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Concept 01 / 07Solutions

Solubility of Solids and Gases in Liquids and Henry's Law

Types of solutions, factors affecting solubility of solids (nature, temperature, Le Chatelier) and of gases (pressure and temperature), Henry's law with its constant and units, applications to carbonated drinks, scuba diving and high altitudes, and limitations.

Figure 1
A solution is a homogeneous mixture of two or more components. The component present in the largest amount is usually the solvent; the others are solutes. Binary solutions can be gaseous (air), liquid (gas in liquid: oxygen in water; liquid in liquid: ethanol in water; solid in liquid: glucose in water) or solid (hydrogen in palladium, amalgams, alloys such as brass). Solubility of a solid in a liquid. "Like dissolves like": polar and ionic solutes dissolve in polar solvents, non-polar solutes in non-polar solvents. A saturated solution is in dynamic equilibrium with undissolved solute. Temperature affects solubility according to Le Chatelier: if dissolution is endothermic (ΔsolH>0\Delta_{sol}H > 0), solubility rises with temperature (most salts, KNO3\mathrm{KNO_{3}}); if exothermic, it falls (such as Ce2(SO4)3\mathrm{Ce_{2}(SO_{4})_{3}}, Li2CO3\mathrm{Li_{2}CO_{3}}). Pressure has little effect, since solids and liquids are nearly incompressible. Solubility of a gas in a liquid. Increases greatly with pressure and decreases with temperature (dissolution of gases is exothermic, and gas molecules escape more easily when hot, which is why aquatic life prefers cold water and why bubbles form in water before it boils). Henry's law. At constant temperature, the solubility of a gas in a liquid is directly proportional to its partial pressure above the liquid. Stated with mole fraction xx of the gas in solution: p=KHxp = K_{H}x (see figure). KHK_{H} (Henry's constant, unit of pressure, often kbar) depends on the gas, solvent and temperature. A larger KHK_{H} means lower solubility; KHK_{H} increases with temperature. At about 293293 to 298 K298\,\mathrm{K} in water: H2\mathrm{H_{2}} 69.2 kbar69.2\,\mathrm{kbar}, N2\mathrm{N_{2}} 76.576.5, O2\mathrm{O_{2}} 34.934.9, CO2\mathrm{CO_{2}} 1.671.67, CH4\mathrm{CH_{4}} 0.41 kbar0.41\,\mathrm{kbar}. Applications. Soft drinks and soda water are bottled under high CO2\mathrm{CO_{2}} pressure; opening the bottle lowers the pressure and the gas bubbles out. Scuba divers breathing compressed air at depth dissolve more N2\mathrm{N_{2}} in blood; on rapid ascent it forms bubbles that block capillaries (the bends). Tanks are filled with air diluted by helium (about 11.7%11.7\% He, 56.2%56.2\% N2\mathrm{N_{2}}, 32.1%32.1\% O2\mathrm{O_{2}}), since helium is much less soluble. At high altitudes the low partial pressure of oxygen reduces its concentration in blood, causing anoxia (weakness, inability to think clearly). Limitations. Henry's law holds for dilute solutions, low pressures, moderate temperatures, and gases that do not react with or dissociate in the solvent (NH3\mathrm{NH_{3}} and HCl in water deviate strongly). Key Takeaways: - Solid solubility follows Le Chatelier with ΔsolH\Delta_{sol}H; pressure has little effect. - Gas solubility increases with pressure and decreases with temperature. - p=KHxp = K_{H}x; higher KHK_{H} means lower solubility. - Applications: carbonated drinks, bends, altitude anoxia. - Pitfall: in p=KHxp = K_{H}x, a large KHK_{H} means a gas is less soluble, not more. Brain-Teaser 1: N2\mathrm{N_{2}} at a partial pressure of 0.987 bar0.987\,\mathrm{bar} is bubbled through water at 293 K293\,\mathrm{K} (KH=76.48 kbarK_{H} = 76.48\,\mathrm{kbar}). How many millimoles of N2\mathrm{N_{2}} dissolve in 1 L1\,\mathrm{L} of water? Solution: x=0.98776480=1.29×10−5x = \frac{0.987}{76480} = 1.29 \times 10^{-5}. 1 L1\,\mathrm{L} of water is 55.5 mol55.5\,\mathrm{mol}, and x≈n55.5x \approx \frac{n}{55.5}, so n=1.29×10−5×55.5=7.16×10−4 mol=0.716 mmoln = 1.29 \times 10^{-5} \times 55.5 = 7.16 \times 10^{-4}\,\mathrm{mol} = 0.716\,\mathrm{mmol}. Brain-Teaser 2: The solubility of H2S\mathrm{H_{2}S} in water at STP (0.987 bar0.987\,\mathrm{bar}) is 0.195 m0.195\,\mathrm{m}. Calculate Henry's constant. Solution: In 1 kg1\,\mathrm{kg} of water (55.5 mol55.5\,\mathrm{mol}): x=0.1950.195+55.5=3.5×10−3x = \frac{0.195}{0.195 + 55.5} = 3.5 \times 10^{-3}. KH=px=0.9873.5×10−3=282 barK_{H} = \frac{p}{x} = \frac{0.987}{3.5 \times 10^{-3}} = 282\,\mathrm{bar}. Its KHK_{H} is far smaller than that of N2\mathrm{N_{2}} (76,480 bar76{,}480\,\mathrm{bar}), so H2S\mathrm{H_{2}S} is much more soluble. Brain-Teaser 3: Why do fish struggle in warm, still water on a summer afternoon? Solution: Dissolving oxygen is exothermic, so its solubility falls as temperature rises (KHK_{H} increases). Warm water holds less dissolved oxygen, while the fish's metabolic need rises, so they may suffocate.