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Química Geral · Liquids, solids and solutions

Propriedades coligativas

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O que você vai aprender

  • Apply Raoult's law to the vapor pressure of a solution with a nonvolatile solute and to the total pressure and vapor composition above an ideal mixture of two volatile liquids.
  • Calculate freezing-point depression and boiling-point elevation with ΔT = i·K·m in water and in other solvents.
  • Calculate osmotic pressure with Π = iMRT and predict which way water moves across a semipermeable membrane.
  • Distinguish ideal from measured van 't Hoff factors and explain ion pairing, partial ionization and association.
  • Determine a molar mass from freezing-point, boiling-point or osmotic-pressure data.
  • Rank solutions by their colligative effects from the concentration of dissolved particles.

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Colligative properties count particles

A colligative property depends on how many solute particles are dissolved in a given amount of solvent, not on what those particles are. There are four: vapor-pressure lowering, boiling-point elevation, freezing-point depression and osmotic pressure. The particle concentration is i times the formula concentration, where the van 't Hoff factor i is the number of particles one formula unit gives in solution. Glucose (drawn) stays as whole molecules, so i = 1; NaCl separates into Na⁺ and Cl⁻, so ideally i = 2. A 0.10 m glucose solution therefore holds 0.10 mol of particles per kilogram of water and a 0.10 m NaCl solution ideally 0.20 mol. Freezing and boiling points use molality, m = moles of solute ÷ kilograms of solvent, because it does not change with temperature; osmotic pressure uses molarity. The working assumption throughout is an ideal, dilute solution.

Raoult's law and vapor-pressure lowering

In an ideal solution each volatile component has a partial vapor pressure P = χ·P°, its mole fraction times the vapor pressure of the pure liquid. A nonvolatile solute contributes nothing to the vapor, so the vapor pressure of the solution is P = χ(solvent)·P°(solvent), and the lowering is ΔP = P° − P = χ(solute)·P°. Example: water at 25 °C (P° = 23.76 torr) with χ(water) = 0.900 has P = 0.900 × 23.76 = 21.38 torr, a lowering of 2.38 torr. For an electrolyte, count moles of ions, not moles of formula units, when you form the mole fractions. The reason is entropy: the solution is more disordered than the pure solvent, so the solvent has less tendency to escape into the vapor.

Mixtures of two volatile liquids

When both components are volatile, each obeys Raoult's law and the total pressure is the sum (Dalton's law): P(total) = χA·P°A + χB·P°B. Example: χA = 0.600 with P°A = 100. torr and χB = 0.400 with P°B = 40.0 torr give PA = 60.0 torr, PB = 16.0 torr and P(total) = 76.0 torr. The vapor composition is yA = PA/P(total) = 60.0/76.0 = 0.789, richer in A than the liquid (0.600): the vapor is always enriched in the more volatile component, which is why distillation separates liquids. Real mixtures deviate. If A–B attractions are stronger than the average of A–A and B–B, the vapor pressure falls below the Raoult prediction (negative deviation) and mixing releases heat; acetone and chloroform (drawn) are the classic case, because the C–H of chloroform hydrogen-bonds to the carbonyl oxygen. If A–B attractions are weaker, as for ethanol and hexane, the vapor pressure lies above the prediction (positive deviation) and mixing absorbs heat.

Freezing-point depression and boiling-point elevation

A nonvolatile solute raises the boiling point, ΔTb = i·Kb·m, and lowers the freezing point, ΔTf = i·Kf·m. Both ΔT values are written as positive numbers: add ΔTb to the normal boiling point and subtract ΔTf from the normal freezing point. For water Kf = 1.86 °C·kg/mol and Kb = 0.512 °C·kg/mol; for benzene Kf = 5.12 °C·kg/mol, with a normal freezing point of 5.50 °C. Example: 0.200 m KBr (drawn, ideal i = 2) has ΔTf = 2 × 1.86 × 0.200 = 0.744 °C, so it freezes at −0.744 °C, and ΔTb = 2 × 0.512 × 0.200 = 0.205 °C, so it boils at 100.205 °C. Because i and m are the same for both, ΔTb/ΔTf = Kb/Kf for any one solution. The freezing point drops because the solute stays in the liquid while the crystals that form are pure solvent: the solute lowers the escaping tendency of the liquid but not of the solid, so solid and liquid balance only at a lower temperature. As solvent freezes out, the remaining solution becomes more concentrated, so the temperature of a freezing solution keeps falling slowly instead of holding flat.

