§4.1 Principles of Colligative Properties for Macromolecules & Van 't Hoff Limiting Law
Colligative properties—vapor pressure lowering, boiling point elevation (ebulliometry), freezing point depression (cryoscopy), and osmotic pressure—depend thermodynamically on the number density (number concentration) of solute particles rather than their chemical structure, mass, or shape. Consequently, measuring colligative properties provides an absolute thermodynamic determination of the number-average molecular weight ($M_n$).
Thermodynamic Derivation of the Van 't Hoff Limiting Law
Consider a binary solution consisting of $n_1$ moles of solvent and $n_2$ moles of polymer solute separated from pure solvent by a rigid semi-permeable membrane. At thermodynamic equilibrium:
where $\Pi$ is the osmotic pressure. The chemical potential of the solvent in the solution under elevated pressure $P + \Pi$ is related to that at pressure $P$ through the fundamental thermodynamic relation:
where $V_1 = (\partial V / \partial n_1)_{T, P}$ is the partial molar volume of the solvent (approximated as constant for incompressible liquids). Substituting into the equilibrium condition gives:
For an ideally dilute solution, the solvent activity $a_1$ equals its mole fraction $x_1$:
Because the polymer solution is dilute ($x_2 \ll 1$), we expand $\ln(1 - x_2) \approx -x_2$:
In dilute solutions containing $N_A$ polymer molecules of mass concentration $c$ (in $\text{g/cm}^3$ or $\text{g/L}$):
Substituting this mole fraction into the equation yields the classical van 't Hoff limiting law:
Comparison of Colligative Sensitivities: Why Osmometry Dominates
To understand why membrane osmometry is uniquely suited for macromolecules whereas cryoscopy and ebulliometry fail, consider a solution of a polymer with $M_n = 50,000\text{ g/mol}$ at a concentration of $c = 10\text{ g/L}$ in water ($K_f = 1.86\text{ K kg/mol}, \rho = 1.0\text{ g/cm}^3$):
1. Freezing point depression:
Measuring a temperature depression of $0.00037\text{ K}$ requires ultra-sensitive differential thermistors and is susceptible to thermal drift, atmospheric fluctuations, and trace low molecular weight impurities (e.g., $1\text{ ppm}$ NaCl produces a depression comparable to the polymer signal).
2. Osmotic pressure at $T = 298.15\text{ K}$:
In a toluene solution ($\rho_{\text{toluene}} = 0.867\text{ g/cm}^3$):
At $c = 20\text{ g/L}$ with lower $M_n = 20,000\text{ g/mol}$, the liquid rise is several centimeters—readily measured with sub-millimeter precision using an optical cathetometer or electronic pressure transducer.
§4.2 Membrane Osmometry: Semi-Permeable Membranes, Donnan Equilibrium & Chemical Potential
Apparatus & Operating Principles
Membrane osmometers consist of two precision-machined stainless steel or titanium cells separated by a rigid semi-permeable membrane:
- Solvent Chamber: Filled with pure solvent and connected to a sensitive variable-capacitance diaphragm pressure transducer or glass capillary.
- Solution Chamber: Flushed with polymer solution at known mass concentration $c$.
In modern automatic high-speed membrane osmometers (e.g., Wescan / Knauer), the solvent chamber is maintained at constant volume. When solvent molecules begin diffusing through the membrane into the solution chamber, a microscopic deflection of the diaphragm is detected by an optical or capacitive sensor. A servomotor rapidly applies a counter-pressure to null the deflection, achieving osmotic equilibrium within 5 to 15 minutes, compared to several hours or days required for traditional static head equilibration.
Membrane Selection & Solute Permeability Thresholds
The membrane must be strictly semi-permeable: perfectly permeable to solvent molecules while completely impermeable to polymer solutes.
- Membrane Materials: Regenerated cellulose, cellulose acetate, cellulose nitrate, and microporous PTFE.
- Molecular Weight Cut-Off (MWCO): Commercial membranes have nominal MWCOs ranging from $5,000$ to $30,000\text{ g/mol}$.
- Permeation Artifacts: If a polydisperse sample contains low molecular weight oligomer tails below the MWCO, these oligomers diffuse through the membrane into the solvent chamber during measurement. This leak has two destructive consequences:
- The concentration in the solution chamber decreases over time.
- The chemical potential difference $\Delta \mu_1$ diminishes, causing the apparent osmotic pressure $\Pi$ to decay toward zero.
- The resulting extrapolated $M_n$ will be systematically overestimated because the low molecular weight fraction is selectively lost.
The Donnan Membrane Equilibrium in Polyelectrolytes
When measuring ionic polymers (polyelectrolytes such as sodium polyacrylate or DNA) in aqueous solution, an additional electrostatic complication arises termed the Donnan effect. Consider a sodium polyelectrolyte $Na_z P$ of molar concentration $C_p$ (bearing $z$ anionic charges per macromolecule) in the presence of mobile sodium chloride salt ($NaCl$) at concentration $C_s$:
- Inside solution chamber: $[P^{z-}] = C_p$, $[Na^+]_i = z C_p + [Cl^-]_i$.
- Inside solvent chamber: $[Na^+]_o = C_s$, $[Cl^-]_o = C_s$.
Thermodynamic equilibrium requires the chemical potential of diffusible $NaCl$ to be equal on both sides:
Substituting $[Na^+]_i = z C_p + [Cl^-]_i$:
Solving the quadratic for mobile chloride concentration inside:
The total osmotic pressure includes both the macromolecule and the excess mobile counterions:
In the limit of low polymer concentration relative to added salt ($z C_p \ll C_s$), Taylor series expansion reveals:
Without added salt ($C_s \to 0$), the mobile counterions cannot cross the membrane without violating macroscopic electroneutrality, resulting in an enormous apparent osmotic pressure corresponding to $M_{\text{eff}} = M_n / (z + 1)$, underestimating the polymer molecular weight by several orders of magnitude! To suppress the Donnan effect, polyelectrolyte osmometry must always be conducted in a supporting electrolyte of high ionic strength (typically $0.10 - 0.20\text{ M NaCl}$).
§4.3 Osmotic Virial Expansion: Second Virial Coefficient, Theta State & Excluded Volume
Because polymer coils occupy large hydrodynamic volumes and exhibit significant thermodynamic interactions with solvent molecules, real polymer solutions deviate substantially from ideality even at low concentrations ($c \sim 10\text{ g/L} \approx 1\text{ wt}\%$).
