If samples (lignosulfonates or precipitated lignins) contain high content of ash, does the ash have a negative effect on the determination of UV spectroscopy? Are more purified samples are better?
On the other hand, the ash that is in addition to finding reflects the purity and amount of inorganic chemistry (Ca, Na, ..) which is incorporated in the compound (example: lignosulfonates C20H24CaO10S2).
Purification of samples results in the changes in the content of OH groups.
Example of Samples and content of ash:
Vanisperse 34.2 %
Boresperse N 23.3%
Orzan 10.8%
Precipiataed Kraft lignin 3.9%
Precipitated lignin II. 0.4%
UV spectroscopy
After being dried overnight at 80°C, precisely weighted amount (about 5 mg) of the lignin was dissolved in 5 mL of dioxane and 5 mL of 0.2 M NaOH. Some of the solutions were not quite clear and these were filtered using a 0.45 m PVDF membrane filter. From each lignin solution, 2 mL was further diluted to 25 mL using either a pH = 6 buffer solution (citrate — NaOH, Merck), and 0.2 M NaOH solution.19 This gave each solution a final concentration of lignin of about 0.08 g/L. UV-Vis spectra were recorded on a CECIL spectrophotometer in the absorption region 200 - 450 nm, scan speed 5 nm/s and resolution 1 nm. Lignin solution with the pH = 6 was used as a reference and the alkaline solutions measured against it. From the difference spectra, the absorbance values at 300 and 350 nm, measured against the solution containing 0.2 M NaOH were recorded. According to an original work of Gartner et al.six structural types of phenolic structures exist (Fig. 1). Maxima at 300 nm and 350-360 nm are assigned to unconjugated phenolic structures (I and III) and that at 350-370 nm to conjugated structures (II and IV). According to Zakis et al. the maximum at 360 nm is attributed only to IIa and IVa types of phenolic structures in lignin.
a) Non-conjugated phenolic structures (I + III)
OH(I + III) = {(0.250 x A300 nm(NaOH) + 0.0595 xA350 nm(NaOH)} x1/(cxd) (eq. 1)
b) Conjugated phenolic structures (II + IV)
OH(II + IV) = {0.0476 xA350 nm(NaOH)} x 1/(cxd) (eq. 2)
c) Total amount of phenolic hydroxyl groups
OH(I + II + III + IV) = {0.250 x A300 nm(NaOH) + 0.107 x A350 nm (NaOH)} x 1/(cxd) (eq. 3)
where Aλ – absorbance at a given wavelength divided by the corresponding molar absorptivity, c – mass concentration in g/L; d – pathlength through the sample in cm.