Colligative Properties of Solutions
General properties of solutions include vapour-pressure lowering, boiling-point elevation, freezing-point depression, and osmotic pressure. These properties are commonly referred to as colligative, or collective, properties. Colligative properties depend only on the number of solute molecules present, not on the size or molar mass of the molecules.
Consider a solution that contains a solvent 1 and a nonvolatile solute 2, such as a solution of sucrose in water. Because the solution is ideal dilute, Raoult’s law applies:
P = x1P1*
Because x1 = 1 – x2, the equation above becomes
P1 = (1 – x2) P1*
Rearranging this equation gives
P1* – P = ΔP = x2 P1*
where ΔP, the decrease in vapour pressure from that of the pure solvent, is directly
proportional to the mole fraction of the solute.
The vapour pressure of a solution falls in the presence of a solute. A convincing explanation is provided by the entropy effect. When a solvent evaporates, the entropy of the universe increases because the entropy of any substance in the gaseous state is greater than that in the liquid state (at the same temperature). The solution process itself is accompanied by an increase in entropy. This result means that there is an extra degree of randomness, or disorder, in a solution that was not present in the pure solvent. Therefore, the evaporation of solvent from a solution will result in a smaller increase in entropy. Consequently, the solvent has less of a tendency to leave the solution, and the solution will have a lower vapour pressure than the pure solvent.
The boiling point of a solution is the temperature at which its vapour pressure is equal to the external pressure.
For a solution containing a nonvolatile solute, the boiling-point elevation originates in the change in the chemical potential of the solvent due to the presence of the solute. The chemical potential of the solvent in a solution is less than the chemical potential of the pure solvent by an amount equal to RT ln x1.
How this change affects the boiling point of the solution can be seen from Figure. The solid lines refer to the pure solvent. Because the solute is nonvolatile, it does not vaporize; therefore, the curve for the vapour phase is the same as that for the pure vapour. On the other hand, because the liquid contains a solute, the chemical potential of the solvent decreases (the dashed curve). The points where the curve for the vapour intersects the curves for the liquids (pure and solution) correspond to the boiling points of the pure solvent and the solution, respectively. We see that the boiling point of the solution (Tb′) is higher than that of the pure solvent (Tb).
We now turn to a quantitative treatment of the boiling-point elevation phenomenon. At the boiling point, the solvent vapour is in equilibrium with the solvent in solution, so that
µ1 (g) = µ1(l) = µ1*(l) + RT ln x1
or Δµ1 = µ1 (g) – µ1* (l) + RT ln x1 …….(1)
where Δµ1 is the Gibbs energy change associated with the evaporation of 1 mole of solvent from the solution at temperature T, its boiling point. Thus, we can write Δµ1 = Δvap G’. Dividing Equation (1) by T, we obtain
ΔvapG’/T = µ1 (g) – µ1* (l)/T = R ln x1
From the Gibbs–Helmholz equation, we write
d(ΔG/T)/dT = – ∆H/T2 (at constant P)
or d(ΔvapG’/T)/dT = – ΔvapH’/T2 = R d(ln x1)/dT
where ΔvapH’ is the molar enthalpy of vaporization of the solvent from the solution. Because the solution is dilute, ΔvapH’ is taken to be the same as the molar enthalpy of vaporization of the pure solvent. Rearranging the last equation gives
d(ln x1) = – ΔvapH’/RT2 dT ………… (2)
To find the relationship between x1 and T, we integrate Equation (2) between the limits Tb′ and Tb, the boiling points of the solution and pure solvent, respectively. Because the mole fraction of the solvent is x1 at Tb′ and 1 at Tb, we write
ln1 ∫ln x1 d ln x1 = Tb∫Tb′ – ΔvapH’/RT2 dT
or ln x1 = ΔvapH’/R (1/Tb’ – 1/Tb)
= – ΔvapH’/R (Tb’ – Tb)/Tb’Tb
= – ΔvapH’/R ΔT/Tb2 ……….(3)
where ΔT = Tb’ – Tb. Two assumptions were used to obtain Equation (3), both of which are based on the fact that Tb′ and Tb differ only by a small amount (a few degrees). First, we assumed ΔvapH’ to be temperature independent and second, Tb′ ≈ Tb, so that Tb′Tb ≈ Tb2.
