4.4 Concentration Units, Henry's Law & Colligative Properties

Key Takeaways

  • Henry's Law (C = k_H P) dictates that gas solubility in a liquid increases with rising partial pressure, whereas gas solubility decreases as temperature increases because dissolution of gas is exothermic.

  • Colligative properties depend solely on the concentration of solute particles (i m or i M); solute addition lowers solvent vapor pressure (Raoult's law), elevates boiling point (Delta Tb = i Kb m), depresses freezing point (Delta Tf = i Kf m), and generates osmotic pressure (Pi = i M R T).

  • Osmotic pressure (Π = iMRT) determines how erythrocytes respond to an infused fluid: isotonic saline leaves cell volume unchanged, hypotonic fluid drives water inward toward haemolysis, and hypertonic fluid draws water out and causes crenation.

Last updated: September 2026

4.4 Concentration Units, Henry's Law & Colligative Properties

5. Solution Concentration Units and Henry's Law

Concentration Definitions

  • Molarity (MM): M=moles of soluteliters of solution(mol/L)M = \frac{\text{moles of solute}}{\text{liters of solution}} \quad (\text{mol/L}) Temperature Dependent: Because liquid volume expands with increasing temperature, molarity decreases as temperature rises.
  • Molality (mm): m=moles of solutekilograms of solvent(mol/kg)m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \quad (\text{mol/kg}) Temperature Independent: Mass does not alter with thermal expansion. Essential for colligative property calculations.
  • Mole Fraction (XAX_A): XA=nAntotal=nAnA+nB+…X_A = \frac{n_A}{n_{\text{total}}} = \frac{n_A}{n_A + n_B + \dots} Dimensionless and Temperature Independent.
  • Mass Percent (w/w%): Mass %=(mass of solutetotal mass of solution)×100%\text{Mass }\% = \left(\frac{\text{mass of solute}}{\text{total mass of solution}}\right) \times 100\%

Henry's Law of Gas Solubility

Henry's law states that at a constant temperature, the solubility of a gas (CC) in a liquid is directly proportional to the partial pressure (PP) of that gas above the liquid:

C=kHPC = k_H P

Where kHk_H is Henry's law constant (specific to the gas, solvent, and temperature).

  • Temperature Dependence of Gas Solubility: Dissolving a gas into a liquid brings free gas molecules into an ordered condensed phase, which is an exothermic process (ΔHsoln<0\Delta H_{\text{soln}} < 0). By Le Chatelier's principle: Gas(g)+Solvent(l)⇌Dissolved Gas(aq)+Heat\text{Gas}_{(g)} + \text{Solvent}_{(l)} \rightleftharpoons \text{Dissolved Gas}_{(aq)} + \text{Heat} Increasing temperature shifts the equilibrium in reverse, decreasing gas solubility. Warm water holds significantly less dissolved oxygen than cold water.
  • Clinical Relevance: In hyperbaric oxygen chambers, raising PO2P_{\text{O}_2} forces more dissolved oxygen directly into blood plasma independently of hemoglobin saturation. Conversely, during rapid scuba ascent, dropping ambient hydrostatic pressure decreases nitrogen solubility, causing dissolved N2\text{N}_2 to precipitate out of blood as painful gas bubbles (decompression sickness / "the bends").

6. Colligative Properties and the van 't Hoff Factor (ii)

Colligative properties depend exclusively on the total number (concentration) of dissolved solute particles in solution, regardless of their chemical identity, mass, or size.

The van 't Hoff Factor (ii)

The van 't Hoff factor accounts for the dissociation of ionic electrolytes into multiple particle species:

∋=moles of particles in solutionmoles of solute formula units dissolved\ni = \frac{\text{moles of particles in solution}}{\text{moles of solute formula units dissolved}}
  • Nonelectrolytes (glucose, sucrose, urea): i=1i = 1.
  • Strong Electrolytes (ideal dissociation):
    • NaCl→Na++Cl−  ⟹  i=2\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \implies i = 2
    • CaCl2→Ca2++2Cl−  ⟹  i=3\text{CaCl}_2 \rightarrow \text{Ca}^{2+} + 2\text{Cl}^- \implies i = 3
    • Al(NO3)3→Al3++3NO3−  ⟹  i=4\text{Al(NO}_3)_3 \rightarrow \text{Al}^{3+} + 3\text{NO}_3^- \implies i = 4
  • Non-ideal behavior: Real van 't Hoff factors are slightly less than whole integers at high concentrations due to transient electrostatic ion pairing.

