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

  1. 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
  2. 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
  3. 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=A450450=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


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