User:Javier Vinals Camallonga/Notebook/Javier Vinals notebook/2013/09/02

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(Citrate-AuNP)
(UV-Vis)
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According to [http://pubs.acs.org/doi/suppl/10.1021/ac0702084/suppl_file/ac0702084si20070321_014144.pdf this reference], the diameter of the gold particles was less or equal to 14nm, since 0.158/A450=0.158/0.098=14nm, which probably means that a mistake while adding the gold was made.
According to [http://pubs.acs.org/doi/suppl/10.1021/ac0702084/suppl_file/ac0702084si20070321_014144.pdf this reference], the diameter of the gold particles was less or equal to 14nm, since 0.158/A450=0.158/0.098=14nm, which probably means that a mistake while adding the gold was made.
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[[User:Matt Hartings|Matt Hartings]] Why is this analysis wrong?? This is a curious statement. Also, what is the final concentration of both the gold nanoparticles AND the concentration of gold (different things).

Revision as of 17:44, 12 September 2013

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Objective

Synthesize 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-AuNP

This procedure was taken from the following reference and has been used by our previous two Experimental Biological Chemistry groups.

  1. Add 1mL of the (~2.5mM -note the exact concentration) gold (HAuCl4) solution to a 10mL volumetric flask
  2. Add an appropriate amount of BSA solution so that the final concentration of gold is 90X that of BSA
                   - Through calculation, we obtained that 0.001809 L was the necessary amount of BSA solution needed to add, so that the final concentration of gold is 90X that of BSA. Calculations done: First we find the amount of moles of BSA necessary for the reaction. [(2.54*10^-3 moles/L of HAuCL4)*(1L/1000mL)*1mL]/90=2.82 *10^-8 moles of BSA. Now we find the volume of BSA necessary to add to the reaction: (2.82 *10^-8 moles BSA) *(1L/15.6 *10^-6 moles)= 0.001809 L of BSA or 1.8 mL of BSA

  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
                  - After we took the solution out of the oven, we observed that some nanoparticles had formed, due to an error during the process fo the experiment. Thus it was discarded away. 
  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

Citrate-AuNP

This procedure is used by Dr. Miller in her research lab - Allison Alix's notebook and is taken from this reference and can be analyzed according to thisthis reference.

  1. Take 50mL of the HAuCl4 solution from the 250mL volumetric flask. Note the concentration
       - 50.5 ml of 0.249 HAuCl4 solution
  1. Heat this solution to boiling while stirring
  2. Add 3mL 1.5mL of 1% (w/v) sodium citrate
       - It was observed that the solution turned purple at the beggining after adding the sodium citrate, and then it turned red after 4 minutes.
  1. Boil solution for another 40 minutes
       - During this part of the experiment at 30 minutes we have to add more water,5 mls of water, since a lot of the solution had evaporated.
  1. Cool to room temperature and measure the volume
      - After we let it cool for 10 minutes, we measured the new volume, which is 25 mls
  1. Determine the final concentration of gold and citrate
     - In order to determine the final concentration, we first found out the initial amount of moles of citrate and gold.
         Gold: 0.05 L* 2.49 *10^-4 M= 1.245 *10^-5 moles of Au.
         Citrate: (0.1010 g/0.01 L)*(1 mol/294 grams)= 0.034 M
                     0.034 M* 0.0015 L= 5.15 *10^-5 moles of citrate.
    - Final concentration
         Gold: 1.245 *10^-5 moles of Au/ 0.025 L= 4.98 *10^-4 M
         Citrate: 5.15*10^-5 moles of citrate/0.025 L= 2.06 *10^-3 M



Note: According to Allison, and Madeline will corroborate, we don't need to reflux the 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

Gold solution (HAuCl4·3H2O) 0.0245g in 0.2505mL water → 0.249mM

Sodium Citrate (Na3C6H5O7·2H2O) 0.1010g in 10.0mL → 1.01%

UV-Vis

"Figure 1"

Image:New Document20130902133945410.pdf


To obtain a UV-Vis sample, we diluted to 1/10 de solution, and then we used 1mL of this solution to run the UV-Vis.

For analyzing the data we used this reference

We obtained that the maximum absorbance was 0.158 at 522-525nm.

According to this reference, the diameter of the gold particles was less or equal to 14nm, since 0.158/A450=0.158/0.098=14nm, which probably means that a mistake while adding the gold was made.

Matt Hartings Why is this analysis wrong?? This is a curious statement. Also, what is the final concentration of both the gold nanoparticles AND the concentration of gold (different things).

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