User:Alicia Rasines Mazo/Notebook/CHEM-581 Experimental Chemistry I/2014/10/03

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Tasks for October 3

  • To begin new film synthesis of PVA-polymer and 10% PVA-NaMT clay film
  • To prepare ionic modified clays for choline chloride and THP bromide
  • To measure leftover fluorescence from solutions with films

New film synthesis

1.0011 g of 22,000 MW PVA were added to 7 mL of distilled water <br.> 1,0043 f of 22,00 MW PVA and 0.1104 h of NAMt were added to 7 mL distilled water <br.> Stirred gently at approx. 80ºC until all crystals were dissolved <br.> Added 0.5 mL 0.8%wt glutaraldehyde <br.> Stirred gently for 3 min to ensure homegenity <br.> Poured into teflon pan

Preparation of Ionic Modified Clays

NaMT cation exchange Assuming 92 miliequivalents (meq) of Na+ per 100g of montmorillonite.<br.>

  • Tributylhexadecylphhosphonium bromide exchange
    1. Add 1.12g of Sodium Montmorillonite (NaMT) to a scintillation vial containing 20mL of 50:50 HPLC grade water:ethanol.
      • (1.12 g NaMT)×(92×10-3 eq/100 g NaMT)×(507.65 g/mol tributylhexadecylphhosphonium bromide)=0.523 g tributylhexadecylphosphonium bromide
    2. Add 0.523g of tributylhexadecylphosphonium bromide to exchange 100% of the Na from the clay
    3. Cap the vial
    4. Stir for 1 week
    5. Vacuum filter using 0.2uM nylon filter paper
    6. Dry in oven at 90C overnight
    7. Grind with a mortar and pestle
    8. Store in a dissecator

Choline chloride exchange

    1. Add 1.12g of Sodium Montmorillonite (NaMT) to a scintillation vial containing 20mL of 50:50 HPLC grade water:ethanol.
    2. (1.12 g NaMT)×(92×10-3 eq/100 g NaMT)×(139.62 g/mol choline chloride)=0.144 g choline chloride
    3. Add 0.144 g of choline chloride to exchange 100% of the Na from the clay
    4. Cap the vial
    5. Stir for 1 week
    6. Vacuum filter using 0.2uM nylon filter paper
    7. Dry in oven at 90C overnight
    8. Grind with a mortar and pestle
    9. Store in a dessicator

Dilutions and Fluoresence Measurements of R6G Solutions

  • PVA film
    • Assuming original concentration was 20μM, it was diluted to 0.1μM (50μL in 10mL)
    • Assuming original concentration was 10μM, it was diluted to 0.1μM (50μL in 10mL)
    • Assuming original concentration was 1μM, it was diluted to 0.5μM (5mL in 10mL)
  • PVA Clay film
    • Assuming original concentration was 20μM, it was diluted to 0.1μM (50μL in 10mL)
    • Assuming original concentration was 10μM, it was diluted to 0.25μM (250μL in 10mL)
    • Assuming original concentration was 1μM, it was diluted to 0.5μM (5mL in 10mL)

UV-Vis Data

<br.> Using calibration curved from Oct.1 at 521 nm:

  • y=0.0464x+0.006
    • Diluted '1μM' PVA clay solution 0.024 absorbance:
      • x=0.3879μM
      • Since it was diluted two-fold, leftover concentration of supposed 1μM R6G solution after removing PVA-Clay film was 0.776μM.
    • Diluted '1μM' PVA solution with 0.031 absorbance:
      • x=0.53879μM
      • Since it was diluted two-fold, leftover concentration of supposed 1μM R6G solution after removing PVA film was 1.076μM.
    • Diluted '10μM' PVA-Clay solution with 0.013 absorbance:
      • Since it was diluted 100-fold, leftover concentration of dye solution was 15.09μM.
    • Diluted '10μM' PVA solution with 0.031 absorbance:
      • Since it was diluted 100-fold, leftover concentration of dye solution was 53.88μM.
    • Diluted '20μM' PVA-Clay solution with 0.015 absorbance:
      • Since it was diluted 200-fold, leftover concentration of dye solution was 38.79μM.
    • Diluted '20μM' PVA solution with 0.065 absorbance:
      • Since it was diluted 200-fold, leftover concentration of dye solution was 254.3μM.

Note: when the solutions were prepared on Sept. 24, the stock concentration of R6G dye was incorrectly determined to be 159μM. Hence why our final concentration values are higher than the initial values. The initial stock concentration used on that date to make the soak solutions is to be determined next week

Fluorescence data

Using calibration curved from Oct.1:

  • y=41351x−2198.8
    • Diluted '1μM' PVA clay solution had an integrated peak intensity of 14411.41:
      • x=0.4017μM
      • Since it was diluted two-fold, leftover concentration of supposed 1μM R6G solution after removing PVA-Clay film was 0.8034μM.
    • Diluted '1μM' PVA solution with integrated peak intensity of 29332.226:
      • x=0.7625μM
      • Since it was diluted two-fold, leftover concentration of supposed 1μM R6G solution after removing PVA film was 1.525μM.
    • Diluted '10μM' PVA-Clay solution with integrated intensity=4091.6224:
      • Since it was diluted 100-fold, leftover concentration of dye solution was 15.21μM.
    • Diluted '10μM' PVA solution with integrated intensity=14043.666:
      • Since it was diluted 100-fold, leftover concentration of dye solution was 39.28μM.
    • Diluted '20μM' PVA-Clay solution with integrated intensity=2000.3547:
      • Since it was diluted 200-fold, leftover concentration of dye solution was 20.31μM.
    • Diluted '20μM' PVA solution with integrated intensity=44923.8:
      • Since it was diluted 200-fold, leftover concentration of dye solution was 227.9μM.