Solutions

A solution is a homogeneous mixture of two or more chemically non-reacting substances. The component present in a larger amount is called the solvent, and the component present in a smaller amount is the solute. Depending on the physical state of the solute and solvent, solutions can be classified into nine types: gas-in-gas (air), gas-in-liquid (CO₂ in water, i.e. soda water), gas-in-solid (H₂ absorbed in palladium), liquid-in-gas (water vapour in air, i.e. humidity), liquid-in-liquid (ethanol in water), liquid-in-solid (mercury in gold, an amalgam), solid-in-gas (camphor vapour in air), solid-in-liquid (sugar or salt in water), and solid-in-solid (alloys such as brass, a solution of zinc in copper).

In this chapter we focus mainly on solid-in-liquid and liquid-in-liquid binary solutions (two components). Because the ratio of solute to solvent can vary continuously, chemists use precise quantitative terms to describe composition rather than vague words like "concentrated" or "dilute".

The commonly used concentration terms are:

• Mass percentage (w/w %) — grams of solute per 100 g of solution.
• Volume percentage (v/v %) — used when both components are liquids; mL of solute per 100 mL of solution.
• Mass by volume percentage (w/v %) — grams of solute per 100 mL of solution; common in medicine and pharmacy.
• Parts per million (ppm) — used for very dilute solutions such as trace pollutants in water or air.
• Mole fraction (x) — ratio of moles of one component to the total moles of all components; mole fractions of all components of a solution always add up to 1.
• Molarity (M) — moles of solute per litre of solution. Molarity changes slightly with temperature because volume expands or contracts on heating/cooling.
• Molality (m) — moles of solute per kilogram of solvent. Because it is based on mass, not volume, molality is independent of temperature — this is why it is preferred for precise colligative-property work.

Worked example: 10 g of glucose (C₆H₁₂O₆, molar mass 180 g mol⁻¹) is dissolved in 200 g of water. Find the molality and the mole fraction of glucose.

1. Moles of glucose = 10 g ÷ 180 g mol⁻¹ = 0.0556 mol.
2. Mass of solvent (water) = 200 g = 0.200 kg.
3. Molality = 0.0556 mol ÷ 0.200 kg = 0.278 mol kg⁻¹.
4. Moles of water = 200 g ÷ 18 g mol⁻¹ = 11.11 mol.
5. Mole fraction of glucose = 0.0556 ÷ (0.0556 + 11.11) = 0.0050.

The solubility of a substance is the maximum amount of it that can dissolve in a specified amount of solvent at a given temperature, producing a saturated solution in dynamic equilibrium with any undissolved solute. For a solid dissolved in a liquid, the general rule is "like dissolves like": polar solutes (e.g. sugar, salts) dissolve readily in polar solvents (e.g. water), while non-polar solutes (e.g. naphthalene) dissolve better in non-polar solvents (e.g. benzene). Dissolution of most solids in liquids is accompanied by a small enthalpy and entropy change; solubility of most solids increases with temperature if the dissolution process is endothermic (as it is for most solid solutes), though there are exceptions.

For a gas dissolved in a liquid, solubility depends strongly on both temperature and pressure. Increasing temperature generally decreases gas solubility (this is why dissolved oxygen levels fall in warm water, harming aquatic life), while increasing pressure increases gas solubility. This pressure dependence is quantified by Henry's law, which states that the partial pressure of a gas in the vapour phase (p) is directly proportional to the mole fraction of the gas dissolved in the solution (x):

p = K_H x, where K_H is the Henry's law constant, specific to a gas-solvent pair at a given temperature. A higher K_H at a given pressure means lower solubility, since x = p/K_H becomes smaller as K_H increases.

Important applications of Henry's law include: (i) carbonated (soda) drinks are bottled under high CO₂ pressure to increase its solubility, and fizzing occurs when the bottle is opened and pressure drops; (ii) scuba divers breathing compressed air can develop excess dissolved N₂ in blood; if they ascend too quickly, this nitrogen escapes as bubbles causing decompression sickness ("the bends") — divers are advised to ascend slowly or use oxygen-helium mixtures; (iii) at high altitudes, the low partial pressure of O₂ lowers dissolved oxygen in blood, causing altitude/anoxia sickness in climbers and aviators.

Worked example: The Henry's law constant for N₂ gas dissolved in water at 293 K is 76,480 bar. Calculate the mole fraction of N₂ in water when the partial pressure of N₂ over the solution is 0.987 bar.

