User:Daniel-Mario Larco/Notebook/AU Biodesign Lab - 09/03/2013/2013/08/28

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  * In this experiment the stock solution had a concentration of 2.54mM
  1. Add an appropriate amount of BSA solution so that the final concentration of gold is 90X that of BSA.
  * In this experiment the appropriate volume of BSA needed was calculated and found to be 0.001809 L (1.809mL).
  1. Add deionized water up to 10mL
  2. Transfer solution to a test tube and cap with aluminum foil
  3. Heat in oven at 80C for 3 hours
  * When the test tube was removed from the oven after 3 hours, it was observed that filaments had formed which is an indication that the BSA/Gold ratio was not correct. This was most likely due to an error when transferring the required volumes of both solution to the volumetric flask.
  1. Transfer solution to a plastic falcon tube (with blue cap)

Stock solutions made Gold solution (HAuCl4·3H2O) 0.0100g in 0.0100mL water → 2.54mM BSA solution 0.0104g BSA (MW = 66776g/mol) in 0.0100mL water → 15.6μM



 - 50.5mL of the 0.249 mM HAuCl4 solution 
  1. Heat this solution to boiling while stirring
  2. Add 3mL 1.5mL of 1% (w/v) sodium citrate
 - Almost instantly after adding the sodium citrate the solution turned purple and gradually changed to a red/burgundy

color after about 4 minutes.

  1. Boil solution for another 40 minutes
 - After about 30 minutes of boiling, about 5mL of deionized H2O were added to the solution to replace the

H2O lost by evaporation.

  1. Cool to room temperature and measure the volume
 - Once cooled, the volume of the solution was measured to be 25mL
  1. Determine the final concentration of gold and citrate
 - Determining the final concentration of gold and citrate required us to calculate the initial amounts of gold and

citrate in the initial solution.

      Gold: 0.05L * 2.49 *10^-4 M = 1.245*10-5 moles of Au.
   Citrate: 0.0015L * (0.101g/0.01L) * 1mol citrate/294g = 5.15*10-5 moles of citrate
 - Final Concentrations 
      Gold: 1.245*10-5 / 0.025L = 4.98*10-4 M
   Citrate: 5.15*10-5 / 0.025L  = 2.06*10-3 M


Stock solutions made



  1. A sample was made up in order to run the UV-Vis, which was 1/10 ratio between the citrate sample and water. A cuvette was

filled with deionized water and was put in the UV-Vis before filling the same cuvette with our diluted sample in order to subtract errors from the actual data.

The data was analyzed using thisthis reference

This is a graph of the results of the UV-Vis of the sample.

The maximum recorded was of 0.158 and this occured between 522nm and 525nm.

Analyzing this maximum absorbency allowed us to

determine that the size of the gold nanoparticles was about 14nm, which allowed us to calculate the

concentration of gold nanoparticles in the solution. The concentration was found using the equation


ε450 = 1.76E+08

A450 = 0.158

c= 8.98E-10 M*cm

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