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# Henry's law

William Henry investigated the relationship between pressure and solubility of a gaseous solute in a particular solvent.

Henry's law

William Henry investigated the relationship between pressure and solubility of a gaseous solute in a particular solvent. According to him, ŌĆ£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 concentrationsŌĆØ. This statement is known as HenryŌĆÖs law.

HenryŌĆÖs law can be expressed as,

psolute ╬▒ xsolute in solution (9.1)

psolute = KHxsolute in solution (9.2)

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 ŌĆśKHŌĆÖ 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 HenryŌĆÖs law

i. HenryŌĆÖs law is applicable at moderate temperature and pressure only.

ii. Only the less soluble gases obeys HenryŌĆÖs law

iii. The gases reacting with the solvent do not obey HenryŌĆÖs law. For example, ammonia or HCl reacts with water and hence does not obey this law.

NH3+ H2O Ōćå NH4+ + OHŌĆō

iv. The gases obeying HenryŌĆÖs 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 = KH x1------------(1)

At pressure 6.0 atm,

p2 = KH x2------------(2)

Dividing equation (1) by (2)

From equation     p1/p2 = x1/x2

1.5/6.0 = 0.24/x2

Therefore x2 = 0.24 x 6.0/1.5 = 0.96 g/L

### Why the carbonated drinks are stored in a pressurized container?

We all know that the carbonated beverages contain carbon dioxide dissolved in them. To dissolve the carbon dioxide in these drinks, the CO2 gas is bubbled through them under high pressure. These containers are sealed to maintain the pressure. When we open these containers at atmospheric pressure, the pressure of the CO2 drops to the atmospheric level and hence bubbles of CO2 rapidly escape from the solution and show effervescence. The burst of bubbles is even more noticeable, if the soda bottle is in warm condition.

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