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4.4 Concentration Units, Henry's Law & Colligative Properties
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8 Biochemistry: Amino Acids, Peptides & Enzymes
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10 Molecular Biology: Nucleic Acids & Genetic Technologies
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14 Human Physiology: Maintenance, Defense & Locomotion
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19.1 Non-Associative Learning & Classical Conditioning 19.2 Operant Conditioning, Reinforcement Schedules & Observational Learning 19.3 Self-Concept & Lifespan Developmental Theories 19.4 Kohlberg's Moral Development & Sociocultural Frameworks 19.5 Social Influence: Conformity, Obedience, Group Dynamics & Prosocial Behavior 19.6 Social Cognition, Attribution Theory, Prejudices, Stereotypes & Discrimination
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4.4 Concentration Units, Henry's Law & Colligative Properties
Key Takeaways Explain This Topic AI
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 ( M M M ) : M = moles of solute liters of solution ( mol/L ) M = \frac{\text{moles of solute}}{\text{liters of solution}} \quad (\text{mol/L}) M = liters of solution moles of solute ( mol/L ) Temperature Dependent : Because liquid volume expands with increasing temperature, molarity decreases as temperature rises.
Molality ( m m m ) : m = moles of solute kilograms of solvent ( mol/kg ) m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \quad (\text{mol/kg}) m = kilograms of solvent moles of solute ( mol/kg ) Temperature Independent : Mass does not alter with thermal expansion. Essential for colligative property calculations.
Mole Fraction ( X A X_A X A ) : X A = n A n total = n A n A + n B + … X_A = \frac{n_A}{n_{\text{total}}} = \frac{n_A}{n_A + n_B + \dots} X A = n total n A = n A + n B + … n A Dimensionless and Temperature Independent .
Mass Percent (w/w%) : Mass % = ( mass of solute total mass of solution ) × 100 % \text{Mass }\% = \left(\frac{\text{mass of solute}}{\text{total mass of solution}}\right) \times 100\% Mass % = ( total mass of solution mass of solute ) × 100%
Henry's Law of Gas Solubility
Henry's law states that at a constant temperature, the solubility of a gas ( C C C ) in a liquid is directly proportional to the partial pressure ( P P P ) of that gas above the liquid: C = k H P C = k_H P C = k H P
Where k H k_H k 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 ( Δ H soln < 0 \Delta H_{\text{soln}} < 0 Δ H soln < 0 ). By Le Chatelier's principle: Gas ( g ) + Solvent ( l ) ⇌ Dissolved Gas ( a q ) + Heat \text{Gas}_{(g)} + \text{Solvent}_{(l)} \rightleftharpoons \text{Dissolved Gas}_{(aq)} + \text{Heat} Gas ( g ) + Solvent ( l ) ⇌ Dissolved Gas ( a q ) + 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 P O 2 P_{\text{O}_2} P 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 N 2 \text{N}_2 N 2 to precipitate out of blood as painful gas bubbles ( decompression sickness / "the bends" ).
6. Colligative Properties and the van 't Hoff Factor ( i i i )
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 ( i i i )
The van 't Hoff factor accounts for the dissociation of ionic electrolytes into multiple particle species: ∋ = moles of particles in solution moles of solute formula units dissolved \ni = \frac{\text{moles of particles in solution}}{\text{moles of solute formula units dissolved}} ∋= moles of solute formula units dissolved moles of particles in solution
Nonelectrolytes (glucose, sucrose, urea): i = 1 i = 1 i = 1 .
Strong Electrolytes (ideal dissociation):
NaCl → Na + + Cl − ⟹ i = 2 \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \implies i = 2 NaCl → Na + + Cl − ⟹ i = 2
CaCl 2 → Ca 2 + + 2 Cl − ⟹ i = 3 \text{CaCl}_2 \rightarrow \text{Ca}^{2+} + 2\text{Cl}^- \implies i = 3 CaCl 2 → Ca 2 + + 2 Cl − ⟹ i = 3
Al(NO 3 ) 3 → Al 3 + + 3 NO 3 − ⟹ i = 4 \text{Al(NO}_3)_3 \rightarrow \text{Al}^{3+} + 3\text{NO}_3^- \implies i = 4 Al(NO 3 ) 3 → Al 3 + + 3 NO 3 − ⟹ 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
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: P A = X A P A ∘ P_A = X_A P_A^\circ P A = X A P A ∘
Where P A P_A P A is the vapor pressure of the solution, X A X_A X A is the solvent mole fraction, and P A ∘ P_A^\circ P A ∘ is the vapor pressure of pure solvent. The vapor pressure lowering ( Δ P \Delta P Δ P ) is: Δ P = P A ∘ − P A = X B P A ∘ = ( i ⋅ X solute ) P A ∘ \Delta P = P_A^\circ - P_A = X_B P_A^\circ = (i \cdot X_{\text{solute}}) P_A^\circ Δ P = P A ∘ − P A = X B P A ∘ = ( i ⋅ X solute ) P A ∘
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 ( Δ H mix > 0 \Delta H_{\text{mix}} > 0 Δ H 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 ( Δ H mix < 0 \Delta H_{\text{mix}} < 0 Δ H mix < 0 ).
