User:Alicia Rasines Mazo/Notebook/CHEM-572 Experimental Biological Chemistry II/2015/03/17

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Tasks for March 17 To analyze evaporated filtrate from last aniline polymerization reaction To make new citrate gold nanoparticles To run new aniline polymerization using gold citrate NPs as catalyst Preparation of gold citrate nanoparticles Stock solution preparation 1% wt HAuCl4 Made on 3/3 (Dissolved 0.1201 g HAuCl4 in 10 mL of distilled water 38.8 mM sodium citrate Added 0.11728 g of sodium citrate in 10 mL odd instilled water 0.075% wt NaBH4 0.00358 g NaBH4 In 5 mL of 38.8 mM sodium citrate solution previously prepared Citrate gold nanoparticles Add 90 mL of water to 100 mL volumetric flask Add 1 mL 1% wt HAuCl4 and stir for a minute Add 2 mL of 38.8 mM sodium citrate solution and stir for 1 minute Add 1 mL of 0.075% NaBH4. Stir for 5 min Note the solution turns deep purple gradually Determination of citrate gold nanoparticles concentration Method described by Wolfgang Heiss in 'Determination of Size and Concentration of Gold Nanoparticles using UV/Vis Spectra' Absorbance of gold citrate NPs at maximum peak (539 mm)=0.343 A450=0.262 A539/A450=0.343/0.262= 1.31 From tabulated data, this value corresponds to a diameter of ≈6 nm ε450=1.26×107 Concentration=A450/ε450=0.262/1.26E+07= 2.08×10-8 M Polymerisation of aniline using gold citrate nanoparticles as catalyst 0.511 mL of aniline, 89 mL of distilled water, 3 mL of 1 M HCl, 7 mL of hydrogen peroxide and 14.28 mL of gold citrate nanoparticles were added in that order. The bottom flask was capped with a septum and run under nitrogen atmosphere for 24 h. Because a new gold nanoparticle was used, it was necessary to recalculate the amount of catalyst to be added to achieve a ratio of 1000:1 An:Au 0.5 g aniline= 0.00537 mol, so need 5.37×10-6 moles of gold in total Gold moles = (0.1201 g/ 339.785 g/mol)/10 since only 1 mL of the stock solution was required to make the AUNPs Thus, there were 3.535 ×10-5 moles in 94 mL of AuNP solution To achieve an amount of 5.37×10-6 Moles, 14.28 mL of AuNP solution was required