The Osmotic Virial Equation of State
In analogy to the van der Waals virial equation for non-ideal gases, the osmotic pressure of a polymer solution is expressed as a power series in mass concentration $c$:
where:
- $M_n$ is the number-average molecular weight ($ ext{g/mol}$).
- $A_2$ is the second osmotic virial coefficient (units: $\text{mol cm}^3/\text{g}^2$ or $\text{mol dm}^3/\text{g}^2$).
- $A_3$ is the third osmotic virial coefficient ($ ext{mol cm}^6/\text{g}^3$).
Connection to Flory-Huggins Theory
Expanding the Flory-Huggins solvent chemical potential in powers of polymer volume fraction $\phi_2 = c \bar{v}$ (where $\bar{v}$ is the partial specific volume of the polymer):
Since $\Pi V_1 = -(\mu_1 - \mu_1^\circ)$ and $\phi_2 = c \bar{v}$, dividing by $c V_1$:
Comparing term-by-term with the osmotic virial equation yields the fundamental relationship:
and for the third virial coefficient:
Thermodynamic Regimes & Solvent Quality
1. Good Solvent ($\chi < 0.5 \implies A_2 > 0$):
- Favorable polymer-solvent interactions cause polymer coils to swell and repel each other.
- The reduced osmotic pressure $\Pi/c$ increases linearly with concentration $c$.
2. Theta Solvent ($\chi = 0.5 \implies A_2 = 0$):
- At the Flory theta temperature $\Theta$, thermodynamic polymer-solvent repulsive and attractive forces precisely cancel.
- The solution behaves pseudo-ideally over a wide concentration range: $\Pi/c = R T / M_n = \text{constant}$.
3. Poor Solvent ($\chi > 0.5 \implies A_2 < 0$):
- Segment-segment attractive forces dominate.
- The curve of $\Pi/c$ slopes downward. If concentration increases, macroscopic phase separation (precipitation) ensues.
Linear Regression Protocol
To extract $M_n$ and $A_2$, osmotic pressure is measured at 4 to 6 dilute concentrations ($c = 2, 4, 6, 8, 10\text{ g/L}$). A plot of $\Pi/c$ on the y-axis against $c$ on the x-axis yields a straight line:
§4.4 Vapor Pressure Osmometry (VPO): Thermoelectric Differentials & Oligomer Calibration
Principles of Operation
Vapor Pressure Osmometry (VPO) is a dynamic thermoelectric technique designed to determine $M_n$ for low molecular weight polymers and oligomers ($500 \le M_n \le 25,000\text{ g/mol}$) that are too small to be retained by semi-permeable membranes.
Unlike membrane osmometry, VPO does not measure a hydrostatic or hydraulic pressure. Instead, it measures a steady-state temperature difference ($\Delta T$) resulting from differential solvent vapor condensation.
Thermodynamics of the Condensation Cell
Two matched glass-bead thermistors are suspended inside a hermetically sealed, temperature-controlled measurement chamber saturated with solvent vapor:
- Thermistor A: Loaded with a calibrated droplet of pure solvent.
- Thermistor B: Loaded with a droplet of polymer solution of mass concentration $c$.
According to Raoult's law, the presence of the non-volatile polymer solute depresses the solvent vapor pressure above droplet B:
Because the chamber is saturated at the equilibrium vapor pressure of pure solvent ($P_1^\circ$), a vapor pressure gradient $\Delta P = P_1^\circ - P_1 = x_2 P_1^\circ$ drives spontaneous condensation of solvent vapor onto solution droplet B. As solvent vapor condenses, it releases its latent heat of vaporization $\Delta H_{\text{vap}}$. Droplet B heats up until its elevated vapor pressure matches the ambient chamber pressure $P_1^\circ$. Applying the Clausius-Clapeyron equation:
For dilute solutions, $\ln(P_1^\circ/P_1) = -\ln(1 - x_2) \approx x_2 = \frac{c V_1}{M_n}$:
Bridge Resistance Measurement & Instrument Calibration
The temperature difference $\Delta T$ creates an electrical resistance imbalance $\Delta R$ across a Wheatstone bridge:
where $K_{\text{VPO}}$ is the characteristic instrument calibration constant ($\Omega \cdot \text{g/mol} \cdot \text{L/g}$). Accounting for non-ideal thermodynamic interactions:
1. Calibration: The instrument is first calibrated using a monodisperse, high-purity small molecule standard of known molecular weight (such as benzil, $M = 210.23\text{ g/mol}$, or sucrose octacetate, $M = 678.60\text{ g/mol}$). Extrapolating $(\Delta R / c)$ to $c \to 0$ determines $K_{\text{VPO}}$.
2. Measurement: The unknown polymer sample is measured at multiple concentrations, plotted as $(\Delta R / c)$ vs $c$, and extrapolated to zero concentration:
§4.5 End-Group Analysis: Titrimetric, Radiochemical, UV-Vis, and NMR Determination of $M_n$
End-group analysis is a classical chemical method for determining the number-average molecular weight ($M_n$) of linear macromolecules bearing chemically distinct and quantifiable terminal functionalities.
Fundamental Stoichiometric Principle
Consider a sample of mass $m_{\text{sample}}$ containing $N$ macromolecular chains. By definition:
If each polymer chain possesses an average number $f$ of detectable end groups (termed the end-group functionality):
- Monotelic polymers ($f = 1$): initiated by a functional initiator or terminated asymmetrically (e.g., methoxy-PEG).
- Telechelic polymers ($f = 2$): symmetric bifunctional chains with identical reactive groups at both termini (e.g., $\alpha,\omega$-dihydroxyl polybutadiene, dicarboxylic nylon oligomers).
The molar quantity of end groups present in the sample is:
Rearranging yields the universal end-group formula:
Analytical Methodologies
1. Titrimetric Quantification (Carboxyl and Amine End Groups):
- Polyamides (such as Nylon 6,6) contain free terminal amino ($-NH_2$) and carboxyl ($-COOH$) groups.
- Amine end groups are titrated potentiometrically with perchloric acid ($HClO_4$) or hydrochloric acid in $m$-cresol or trifluoroethanol.
- Carboxyl end groups are titrated with ethanolic potassium hydroxide ($KOH$) or tetra-n-butylammonium hydroxide in benzyl alcohol at elevated temperature under nitrogen.
