User:Daniel-Mario Larco/Notebook/AU Biodesign Lab - 08/28/2013
Biomaterials Design Lab | Main project page |
ObjectiveSynthesize two different sets of gold nanoparticles. In one set, Au3+ is reduced by a protein (bovine serum albumin, BSA) and the synthesized nanoparticle is also surrounded and stabilized by BSA. In the second set, Au3+ is reduced by citrate, and the AuNP is stabilized by citrate in solution. The BSA-AuNPs are purple in color and the citrate-AuNPs are more of a burgundy (reddish) color. BSA-AuNPThis procedure was taken from the following reference and has been used by our previous two Experimental Biological Chemistry groups.
* In this experiment the stock solution had a concentration of 2.54mM
* In this experiment the appropriate volume of BSA needed was calculated and found to be 0.001809 L (1.809mL).
* 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.
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 Citrate-AuNPThis procedure is used by Dr. Miller in her research lab - Allison Alix's notebook and is taken from this reference.
- 50.5mL of the 0.249 mM HAuCl4 solution
- Almost instantly after adding the sodium citrate the solution turned purple and gradually changed to a red/burgundy color after about 4 minutes.
- After about 30 minutes of boiling, about 5mL of deionized H2O were added to the solution to replace the H2O lost by evaporation.
- Once cooled, the volume of the solution was measured to be 25mL
- 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
solution (which would carry out the reaction in a closed system). We should be able to bring the solution to a boil on the bench top. Stock solutions made
UV-Vis
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 The maximum recorded was of 0.158 and this occured between 522nm and 525nm. File:Citrate-AuNP-UV-Vis-DML 2013 28 08.pdf is a graph of the results of the UV-Vis of the sample. Analyzing this maximum absorbency allowed us to determine that the size of the gold nanoparticles was about 14nm. | |
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Matt Hartings Using that reference that you used to determine the size of your nanoparticles, what is the concentration of nanoparticles? Also, you really need to have a note in here about your BSA-AuNP synthesis going wrong (getting fibers instead of a homogeneous solution!)