The Four Colligative Properties

  1. Vapor Pressure Lowering (Raoult's Law): Dissolving a non-volatile solute reduces the mole fraction of volatile solvent at the liquid surface, decreasing the rate of solvent vaporization. For an ideal solution:

    PA=XAPA∘P_A = X_A P_A^\circ

    Where PAP_A is the vapor pressure of the solution, XAX_A is the solvent mole fraction, and PA∘P_A^\circ is the vapor pressure of pure solvent. The vapor pressure lowering (ΔP\Delta P) is:

    ΔP=PA∘−PA=XBPA∘=(i⋅Xsolute)PA∘\Delta P = P_A^\circ - P_A = X_B P_A^\circ = (i \cdot X_{\text{solute}}) P_A^\circ
    • Deviations from Raoult's Law:
      • Positive Deviation: Solute-solvent attractions are weaker than solute-solute or solvent-solvent attractions. Molecules escape more readily into the vapor phase, resulting in a higher observed vapor pressure and endothermic mixing (ΔHmix>0\Delta H_{\text{mix}} > 0, e.g., ethanol and hexane).
      • Negative Deviation: Solute-solvent attractions are stronger than interactions in the pure components (e.g., hydrogen bonding between acetone and chloroform). Molecules are held more tightly in the liquid, resulting in a lower observed vapor pressure and exothermic mixing (ΔHmix<0\Delta H_{\text{mix}} < 0).
  2. Boiling Point Elevation: A liquid boils when its vapor pressure equals atmospheric pressure. Because solute lowers vapor pressure, a higher temperature is required to reach atmospheric pressure:

    ΔTb=iKbm\Delta T_b = i K_b m

    Where KbK_b is the ebullioscopic constant (for water, Kb=0.512∘C⋅kg/molK_b = 0.512^\circ\text{C}\cdot\text{kg/mol}) and mm is molality. The new boiling point is Tb=Tb∘+ΔTbT_b = T_b^\circ + \Delta T_b.

  3. Freezing Point Depression: Solute particles disrupt the orderly packing of solvent molecules into a crystalline solid lattice, requiring lower thermal energy to solidify:

    ΔTf=iKfm\Delta T_f = i K_f m

    Where KfK_f is the cryoscopic constant (for water, Kf=1.86∘C⋅kg/molK_f = 1.86^\circ\text{C}\cdot\text{kg/mol}). The new freezing point is Tf=Tf∘−ΔTfT_f = T_f^\circ - \Delta T_f. Practical applications include salting icy roads and ethylene glycol automotive antifreeze.

  4. Osmotic Pressure (Π\Pi): Osmotic pressure is the exact hydrostatic pressure that must be applied to a solution to prevent the net inward osmotic flow of pure water across a semipermeable membrane:

    Π=iMRT\Pi = i M R T

    Where MM is molarity, R=0.0821 L⋅atm/(mol⋅K)R = 0.0821\text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K}), and TT is absolute temperature in Kelvin.

Colligative PropertyGoverning EquationProportionality Constant (Water)Particle DependenceClinical / Practical Application
Vapor Pressure LoweringΔP=XsolutePA∘\Delta P = X_{\text{solute}} P_A^\circN/Ai×mole fractioni \times \text{mole fraction}Atmospheric distillation columns
Boiling Point ElevationΔTb=iKbm\Delta T_b = i K_b mKb=0.512∘C⋅kg/molK_b = 0.512^\circ\text{C}\cdot\text{kg/mol}i×molalityi \times \text{molality}Autoclave sterilization under pressure
Freezing Point DepressionΔTf=iKfm\Delta T_f = i K_f mKf=1.86∘C⋅kg/molK_f = 1.86^\circ\text{C}\cdot\text{kg/mol}i×molalityi \times \text{molality}Road salting; osmometry molecular mass
Osmotic PressureΠ=iMRT\Pi = i M R TR=0.0821 L⋅atm/(mol⋅K)R = 0.0821\text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K})i×molarityi \times \text{molarity}IV fluid tonicity; renal tubular reabsorption

7. Physiological Applications: Intravenous Fluids, Osmolarity & Erythrocytes

Normal human blood plasma has a physiological osmolarity of approximately 280−300 mOsm/L280 - 300\text{ mOsm/L} (primarily generated by dissolved Na+,Cl−\text{Na}^+, \text{Cl}^-, bicarbonate, glucose, and serum albumin).

Tonicity and Erythrocyte Morphologies

  • Isotonic Solutions: Possess an effective osmolarity equal to intracellular fluid (e.g., 0.9% w/v NaCl0.9\%\text{ w/v NaCl} normal saline ≈308 mOsm/L\approx 308\text{ mOsm/L}, or 5% w/v5\%\text{ w/v} dextrose in water [D5W]). Zero net water flux occurs across the erythrocyte membrane; red blood cells maintain normal biconcave disc morphology.
  • Hypertonic Solutions (>300 mOsm/L> 300\text{ mOsm/L}, e.g., 3.0% NaCl3.0\%\text{ NaCl}): External solute concentration exceeds intracellular concentration. Water rapidly diffuses down its chemical potential gradient out of the cell into the hypertonic extracellular environment, causing erythrocyte cellular shrinkage and wrinkling (crenation).
  • Hypotonic Solutions (<280 mOsm/L< 280\text{ mOsm/L}, e.g., 0.45% NaCl0.45\%\text{ NaCl} or sterile pure water): External solute concentration is lower than intracellular fluid. Water rushes osmotically into the erythrocyte, causing cellular swelling, membrane disruption, and cell rupture (osmotic hemolysis). Intravenous administration of pure water is fatal.

