Solutions

Chemistry refresher A solution is a homogeneous mixture of several components. Consider a two-component solution. One component has a mole fraction nw (the ratio of the number of moles of a component to the total number of moles), and the other component has mole fraction n. The component with a greater mole fraction, let it be nw, is called the solvent. The solvent determines the state of matter of the solution (gas, liquid or solid). The component with the smaller mole fraction, n, is called the solute. In Chapter 5 we considered a cloud droplet as an example of a solution with water as a solvent and the salt as a solute. A solution of a salt in a solvent such as water is saturated when the rates of dissolving and crystallization are equal. In this case there could be some substance in the crystalline form present in the composite system. For example, there might be a salt crystal inside a cloud droplet. The amount of dissolved material (solute) in the saturated solution is called its solubility, which might depend strongly on temperature and weakly on pressure. We are interested in the effect of the dissolved solute on the vapor pressure of the solvent.

Raoult's Law Consider a solution that is in chemical equilibrium. The vapor pressure of each component of the solution is approximately

where ppure is the vapor pressure of a pure ith component and m is the mole fraction of an ith component of the solution. The solution is called ideal when both solvent and solute obey Raoult's Law. Raoult's Law applies when the components of the solution are present in high concentrations. We used Raoult's Law when we considered the equilibrium vapor pressure over a droplet containing dissolved electrolytes.

For solutions at low concentrations the vapor pressure of the solute obeys Henry's Law. According to Henry's Law, the vapor pressure of a solute, p, is a product of the mole fraction of the solute, n, and an empirical tabulated constant, KH, expressed in units of pressure:

Generally, the value of KH increases with increasing temperature. Thus, at the same pressure the mole fraction of a solute decreases with increasing temperature.

When the atmospheric pressure decreases, the partial pressure of a gas decreases, and the molar solubility of a gas decreases. For example, high in the mountains the atmospheric pressure is low; as a result the solubility of oxygen in human blood decreases, which can cause respiration problems. At the opposite end, the higher the pressure, the higher the solubility of gases. You might say the gas is "squeezed" into the solution.

Example 8.11 Calculate the molar solubility of nitrogen dissolved in 1 l of water at 25 °C and atmospheric pressure of 1 atm. Henry's Law constant for nitrogen at 25 °C is 8.68 x 109 Pa. The percentage by volume of N2 in dry air is 78.1. Answer: The partial pressure of N2 at 1 atm is pN2 = 0.781 x 1 atm = 7.91 x

104 Pa. From Henry's Law nN2 = Pn2/Kh = (7.91 x 104Pa)/(8.68 x 109Pa) = 9.1 x 10-6. The mole fraction of nitrogen nN2 = vN2/(vN2 + vH2O) ~ vN2/vH2O since vn2 ^ vH2O. The number of moles of H2O in 1 l is (1000/18) mol. Then, vn2 = (9.1 x 10-6 x 1000/18) mol = 5.05 x 10-4mol. The molar solubility of nitrogen is 5.05 x 10-4 mol l-1. □

Example 8.12 Calculate the molar solubility of CO2 in moles per liter dissolved in water at 25 °C and CO2 pressure of 2.4 atm (pressure used to carbonate soda). Henry's Law constant for CO2 at 25 °C is 1.67 x 108 Pa. Answer: 2.4 atm = 2.43 x 105 Pa. The mole fraction of CO2 according to Henry's Law is ncO2 = PcO2/Kh = (2.43 x 105Pa)/(1.67 x 108Pa). Since there is (1000/18) mol of H2O in 1l, the molar solubility of CO2 is (2.43 x 105Pa x 1000/18 mol l-1)/(1.67 x 108Pa) = 8.1 x 10-2 moll-1. When one opens a bottle of soda, the pressure decreases; as a result the solubility of CO2 decreases, and the

es vapor

e' vapor

pure solvent

solution

Figure 8.7 Notation. (a) Pure solvent in equilibrium with its vapor at pressure es. (b) Solution in equilibrium with the solvent's vapor at pressure e'.

Figure 8.7 Notation. (a) Pure solvent in equilibrium with its vapor at pressure es. (b) Solution in equilibrium with the solvent's vapor at pressure e'.

bubbles of CO2 emerge from the solution. At higher temperature the solubility of CO2 decreases since Henry's constant increases (soda from a fridge sparkles less than soda held at room temperature). □

8.6.1 Molar Gibbs energy of an ideal solution

In this section we will find out how the Gibbs energy of a pure solvent changes when a small amount of solute is added. We will consider liquids that are at equilibrium with their vapors. This means that the Gibbs energy of the vapor is equal to the Gibbs energy of the liquid. Let us denote the equilibrium vapor pressure over a pure solvent as es, and the equilibrium vapor pressure over the solution as e1 (see Figure 8.7). From (8.69) the molar Gibbs energy (the Gibbs energy per mole) of vapor at pressure es is

where es is in atm, G is the molar Gibbs energy6 at pressure p and temperature T and Gt is the molar Gibbs energy at 1 atm and temperature T. Since at equilibrium the molar Gibbs energy of a liquid is equal to that of the vapor, Gvapor = Gliquid, the Gibbs energy of a pure solvent, denoted as Gw, is

