User:Helen L. Slucher/Notebook/CHEM 571/2013/08/28: Difference between revisions

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Citrate AuNP
Citrate AuNP
Peak Wavelength: 518 nm
Peak Wavelength = 518 nm


Absorbance at Peak: .612
Absorbance at Peak = 0.612


In order to determine the concentration, the Absorbance at 450 is known.
Final Concentration
Abs (450): .389
<br>A<sub>450</sub>=0.389
Abs (518)/ Abs (450)= .612/ .389= 1.57
Abs (518)/ Abs (450)= .612/ .389= 1.57
<br>A<sub>518</sub>/A<sub>450</sub>=1.57


Using the data found [http://pubs.acs.org/doi/suppl/10.1021/ac0702084/suppl_file/ac0702084si20070321_014144.pdf here] the value of 1.57 resulted in 12 d/nm and a molar absorptivity of 1*10^8.
Using the data found [http://pubs.acs.org/doi/suppl/10.1021/ac0702084/suppl_file/ac0702084si20070321_014144.pdf here]:
 
<br>1.57 = 12 d/nm and a molar absorptivity of 1*10^8.
Using Beer's law and the newly found value for molar absorptivity, the concentration was determined to be 3.57*10^-9M.
<br>ε<sub>450</sub>=1.09*10<sup>8</sup> M<sup>-1</sup>cm<sup>-1</sup>


<br>[[Image:Absorbtionvswavelength.png|right|]]
<br>[[Image:Absorbtionvswavelength.png|right|]]
Beer's Law gives an equation for the absorption of light and the properties of the solution.
Beer's Law gives an equation for the absorption of light and the properties of the solution.
:<math>\ A=\epsilon bc </math>
:<math>\ A=\epsilon bc </math>
<br>A is absorbance.
<br>ε is molar absorptivity with units of L mol<sup>-1</sup> cm<sup>-1</sup>.
<br>b is the path length of the sample.
<br>c is the concentration with units of mol L<sup>-1</sup>.
<br>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.
<br>A<sub>450</sub>=0.389
<br>A<sub>518</sub>=0.612
<br>A<sub>518</sub>/A<sub>450</sub>=1.57
<br>From table S-1, d/nm=12
<br> d/nm is the diameter of the particle is nm.
<br>ε<sub>450</sub>=1.09*10<sup>8</sup> M<sup>-1</sup>cm<sup>-1</sup>


<br>To find the concentration, the following equation is used; c=A<sub>450</sub>/ε<sub>450</sub>
c=A<sub>450</sub>/ε<sub>450</sub>
<br>c=0.389/1.09*10<sup>8</sup>L mol<sup>-1</sup> cm<sup>-1</sup>
<br>c=3.57*10<sup>-9</sup> mol L<sup>-1</sup>
<br>c=3.5688*10<sup>-9</sup> mol L<sup>-1</sup>


==References==
==References==


[http://openwetware.org/wiki/User:Matt_Hartings/Notebook/AU_Biomaterials_Design_Lab/2013/08/28 Dr. Harting's AU Biomaterials Design Lab]
[http://openwetware.org/wiki/User:Matt_Hartings/Notebook/AU_Biomaterials_Design_Lab/2013/08/28 Dr. Harting's AU Biomaterials Design Lab]

Revision as of 15:43, 16 September 2013

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Overview

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-Au Nanoparticles

  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
  3. Add deionized water up to 10mL
  4. Transfer solution to a test tube and cap with aluminum foil
  5. Heat in oven at 80C for 3 hours
  6. Transfer solution to a plastic falcon tube (with blue cap)

Stock solutions made

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

Citrate-Au Nanoparticles

  1. Take 50mL of the HAuCl4 solution from the 250mL volumetric flask. Note the concentration
  2. Heat this solution to boiling while stirring
  3. Add 3mL 1.5mL of 1% (w/v) sodium citrate
  4. Boil solution for another 40 minutes
  5. Cool to room temperature and measure the volume
  6. Determine the final concentration of gold and citrate

Stock solutions made

  1. Gold solution (HAuCl4·3H2O) 0.0245g in 0.2505mL water → 0.249mM
  2. Sodium Citrate (Na3C6H5O7·2H2O) 0.1010g in 10.0mL → 1.01%

Notes

Citrate AuNP Peak Wavelength = 518 nm

Absorbance at Peak = 0.612

Final Concentration
A450=0.389 Abs (518)/ Abs (450)= .612/ .389= 1.57
A518/A450=1.57

Using the data found here:
1.57 = 12 d/nm and a molar absorptivity of 1*10^8.
ε450=1.09*108 M-1cm-1


Beer's Law gives an equation for the absorption of light and the properties of the solution.

[math]\displaystyle{ \ A=\epsilon bc }[/math]

c=A450/ε450
c=3.57*10-9 mol L-1

References

Dr. Harting's AU Biomaterials Design Lab

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