Osmotic pressure

A semipermeable membrane passes solvent but not solute. Water moves through it from the side with the lower particle concentration to the side with the higher one. The osmotic pressure is the pressure that must be applied to the more concentrated side to stop that flow: Π = i·M·R·T, with M in mol/L, R = 0.08206 L·atm/(mol·K) and T in kelvin. Example: 0.0500 M glucose at 25 °C has Π = 0.0500 × 0.08206 × 298.15 = 1.22 atm. Two solutions with the same particle concentration are isotonic. A red blood cell in a hypotonic solution (fewer particles outside) takes in water, swells and can burst; in a hypertonic solution it loses water and shrinks. Applying more than Π to a salt solution pushes water out through the membrane the other way, which is reverse osmosis. Osmotic pressures are large even for dilute solutions, which makes them the method of choice for measuring the molar masses of proteins and polymers.

The van 't Hoff factor: ideal and real

The ideal i of a strong electrolyte is the number of ions in its formula: 2 for NaCl, 3 for CaCl₂, 4 for K₃PO₄. Measured values are smaller, because oppositely charged ions attract and some travel together as ion pairs that count as one particle. The shortfall grows with ion charge and with concentration: dilute NaCl behaves as if i ≈ 1.9, while MgSO₄ (drawn), with 2+ and 2− ions, falls far below its ideal 2. A measured i comes straight from the data: i = ΔT(measured) ÷ (K·m). A weak acid HA that is ionized to the fraction α gives (1 − α) mol HA, α mol H⁺ and α mol A⁻ per mole, so i = 1 + α. Solutes can also associate: carboxylic acids pair up through hydrogen bonds in benzene, so their measured i approaches 0.5 and their apparent molar mass approaches twice the true value.

Molar mass from colligative data

Run the colligative equation backwards. From a freezing-point or boiling-point change, m = ΔT ÷ (i·K); moles of solute = m × kilograms of solvent; molar mass = grams of solute ÷ moles. Example: 1.20 g of a nonelectrolyte in 30.0 g of benzene lowers the freezing point by 1.28 °C, so m = 1.28 ÷ 5.12 = 0.250 mol/kg, n = 0.250 × 0.0300 = 7.50 × 10⁻³ mol and M = 1.20 ÷ 7.50 × 10⁻³ = 160 g/mol. From osmotic pressure, n = ΠV ÷ RT. If the solute is assumed to be a nonelectrolyte but actually gives i particles, the apparent molar mass comes out i times too small; if it dimerizes, it comes out about twice too large.

Comparing solutions

To rank solutions by any colligative effect, compare the particle concentration i × m (or i × M). The largest value has the lowest freezing point, the highest boiling point, the highest osmotic pressure and the lowest vapor pressure. At equal molality the effect follows i: sucrose (1) < LiCl (2) < MgCl₂ (3). When the molalities differ, multiply before comparing: 0.40 m glucose (0.40) gives a larger effect than 0.15 m NaCl (0.30). To compare equal masses, compare i ÷ molar mass, since moles of particles per gram = i ÷ M.

What to carry forward

Every colligative effect scales with the concentration of dissolved particles, i × m or i × M. Use P = χ·P° for vapor pressure (and add the partial pressures for two volatile liquids), ΔTf = i·Kf·m and ΔTb = i·Kb·m with molality and kilograms of solvent, and Π = iMRT with T in kelvin. Real electrolytes give smaller i than ideal because of ion pairing, weak electrolytes give 1 + α, and associating solutes give less than 1. Work the same equations backwards to get a molar mass.

Problemas de prática (40)

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Preveja o produto 4

Osmotic pressure

  • Which way does water move?Predict the direction of osmosis between an electrolyte and a nonelectrolyte solution by comparing particle concentrations.
  • A red blood cell in pure waterPredict the effect of a hypotonic medium on a cell from the direction of osmosis.

Counting particles and comparing solutions

  • Two beakers under one bell jarPredict mass transfer between pure solvent and a solution sharing one vapor space from their vapor pressures.

Vapor pressure and Raoult's law

  • Composition of the vapor above a mixturePredict which component is enriched in the vapor above an ideal mixture of two volatile liquids.

Classificar 1

  • Which property is colligative?Classify solution properties as colligative or not by whether they depend only on the number of solute particles.

Comparar 5

Counting particles and comparing solutions

  • Boiling points at different molalitiesRank aqueous solutions of different molality and van 't Hoff factor by boiling point using i × m.
  • Best de-icer per gramCompare equal masses of salts by the moles of particles they give, i ÷ molar mass.
  • Highest osmotic pressureIdentify the solution with the highest osmotic pressure by comparing i × M across solutes.
  • Osmotic pressure with a weak electrolyteRank solutions by osmotic pressure when one solute is a partially ionized weak electrolyte.
  • Freezing points at equal molalityRank solutions of equal molality by freezing point using ideal van 't Hoff factors.