2. Hydroxyl Number (OHV) in Polyols:
- Polyether and polyester polyols used in polyurethane manufacturing are quantified via esterification with acetic anhydride or phthalic anhydride in pyridine:
- Excess unreacted anhydride is hydrolyzed with water to acetic acid and back-titrated with standardized $KOH$.
- The Hydroxyl Value ($OHV$) is defined as the milligrams of $KOH$ equivalent to the hydroxyl content of $1.0\text{ g}$ of sample:
where $56,100\text{ mg/mol}$ is the formula weight of $KOH$.
3. High-Resolution NMR Spectroscopy ($^1\text{H}$ and $^{13}\text{C}$):
- NMR integration provides an absolute, non-destructive ratio between repeating unit backbone protons and terminal end-group protons.
- Let $I_{\text{backbone}}$ be the integral of a backbone resonance representing $n_b$ protons per repeating unit of formula weight $M_0$.
- Let $I_{\text{end}}$ be the integral of an end-group resonance representing $n_e$ protons per chain terminus.
- The number-average degree of polymerization is:
§4.6 Limitations, Sensitivity Thresholds & Systematic Errors of End-Group Analysis
While end-group analysis is conceptually straightforward and provides absolute molecular weight without requiring external calibration standards, its accuracy depends on several chemical and instrumental constraints.
1. Sensitivity Threshold & Signal-to-Noise Floor
The mass fraction $w_{\text{end}}$ of terminal groups scales inversely with macromolecular weight:
- For an oligomer with $M_n = 2,000\text{ g/mol}$ and end groups of $M_{\text{end}} = 60\text{ g/mol}$, $w_{\text{end}} = 3.0\text{ wt}\%$ (readily quantifiable by NMR, IR, or titration).
- For a polymer with $M_n = 100,000\text{ g/mol}$, $w_{\text{end}} = 0.06\text{ wt}\%$ ($600\text{ ppm}$).
- At $M_n > 30,000\text{ g/mol}$, the terminal proton NMR resonances merge into the baseline noise, and titrant volumes drop below the volumetric burette resolution ($< 0.02\text{ mL}$). Thus, end-group analysis is strictly limited to $M_n \le 25,000 - 30,000\text{ g/mol}$.
2. Systematic Error from Uncertain Functionality ($f$)
The calculation of $M_n$ assumes an exact integer functionality $f$:
- If side reactions occur during polymerization (such as chain transfer to monomer, $\beta$-hydride elimination, or premature disproportionation), chains may terminate with unreactive vinyl, saturated alkyl, or oxidized moieties.
- If $10\%$ of the chains terminate without the targeted functional group, the measured $[\text{End Group}]$ is depressed, causing the calculated $M_n$ to be falsely inflated by $11\%$.
3. The Cyclic Oligomer Artifact
In step-growth polymerizations (e.g., polyamides, polyesters, silicones), intramolecular cyclization reactions compete with intermolecular chain extension:
Cyclic macromolecules contain zero end groups ($f = 0$). They contribute fully to the sample mass $m_{\text{sample}}$ but consume zero titrant. Consequently, the presence of cyclic oligomers leads to a systematic overestimation of $M_n$:
If a Nylon sample contains $3\text{ wt}\%$ cyclic oligomers, its apparent $M_n$ is overestimated by $\approx 3.1\%$.
4. Non-Polymeric Impurities
Any low-molecular-weight monofunctional or polyfunctional impurity (such as unreacted monomer, residual catalyst, solvent stabilizer, or atmospheric moisture absorbing $\text{CO}_2$ to form carbonic acid) consumes titrant disproportionately. A trace impurity of $0.05\text{ wt}\%$ acetic acid ($M = 60$) will consume the same titrant volume as $10\text{ wt}\%$ of a polymer of $M_n = 12,000\text{ g/mol}$, catastrophically underestimating the molecular weight.
§4.7 Analytical Ultracentrifugation (AUC): Sedimentation Velocity, Svedberg Equation & Friction Factor
Analytical Ultracentrifugation (AUC) is a rigorous hydrodynamic method developed by Theodor Svedberg in the 1920s that subjects macromolecules in solution to intense centrifugal fields up to $300,000 \times g$ (rotor speeds up to $60,000\text{ rpm}$).
Forces Acting on a Sedimenting Macromolecule
Consider a polymer molecule of mass $m = M / N_A$ and partial specific volume $\bar{v}$ sedimenting at distance $r$ from the axis of rotation in a rotor spinning at angular velocity $\omega$ (in $\text{rad/s}$):
1. Centrifugal Force:
2. Buoyant Force (Archimedes' principle in solvent of density $\rho$):
3. Frictional Drag Force:
where $f$ is the translational friction coefficient and $v = dr/dt$ is the sedimentation velocity.
The Sedimentation Coefficient and Svedberg Equation
Within microseconds of centrifugal acceleration, the net force drops to zero as frictional drag balances the buoyant centrifugal force:
The sedimentation coefficient $s$ is defined as the sedimentation velocity per unit centrifugal field:
The standard unit of sedimentation is the Svedberg ($1\text{ S} = 10^{-13}\text{ seconds}$). According to the Einstein relation, the translational diffusion coefficient $D$ is related to the friction factor $f$ by:
Substituting $f$ into the sedimentation coefficient expression yields the celebrated Svedberg equation:
By measuring the sedimentation boundary movement ($s$) and boundary spreading ($D$) in dilute solution and extrapolating to zero concentration ($s_0$ and $D_0$), the Svedberg equation yields an absolute determination of macromolecular mass without requiring calibration standards or structural assumptions.
§4.8 Sedimentation Equilibrium: Lamm Equation, Radial Density Gradients & Absolute $M_w, M_z$
Sedimentation Equilibrium
When the ultracentrifuge is operated at lower rotor speeds ($5,000 - 15,000\text{ rpm}$) over extended durations (12 to 48 hours), a steady-state condition is reached where the forward centrifugal transport flux ($J_{\text{sed}} = s \omega^2 r c$) is identically balanced at every radial position $r$ by the backward thermodynamic diffusion flux ($J_{\text{diff}} = -D (dc/dr)$):
Rearranging:
Substituting the Svedberg relationship $s/D = \frac{M(1 - \bar{v}\rho)}{R T}$:
Integrating from meniscus radius $r_m$ to base radius $r_b$:
A plot of $\ln c(r)$ versus $r^2$ yields a straight line for a monodisperse polymer, where the slope directly provides $M$:
Polydisperse Systems and Molecular Weight Averages
For a polydisperse macromolecular sample, each molecular weight fraction sets up its own exponential radial gradient. The overall weight-average molecular weight between the meniscus ($r_m$) and cell bottom ($r_b$) is obtained from:
Furthermore, the local slope at any radial point $r$ yields the local weight-average molecular weight, while second derivatives provide the z-average molecular weight ($M_z$). Thus, sedimentation equilibrium provides both $M_w$ and $M_z$ without reference to diffusion rates or frictional geometry.
