Wednesday, December 11, 2019

Di-block copolymer analysis: GPC

Gel Permeation Chromatography is a well known method of measuring molecular weight distribution of polymers. For di-block copolymers, a proper calibration curve can be constructed by using Mark-Houwink parameters of homopolymers (see here). Usually, di-block copolymers are obtained by synthesis one block first then "grow" the other block from the end of preceding block. The molecular weight distributions of final di-block copolymer and preceding block are accessible by GPC, the molecular weight distribution of the 2nd block, however, is hard to obtain.

Here I present a simple method to calculate the molecular weight distribution of 2nd block by GPC data of preceding block and final di-block copolymer. Assuming that we already have the molecular weight distribution of the preceding block, $P_1(n)$, it is generally that the length of the 2nd block "grow" from the depends on the preceding chain length $n$, therefore, the pdf of the 2 blocks should be a joint pdf $Q(n, m)=P_1(n)P_2(m, n)$, and the pdf of the final di-block copolymer is straightforward:
$$ P(x)=\int Q(n, x-n) \mathrm{d}n $$
However, it is reasonable to assume that the $P_2(m, n)$ can be expressed as some $f(m)g(n)$ form by dropping some higher order correlations between $m,n$; hence, the $g(n)$ can be considered as a "fixing parameter" on the distribution of $P_1(n)$, the calculation of $P(x)$ is therefore a convolution. The simplest case is $g(n)=1$, which means the $P_1(n)$ and $P_2(m)$ are independent, the growth of 2nd block on the end of the 1st block is not effected by the length of the preceding block; or $g(n)\sim n^{-1}$ by considering the diffusion of the preceding block: small preceding chains tend to "grow" more 2nd block.

The evaluation of pdf of the 2nd block follows 3 steps:

  1. Determine the range of molecular weight: $(x_{min}-n_{max}, x_{max}-n_{min})$, where $x$ is the chain length of the di-block copolymer and $n$ is the length of the preceding block;
  2. Interpolate the distributions obtained from GPC into equally spaced molecular weight, negative part is simply 0;
  3. Deconvolute $P(x)$ with $P_1(n)$.

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