Vapor Pressure of Solutions Chapter 13 Part 3

Vapor Pressure The pressure of the vapor present. Vapor is the liquid molecule in gas form over the liquid surface. Remove the liquid and you have a gas!

Vapor Pressure of Solutions In a closed container at constant temperature an equilibrium vapor pressure is established. The picture on the left indicates that vapor molecules leave a solvent to dilute a solution. Why?

Vapor Pressure of Solutions

The vapor pressure of a liquid is much different in a solution than it is in a pure liquid. The dissolved nonvolatile solute lowers the vapor pressure of a solvent. The solute decreases the number of solvent molecules per unit volume lowering the tendency for the molecules to escape into vapor.

Vapor Pressure of Solutions In a solution that is half nonvolatile solute particles and half solvent, one would expect a vapor pressure of 1/2 the pure solvent, since only half as many molecules can escape. That is what is exhibited by such a solution.

Raoult's Law The common mathematical statement for this behavior is known as Raoult's Law: P soln = X solvent P solvent P soln is the observed vapor pressure of the solution X solvent is the mole fraction P solvent is the vapor pressure of the pure solvent.

Raoult’s Law From Raoult’s law we see clearly that the amount of change in the vapor pressure is dependent on the amount of the nonvolatile solute added to the solution (mole fraction) not the quality of the solute.

Raoult’s Law Raoult’s Law is a linear equation (y=mx+b) A plot of P soln vs mole fraction gives a straight line with a slope equal to P solvent.

Ideal Solutions Liquid-liquid solutions that obey Raoult’s Law are called ideal solutions. For solutions that contain volatile solutions a modified Raoult’s Law. Ptotal = Pa + Pb = XaP°a +XbP°b Pa and Pb are the partial pressure of the two liquids in solution. P°a and P°b are the partial pressure of the pure solvent. X is the mole fraction.

Nonideal Solutions

Problems A solution is prepared by mixing 5.81 g acetone (C 3 H 6 O, molar mass =58.1 g/mol) and 11.9 g chloroform (HCCl 3, molar mass=119.4 g/mol). At 35°C, this solution has a total vapor pressure of 260 torr. Is this an ideal solution? The vapor pressures of pure acetone and pure chloroform at 35°C are 345 and 293 respectively.

Solution Ideal solutions follow Raoult’s Law. Moles of each volatile liquid: 5.81 g acetone/58.1 g/mol = mol ace 11.9 g chloro/119 g/mol= mol chloro Equal number of moles thus mole fraction: Xa = 0.500Xc= Ptotal=(0.500)(345torr) +(0.500)(293 torr) Ptotal= 319 torr expected

Discussion Since the observed vapor pressure is 260 torr and the calculated vapor pressure is 319 torr, this is not an ideal solution. Why is the vapor pressure lower? IMF’s: both molecules have a dipoles and these interactions reduce the molecules tendency to escape. This lowers the vapor pressure more than expected.

Colligative Properties These properties are known as the colligative properties of solutions (collected properties) and they are: vapor pressure lowering boiling point elevation freezing point depression osmotic pressure elevation

Boiling point Elevation Boiling occurs when the vapor pressure of a liquid equals atmospheric pressure. But since the vapor pressure of a solution is always lower than that of the pure solvent, more heat will need to be applied to raise it to atmospheric pressure.

Boiling Point Elevation The quantitative relationship which describes this behavior looks like this: ∆ T b = K b m ∆ T b is the change in the boiling point. K b is the "molal boiling point constant" which is a property of the solvent. m is the molality of the solute in the solution. Why Molality?

Boiling Point Elevation

Freezing Point Depression When you add salt to ice the ice melts and the temperature of the solution drops. The resulting solution will not freeze at 0 °C. Salt water has a lower vapor pressure than pure water. Another way to look at it, is the particles interfere with the water molecules’ ability to form a crystal. The freezing point is depressed.

Freezing Point Depression

∆ T=K f m Where ∆ T is the freezing point depression, the change in freezing point between the pure solvent and the solution. K f is the molal freezing point constant. Values depend on the solvent. m is the molality of the solute in the solution