User:Tamanika Tinsley/Notebook/Chem 581 Biomaterials Design Lab/2014/09/10

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Sept 10 - Ionic liquid modified clays

  • Group Members Andrew Farag Sr (Undergrad) and Michael Bible Jr (Undergrad)
  • Group Name : AMT

Objective

Finish film preparation and begin preparation of ionic liquid modified clays.

Description

We are going to exchange the sodium in the sodium montmorillonite for an organic cation. The procedures are taken from these two sources (1, 2 along with Karlena Brown's thesis. We are working from the starting point that there are 92 miliequivalents (meq) of Na+ per 100g of montmorillonite.

  1. Add 1.1g of Sodium Montmorillonite (NaMT) to a scintillation vial containing 20mL of 50:50 HPLC grade water:ethanol.
  2. Determine the amount of cation to add if using either tributylhexadecylphosphonium bromide or (1-hexadecyl)triphenylphosphonium bromide to exchange 100% of the Na from the clay.
  3. Add this amount of phosphonium ion to the clay solution.
  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


Here's an excellent paper on good graphing procedures.

Notes and Calculations

Today we began the exchange of sodium in Sodium Montmorillonite (NaMT) for an organic cation. We worked with the assumption that there are 92 milliequivalents (meq) of Na+ per 100g of NaMT.

  1. First 20 mL of a 50:50 HPLC grade water:ethanol solution was added to a scintillation vial.
  2. To this, we added 1.1488g of NaMT.
  3. We then added, 0.5349g of tributylhexadecylphosponium bromide to the vial.
  4. The vial was capped and left to stir for a week

To exchange 100% of the Na+ from the clay an equivalent number of moles of tributylhexadecylphosphonium bromide needs to be added to our NaMT.

  • The starting weight of NaMT was 1.1488g which has 1.05x10-3 equivalents of Na+.
    • (1.1488g NaMT)x[(92 meq Na+)/(100g NaMT)]=1.05 milliequivalents of Na+
  • Tributylhexadecylphosphonium bromide has a molecular weight of 507.65g/mol.
  • Thus we determined that we needed 0.533g of Tributylhexadecylphosphonium bromide by the following calculation:
    • (1.05x10-3mols Tbhp bromide)x[(507g Tbhp bromide)/(1 mol Tbhp bromide)] = 0.533g Tributylhexadecylphosphonium bromide.
      • Note, however, the actual mass of Tributylhexadecylphosphonium bromide used was 0.5349g.
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