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: Δ T b = i K b m \Delta T_b = i K_b m Δ T b = i K b m
Where K b K_b K b is the ebullioscopic constant (for water, K b = 0.512 ∘ C ⋅ kg/mol K_b = 0.512^\circ\text{C}\cdot\text{kg/mol} K b = 0.51 2 ∘ C ⋅ kg/mol ) and m m m is molality. The new boiling point is T b = T b ∘ + Δ T b T_b = T_b^\circ + \Delta T_b T b = T b ∘ + Δ T b .
Freezing Point Depression :
Solute particles disrupt the orderly packing of solvent molecules into a crystalline solid lattice, requiring lower thermal energy to solidify: Δ T f = i K f m \Delta T_f = i K_f m Δ T f = i K f m
Where K f K_f K f is the cryoscopic constant (for water, K f = 1.86 ∘ C ⋅ kg/mol K_f = 1.86^\circ\text{C}\cdot\text{kg/mol} K f = 1.8 6 ∘ C ⋅ kg/mol ). The new freezing point is T f = T f ∘ − Δ T f T_f = T_f^\circ - \Delta T_f T f = T f ∘ − Δ T f . Practical applications include salting icy roads and ethylene glycol automotive antifreeze.
Osmotic Pressure ( Π \Pi Π ) :
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 = i M R T \Pi = i M R T Π = i MRT
Where M M M is molarity, R = 0.0821 L ⋅ atm / ( mol ⋅ K ) R = 0.0821\text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K}) R = 0.0821 L ⋅ atm / ( mol ⋅ K ) , and T T T is absolute temperature in Kelvin.
Colligative Property Governing Equation Proportionality Constant (Water) Particle Dependence Clinical / Practical Application
Vapor Pressure Lowering Δ P = X solute P A ∘ \Delta P = X_{\text{solute}} P_A^\circ Δ P = X solute P A ∘ N/A i × mole fraction i \times \text{mole fraction} i × mole fraction Atmospheric distillation columns
Boiling Point Elevation Δ T b = i K b m \Delta T_b = i K_b m Δ T b = i K b m K b = 0.512 ∘ C ⋅ kg/mol K_b = 0.512^\circ\text{C}\cdot\text{kg/mol} K b = 0.51 2 ∘ C ⋅ kg/mol i × molality i \times \text{molality} i × molality Autoclave sterilization under pressure
Freezing Point Depression Δ T f = i K f m \Delta T_f = i K_f m Δ T f = i K f m K f = 1.86 ∘ C ⋅ kg/mol K_f = 1.86^\circ\text{C}\cdot\text{kg/mol} K f = 1.8 6 ∘ C ⋅ kg/mol i × molality i \times \text{molality} i × molality Road salting; osmometry molecular mass
Osmotic Pressure Π = i M R T \Pi = i M R T Π = i MRT R = 0.0821 L ⋅ atm / ( mol ⋅ K ) R = 0.0821\text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K}) R = 0.0821 L ⋅ atm / ( mol ⋅ K ) i × molarity i \times \text{molarity} i × 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/L 280 - 300\text{ mOsm/L} 280 − 300 mOsm/L (primarily generated by dissolved Na + , Cl − \text{Na}^+, \text{Cl}^- Na + , 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 NaCl 0.9\%\text{ w/v NaCl} 0.9% w/v NaCl normal saline ≈ 308 mOsm/L \approx 308\text{ mOsm/L} ≈ 308 mOsm/L , or 5 % w/v 5\%\text{ w/v} 5% 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} > 300 mOsm/L , e.g., 3.0 % NaCl 3.0\%\text{ NaCl} 3.0% 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} < 280 mOsm/L , e.g., 0.45 % NaCl 0.45\%\text{ NaCl} 0.45% 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 ( T f T_f T f ) and boiling point ( T b T_b T b ) of an aqueous solution prepared by dissolving 111 g 111\text{ g} 111 g of calcium chloride ( CaCl 2 \text{CaCl}_2 CaCl 2 , molar mass = 111.0 g/mol = 111.0\text{ g/mol} = 111.0 g/mol ) into 1000 g 1000\text{ g} 1000 g ( 1.00 kg 1.00\text{ kg} 1.00 kg ) of pure distilled water, assuming ideal electrolyte dissociation. ( K f = 1.86 ∘ C ⋅ kg/mol K_f = 1.86^\circ\text{C}\cdot\text{kg/mol} K f = 1.8 6 ∘ C ⋅ kg/mol , K b = 0.512 ∘ C ⋅ kg/mol K_b = 0.512^\circ\text{C}\cdot\text{kg/mol} K b = 0.51 2 ∘ C ⋅ kg/mol ).
