Henry’s Law – Statement, Formula, Constant, Solved Examples
William Henry investigated the relationship between pressure and solubility of a gaseous solute in a particular solvent. According to him, “the partial pressure of the gas in vapour phase (vapour pressure of the solute) is directly proportional to the mole fraction (x) of the gaseous solute in the solution at low concentrations”. This statement is known as Henry’s law.
Henry’s law can be expressed as,
Psolute α xsolute in solution
Psolute = KHxsolute in solution
Here, psolute represents the partial pressure of the gas in vapour state which is commonly called as vapour pressure. xsolute in solution represents the mole fraction of solute in the solution. KH is a empirical constant with the dimensions of pressure.
The value of ‘KH’ depends on the nature of the gaseous solute and solvent. The above equation is a straight line in the form of y=mx. The plot partial pressure of the gas against its mole fraction in a solution will give a straight line as shown in fig 9.3. The slope of the straight line gives the value of KH.
Limitations of Henry’s law
- Henry’s law is applicable at moderate temperature and pressure only.
- Only the less soluble gases obeys Henry’s law
- The gases reacting with the solvent do not obey Henry’s law. For example, ammonia or HCl reacts with water and hence does not obey this law.
NH3 + H2O ⇄ NH+4 + OH–
The gases obeying Henry’s law should not associate or dissociate while dissolving in the solvent.
Example Problem 2:
0.24 g of a gas dissolves in 1 L of water at 1.5 atm pressure. Calculate the amount of dissolved gas when the pressure is raised to 6.0 atm at constant temperature.
psolute = KHxsolute in solution
At pressure 1.5 atm,
P1 = KHx1 …………… (1)
At pressure 6.0 atm,
p2 = KH x2 ……………. (2)
Dividing equation (1) by (2)
p1/p2 = x1/x2
1.5/6.0 = 0.24/x2
Therefore x2 = 0.24 × 6.0/1.5 = 0.96 g/L
Find free online Chemistry Topics covering a broad range of concepts from research institutes around the world.