User:Tamanika Tinsley/Notebook/Chem 581 Biomaterials Design Lab/2014/10/03

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Oct 3 - Film and Clay Prep Group Members Andrew Farag Sr (Undergrad) and Michael Bible Jr (Undergrad) Group Name : AMT Objective To start synthesis of PVA and PVA with Clay films, and a new ion exchanged clay Exchange of Na+ in Sodium Montmorillonite for Organic Cations Using a procedure analogous to the one used on September 10th, two cation exchanges were setup using tributylhexadecylphosponium bromide and choline. 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. To begin, we added 20 mL of a 50:50 HPLC grade water:ethanol solution to two scintillation vials. To one of the vials, we added 1.1078g of NaMT. We then added, 0.5254g of tributylhexadecylphosponium bromide to the vial. The vial was then capped and stirring was begun with the intention to stir the solution for a week. To the other vial, we added 1.1014g of NaMT. We then added, 0.0.1665g of choline to the vial. The vial was then capped and stirring was begun with the intention to stir the solution for a week. Exchanged ion clay was completed by Michael New films were created by Andrew Calculations We wanted to exchange 100% of the Na+ from the clay so we needed to add an equivalent number of moles of tributylhexadecylphosphonium bromide to our NaMT. We started with 1.1078g of NaMT which has 1.02x10-3 equivalents of Na+. (1.1078g NaMT)x[(92 meq Na+)/(100g NaMT)]=1.02 milliequivalents of Na+ Tributylhexadecylphosphonium bromide has a molecular weight of 507.65g/mol. Thus we determined that we needed 0.517g of Tributylhexadecylphosphonium bromide by the following calculation: (1.02x10-3mols Tbhp bromide)x[(507.65g Tbhp bromide)/(1 mol Tbhp bromide)] = 0.517g Tributylhexadecylphosphonium bromide. Note, however, the actual mass of Tributylhexadecylphosphonium bromide used was 0.5254g. We wanted to exchange 100% of the Na+ from the clay so we needed to add an equivalent number of moles of choline to our NaMT. We started with 1.1014g of NaMT which has 1.013x10-3 equivalents of Na+. (1.1014g NaMT)x[(92 meq Na+)/(100g NaMT)]=1.013 milliequivalents of Na+ Choline has a molecular weight of 139.63g/mol. Thus we determined that we needed 0.141g of choline by the following calculation: (1.013x10-3mols choline)x[(139.63g choline)/(1 mol Tbhp bromide)] = 0.141g choline. Note, however, the actual mass of choline used was 0.1665g. Calculations were completed by Michael