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Solutions - Solubility of Gases in Liquids (Henry's Law)

Grade 12CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

The solubility of a gas in a liquid is significantly affected by pressure and temperature. The process of dissolution is generally exothermic, meaning solubility decreases as temperature increases according to Le Chatelier's Principle.

Henry's Law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of the liquid or solution.

The most common form of Henry's Law states: The partial pressure of the gas in the vapour phase (pp) is proportional to the mole fraction of the gas (xx) in the solution: p=KHxp = K_H \cdot x.

Henry's Law Constant (KHK_H) is a function of the nature of the gas. At a given pressure, the higher the value of KHK_H, the lower is the solubility of the gas in the liquid.

KHK_H values increase with an increase in temperature, which explains why aquatic species are more comfortable in cold water than in warm water (due to higher dissolved O2O_2).

Applications of Henry's Law include: 1. Increasing solubility of CO2CO_2 in soft drinks by sealing the bottle under high pressure. 2. Using Helium to dilute air in scuba tanks to avoid 'bends' caused by N2N_2 bubbles. 3. Explaining Anoxia at high altitudes where low partial pressure of O2O_2 leads to low oxygen concentrations in the blood.

📐Formulae

p=KHxp = K_H \cdot x

xgas=ngasngas+nsolventx_{gas} = \frac{n_{gas}}{n_{gas} + n_{solvent}}

Solubilityp\text{Solubility} \propto p

💡Examples

Problem 1:

If N2N_2 gas is bubbled through water at 293 K293\text{ K}, how many millimoles of N2N_2 gas would dissolve in 1 litre1\text{ litre} of water? Assume that N2N_2 exerts a partial pressure of 0.987 bar0.987\text{ bar}. Given that Henry's law constant for N2N_2 at 293 K293\text{ K} is 76.48 kbar76.48\text{ kbar}.

Solution:

  1. Convert KHK_H to the same unit as pressure: KH=76.48 kbar=76,480 barK_H = 76.48\text{ kbar} = 76,480\text{ bar}.
  2. Use Henry's Law: x=pKH=0.98776480=1.29×105x = \frac{p}{K_H} = \frac{0.987}{76480} = 1.29 \times 10^{-5}.
  3. In 1 L1\text{ L} of water, nH2O=1000 g18 g/mol=55.5 moln_{H_2O} = \frac{1000\text{ g}}{18\text{ g/mol}} = 55.5\text{ mol}.
  4. Since nN2nH2On_{N_2} \ll n_{H_2O}, mole fraction xnN2nH2Ox \approx \frac{n_{N_2}}{n_{H_2O}}.
  5. nN2=xnH2O=1.29×105×55.5=7.16×104 moln_{N_2} = x \cdot n_{H_2O} = 1.29 \times 10^{-5} \times 55.5 = 7.16 \times 10^{-4}\text{ mol}.
  6. Convert to millimoles: 7.16×104×1000=0.716 mmol7.16 \times 10^{-4} \times 1000 = 0.716\text{ mmol}.

Explanation:

The problem applies p=KHxp = K_H \cdot x to find the mole fraction of the gas, then relates that mole fraction to the total number of moles of the solvent (water) to find the amount of dissolved gas.

Solubility of Gases in Liquids (Henry's Law) Revision - Class 12 Chemistry CBSE