When a solute is added, the molar Gibbs energy of the solvent, G', which is at equilibrium with its vapor at pressure e', is

6 The molar Gibbs energy is often called the chemical potential.

Subtracting (8.80) from (8.81), we obtain the difference between the molar Gibbs energy of the solvent in the solution and the pure solvent:

From Raoult's Law, e'/es = nw, where nw is the mole fraction of solvent (e.g. water) in the solution. Then,

We will consider a dilute solution, when the mole fraction of the solute, n, is much smaller than that of the solvent, n ^ nw. Since n + nw = 1, we rewrite (8.83) as

Taking into account that for n ^ 1, the logarithmic function can be written as ln(1 - n) ~ -n, we obtain

We will use (8.85) in the next section to find the temperature at which the solution boils and freezes.

8.6.2 The elevation of the boiling point and the lowering of the freezing point of a solution

When some solute is dissolved in a pure solvent, the boiling and freezing points of the solution are not the same as for the pure solvent. We will show that the change is proportional to the amount of solute. We will consider nonvolatile solutes (for example, a salt). In this case the vapor of the solute is a pure gas.

Example 8.13 Show that the addition of a dissolved solute in a solution will elevate the boiling point compared to the boiling point of pure solvent. Answer: The boiling point is the temperature at which the saturated vapor pressure of a liquid is the same as the atmospheric pressure. We consider two cases.

Case 1: equilibrium between a pure solvent and its vapor (see Figure 8.8a). The pure solvent boils at temperature T0. The vapor pressure is 1 atm. At equilibrium, the molar Gibbs energies of the pure solvent, denoted below as Gw, and its vapor, denoted as Gv, are equal to each other:

vapor pure solvent vapor solution vapor solution

Figure 8.8 Notation. (a) Molar Gibbs energy for the vapor over the pure solvent, Gv (T0, p), and molar Gibbs energy for the pure solvent, Gw (T0, p). (b) Molar Gibbs energy of solvent vapor over the solution, Gv (T,p), and molar Gibbs energy of the solvent, G (T, p).

Case 2: there is an amount n of nonvolatile solute in the solvent. The solution is at equilibrium with the solvent's vapor (see Figure 8.8b). The vapor pressure at the boiling point is 1 atm, the same as in case 1, but the boiling point differs. Assume that the solution boils at a temperature T = T0 + AT where AT ^ T0. From (8.85) the molar Gibbs energy of the solvent in the solution is G' = Gw — R*Tn. At equilibrium, the molar Gibbs energy of the solvent in the solution, G', is equal to the molar Gibbs energy of the vapor, Gv:

Since AT ^ T0 we can expand both sides of (8.88) in Taylor's series retaining only the linear term:

Substituting these expansions in (8.88) and taking into account (8.86), we obtain

Here we also replaced Tn by T0n since n has a small value proportional to AT.

As shown in Section 4.8, the temperature derivative of the Gibbs energy at a constant pressure is minus the entropy. So, and d Gw d T

where sw is the entropy of 1 mol of the solvent and sv is the entropy of 1 mol of the solvent's vapor. Then, from (8.91) we obtain

R*Ton

Multiplying and dividing the right-hand side of the last equation by T0 and taking into account that (sv — sw)T0 is the amount of heat required to evaporate 1 mol of the solvent at the boiling point, in other words (sv — sw)T0 = AvapHwe obtain the final formula for the elevation of the boiling point AT:

Since AT is positive, the boiling point of the solution is higher than that of the pure solvent. The change in the boiling point is proportional to the amount of solute n. □

Example 8.14 What is the change in the boiling temperature of 11 of water with 15 g of NaCl dissolved in it?

Answer: The molecular weight of NaCl is 58.44 g mol-1. Substituting in (8.96) n —

(15/58.44)/(1000/18),R* — 8.31 J K-1 mol-1, To — 373 K, Avap kJ mol-1, we obtain AT — o.13 K.

Example 8.15 Calculate the change in the freezing point of the solution. Answer: We assume that only pure solvent is frozen, while the solute remains in the solution. Then the calculation of the freezing point of the solution is similar to the calculation of the boiling point except we have fusion instead of vaporization. AfusH is the heat released when 1 mol of the solvent is frozen at a temperature To;

this heat is negative. Substituting AfasH° in (8.96), we find that the freezing point of a solution is lower than that of the pure solvent. The difference AT is:

R*T02n

AfusH |

The decrease in the freezing point is proportional to the amount of solute. □

Example 8.16 What is the change in the freezing temperature of 11 of water with 15 g of NaCl dissolved in it?

Answer: |AfusH°| for water is 6.008 kJ mol-1, T0 = 273 K, n = (15/58.44)/ (1000/18), R* = 8.31 J K-1 mol-1. According to (8.97) the change in freezing temperature AT = -0.48 K. □