Mark-Houwink Equation: Polymer Molar Mass
The Mark-Houwink equation, often called the Mark-Houwink-Sakurada equation, connects the dilute solution viscosity of a macromolecule directly to its molecular mass. In competitive postgraduate examinations such as CSIR NET, GATE, IIT-JAM, and CUET-PG, polymer chemistry is an essential scoring domain. Questions regularly test the determination of the viscosity-average molar mass, the interpretation of the exponent parameter across different solvent regimes, and graphical extrapolation of relative, specific, and reduced viscosities. Mastering the mathematical progression from efflux time measurements to final molecular weight allows students to tackle both multi-part objective problems and numerical response questions with speed and precision.
Physical Principles and the Dilute Solution Viscosity Concept
When high molecular weight polymer chains dissolve in a liquid solvent, the overall resistance to fluid flow increases substantially. As a macromolecule moves through the solvent streamlines, it creates hydrodynamic drag that dissipates mechanical energy. At very high dilutions, polymer coils are separated by vast expanses of pure solvent. Under these conditions, intermolecular chain entanglements vanish entirely, ensuring that any increase in viscosity arises solely from isolated solute-solvent friction.
Experimental measurements begin by recording the flow time of pure solvent, denoted as t₀, through a capillary viscometer such as an Ostwald or Ubbelohde viscometer. The flow time of the polymer solution at concentration c is recorded as t. The ratio of solution viscosity to pure solvent viscosity gives the relative viscosity η_rel. The fractional increase in viscosity above that of the neat solvent is defined as the specific viscosity η_sp. When specific viscosity is divided by the mass concentration c, it yields the reduced viscosity, while the natural logarithm of relative viscosity divided by concentration gives the inherent viscosity. Extrapolating both quantities to zero concentration yields the intrinsic viscosity [η], which carries dimensions of reciprocal concentration.
Primary Mark-Houwink Relationship:
[η] = K · (M_v)^a
Logarithmic Form for Parameter Evaluation:
ln([η]) = ln(K) + a · ln(M_v) or log₁₀([η]) = log₁₀(K) + a · log₁₀(M_v)
Variables, Units, and Physical Dimensions:
• [η] = Intrinsic viscosity, expressed in mL/g or dL/g (dimensions: [M⁻¹ L³]).
• M_v = Viscosity-average molar mass of the polymer sample, in g/mol.
• K = Mark-Houwink empirical constant, characteristic of the polymer-solvent-temperature system (units depend on a, typically mL/g or dL/g).
• a = Mark-Houwink exponent reflecting chain conformation and solvent quality (dimensionless number, typically ranging from 0.5 to 0.8 for flexible coils).
Chain Conformation and the Significance of Exponent a
The exponent a in the Mark-Houwink relationship provides direct insight into the conformational shape of the polymer chain in the given solvent medium:
1. Spherical Globules (a = 0): For compact, non-draining hard spheres described by the Einstein viscosity law, intrinsic viscosity is independent of molecular weight, yielding an exponent of zero.
2. Theta Solvents (a = 0.5): In a Flory theta solvent at the exact theta temperature, polymer segment-segment attractions precisely balance segment-solvent interactions. The unperturbed random coil obeys ideal Gaussian chain statistics, and the hydrodynamic volume scales with the square root of molar mass.
3. Thermodynamically Good Solvents (a = 0.65 to 0.80): Favourable solvent-polymer contacts lead to chain swelling driven by excluded volume effects. The coil expands into an open conformation, causing the intrinsic viscosity to scale with a higher power of molecular mass.
4. Rigid Rods and Helices (a = 1.0 to 2.0): Extended, stiff rod-like macromolecules exhibit steep increases in hydrodynamic resistance as the rod length increases, driving the exponent toward or above unity.
Exam details
- Source: practice problem written for this article (not a past-paper question)
- Relevant exams: CSIR NET Chemical Sciences, GATE Chemistry, IIT-JAM Chemistry, CUET-PG
- Topic: Polymer Chemistry and Solution Viscometry
- Difficulty: Advanced undergraduate to postgraduate
Question: A solution of monodisperse polystyrene in toluene at 25 °C has an intrinsic viscosity of 80.0 mL/g. The Mark-Houwink parameters for this system are K = 0.0160 mL/g and a = 0.700. Calculate the viscosity-average molar mass of the polystyrene sample.
