Most of the gases dissolve in water or some other liquids to a greater or less extent. In a gas the molecules arc far apart. After dissolution in a liquid solvent. the molecules of the gas are much closer. it is JUSt like saying that before a gas dissolves in liquid. it must be condensed to g1ve a liquid. The condensatiOn of a gas IS an cxothem1ic process. l11e enthalpy of condensatiOn is larger than the enthalpy of solution. Thus the dissolution of a gas is an exothermic process (i c. heat is evolved). The solubility of a gas in a liquid is measured in tcm1s of absorption coefficient or Bunsen coefficient. This coefficient has been named after the scientist. Bunsen. who introduced it. It is denoted by ac. it is defined as the volume of a gas at standard temperature and pressure (273. 15K and 1.013 x I()'< Pa) dissolved by unit volume of the solvent at the temperature of the experiment and under a pressure of
1.013 x 105 Pa. The absorption coefficients of some gases are given in Table 2.1.
Table 2.1: Absorption Coefficients at 293 K
Solvent Carbon (IV) Oxide Hydrogen Oxgyen Nitrogen
Water Ethanol Benzene
0.88 3.00
---0.018 0.081 0.060
O.D28 0.142 0.165
0.015 0.130 0.105
3.2.1 Factors affecting Solubility of Gases The solubility of a gas in a liquid depends upon:
• Temperature
• Pressure. and
• Nature of the gas and the solvent.
We will consider each of these factors separately.
1. Effect of temperature
The dissolution of a gas in a liquid is an exothermic process. Hence,_ according to Le Chatelier 's principle. the solubility of a gas in a liquid decreases with rise in temperature. This behaviour can be seen when bubbles of dissolved air escape on heating water below 373 K.
2. Effect of pressure- Henry's law
The solubility of solids in liquids is not so mucll affected by pressure, as the volume changes are not so htgh, But the solubility of a gas inliquid varies considerably with pressure. In1803, the English Chemist.
Henry found that the solubility of a gas increases as the gas pressure is increased at a given temperature.
He proposed the generalization that is known as Henry's law. It may be stated as fullows:
119
At constant temperature, the partial pressure of a gas over a solution is directly proportional to the concentration of the gas in· the solution.
For example. doubling the pressure of oxygen, doubles the amount of oxygen that "ill dissolve in a g1ven amount of the solvent
Mathematically. Henry's law is expressed as follows:
p = Kx . . ., . ,,,3. I
Where pis the partial pressure of the gas over the solution and xis the solubility of the gas in terms of its mole fraction in the solution: K is a constant. characteristic of the specific combination of the solvent and the gas.
It is called Henry's law constant. The partial pressure is expressed in tenns of Pascal (Pa).
From Eq. 3. I. the Henry's Law constant is g•ven as
' '' '''' ''''''' ... 3.2
K has the dimensions of pressure. K is not constant as expected from Eq. 3.2 because of non ideality of the solution. Its value is obtained by plotting the rat1o pix vs x and extrapolating to x = 0. Such a graph is shm\n in Fig2.2.
Pix
0
0
Fig Z.2: Et·a/ua/ion of Henry:, law consla171
Henry's law consta01t for some ga>es are given m Table 2.2:
Table 2.2: Henry's Law Constants at 298 K
Gas K/10' Pa
In water
In benzene
N2
o, o, co co,
7.12 8 68 4.40 5.80
0.37 0.24
-
0.16
017
I
0.01It has been found that Henry's law is followed most closely by dilute solutions of gases that do not react with the solvent. Thus the law is valid for the solubility of hydrogen chloride and ammonia in water. Hydrogen chloride ionizes in water and ammonia enters into chemical combination with water.
H,O + HCI "'"===;;;:o H,Oe + CI
H,O + NH3 ----+ NH.,OH .!====;;;: NH; + OH 3, Nature of the gas and the solvent
Generally, gases. which react chemically with the solvent are more soluble in it than in other solvents. For example, hydrogen chloride gas is more soluble in water than in benzene. Gases, which can be easily liquefied.
are more soluble in common solvents.
Example 2
The Henry's law constant for 0, is 4.40 x 109 Pa. Calculate the molarity of oxygen in water at 298 K. The partial pressure of oxygen over the solution is 1.00 x 10' Pa. Assume that 1.00 dm 3 of the aqueous solution weighs 1.00 kg.
Solution
K = 4.40 X 109 Pa P= I.OOx 10' Pa Mole fraction of oxygen,
of02 is negligible in comparison to the number of moles ofH,O.
then Xo1 = no2 + nlt:!u
Substituting the values in Eq. 3.1, we get
.1; -3
n0,= 1.26xiO.moldm
3 3 3
In other words. the solubility of oxygen in water at 298 K = 1.26 X 10- mol dm- solution conta,ins 1.26 x I
o·'
mol of oxygen.Exercise 2
Why can fish not live in warm water'>
Smce j kg (or I dm ) of the
4.0 Conclusion
In this Unit. we have expressed a solution as a homogenous mixture of two or more substances. the three phases at which solutions are fanned namelv solid. liquid and gaseous arc also discussed. Solutions can be divided mto nme types depending upon the physical state of the solute and the solutiOn The compositiOn of the solution can be measured Pressure has onlv a small effect on the solubility of solids m liquids. The solubility of a gas varies \VIth pressure. All these two phenomena are also d1scussed m th1s Unll
5.0 Summary
We can summarise what we have learnt as follows
• A phase is a homogeneous part of a system which is susceptible to change on certain conditions
• Different solutions ex1st but can be d1fferentiated mto solid solution. liquid solution and gaseous solutiOns
• The solubility of a gas in a liquid depends upon temperature. pressure and nature of the gas and the solvent
6.0 References and Other Resources
Goldberg and Dillard. C. (I '174) C allege Chemistry 3rd Revised Edition New York Macmillan Publishmg Co
IGNOU ( 1993) Solution and Phase Equilismg- Phvsical Chcmistrv 2 CHEM -04 New Delhi.