Solution :
Calculate moles of solute: n = 111 g 111.0 g/mol = 1.00 mol CaCl 2 n = \frac{111\text{ g}}{111.0\text{ g/mol}} = 1.00\text{ mol } \text{CaCl}_2 n = 111.0 g/mol 111 g = 1.00 mol CaCl 2
Calculate molality ( m m m ): m = 1.00 mol 1.00 kg solvent = 1.00 m m = \frac{1.00\text{ mol}}{1.00\text{ kg solvent}} = 1.00\text{ m} m = 1.00 kg solvent 1.00 mol = 1.00 m
Determine van 't Hoff factor ( i i i ): CaCl 2 ( s ) → Ca 2 + ( a q ) + 2 Cl − ( a q ) ⟹ i = 1 + 2 = 3 \text{CaCl}_2(s) \rightarrow \text{Ca}^{2+}(aq) + 2\text{Cl}^-(aq) \implies i = 1 + 2 = 3 CaCl 2 ( s ) → Ca 2 + ( a q ) + 2 Cl − ( a q ) ⟹ i = 1 + 2 = 3
Calculate freezing point depression ( Δ T f \Delta T_f Δ T f ): Δ T f = i K f m = ( 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} Δ T f = i K f m = ( 3 ) ( 1.8 6 ∘ C ⋅ kg/mol ) ( 1.00 mol/kg ) = 5.5 8 ∘ C T f = 0.00 ∘ C − 5.58 ∘ C = − 5.58 ∘ C T_f = 0.00^\circ\text{C} - 5.58^\circ\text{C} = -5.58^\circ\text{C} T f = 0.0 0 ∘ C − 5.5 8 ∘ C = − 5.5 8 ∘ C
Calculate boiling point elevation ( Δ T b \Delta T_b Δ T b ): Δ T b = i K b m = ( 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} Δ T b = i K b m = ( 3 ) ( 0.51 2 ∘ C ⋅ kg/mol ) ( 1.00 mol/kg ) = 1.53 6 ∘ C T b = 100.00 ∘ C + 1.54 ∘ C = 101.54 ∘ C T_b = 100.00^\circ\text{C} + 1.54^\circ\text{C} = 101.54^\circ\text{C} T b = 100.0 0 ∘ C + 1.5 4 ∘ C = 101.5 4 ∘ 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} ≈ 66 , 500 g/mol ) at a concentration of 40.0 g/L 40.0\text{ g/L} 40.0 g/L at normal body temperature ( 37.0 ∘ C = 310.15 K 37.0^\circ\text{C} = 310.15\text{ K} 37. 0 ∘ C = 310.15 K ). Assuming albumin behaves as a monomeric nonelectrolyte ( i = 1 i = 1 i = 1 ), calculate the osmotic pressure (oncotic pressure) in atmospheres and in mmHg \text{mmHg} mmHg ( 1 atm = 760 mmHg 1\text{ atm} = 760\text{ mmHg} 1 atm = 760 mmHg ).
Solution :
Calculate the molar concentration ( M M M ): M = 40.0 g/L 66 , 500 g/mol = 6.015 × 10 − 4 M M = \frac{40.0\text{ g/L}}{66,500\text{ g/mol}} = 6.015 \times 10^{-4}\text{ M} M = 66 , 500 g/mol 40.0 g/L = 6.015 × 1 0 − 4 M
Apply the van 't Hoff osmotic pressure equation ( Π = i M R T \Pi = i M R T Π = i MRT ): 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 ) ( 6.015 × 1 0 − 4 mol/L ) ( 0.0821 L ⋅ atm / ( mol ⋅ K )) ( 310.15 K ) Π = 1.532 × 10 − 2 atm \Pi = 1.532 \times 10^{-2}\text{ atm} Π = 1.532 × 1 0 − 2 atm
Convert to millimeters of mercury ( mmHg \text{mmHg} 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} Π = ( 1.532 × 1 0 − 2 atm ) × 760 mmHg/atm ≈ 11.6 mmHg This oncotic pressure acts across capillary walls to pull fluid back from the interstitial space into the vascular lumen.