Equation (3) gives the boiling-point elevation, ΔT, in terms of the concentration of the solvent (x1). By custom, however, we express the concentration in terms of the amount of solute present, so we write
ln x1 = ln (1 – x2) = – ΔvapH’/R ΔT/Tb2
where, ln (1 – x2) = – x2
We now have ΔT = RTb2 x2/ ΔvapH’
To convert the mole fraction x2 into a more practical concentration unit, such as molality (m2), we write
x2 = n2/n1 + n2 ≈ n2/n1 = n2 x ℳ1 /w1 (n1 >> n2)
where w1 is the mass of the solvent in kg and ℳ1 is the molar mass of the solvent in kg mol-1, respectively. Because n2/w1 gives the molality of the solution, m2, it follows that x2 = ℳ1m2 and thus
ΔT = RTb2ℳ1m2/ ΔvapH’ ……….(4)
Note that all the quantities in the first term on the right of Equation (4) are constants for a given solvent, and so we have
Kb = RTb2ℳ1/ ΔvapH’ ………..(5)
where Kb is called the molal boiling-point-elevation constant. The units of Kb are K mol-1 kg. Finally,
ΔT = Kb m2 …………(6)
The advantage of using molality is that it is independent of temperature and thus is suitable for boiling-point elevation studies.
The figure below shows the phase diagrams of pure water and an aqueous solution. Upon the addition of a nonvolatile solute, the vapour pressure of the solution decreases at every temperature. Consequently, the boiling point of the solution at 1 atm will be greater than 373.15 K.
The thermodynamic analysis of freezing-point depression is similar to that of boiling-point elevation. If we assume that when a solution freezes, the solid that separates from the solution contains only the solvent, then the curve for the chemical potential of the solid does not change. Consequently, the solid curve for the solid and the dashed curve for the solvent in solution now intersect at a point (Tf′)below the freezing point of the pure solvent (Tf). By following the same procedure as that for the boiling-point elevation, we can show that the drop in freezing point ΔT (i.e., Tf – Tf′, where Tf and Tf′ are the freezing points of the pure solvent and solution, respectively) is
ΔT = Kf m2 ………..(7)
where Kf is the molal freezing-point-depression constant given by
Kf = RTf2ℳ1/ΔfusH’ …………(8)
where ΔfusH’ is the molar enthalpy of fusion of the solvent.
At 1 atm, the freezing point of the solution lies at the intersection point of the dashed curve (between the solid and liquid phases) and the horizontal line at 1 atm. It is interesting that whereas the solute must be nonvolatile in the boiling-point-elevation case, no such restriction applies to lowering the freezing point. A proof of this statement is the use of ethanol (b.p. 5 351.65 K) as an antifreeze.
The phenomenon of osmosis is illustrated in Figure. The left compartment of the apparatus contains pure solvent; the right compartment contains a solution. The two compartments are separated by a semipermeable membrane (e.g., a cellophane membrane), one that permits the solvent molecules to pass through but does not permit the movement of solute molecules from right to left. Practically speaking, then, this system has two different phases. At equilibrium, the height of the solution in the tube on the right is greater than that of the pure solvent in the left tube by h. This excess hydrostatic pressure is called the osmotic pressure.
We can now derive an expression for osmotic pressure as follows.