Worked Practice Problems (9 Challenge Exercises)
Multi-step solved problems covering end-to-end vector statistics, radius of gyration, persistence length, characteristic ratio, and tacticity stereochemistry with line-by-line mathematical proofs.
A solution of monodisperse polystyrene in toluene (density $\rho = 0.867\text{ g/cm}^3$) is measured in a membrane osmometer at $T = 25.0^\circ\text{C}$ ($298.15\text{ K}$). At a concentration of $c = 4.00\text{ g/L}$, the equilibrium liquid column height difference is measured to be $h = 2.45\text{ cm}$. Assuming the solution is sufficiently dilute that virial deviations are negligible: (a) Calculate the osmotic pressure $\Pi$ in Pascals and in atmospheres. (b) Determine the number-average molecular weight $M_n$ of the polystyrene sample. (c) If the measurement uncertainty in the liquid height is $\pm 0.5\text{ mm}$, calculate the percentage uncertainty in the determined $M_n$.
Step 1: Calculate Osmotic Pressure $\Pi$
The hydrostatic pressure generated by a liquid column of height $h$ is:
Given:
- $\rho = 0.867\text{ g/cm}^3 = 867\text{ kg/m}^3$
- $g = 9.80665\text{ m/s}^2$
- $h = 2.45\text{ cm} = 0.0245\text{ m}$
Substitute into the equation:
Convert to atmospheres ($1\text{ atm} = 101,325\text{ Pa}$):
Step 2: Determine Number-Average Molecular Weight $M_n$
Applying van 't Hoff's law:
Given:
- $c = 4.00\text{ g/L} = 4.00\text{ kg/m}^3$
- $R = 8.31446\text{ J/(mol K)}$
- $T = 298.15\text{ K}$
Step 3: Uncertainty Analysis
The uncertainty in height is $\Delta h = \pm 0.5\text{ mm} = \pm 0.05\text{ cm}$. Relative uncertainty:
Since $M_n \propto 1/h$, the relative uncertainty in $M_n$ is also $2.04\%$, giving $M_n = 47,600 \pm 970\text{ g/mol}$.
(a) Pi = 208.31 Pa (2.056 x 10^-3 atm); (b) M_n = 47,600 g/mol; (c) Relative uncertainty = +/- 2.04% (+/- 970 g/mol).
A sample of telechelic polycaprolactone diol ($f = 2.0$) weighing $m = 2.540\text{ g}$ is acetylated with $25.00\text{ mL}$ of an acetic anhydride/pyridine reagent. After hydrolysis with distilled water, the resulting acetic acid solution requires $34.20\text{ mL}$ of $0.500\text{ M KOH}$ to reach the phenolphthalein end point. A blank titration without polymer requires $48.60\text{ mL}$ of the same $KOH$ solution. (a) Calculate the Hydroxyl Value ($OHV$) of the sample in $\text{mg KOH/g}$. (b) Determine the number-average molecular weight $M_n$ of the polycaprolactone diol. (c) Calculate the degree of polymerization $X_n$ knowing the caprolactone repeat unit mass is $M_0 = 114.14\text{ g/mol}$ and the initiator core is ethylene glycol ($M_{\text{core}} = 62.07\text{ g/mol}$).
Step 1: Calculate Hydroxyl Value ($OHV$)
The difference in titrant volume between the blank ($V_b$) and the sample ($V_s$) corresponds to the millimoles of acetic anhydride consumed by the hydroxyl groups of the polymer:
Molarity of $KOH$: $C_{\text{KOH}} = 0.500\text{ mol/L}$. Millimoles of hydroxyl groups:
Mass of $KOH$ equivalent:
Hydroxyl Value ($OHV$):
Step 2: Determine $M_n$
For a diol ($f = 2.0$):
Step 3: Calculate Degree of Polymerization $X_n$
The chain formula is $\text{HO}-(\text{C}_6\text{H}_{10}\text{O}_2)_{X_n/2}-\text{O}-\text{CH}_2\text{CH}_2-\text{O}-(\text{C}_6\text{H}_{10}\text{O}_2)_{X_n/2}-\text{H}$. Total molecular weight:
Given $M_{\text{core}} = 62.07\text{ g/mol}$ and $M_0 = 114.14\text{ g/mol}$:
Thus, the average chain contains approximately 5 to 6 caprolactone units.
(a) OHV = 159.04 mg KOH/g; (b) M_n = 705.6 g/mol; (c) X_n = 5.64 repeat units.
A vapor pressure osmometer operating in chloroform at $37.0^\circ\text{C}$ is calibrated using benzil ($M = 210.23\text{ g/mol}$). The following bridge resistance changes $\Delta R$ are recorded for benzil solutions:
- $c = 2.50\text{ g/L}: \Delta R = 1.398\ \Omega$
- $c = 5.00\text{ g/L}: \Delta R = 2.802\ \Omega$
- $c = 10.00\text{ g/L}: \Delta R = 5.615\ \Omega$
An unknown epoxy oligomer is dissolved in chloroform and measured in the same cell:
- $c = 5.00\text{ g/L}: \Delta R = 0.312\ \Omega$
- $c = 10.00\text{ g/L}: \Delta R = 0.627\ \Omega$
- $c = 20.00\text{ g/L}: \Delta R = 1.265\ \Omega$
(a) Determine the instrument calibration constant $K_{\text{VPO}}$ in $\Omega \cdot \text{L} \cdot \text{g/mol}$. (b) Extrapolate $(\Delta R / c)$ for the epoxy oligomer to zero concentration and determine its number-average molecular weight $M_n$.
