User:Zachary I. Mendel/Notebook/Zacks Notebook/2013/08/28
|Project name|| Main project page|
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.
Start by placing all of your materials into a volumetric flask so that you know the exact volumes and exact amounts of what you have added. I will have prepared the initial protein and gold solutions for you (this one time only). In the future, you will be expected to make most of the starting solutions.
This procedure was taken from the following reference and has been used by our previous two Experimental Biological Chemistry groups.
- Add 1mL of the 2.54mM gold (HAuCl4) solution to a 10mL volumetric flask
- Add an appropriate amount of BSA solution so that the final concentration of gold is 90X that of BSA
- Calculations = ((0.00254 moles/L Gold)* (1L/1000ml)* (1ml)) / 90 = moles BSA = 2.82*10^-8 moles BSA [BSA] = 15.6μM - (2.82*10^-8 moles) * (1L/15.6*10^-6 moles) = 0.00180L or 1.8mL BSA to add to solution
- Add deionized water up to 10mL
- Transfer solution to a test tube and cap with aluminum foil
- Heat in oven at 80C for 3 hours
- 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
- According to the graph the peak absorbance is around 520nm which corresponds to an particle size less than 32nm.
- Take 50mL of the HAuCl4 solution from the 250mL volumetric flask. Note the concentration
- Heat this solution to boiling while stirring
3mL1.5mL of 1% (w/v) sodium citrate
- During this part the solution turned purple and later red
- Boil solution for another 40 minutes
- Note = A lot of our water evaporated so we added more water here
- Cool to room temperature and measure the volume
- Final volume was determined to be
- Determine the final concentration of gold and citrate
- First had to calculate moles of Au and Citrate - Au - 0.249mM solution = 0.249moles/mL * 50mL solution = 12.45 moles Au - Sodium Citrate - (0.1010g/10mL) * (1000mL/1L) / (258.06g/mol) = 0.03913 moles Sodium Citrate - Final Concentrations - Au = 1.3*10^-3 M - Sodium Citrate = 5.0*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 (Na3
C6H5O7·2H2O) 0.1010g in 10.0mL → 1.01%
Matt Hartings Two things. 1) What is the final concentration of gold in your citrate-AuNP samples (based on your initial conc and final volume)? And 2) Use the reference that I gave in order to determine the size and concentration of the gold nanoparticles. (Note: gold and gold nanoparticles are different things!)
- The maximum absorbance of the citrate-AuNP solution was 0.405 at 520nm. At 450 the absorbance was determined to be 0.259. According to Hais et al. article, the max absorbance (0.405) was divided by 0.259. This gave a vlaue of 1.56. This corresponded to a diameter size of 12nm.
- A 12nm diameter np corresponds to an epsilon 450 value of 1.09E08. Using the equation C=A450/(epsilon)450 the concentration can be calculated. This gave us a value of 2.38E-9 M solution.
- Final volume of citrate-AuNP was 9.5ml
1. The final concentration of gold in the citrate-AuNP samples
2. Size and concentration of gold nanoparticles.