Calcular 20

Van 't Hoff factor and molar mass

  • Apparent molar mass of benzoic acid in benzeneCalculate an apparent molar mass from freezing-point data and recognize when it signals association of the solute.
  • Ideal van 't Hoff factor of aluminum sulfateFind the ideal van 't Hoff factor of a strong electrolyte that contains a polyatomic ion.
  • Measured van 't Hoff factor of NaClCalculate an experimental van 't Hoff factor from a measured freezing-point depression.
  • Molar mass from boiling-point elevationDetermine the molar mass of a nonvolatile nonelectrolyte from the boiling-point elevation of its solution.
  • Molar mass from freezing point in cyclohexaneDetermine the molar mass of a nonelectrolyte from the freezing-point depression it causes in an organic solvent.
  • Percent ionization from a freezing pointFind the percent ionization of a weak acid from its measured van 't Hoff factor, i = 1 + α.

Vapor pressure and Raoult's law

  • Vapor-pressure lowering in ethanolCalculate the vapor-pressure lowering of a solvent from the mole fraction of a nonvolatile solute.
  • Vapor pressure of a glucose solutionFind the vapor pressure of an aqueous solution from the masses of a nonvolatile solute and water using Raoult's law.
  • Total vapor pressure of hexane and heptaneCalculate the total vapor pressure above an ideal mixture of two volatile liquids from Raoult's and Dalton's laws.
  • Vapor pressure of a salt solutionApply Raoult's law to a strong electrolyte solution by counting moles of ions in the mole fraction.
  • Vapor composition from the total pressureWork back from the total vapor pressure of an ideal binary mixture to the liquid composition, then find the vapor composition.

Freezing and boiling points

  • From freezing point to boiling pointUse one colligative measurement to predict another for the same solution through the ratio Kb/Kf.
  • Freezing point of a glucose solutionCalculate the freezing point of an aqueous nonelectrolyte solution from its molality.
  • Freezing point from molarity and densityConvert molarity to molality with the solution density before applying the freezing-point equation.
  • Freezing point of naphthalene in benzeneCalculate the freezing point of a solution in a non-aqueous solvent from its Kf and freezing point.
  • Find Kf of an unknown solventDetermine a solvent's freezing-point constant from the depression caused by a known solute.
  • Boiling point of a urea solutionCalculate the boiling point of an aqueous solution from the mass of a nonvolatile nonelectrolyte and the mass of water.

Osmotic pressure

  • Osmotic pressure of dilute glucoseCalculate the osmotic pressure of a dilute nonelectrolyte solution with Π = MRT.
  • Osmotic pressure of a mixed solutionAdd the particle contributions of two solutes to find the osmotic pressure of a mixture.
  • From osmotic pressure to freezing pointConvert an osmotic-pressure measurement into a predicted freezing point for the same dilute solution.

Encontre o erro 2

Van 't Hoff factor and molar mass

  • A molar mass that is too smallDiagnose an apparent molar mass that is too small because a strong electrolyte was treated as a nonelectrolyte.

Freezing and boiling points

  • Find the error in a freezing-point calculationDiagnose a freezing-point calculation that treats a strong electrolyte as a nonelectrolyte.

Interprete as evidências 2

Van 't Hoff factor and molar mass

  • Acetic acid in benzene looks twice as heavyInterpret an apparent molar mass that is double the formula mass as association of the solute.

Freezing and boiling points

  • Reading a solution's cooling curveInterpret the sloping freezing region of a solution's cooling curve.

Planeje uma síntese 2

Freezing and boiling points

  • How much antifreeze?Plan the mass of a nonelectrolyte antifreeze needed to reach a target freezing point.

Osmotic pressure

  • Glucose solution isotonic with salinePlan a nonelectrolyte solution that is isotonic with a given electrolyte solution.

Explique por quê 4

Vapor pressure and Raoult's law

  • Acetone and chloroform deviate from Raoult's lawExplain a negative deviation from Raoult's law in terms of the attractions between unlike molecules.
  • Why a solute lowers vapor pressureExplain vapor-pressure lowering by a nonvolatile solute in terms of the solvent's mole fraction.

Freezing and boiling points

  • Why a solute lowers the freezing pointExplain freezing-point depression by the exclusion of solute from the solid solvent.

Van 't Hoff factor and molar mass

  • Why MgSO₄ falls short of i = 2Explain why a 2+/2− salt departs further from its ideal van 't Hoff factor than a 1+/1− salt.

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