Synthesize two different sets of gold nanoparticles. In one set, Au^{3+} is reduced by a protein (bovine serum albumin, BSA) and the synthesized nanoparticle is also surrounded and stabilized by BSA. In the second set, Au^{3+} 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
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 1.2mL of the (~2.54mM -note the exact concentration) gold (HAuCl_{4}) 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
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 (HAuCl_{4}·3H_{2}O) 0.0100g in 0.0100mL water → 2.54mM
BSA solution 0.0104g BSA (MW = 66776g/mol) in 0.0100mL water → 15.6μM
Take 50mL of the HAuCl_{4} solution from the 250mL volumetric flask. Note the concentration
Heat this solution to boiling while stirring
Add 3mL 1.5mL of 1% (w/v) sodium citrate
Boil solution for another 40 minutes
Cool to room temperature and measure the volume
Determine the final concentration of gold and citrate
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 (HAuCl_{4}·3H_{2}O) 0.0245g in 0.2505mL water → 0.249mM
Sodium Citrate (Na_{3}C_{6}H_{5}O_{7}·2H_{2}O) 0.1010g in 10.0mL → 1.01%
Calculations
Beer's Law gives an equation for the absorption of light and the properties of the solution.
[math]\displaystyle{ \ A=\epsilon bc }[/math]
A is absorbance.
ε is molar absorptivity with units of L mol^{-1} cm^{-1}.
b is the path length of the sample.
c is the concentration with units of mol L^{-1}.
From the Absorption vs Wave Length Graph, the peak is at a wave length of 518nm and absorbance is 0.612. As the peak is at a wave length less than 520nm, table S-1 from the reference is used.
A_{450}=0.389
A_{518}=0.612
A_{518}/A_{450}=1.57
From table S-1, d/nm=12
d/nm is the diameter of the particle is nm.
ε_{450}=1.09*10^{8} M^{-1}cm^{-1}
To find the concentration, the following equation is used; c=A_{450}/ε_{450} c=0.389/1.09*10^{8}L mol^{-1} cm^{-1} c=3.5688*10^{-9} mol L^{-1}
Matt Hartings Based on your final volume (of the citrate-AuNPs), what do you estimate the final concentration of gold (not gold nano particles) to be?
the initial concentration of gold solution is 2.54 mM in 0.01 mL of water. Considering the final volume of Citrate-AuNP is 26 mL, the concentration of gold is 9.769x10^{-4}mM.