Let µ1L and µ1R be the chemical potential of the solvent in the left and right compartments, respectively. Initially, before equilibrium is established, we have
µ1L = µ1* + RT ln x1
= µ1* (x1 = 1)
µ1R = µ1* + RT ln x1 (x1 < 1)
µ1L = µ1* > µ1R = µ1* + RT ln x1
Note that µ1L is the same as the standard chemical potential for the pure solvent, µ1*, and the inequality sign denotes that RT ln x1 is a negative quantity. Consequently, more solvent molecules, on the average, will move from left to right across the membrane. The process is spontaneous because the dilution of the solution in the right compartment by solvent leads to a decrease in the Gibbs energy and an increase in entropy. Equilibrium is finally reached when the ﬂow of solvent is exactly balanced by the hydrostatic pressure difference in the two side tubes. This extra pressure increases the chemical potential of the solvent in solution, µ1R.
We know that
(∂G/∂P)T = V
We can write a similar equation for the variation of the chemical potential with pressure at constant temperature. Thus, for the solvent component in the right compartment,
(∂µ1R/∂P)T = V1’ ……………(9)
where V1’ is the partial molar volume of the solvent. For a dilute solution, V1’ is approximately equal to V’, the molar volume of the pure solvent. The increase in the chemical potential of the solvent in the solution compartment (Δ µ1R) when the pressure increases from P, the external atmospheric pressure to (P + Π) is given by
Δ µ1R = P∫P +Π V’dP = V’Π
Note that V’ is treated as a constant because the volume of a liquid changes little with pressure. Π represents the osmotic pressure. The term osmotic pressure of a solution refers to the pressure that must be applied to the solution to increase the chemical potential of the solvent to the value of its pure liquid under atmospheric pressure.
At equilibrium, the following relations must hold:
µ1L = µ1R = µ1* + RT ln x1 + ΠV’
Because µ1L = µ1*, we have
ΠV’ = – RT ln x1 …………(10)
To relate Π to the concentration of the solute, we take the following steps. From the procedure employed for boiling-point elevation:
– ln x1 = – ln (1 – x2)
≈ x2 (x2 << 1)
x2 = n2/n1 + n2 ≈ n2/n1 (n1 >> n2)
where n1 and n2 are the number of moles of solvent and solute, respectively. Equation (10) now becomes
ΠV’ = RT x2
= RT (n2/n1) ………….(11)
Substituting V’= V/n1 into Equation (11), we get
ΠV = n2RT
If V is in litres, then
Π = (n2/V)RT
= MRT …………(12)
where M is the molarity of the solution. Note that molarity is a convenient concentration unit here because osmotic pressure measurements are normally made at a constant temperature. Alternatively, we can rewrite Equation (12) as
Π = (c2/ℳ2 ) RT …………….(13)
Or Π/ c2 = RT/ ℳ2 ……………(14)
where c2 is the concentration of the solute in g L-1 of the solution and ℳ2 is the molar mass of the solute. Equation (14) provides a way to determine molar masses of compounds from osmotic pressure measurements.
For a nonideal solution, the osmotic pressure at any concentration, c2, is given by
Π/ c2 = RT/ ℳ2 (1 + Bc2 + Cc22 + Dc23 + …) …….. (15)
where B, C, and D are called the second, third, and fourth virial coefficients, respectively. The magnitude of the virial coefficients is such that B >> C >> D. In dilute solutions, we need be concerned only with the second virial coefficient. For an ideal solution, the second and higher virial coefficients are all equal to zero, so Equation (15) reduces to Equation (14).
If two solutions are of equal concentration, and hence have the same osmotic pressure, they are said to be isotonic. For two solutions of unequal osmotic pressures, the more concentrated solution is said to be hypertonic, and the less concentrated solution is said to be hypotonic.
A related phenomenon to osmosis is called reverse osmosis. If we apply a pressure greater than the equilibrium osmotic pressure to the solution compartment, the pure solvent will ﬂow from the solution to the solvent compartment. This reversal of the osmotic process results in the unmixing of the solution components. An important application of reverse osmosis is the desalination of water.