Solution:
Step 1: Write down the governing equation and identify the given experimental data:
[η] = K · (M_v)^a
[η] = 80.0 mL/g
K = 0.0160 mL/g
a = 0.700
Step 2: Rearrange the formula to isolate the molar mass term (M_v)^a:
(M_v)^a = [η] / K
(M_v)^0.700 = 80.0 / 0.0160
Step 3: Perform the exact division:
80.0 / 0.0160 = 80.0 / (1.60 × 10⁻²) = 5000
Step 4: Solve for M_v by raising both sides to the reciprocal power 1/0.700 = 10/7 ≈ 1.42857:
M_v = (5000)^(1 / 0.700) = (5000)^(1.428571)
Step 5: Apply logarithms to evaluate the exponentiation step by step:
log₁₀(5000) = log₁₀(5 × 10³) = 3 + log₁₀(5) = 3 + 0.698970 = 3.698970
log₁₀(M_v) = 1.428571 × 3.698970 = 5.284243
Step 6: Take the inverse logarithm (antilogarithm):
M_v = 10^(5.284243) = 10^(0.284243) × 10⁵
Since 10^(0.284243) ≈ 1.92415, we obtain:
M_v = 1.92415 × 10⁵ g/mol = 192415 g/mol ≈ 1.92 × 10⁵ g/mol
Step 7: Verification by the second route (substituting M_v back into [η] = K · M_v^a):
[η] = 0.0160 × (192415)^0.700
ln(192415) = 12.1674
0.700 × 12.1674 = 8.51718
exp(8.51718) = 5000.0
[η] = 0.0160 × 5000.0 = 80.0 mL/g. The result matches perfectly.
Answer: 1.92 × 10⁵ g/mol (or 192400 g/mol)
Question: Two poly(methyl methacrylate) samples, Sample A and Sample B, dissolved in acetone have viscosity-average molar masses of 64000 g/mol and 216000 g/mol, respectively. If the intrinsic viscosity of Sample A is 32.0 mL/g, and the Mark-Houwink exponent a is 0.600, find the intrinsic viscosity of Sample B in the same solvent at the same temperature.
Solution:
Step 1: State the Mark-Houwink relationship for both samples in identical conditions:
[η]_A = K · (M_A)^a and [η]_B = K · (M_B)^a
Step 2: Formulate the ratio [η]_B / [η]_A to eliminate the constant K:
[η]_B / [η]_A = (M_B / M_A)^a
Step 3: Insert the given values into the ratio:
M_B / M_A = 216000 / 64000 = 216 / 64 = 27 / 8 = 3.375
Step 4: Notice the fractional power simplification: 27 / 8 = (3/2)³ = (1.5)³
Given a = 0.600 = 3/5:
(M_B / M_A)^a = (27 / 8)^(3/5) = ((3/2)³)^(3/5) = (1.5)^(9/5) = (1.5)^1.8
Step 5: Compute using base-10 logarithms:
log₁₀(3.375) = 0.528274
a × log₁₀(3.375) = 0.600 × 0.528274 = 0.316964
10^(0.316964) = 2.07474
Step 6: Multiply by [η]_A to calculate [η]_B:
[η]_B = 32.0 mL/g × 2.07474 = 66.39 mL/g
Step 7: Verification by calculating K independently:
K = [η]_A / (M_A)^0.600 = 32.0 / (64000)^0.600
(64000)^0.600 = (6.4 × 10⁴)^0.600 = (10^4.80618)^0.600 = 10^2.88371 = 765.08
K = 32.0 / 765.08 = 0.0418257 mL/g
[η]_B = 0.0418257 × (216000)^0.600
(216000)^0.600 = (10^5.33445)^0.600 = 10^3.20067 = 1587.35
[η]_B = 0.0418257 × 1587.35 = 66.39 mL/g. Both routes converge.