8. Worked Numerical Examples

Example 1: Colligative Shift for an Electrolyte Solution

Calculate the freezing point (TfT_f) and boiling point (TbT_b) of an aqueous solution prepared by dissolving 111 g111\text{ g} of calcium chloride (CaCl2\text{CaCl}_2, molar mass =111.0 g/mol= 111.0\text{ g/mol}) into 1000 g1000\text{ g} (1.00 kg1.00\text{ kg}) of pure distilled water, assuming ideal electrolyte dissociation. (Kf=1.86∘C⋅kg/molK_f = 1.86^\circ\text{C}\cdot\text{kg/mol}, Kb=0.512∘C⋅kg/molK_b = 0.512^\circ\text{C}\cdot\text{kg/mol}).

Solution:

  1. Calculate moles of solute: n=111 g111.0 g/mol=1.00 mol CaCl2n = \frac{111\text{ g}}{111.0\text{ g/mol}} = 1.00\text{ mol } \text{CaCl}_2
  2. Calculate molality (mm): m=1.00 mol1.00 kg solvent=1.00 mm = \frac{1.00\text{ mol}}{1.00\text{ kg solvent}} = 1.00\text{ m}
  3. Determine van 't Hoff factor (ii): CaCl2(s)→Ca2+(aq)+2Cl−(aq)  ⟹  i=1+2=3\text{CaCl}_2(s) \rightarrow \text{Ca}^{2+}(aq) + 2\text{Cl}^-(aq) \implies i = 1 + 2 = 3
  4. Calculate freezing point depression (ΔTf\Delta T_f): ΔTf=iKfm=(3)(1.86∘C⋅kg/mol)(1.00 mol/kg)=5.58∘C\Delta T_f = i K_f m = (3)(1.86^\circ\text{C}\cdot\text{kg/mol})(1.00\text{ mol/kg}) = 5.58^\circ\text{C} Tf=0.00∘C−5.58∘C=−5.58∘CT_f = 0.00^\circ\text{C} - 5.58^\circ\text{C} = -5.58^\circ\text{C}
  5. Calculate boiling point elevation (ΔTb\Delta T_b): ΔTb=iKbm=(3)(0.512∘C⋅kg/mol)(1.00 mol/kg)=1.536∘C\Delta T_b = i K_b m = (3)(0.512^\circ\text{C}\cdot\text{kg/mol})(1.00\text{ mol/kg}) = 1.536^\circ\text{C} Tb=100.00∘C+1.54∘C=101.54∘CT_b = 100.00^\circ\text{C} + 1.54^\circ\text{C} = 101.54^\circ\text{C}

Example 2: Osmotic Pressure of a Protein Solution

A physician analyzes an artificial colloidal plasma expander containing human serum albumin (molar mass ≈66,500 g/mol\approx 66,500\text{ g/mol}) at a concentration of 40.0 g/L40.0\text{ g/L} at normal body temperature (37.0∘C=310.15 K37.0^\circ\text{C} = 310.15\text{ K}). Assuming albumin behaves as a monomeric nonelectrolyte (i=1i = 1), calculate the osmotic pressure (oncotic pressure) in atmospheres and in mmHg\text{mmHg} (1 atm=760 mmHg1\text{ atm} = 760\text{ mmHg}).

Solution:

  1. Calculate the molar concentration (MM): M=40.0 g/L66,500 g/mol=6.015×10−4 MM = \frac{40.0\text{ g/L}}{66,500\text{ g/mol}} = 6.015 \times 10^{-4}\text{ M}
  2. Apply the van 't Hoff osmotic pressure equation (Π=iMRT\Pi = i M R T): Π=(1)(6.015×10−4 mol/L)(0.0821 L⋅atm/(mol⋅K))(310.15 K)\Pi = (1)(6.015 \times 10^{-4}\text{ mol/L})(0.0821\text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K}))(310.15\text{ K}) Π=1.532×10−2 atm\Pi = 1.532 \times 10^{-2}\text{ atm}
  3. Convert to millimeters of mercury (mmHg\text{mmHg}): Π=(1.532×10−2 atm)×760 mmHg/atm≈11.6 mmHg\Pi = (1.532 \times 10^{-2}\text{ atm}) \times 760\text{ mmHg/atm} \approx 11.6\text{ mmHg} This oncotic pressure acts across capillary walls to pull fluid back from the interstitial space into the vascular lumen.
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Thermodynamics of Phase Transitions and Colligative Solution Mechanics
Test Your Knowledge

At 25 degrees Celsius, pure water has a vapor pressure of 23.8 mmHg. If 2.0 moles of a non-volatile nonelectrolyte solute are completely dissolved in 8.0 moles of water, what is the vapor pressure of the resulting solution according to Raoult's law?

A

4.76 mmHg

B

23.8 mmHg

C

29.75 mmHg

D

19.04 mmHg

Test Your Knowledge

Which of the following four aqueous solutions will exhibit the lowest (most depressed) freezing point, assuming complete electrolyte dissociation?

A

0.10 m C6H12O6 (glucose)

B

0.10 m AlCl3

C

0.10 m CaCl2

D

0.10 m NaCl

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