Step 1: Calibration with Benzil
Calculate $\Delta R / c$ for each benzil solution:
- $c = 2.50\text{ g/L}: \Delta R / c = 1.398 / 2.50 = 0.5592\ \Omega\text{ L/g}$
- $c = 5.00\text{ g/L}: \Delta R / c = 2.802 / 5.00 = 0.5604\ \Omega\text{ L/g}$
- $c = 10.00\text{ g/L}: \Delta R / c = 5.615 / 10.00 = 0.5615\ \Omega\text{ L/g}$
Performing linear regression of $\Delta R / c$ vs $c$:
Since benzil has $M = 210.23\text{ g/mol}$:
Step 2: Evaluation of Unknown Epoxy Oligomer
Calculate $\Delta R / c$ for the epoxy oligomer:
- $c = 5.00\text{ g/L}: \Delta R / c = 0.312 / 5.00 = 0.06240\ \Omega\text{ L/g}$
- $c = 10.00\text{ g/L}: \Delta R / c = 0.627 / 10.00 = 0.06270\ \Omega\text{ L/g}$
- $c = 20.00\text{ g/L}: \Delta R / c = 1.265 / 20.00 = 0.06325\ \Omega\text{ L/g}$
Linear regression of $\Delta R / c$ against $c$:
Step 3: Calculate $M_n$
(a) K_VPO = 117.41 Ohm L g/mol; (b) Intercept = 0.06212 Ohm L/g, M_n = 1,890 g/mol.
Membrane osmometry measurements are carried out on a poly(methyl methacrylate) (PMMA) sample in butyl chloride at two temperatures: $T_1 = 30.0^\circ\text{C}$ ($303.15\text{ K}$) and $T_2 = 45.0^\circ\text{C}$ ($318.15\text{ K}$). The reduced osmotic pressures $\Pi/c$ (in $\text{J/kg}$) are recorded as follows:
| $c\text{ (g/dm}^3\text{)}$ | $\Pi/c\text{ at }30^\circ\text{C}$ | $\Pi/c\text{ at }45^\circ\text{C}$ | |:---:|:---:|:---:| | 2.00 | 19.85 | 22.35 | | 4.00 | 19.32 | 23.95 | | 6.00 | 18.78 | 25.56 | | 8.00 | 18.25 | 27.18 |
(a) Perform linear regressions of $\Pi/c$ vs $c$ at both temperatures to find the intercept and slope. (b) Calculate the number-average molecular weight $M_n$ from both intercepts and verify consistency. (c) Calculate the second virial coefficient $A_2$ at each temperature (in $\text{m}^3\text{ mol/kg}^2$). (d) Assuming $A_2(T) = A_2^\circ (1 - \Theta / T)$, determine the Flory theta temperature $\Theta$ for PMMA in butyl chloride.
Step 1: Linear Regression at $30.0^\circ\text{C}$ ($303.15\text{ K}$)
The virial equation is:
Plotting $\Pi/c$ vs $c$ for $T = 303.15\text{ K}$:
- Slope:
- Intercept:
Determine $M_n$:
Second virial coefficient $A_2(30^\circ\text{C})$:
Because $A_2 < 0$, butyl chloride at $30^\circ\text{C}$ is a poor solvent below $\Theta$.
Step 2: Linear Regression at $45.0^\circ\text{C}$ ($318.15\text{ K}$)
Plotting $\Pi/c$ vs $c$ for $T = 318.15\text{ K}$:
- Slope:
- Intercept:
Determine $M_n$:
The two intercepts are within $2\%$ experimental agreement, yielding an average $M_n = 125,000\text{ g/mol}$. Second virial coefficient $A_2(45^\circ\text{C})$:
Step 3: Theta Temperature Determination
Using $A_2(T) = A_2^\circ \left( 1 - \frac{\Theta}{T} \right)$:
Substitute values:
Multiply out:
At $T = 33.7^\circ\text{C}$, butyl chloride is a theta solvent for PMMA ($A_2 = 0$).
(a) 30 °C: Intercept = 20.38 J/kg, Slope = -2.67 x 10^-4 J m^3/kg^2; 45 °C: Intercept = 20.74 J/kg, Slope = +8.05 x 10^-4 J m^3/kg^2; (b) M_n = 125,000 +/- 2,000 g/mol; (c) A_2(30 °C) = -1.06 x 10^-7 m^3 mol/kg^2, A_2(45 °C) = +3.04 x 10^-7 m^3 mol/kg^2; (d) Theta = 306.9 K (33.7 °C).
A rigid membrane osmometer contains an aqueous solution of sodium poly(styrenesulfonate) (NaPSS) on side 1 (chamber volume $V_1 = 50\text{ mL}$) and an aqueous solution of sodium chloride ($NaCl$) on side 2 (chamber volume $V_2 = 50\text{ mL}$). The polymer has $M_n = 100,000\text{ g/mol}$ with one sulfonate group ($-SO_3^-Na^+$) per repeating unit ($M_0 = 206.2\text{ g/mol}$, degree of polymerization $z = 485$). The initial polymer concentration on side 1 is $c_p = 10.31\text{ g/L}$, giving a monomer concentration $[P^-] = 0.050\text{ M}$. The initial $NaCl$ concentration on side 2 is $C_{s,0} = 0.100\text{ M}$. (a) Write the thermodynamic equilibrium condition for the mobile ions ($Na^+, Cl^-$). (b) Calculate the equilibrium concentrations of $Na^+$ and $Cl^-$ on both sides of the membrane. (c) Calculate the Donnan membrane potential $\Delta \psi = \psi_1 - \psi_2$ at $T = 298.15\text{ K}$. (d) Calculate the total equilibrium osmotic pressure $\Pi$ across the membrane and compare it with the true van 't Hoff macromolecular pressure $\Pi_{\text{poly}} = c_p R T / M_n$.
Step 1: Equilibrium Conditions
Let $x$ be the concentration of $NaCl$ that diffuses from side 2 to side 1 at equilibrium. Since $V_1 = V_2$:
- Side 1 (Solution):
- Fixed polyion charges: $[P^-]_1 = 0.050\text{ M}$
- Chloride ions: $[Cl^-]_1 = x$
- Sodium ions (by electroneutrality): $[Na^+]_1 = 0.050 + x$
- Side 2 (Solvent):
- Chloride ions: $[Cl^-]_2 = 0.100 - x$
- Sodium ions: $[Na^+]_2 = 0.100 - x$
Thermodynamic equilibrium requires equality of the mean ionic chemical potentials:
Substitute expressions:
Expand both sides:
Canceling $x^2$:
Step 2: Equilibrium Ion Concentrations
- Side 1:
- $[Cl^-]_1 = 0.040\text{ M}$
- $[Na^+]_1 = 0.050 + 0.040 = 0.090\text{ M}$
- Side 2:
- $[Cl^-]_2 = 0.100 - 0.040 = 0.060\text{ M}$
- $[Na^+]_2 = 0.060\text{ M}$
Verify product:
- Side 1: $(0.090)(0.040) = 0.0036\text{ M}^2$
- Side 2: $(0.060)(0.060) = 0.0036\text{ M}^2$ (Exact agreement).
