the 1.0 M solution (0.42 torr). To verify this trend, we can try several solutions and plot the vapor pressure as a function of the concentration of the glucose solution. The result is shown in Figure 1, where the vapor pressure appears to decrease in a linear fashion as we increase the concentration of glucose.

185 It isn't obvious just from looking, but the graph in Figure 1 is not exactly a straight line. It curves slightly. It isn't obvious either that we can get a better straight line by changing the way in which the concentration of the glucose is measured. Instead of using the molarity (moles per liter), we can measure the mole fraction of the glucose, dened by:

Xglucose=nglucose/(nglucose+nwater)

Note that the mole fraction is, as the name suggests, the fraction of the total number of moles of substance that is glucose. If we plot the vapor pressure of the glucose solutions as a function of the mole fraction of glucose, we get the graph in Figure 2, which essentially turns out to be a perfectly straight line.

Figure 18.2

Based on the graph in Figure 2, we can say that the vapor pressure of the glucose solution is a linear function of the mole fraction of the glucose in the solution. With a little work, we can show that the line in Figure 2 is described by the equation:

Pvap=P*vap*(1-Xglucose)

where Pvap* is the vapor pressure of the pure water. This means that it is only the number of moles of

glucose that aects the vapor pressure of the solution. If we look back at the denition of mole fraction, we can see that the mole fraction of glucose plus the mole fraction of water must equal 1. This means that Xglucose + Xwater = 1, so we can rewrite the above equation as:

Pvap=P*vap*Xwater

This is a somewhat unexpected result: glucose seems to be absent from the equation! In fact, it is present, but only in the expression for the mole fraction of water

We should check this result by comparing it to other aqueous solutions of nonvolatile solutes. Ethylene glycol, commonly used in antifreeze solutions, is soluble in water but has a very small vapor pressure at 25 ºC, less than 0.1 torr. When we prepare ethylene glycol solutions with molarities in the range of 1.0 M to 3.0 M and measure the vapor pressures as before, we discover a remarkable result. The vapor pressures follow the same graphs as shown in Figures 1 and 2. There is no discernible dierence within the accuracy of our measurements. This means that, in our equation Pvap=P*vap*Xwater, the vapor pressure of the aqueous

solution does not depend on the identity of the nonvolatile solute. It depends only on the number of moles of the nonvolatile solute. This is not at all an intuitive result. However, it is quite general, and these experimental results and the equation above both go by the name Raoult's Law. Raoult's measurements

186 CHAPTER 18. PHASE EQUILIBRIUM IN SOLUTIONS showed that this equation works for almost all solvent and nonvolatile solute combinations, provided only that the concentration of the solute is not too high.

Given this surprising result, we need to generate a model which can account for it. Before doing so, we will consider an additional aspect of our observation. Imagine that we start with pure liquid water in equilibrium with its vapor at the normal boiling point. This means that the temperature is 100 ºC, and the applied pressure and the vapor pressure are both 1 atm. Now imagine that we add some amount of glucose to the water. According to Raoult's law, the vapor pressure of the solution we just formed must be less than the vapor pressure of the pure water. Therefore, the vapor pressure is less than 1 atm, which is less than the applied pressure. Since the applied pressure is greater, all of the vapor must now condense into the liquid, and we no longer have liquid-vapor equilibrium. The solution is not at its boiling point, even though the temperature is 100 ºC. The addition of the nonvolatile solute has disrupted the liquid-vapor equilibrium.

How can we restore the equilibrium? There are two ways. The obvious way would be to lower the applied pressure to the vapor pressure of the solution. A less obvious way would be to increase the temperature without changing the applied pressure. To see this, remember that Raoult's law gives the vapor pressure of the solution in terms of the vapor pressure of the pure liquid: Pvap=P*vap*Xwater. If we need Pvap to be 1

atm even though Xwater is less than 1, we can increase Pvap* by increasing the temperature. This means

that we can nd a normal boiling point for the solution, but it will be at a higher temperature than the boiling point of the pure liquid at the same applied pressure. The boiling point is elevated by the presence of the nonvolatile solute. The amount by which the boiling point changes is typically quite small, about 0.5 ºC for a 1 M solution, but it is easily observable.