Answer: 66.4 mL/g
Question: In a dilute solution viscometry experiment on poly(vinyl chloride) in cyclohexanone at 30 °C, the intrinsic viscosity values obtained for two fractions with molar masses of 25000 g/mol and 160000 g/mol are 20.0 mL/g and 80.0 mL/g, respectively. Determine the Mark-Houwink constants K and a for this polymer-solvent pair.
Solution:
Step 1: Write the logarithmic equations for the two fractions:
Fraction 1: ln(20.0) = ln(K) + a · ln(25000)
Fraction 2: ln(80.0) = ln(K) + a · ln(160000)
Step 2: Subtract equation 1 from equation 2 to isolate a:
ln(80.0) - ln(20.0) = a · [ln(160000) - ln(25000)]
ln(80.0 / 20.0) = a · ln(160000 / 25000)
Step 3: Simplify the numerical quotients inside the natural logarithms:
80.0 / 20.0 = 4.000
160000 / 25000 = 160 / 25 = 6.400
Step 4: Compute the logarithms and solve for exponent a:
ln(4.000) = 1.386294
ln(6.400) = 1.856298
a = 1.386294 / 1.856298 = 0.746806 ≈ 0.747
Step 5: Solve for ln(K) and K using Fraction 1 data:
ln(K) = ln(20.0) - a · ln(25000)
ln(20.0) = 2.995732
ln(25000) = 10.126631
a · ln(25000) = 0.746806 × 10.126631 = 7.562629
ln(K) = 2.995732 - 7.562629 = -4.566897
K = exp(-4.566897) = 0.010390 mL/g ≈ 1.04 × 10⁻² mL/g
Step 6: Verify using Fraction 2:
ln(K) = ln(80.0) - 0.746806 × ln(160000) = 4.382027 - 0.746806 × 11.982929 = 4.382027 - 8.948924 = -4.566897
K = exp(-4.566897) = 0.0104 mL/g. Both equations give identical constants.
Answer: a = 0.747, K = 1.04 × 10⁻² mL/g
Common Student Mistakes
1. Unit Inconsistency in Intrinsic Viscosity: Intrinsic viscosity is reported in different units across textbooks, commonly mL/g, dL/g, or cm³/g (note: 1 dL/g = 100 mL/g = 100 cm³/g). Using a value of K calibrated for dL/g alongside an intrinsic viscosity measured in mL/g introduces an error factor of 100 or more into the calculated molecular weight.
2. Confusing Viscosity Averages with Number or Weight Averages: The viscosity-average molar mass M_v lies strictly between the number-average molar mass M_n and the weight-average molar mass M_w (M_n < M_v ≤ M_w for polydisperse samples when 0 < a < 1). Only for a perfectly monodisperse sample where polydispersity index equals unity do M_n, M_v, and M_w become equal.
3. Extrapolation Errors in Determining [η]: Students often equate reduced viscosity η_sp/c at a finite single concentration to the intrinsic viscosity [η]. Intrinsic viscosity requires extrapolation to zero concentration using the Huggins equation (η_sp/c = [η] + k_H [η]² c) or the Kraemer equation (ln(η_rel)/c = [η] - k_K [η]² c).
4. Algebraic Inversion of Exponents: In solving for M_v from (M_v)^a = [η]/K, examinees frequently multiply by a rather than raising to the power 1/a, leading to dimensional failure.
| Target Examination | Question Format and Assessment Scope | Key Focus Concepts |
|---|---|---|
| CSIR NET Chemical Sciences | Objective questions in Part B and advanced numerical problems in Part C | Evaluation of M_v, Huggins/Kraemer plots, conformational scaling regimes for exponent a |
| GATE Chemistry | Numerical Answer Type (NAT) and multiple-choice questions | Exact computation of molar mass from intrinsic viscosity, parameter extraction via log-log plots |
| IIT-JAM Chemistry | Multiple-choice and numerical answer questions in physical chemistry | Viscosity definitions (relative, specific, reduced), basic ratio calculations of molecular weight |
| CUET-PG Chemistry | Single-correct multiple-choice questions | Qualitative meaning of theta condition (a = 0.5), order of molar mass averages (M_n < M_v < M_w) |
Reaction mechanisms stick only with practice: attempt previous-year CSIR-NET questions and a timed quiz.
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