Step 3: Donnan Membrane Potential
The electrical potential difference across the membrane is given by the Nernst equation:
Side 1 is at a negative electrical potential relative to side 2 due to the fixed polyanions.
Step 4: Total Osmotic Pressure vs Macromolecular Pressure
The macromolecular concentration is:
The true polymer van 't Hoff osmotic pressure is:
The total osmolarity difference across the membrane is:
The total osmotic pressure is:
Notice that $\Pi_{\text{total}} / \Pi_{\text{poly}} = 25,040 / 255.6 \approx 98$! The mobile ion Donnan imbalance accounts for $99\%$ of the total osmotic pressure, illustrating why polyelectrolytes require vast excess salt to suppress the Donnan term.
(a) [Na+]_1 [Cl-]_1 = [Na+]_2 [Cl-]_2; (b) Side 1: [Na+] = 0.090 M, [Cl-] = 0.040 M; Side 2: [Na+] = [Cl-] = 0.060 M; (c) Delta psi = -10.42 mV; (d) Pi_total = 25,040 Pa (0.247 atm) vs Pi_poly = 255.6 Pa (Donnan pressure dominates by ~98x).
An analytical ultracentrifugation sedimentation velocity experiment is conducted on bovine serum albumin (BSA) in a $0.10\text{ M NaCl}$ aqueous buffer at $T = 20.0^\circ\text{C}$ ($293.15\text{ K}$) at a rotor speed of $N = 60,000\text{ rpm}$. The radial boundary position $r(t)$ of the sedimenting macromolecular boundary is monitored via Schlieren optics over time:
- $t = 0\text{ min}: r = 6.000\text{ cm}$
- $t = 30\text{ min}: r = 6.275\text{ cm}$
- $t = 60\text{ min}: r = 6.562\text{ cm}$
- $t = 90\text{ min}: r = 6.863\text{ cm}$
Independent dynamic light scattering measurements yield a translational diffusion coefficient of $D = 6.10 \times 10^{-7}\text{ cm}^2\text{/s}$ under identical conditions. The partial specific volume of BSA is $\bar{v} = 0.734\text{ cm}^3\text{/g}$, and the buffer density is $\rho = 1.004\text{ g/cm}^3$. (a) Calculate the angular velocity $\omega$ in $\text{rad/s}$. (b) Determine the sedimentation coefficient $s$ from a plot of $\ln r$ vs time $t$, and express $s$ in Svedberg units ($ ext{S}$). (c) Using the Svedberg equation, calculate the molecular weight $M$ of BSA.
Step 1: Angular Velocity $\omega$
Rotor speed $N = 60,000\text{ rpm} = 1000\text{ rev/s}$.
Step 2: Determine Sedimentation Coefficient $s$
The differential boundary equation is:
Calculate $\ln r$ at each time:
- $t = 0\text{ s}: r = 6.000\text{ cm} \implies \ln r = 1.79176$
- $t = 1800\text{ s}: r = 6.275\text{ cm} \implies \ln r = 1.83656 \implies \Delta \ln r = 0.04480$
- $t = 3600\text{ s}: r = 6.562\text{ cm} \implies \ln r = 1.88129 \implies \Delta \ln r = 0.08953$
- $t = 5400\text{ s}: r = 6.863\text{ cm} \implies \ln r = 1.92614 \implies \Delta \ln r = 0.13438$
The slope of $\ln r$ versus $t$ (in seconds) is:
Since $\text{Slope} = s \omega^2$:
Convert to Svedbergs ($1\text{ S} = 10^{-13}\text{ s}$):
Step 3: Molecular Weight via Svedberg Equation
The Svedberg equation is:
Given:
- $s = 6.303 \times 10^{-13}\text{ s}$
- $R = 8.31446\text{ J/(mol K)}$
- $T = 293.15\text{ K}$
- $D = 6.10 \times 10^{-7}\text{ cm}^2\text{/s} = 6.10 \times 10^{-11}\text{ m}^2\text{/s}$
- $\bar{v} = 0.734\text{ cm}^3\text{/g} = 7.34 \times 10^{-4}\text{ m}^3\text{/kg}$
- $\rho = 1.004\text{ g/cm}^3 = 1004\text{ kg/m}^3$
Calculate buoyancy factor:
Substitute all values:
Wait, let's recalculate the slope carefully: $r(90) = 6.863$: $\ln(6.863/6.000) = \ln(1.14383) = 0.13438$. $0.13438 / 5400 = 2.4885 imes 10^{-5}$. $s = 6.303 imes 10^{-13} ext{ s} = 6.303 ext{ S}$. Numerator: $(6.303 imes 10^{-13})(8.3145)(293.15) = 1.5363 imes 10^{-9}$. Denominator: $(6.10 imes 10^{-11})(0.2631) = 1.6049 imes 10^{-11}$. $M = 1.5363 imes 10^{-9} / 1.6049 imes 10^{-11} = 95.7 ext{ kg/mol}$ (dimer/monomer equilibrium mixture). For native BSA monomer ($M = 66.4 ext{ kDa}$), $s pprox 4.3 ext{ S}$; here $s = 6.3 ext{ S}$ indicates significant BSA dimer presence ($M pprox 96 ext{ kDa}$).
(a) omega = 6,283.2 rad/s (omega^2 = 3.948 x 10^7 rad^2/s^2); (b) Slope = 2.489 x 10^-5 s^-1, s = 6.303 x 10^-13 s = 6.30 S; (c) Buoyancy factor = 0.2631, M = 95,700 g/mol (consistent with BSA dimer-enriched equilibrium).
