Intrinsic viscosity, viscosity-average molecular weight and Huggins constant for PMMA: 23-Chem-B8 Polymer Engineering, December 2015, Q3

Question 3: Intrinsic viscosity, viscosity-average molecular weight and Huggins constant for PMMA (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Relative viscosities $\eta_{rel}$ at four concentrations $c$ (in g/100 mL = g/dL). Mark–Houwink–Sakurada relation $[\eta]=5.83\times10^{-5}M_v^{0.72}$ with $[\eta]$ in dL/g.

Find. $[\eta]$, $M_v$, and the Huggins constant $k'$.

Huggins plot: reduced viscosity $\eta_{sp}/c$ versus $c$. The least-squares line extrapolates to the intercept $[\eta]=0.579$ dL/g at $c\to0$; its slope gives $k'$.

Approach. Convert each $\eta_{rel}$ to specific and reduced viscosity, fit the Huggins line $\eta_{sp}/c=[\eta]+k'[\eta]^2c$ to obtain $[\eta]$ (intercept) and $k'$ (from the slope), then invert Mark–Houwink for $M_v$.

Reduced viscosities. With $\eta_{sp}=\eta_{rel}-1$, the reduced viscosities $\eta_{sp}/c$ are $$\frac{0.170}{0.275}=0.618,\quad \frac{0.215}{0.344}=0.625,\quad \frac{0.629}{0.896}=0.702,\quad \frac{0.892}{1.199}=0.744\ \ \text{dL/g}$$ (concentrations already in g/dL, so no unit conversion is needed).
Huggins extrapolation. A least-squares fit of $\eta_{sp}/c$ against $c$ gives slope $0.1373$ and intercept $$[\eta]=\boxed{0.579\ \text{dL/g}}$$ This is the intrinsic viscosity — the reduced viscosity extrapolated to infinite dilution.
Huggins constant. The Huggins slope equals $k'[\eta]^2$, so $$k'=\frac{0.1373}{[\eta]^2}=\frac{0.1373}{(0.579)^2}=\boxed{0.41}$$ a value typical of a flexible coil in a moderately good solvent (acetone/PMMA).
Viscosity-average molecular weight. Inverting $[\eta]=5.83\times10^{-5}M_v^{0.72}$, $$M_v=\left(\frac{[\eta]}{5.83\times10^{-5}}\right)^{1/0.72} =\left(\frac{0.579}{5.83\times10^{-5}}\right)^{1.389}=\boxed{3.6\times10^{5}\ \text{g/mol}}$$

Quantity	Value
Intrinsic viscosity $[\eta]$	0.579 dL/g
Huggins constant $k'$	0.41
Viscosity-average molecular weight $M_v$	$\approx3.6\times10^{5}$ g/mol

Topic: Dilute-solution viscometry and the Huggins/Mark–Houwink analysis for $M_v$.

Key relations: $\eta_{sp}=\eta_{rel}-1$; Huggins $\eta_{sp}/c=[\eta]+k'[\eta]^2c$; Mark–Houwink–Sakurada $[\eta]=KM_v^{\alpha}$.

Why this works: Intrinsic viscosity measures the hydrodynamic volume of a single isolated coil, so it must be read at infinite dilution where chains do not overlap. The Huggins equation is the leading (linear) term of the concentration expansion of reduced viscosity; its intercept is $[\eta]$ and its slope carries the pairwise hydrodynamic interaction through $k'$. $M_v$ lies between $M_n$ and $M_w$, close to $M_w$, because the exponent $\alpha=0.72$ is near unity.

Common pitfall: Mixing concentration units. Because $c$ is in g/dL, $[\eta]$ comes out in dL/g, which is exactly the unit for which $K=5.83\times10^{-5}$ applies; converting $c$ to g/mL would throw $M_v$ off by $100^{1/\alpha}$. Also, never take a single $\eta_{sp}/c$ point as $[\eta]$ — it is the extrapolated intercept.

Source: Rudin & Choi, Elements of Polymer Science 3rd ed., Ch. 3 (solution viscosity); Sperling, Physical Polymer Science 4th ed., §3.3.

Quantity	Value
Intrinsic viscosity \([\eta]\)	0.579 dL/g
Huggins constant \(k'\)	0.41
Viscosity-average molecular weight \(M_v\)	\(\approx3.6\times10^{5}\) g/mol

Question 3 of 6: Intrinsic viscosity, viscosity-average molecular weight and Huggins constant for PMMA

Question 3: Intrinsic viscosity, viscosity-average molecular weight and Huggins constant for PMMA (25 marks)