1. By Henry's law, x(N₂) = p / K_H.
2. x(N₂) = 0.987 bar ÷ 76,480 bar = 1.29 × 10⁻⁵.

For a solution of two volatile liquids (A and B), French chemist Raoult's law states that the partial vapour pressure of each component in the vapour phase is directly proportional to its mole fraction in the liquid solution: p_A = p°_A x_A and p_B = p°_B x_B, where p°_A and p°_B are the vapour pressures of pure A and pure B at that temperature. By Dalton's law of partial pressures, the total vapour pressure of the solution is p_total = p_A + p_B = p°_A x_A + p°_B x_B. A plot of p_total versus mole fraction is linear for an ideal solution, lying between the two pure-component vapour pressures.

Note that the composition of the vapour above the solution is generally different from the composition of the liquid, because the more volatile component (higher p°) contributes proportionally more to the vapour. The mole fraction of a component in the vapour phase, y_A, is found from y_A = p_A/p_total (Dalton's law applied to the vapour).

For a solution of a non-volatile solute in a volatile solvent (e.g. sugar in water), only the solvent contributes to the vapour pressure. Raoult's law becomes a special/limiting case: p₁ = p°₁ x₁, and since x₁ is always less than 1, the vapour pressure of the solution p₁ is always lower than that of the pure solvent p°₁. This lowering of vapour pressure is the origin of all colligative properties. Raoult's law can also be seen as a special case of Henry's law where the Henry's law constant K_H becomes equal to p°₁, the vapour pressure of the pure liquid.

Worked example: Liquids A and B form an ideal solution. At a given temperature, p°_A = 450 mm Hg and p°_B = 700 mm Hg. If the mole fraction of A in the liquid is 0.25 (so x_B = 0.75), find the total vapour pressure and the composition of the vapour.

1. p_A = p°_A x_A = 450 × 0.25 = 112.5 mm Hg.
2. p_B = p°_B x_B = 700 × 0.75 = 525 mm Hg.
3. p_total = 112.5 + 525 = 637.5 mm Hg.
4. Vapour mole fraction of A: y_A = p_A/p_total = 112.5/637.5 = 0.1765.
5. y_B = 1 − 0.1765 = 0.8235 (vapour is richer in the more volatile component B, as expected).

An ideal solution is one that obeys Raoult's law at every concentration and temperature. In such a solution, solute-solute, solvent-solvent, and solute-solvent intermolecular interactions are essentially of the same strength, so molecules experience nearly the same environment before and after mixing. Consequently, ideal solutions form with no enthalpy change on mixing (ΔmixH = 0) and no volume change on mixing (ΔmixV = 0). Close approximations to ideal behaviour include n-hexane and n-heptane, or benzene and toluene, whose molecules are structurally very similar.

Most real solutions show deviations from Raoult's law, called non-ideal solutions, classified as follows:

• Positive deviation: occurs when A–B interactions are weaker than A–A and B–B interactions. Molecules escape into the vapour phase more easily than Raoult's law predicts, so the observed vapour pressure is higher than the calculated ideal value, and ΔmixH is positive (mixing is endothermic) with ΔmixV positive (slight expansion). Examples: ethanol and acetone; acetone and carbon disulphide.
• Negative deviation: occurs when A–B interactions are stronger than A–A and B–B interactions (e.g. through hydrogen bonding), so molecules are held back from escaping into the vapour, giving lower-than-ideal vapour pressure. ΔmixH is negative (mixing is exothermic) and ΔmixV is negative (slight contraction). Examples: chloroform and acetone; phenol and aniline.

When deviations are large enough, the vapour pressure–composition curve develops a maximum or minimum, and the solution at that composition boils at a constant temperature without change in composition — such mixtures are called azeotropes. A solution showing large positive deviation forms a minimum boiling azeotrope (e.g. 95% ethanol + 5% water by mass, boiling at 351.1 K, which is why ethanol cannot be purified above this concentration by simple distillation). A solution showing large negative deviation forms a maximum boiling azeotrope (e.g. a mixture of about 68% nitric acid and 32% water boiling at 393.5 K).

Colligative properties are properties of a dilute solution containing a non-volatile solute that depend only on the number of solute particles relative to the total number of particles present, and not on their chemical nature. The four classical colligative properties are: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure (covered in the next section). Because they depend purely on particle count, colligative properties provide a convenient experimental route to determine the molar mass of an unknown non-volatile, non-electrolyte solute.

1. Relative lowering of vapour pressure: From Raoult's law, p₁ = p₁° x₁ = p₁°(1 − x₂), which rearranges to (p₁° − p₁)/p₁° = x₂. For a dilute solution, x₂ = n₂/(n₁ + n₂) ≈ n₂/n₁, so the relative lowering of vapour pressure equals the mole fraction of solute, and can be used to find its molar mass M₂ if the mass of solute (w₂) and solvent (w₁) and molar mass of solvent (M₁) are known.