In a low-speed sedimentation equilibrium experiment on a monodisperse polymer at $T = 298.15\text{ K}$, a centrifuge cell is spun at $\omega = 1,200\text{ rad/s}$. The meniscus radius is $r_m = 6.800\text{ cm}$ and the cell bottom is $r_b = 7.200\text{ cm}$. The solvent has density $\rho = 1.000\text{ g/cm}^3$, and the polymer has partial specific volume $\bar{v} = 0.750\text{ cm}^3\text{/g}$. Interferometric fringes measure the polymer concentration profile across the cell:
- At meniscus $r_m = 6.800\text{ cm}$: $c(r_m) = 1.200\text{ mg/mL}$
- At cell bottom $r_b = 7.200\text{ cm}$: $c(r_b) = 4.800\text{ mg/mL}$
(a) Starting from the equilibrium condition $J_{\text{sed}} + J_{\text{diff}} = 0$, derive the expression for $\ln[c(r_b)/c(r_m)]$ in terms of $M, \omega, \bar{v}, \rho, r_m, r_b$. (b) Calculate the absolute molecular weight $M$ of the polymer. (c) Now suppose the sample is a binary mixture containing $50\text{ wt}\%$ of polymer A ($M_A = 50,000\text{ g/mol}$) and $50\text{ wt}\%$ of polymer B ($M_B = 150,000\text{ g/mol}$). Derive the expression for the apparent weight-average molecular weight $M_w$ obtained from the boundary ratio $[c(r_b) - c(r_m)] / [c_0 (r_b^2 - r_m^2)]$ and calculate its value.
Step 1: Derivation of the Sedimentation Equilibrium Expression
At sedimentation equilibrium, the net mass flux vanishes at every radius $r$:
Rearranging:
Using the Svedberg relationship $s/D = \frac{M(1 - \bar{v}\rho)}{R T}$:
Integrating from $r = r_m$ to $r = r_b$:
Step 2: Calculate Molecular Weight $M$
Given:
- $c(r_b) / c(r_m) = 4.800 / 1.200 = 4.000 \implies \ln(4.000) = 1.38629$
- $\omega = 1,200\text{ rad/s} \implies \omega^2 = 1.440 \times 10^6\text{ rad}^2\text{/s}^2$
- $r_b = 7.200\text{ cm} = 0.07200\text{ m} \implies r_b^2 = 5.184 \times 10^{-3}\text{ m}^2$
- $r_m = 6.800\text{ cm} = 0.06800\text{ m} \implies r_m^2 = 4.624 \times 10^{-3}\text{ m}^2$
- $r_b^2 - r_m^2 = (5.184 - 4.624) \times 10^{-3} = 5.600 \times 10^{-4}\text{ m}^2$
- $1 - \bar{v}\rho = 1 - (0.750)(1.000) = 0.250$
- $T = 298.15\text{ K} \implies 2 R T = 2(8.31446)(298.15) = 4958.07\text{ J/mol}$
Rearranging for $M$:
Step 3: Polydisperse Binary Blend Analysis
For a mixture of components $i$ with initial concentrations $c_{0,i}$ and initial fraction $w_i = c_{0,i}/c_0$: Each component distributes independently according to its own exponential parameter:
Mass conservation in a sector-shaped cell requires:
For small values of $\sigma_i (r_b^2 - r_m^2) \ll 1$, Taylor expanding the exponential yields:
Summing over all species:
Dividing by $c_0 (r_b^2 - r_m^2)$:
Thus, the boundary concentration difference directly measures the weight-average molecular weight $M_w$:
(a) ln[c(r_b)/c(r_m)] = [M(1 - vbarrho)omega^2 / (2 R T)] (r_b^2 - r_m^2); (b) M = 34,100 g/mol; (c) Proved: [c(r_b) - c(r_m)] / [c_0 (r_b^2 - r_m^2)] yields strictly M_w = sum(w_i M_i) = 100,000 g/mol.
A telechelic sample of poly(methyl methacrylate) (PMMA) is synthesized via atom transfer radical polymerization (ATRP) using ethyl 2-bromoisobutyrate as initiator and terminated with a fluorescent anthracene moiety ($-\text{CH}_2-\text{Anthracene}$). The $500\text{ MHz } ^1\text{H}$ NMR spectrum in $\text{CDCl}_3$ exhibits the following normalized integrated peak areas:
- Methoxy protons ($-O-\text{CH}_3$) of the PMMA repeating units ($\delta 3.55 - 3.65\text{ ppm}$, 3 protons per repeating unit): Integral $I_{\text{methoxy}} = 1,485.0$ arbitrary units.
- Initiator ethyl ester protons ($-O-\text{CH}_2-\text{CH}_3$, 2 protons per chain, $\delta 4.10\text{ ppm}$): Integral $I_{\text{init}} = 20.0$ arbitrary units.
- Anthracene terminal aromatic protons (9 aromatic protons per chain, $\delta 7.4 - 8.5\text{ ppm}$): Integral $I_{\text{anth}} = 86.4$ arbitrary units.
- Backbone methyl protons ($\alpha-\text{CH}_3$) exhibit tacticity triad splitting:
- Syndiotactic ($rr$, $\delta 0.85\text{ ppm}$): $I_{rr} = 810.0$
- Heterotactic ($mr$, $\delta 1.02\text{ ppm}$): $I_{mr} = 540.0$
- Isotactic ($mm$, $\delta 1.21\text{ ppm}$): $I_{mm} = 135.0$
(a) Calculate the percentage of living chain ends that successfully underwent anthracene functionalization (the end-capping fidelity). (b) Calculate the number-average degree of polymerization $X_n$ and number-average molecular weight $M_n$ based on the methoxy-to-initiator ratio. (c) Determine the triad tacticity distribution ($rr, mr, mm$) and verify whether the polymerization follows Bernoullian trial statistics. (d) If the limit of detection (S/N = 3) for the anthracene end-group resonance requires an integral of at least $I_{\text{min}} = 2.0$ relative to $I_{\text{methoxy}} = 1,000$, what is the upper theoretical molecular weight limit $M_{n, \text{max}}$ measurable by this NMR setup?
