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Solutions - Solubility of gases in liquids

Grade 12ICSEChemistry

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

🔑Concepts

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The solubility of a gas in a liquid is the maximum amount of gas that can be dissolved in a specific volume of liquid at a given temperature and pressure.

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Nature of Gas and Solvent: Polar gases like HClHCl and NH3NH_3 are highly soluble in polar solvents like H2OH_2O, while non-polar gases like O2O_2 and N2N_2 show low solubility in water.

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Effect of Temperature: The dissolution of a gas in a liquid is typically an exothermic process (ΔHsoln<0\Delta H_{soln} < 0). According to Le Chatelier's Principle, the solubility of gases in liquids decreases with an increase in temperature.

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Effect of Pressure (Henry's Law): 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.

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Henry's Law Constant (KHK_H): The value of KHK_H depends on 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.

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Biological Application (Anoxia): At high altitudes, the partial pressure of O2O_2 is less than that at ground level. This leads to low concentrations of oxygen in the blood and tissues of climbers, causing a condition known as Anoxia.

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Scuba Diving: To avoid 'bends' (painful formation of N2N_2 bubbles in the blood), scuba divers use tanks filled with air diluted with Helium (11.7% He,56.2% N2, and 32.1% O211.7\% \text{ } He, 56.2\% \text{ } N_2, \text{ and } 32.1\% \text{ } O_2).

📐Formulae

p=KH⋅χp = K_H \cdot \chi

χgas=ngasngas+nsolvent\chi_{gas} = \frac{n_{gas}}{n_{gas} + n_{solvent}}

m=k⋅pm = k \cdot p

💡Examples

Problem 1:

Calculate the solubility of N2N_2 gas in water at 293 K293\text{ K} if the partial pressure of N2N_2 is 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 units as pressure: KH=76.48×103 bar=76480 barK_H = 76.48 \times 10^3\text{ bar} = 76480\text{ bar}.
  2. Use Henry's Law: χN2=pN2KH=0.98776480≈1.29×10−5\chi_{N_2} = \frac{p_{N_2}}{K_H} = \frac{0.987}{76480} \approx 1.29 \times 10^{-5}.
  3. Since 1 liter of water contains 55.5 moles55.5\text{ moles} (1000/181000/18), and nN2n_{N_2} is very small: χN2≈nN2nH2O\chi_{N_2} \approx \frac{n_{N_2}}{n_{H_2O}}.
  4. nN2=1.29×10−5×55.5=7.16×10−4 molesn_{N_2} = 1.29 \times 10^{-5} \times 55.5 = 7.16 \times 10^{-4}\text{ moles}.
  5. In millimoles: 0.716 mmol0.716\text{ mmol}.

Explanation:

The mole fraction χ\chi is calculated first using the ratio of partial pressure to the Henry's constant. The total number of moles is then used to find the specific amount of gas dissolved in a unit volume of solvent.

Problem 2:

Why do aquatic species feel more comfortable in cold water than in warm water?

Solution:

Solubility of gases like O2O_2 in water decreases with an increase in temperature because the dissolution process is exothermic (ΔH<0\Delta H < 0).

Explanation:

In cold water, the solubility of Oxygen (O2O_2) is higher compared to warm water. Therefore, more dissolved oxygen is available for the aquatic animals to breathe, making them more comfortable.

Solubility of gases in liquids - Revision Notes & Key Formulas | ICSE Class 12 Chemistry