2. Elevation of boiling point (ΔTb): Since a non-volatile solute lowers the solvent's vapour pressure, a higher temperature is needed for the vapour pressure to reach atmospheric pressure and boil — hence the boiling point rises. The elevation is directly proportional to the molal concentration: ΔTb = Kb m, where Kb is the molal elevation constant (ebullioscopic constant) of the solvent, a solvent-specific property with unit K kg mol⁻¹.

3. Depression of freezing point (ΔTf): Similarly, a solution freezes at a lower temperature than the pure solvent because the solute lowers the vapour pressure of the liquid phase, shifting the solid-liquid equilibrium to a lower temperature. ΔTf = Kf m, where Kf is the molal depression constant (cryoscopic constant) of the solvent. Freezing-point depression is widely used in practice, e.g. ethylene glycol added to car radiators as antifreeze, and salt spread on icy roads to lower the freezing point of water.

Worked example (boiling point elevation and molar mass): 1.5 g of a non-volatile, non-electrolyte solute is dissolved in 50 g of a solvent with Kb = 2.53 K kg mol⁻¹, producing a boiling point elevation of 0.6325 K. Find the molar mass of the solute.

1. ΔTb = Kb × m, so molality m = ΔTb / Kb = 0.6325 / 2.53 = 0.250 mol kg⁻¹.
2. Moles of solute in 50 g (0.050 kg) of solvent = m × mass of solvent (kg) = 0.250 × 0.050 = 0.0125 mol.
3. Molar mass = mass of solute / moles = 1.5 g / 0.0125 mol = 120 g mol⁻¹.

This calculation is usually written directly using the formula M₂ = (1000 × Kb × w₂)/(ΔTb × w₁), where w₁ and w₂ are the masses (in grams) of solvent and solute respectively; an analogous formula, M₂ = (1000 × Kf × w₂)/(ΔTf × w₁), applies for freezing-point depression data.

Osmosis is the spontaneous flow of solvent molecules through a semipermeable membrane (which allows only solvent, not solute, to pass) from a region of pure solvent (or lower solute concentration) into a solution (or higher solute concentration), until equilibrium is reached. The minimum external pressure that must be applied to the solution side to just stop this net flow of solvent is called the osmotic pressure (π) of the solution. Osmotic pressure is a colligative property because, like the others, it depends only on the number of solute particles per unit volume, not their identity. For dilute solutions, osmotic pressure obeys a van't Hoff equation directly analogous to the ideal gas equation: πV = nRT, or π = CRT, where C is the molar concentration of the solution.

Osmotic pressure measurements are especially useful for biological macromolecules (e.g. proteins) because even a small molar concentration produces an easily measurable π, unlike the very small ΔTb or ΔTf these large, dilute solutes would produce. Two solutions with the same osmotic pressure at a given temperature are called isotonic solutions — this concept is critical in medicine (e.g. IV fluids and the fluid inside blood cells must be isotonic to avoid cells swelling/bursting or shrinking). If external pressure greater than the osmotic pressure is applied to the solution side, solvent flows in the reverse direction, from solution to pure solvent — this is reverse osmosis, the principle used in desalination of sea water using a semipermeable membrane, typically cellulose acetate.

Colligative property formulas assume the solute neither dissociates nor associates in solution. Many real solutes deviate from this — electrolytes like NaCl dissociate into ions (increasing the effective particle count), while some solutes like carboxylic acids can associate into dimers in non-polar solvents (decreasing the effective particle count). This causes an experimentally measured colligative property (and hence molar mass) to differ from the value calculated assuming no dissociation/association — called an abnormal molar mass. Jacobus van't Hoff introduced a correction factor, the van't Hoff factor (i), defined as: i = (normal, i.e. calculated molar mass) / (observed, i.e. abnormal molar mass) = (observed colligative property) / (calculated colligative property assuming no dissociation/association). All colligative property equations are modified by including i: ΔTb = i Kb m, ΔTf = i Kf m, π = i C R T, and (p₁°−p₁)/p₁° = i x₂. For a solute dissociating into n ions with degree of dissociation α, i = 1 + (n−1)α (so i is greater than 1); for a solute associating with n molecules combining and degree of association α, i = 1 + (1/n − 1)α (so i is less than 1).

Worked example: Calculate the osmotic pressure at 300 K of (a) a 0.10 M glucose solution and (b) a 0.10 M NaCl solution, assuming NaCl dissociates completely (i = 2). (R = 0.0821 L atm K⁻¹ mol⁻¹)

1. Glucose (non-electrolyte, i = 1): π = CRT = 0.10 × 0.0821 × 300 = 2.463 atm.
2. NaCl (i = 2): π = i × C R T = 2 × 0.10 × 0.0821 × 300 = 4.926 atm — exactly double the glucose value, because NaCl produces twice as many particles (Na⁺ and Cl⁻) per formula unit.