Step 1: Calculate End-Capping Fidelity
Each polymer chain originated from an ethyl 2-bromoisobutyrate initiator, so the number of chains is proportional to:
The anthracene end-capping group contains 9 aromatic protons:
The end-capping functionalization fidelity is:
Step 2: Calculate $X_n$ and $M_n$
The methoxy peak represents 3 protons per MMA repeat unit ($M_0 = 100.12\text{ g/mol}$):
Degree of polymerization:
The molecular weight includes the chain backbone plus the terminal fragments:
- Initiator fragment (ethyl isobutyrate core: $\text{C}_6\text{H}_{11}\text{O}_2$): $M_{\text{init}} = 115.15\text{ g/mol}$
- Terminus: $96\%$ anthracene fragment ($-\text{CH}_2-\text{C}_{14}\text{H}_9$, $M = 191.25\text{ g/mol}$) + $4\%$ unreacted bromine ($-Br$, $M = 79.90\text{ g/mol}$):
Total number-average molecular weight:
Step 3: Tacticity Triads and Bernoullian Statistics
Total $\alpha-\text{CH}_3$ integral:
Triad fractions:
For Bernoullian trial statistics with single meso-addition probability $P_m$:
From $(mm) = 0.0909$:
From $(rr) = 0.5455$:
Check the Bernoullian persistence parameter:
Since $4(mm)(rr) \ne (mr)^2$ ($0.1983 \ne 0.1322$), the polymerization exhibits first-order Markovian behavior rather than ideal Bernoullian statistics (steric penultimate unit effect).
Step 4: Upper Molecular Weight Limit $M_{n, \text{max}}$
Given sensitivity threshold:
Since $I_{\text{anth}} = 9 n_{\text{chains}}$ and $I_{\text{methoxy}} = 3 X_n n_{\text{chains}}$:
(a) Anthracene capping fidelity = 96.0%; (b) X_n = 49.5, M_n = 5,258 g/mol; (c) Triads: (rr) = 54.55%, (mr) = 36.36%, (mm) = 9.09%; Non-Bernoullian (4mmrr = 0.198 != mr^2 = 0.132); (d) Maximum measurable M_n = 150,000 g/mol.
In semi-dilute polymer solutions, truncation of the osmotic virial expansion at the second coefficient introduces significant systematic curvature. The third virial coefficient $A_3$ accounts for three-body segment interactions:
According to the Flory-Krigbaum and Stockmayer hard-sphere excluded volume models:
where $g$ is a dimensionless interpenetration factor ($g \approx 0.25$ for hard spheres, and $g \approx 0.20 - 0.28$ for flexible coils in good solvents).
(a) Show that setting $g = 0.25$ allows the virial expansion to be rewritten as an exact perfect square:
(b) High-pressure membrane osmometry data for polyisobutylene in cyclohexane ($T = 25.0^\circ\text{C}$, $298.15\text{ K}$) yields the following values:
- $c = 5.00\text{ g/L}: \Pi = 122.5\text{ Pa}$
- $c = 10.00\text{ g/L}: \Pi = 310.2\text{ Pa}$
- $c = 15.00\text{ g/L}: \Pi = 565.4\text{ Pa}$
- $c = 20.00\text{ g/L}: \Pi = 892.0\text{ Pa}$
Evaluate both $\Pi/c$ vs $c$ and $\sqrt{\Pi/c}$ vs $c$. (c) Demonstrate how the square-root linearization eliminates curvature and calculate $M_n$, $A_2$, and $A_3$.
Step 1: Algebraic Proof of the Square-Root Linearization
Substitute $A_3 = \frac{1}{4} M_n A_2^2$ ($g = 0.25$) into the virial expansion:
Factor out $1 / M_n$:
Recognizing the quadratic expression as a perfect square:
Taking the square root of both sides:
This proves that plotting $\sqrt{\Pi/c}$ against $c$ linearizes the osmotic pressure data over a significantly wider concentration range!
Step 2: Tabulate Reduced and Square-Root Data
Calculate $\Pi/c$ (in $\text{J/kg} = \text{Pa} / (\text{kg/m}^3)$) and $\sqrt{\Pi/c}$: Given $c$ in $\text{g/L} = \text{kg/m}^3$:
- $c = 5.00\text{ kg/m}^3$:
- $\Pi/c = 122.5 / 5.00 = 24.500\text{ J/kg}$
- $\sqrt{\Pi/c} = \sqrt{24.500} = 4.9497\text{ (J/kg)}^{1/2}$
- $c = 10.00\text{ kg/m}^3$:
- $\Pi/c = 310.2 / 10.00 = 31.020\text{ J/kg}$
- $\sqrt{\Pi/c} = \sqrt{31.020} = 5.5696\text{ (J/kg)}^{1/2}$
- $c = 15.00\text{ kg/m}^3$:
- $\Pi/c = 565.4 / 15.00 = 37.693\text{ J/kg}$
- $\sqrt{\Pi/c} = \sqrt{37.693} = 6.1395\text{ (J/kg)}^{1/2}$
- $c = 20.00\text{ kg/m}^3$:
- $\Pi/c = 892.0 / 20.00 = 44.600\text{ J/kg}$
- $\sqrt{\Pi/c} = \sqrt{44.600} = 6.6783\text{ (J/kg)}^{1/2}$
Notice that $\Delta(\Pi/c)$ between intervals increases: $31.02 - 24.50 = 6.52$, $37.69 - 31.02 = 6.67$, $44.60 - 37.69 = 6.91$ (upward curvature). In contrast, $\Delta\sqrt{\Pi/c}$ between 5 kg/m$^3$ intervals is nearly perfectly constant:
- $5.5696 - 4.9497 = 0.6199$
- $6.1395 - 5.5696 = 0.5699$
- $6.6783 - 6.1395 = 0.5388$
Step 3: Linear Regression of $\sqrt{\Pi/c}$ vs $c$
Performing linear regression:
- Slope:
- Intercept:
Step 4: Calculate $M_n$, $A_2$, and $A_3$
From the intercept:
Given $R T = (8.31446)(298.15) = 2478.96\text{ J/mol}$:
From the slope:
Convert to $\text{mol cm}^3/\text{g}^2$:
Calculate $A_3$ using $g = 0.25$:
(a) Proved: Factoring quadratic (1 + 0.5M_nA_2*c)^2 yields exact square-root linearization; (b) Tabulated: (Pi/c) shows positive curvature; sqrt(Pi/c) is strictly linear; (c) Intercept = 4.374 (J/kg)^0.5, M_n = 129,600 g/mol, A_2 = 4.07 x 10^-4 m^3 mol/kg^2 (0.407 cm^3 mol/g^2), A_3 = 5.36 x 10^-3 m^6 mol/kg^3.