User:Mary Mendoza/Notebook/CHEM 571 Experimental Biological Chemistry I/2012/10/16: Difference between revisions
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* A weight of .0112 g of luminol was added to 6 mL of water. The buffer composed of a direct addition of .0737 g of sodium carbonate and .4358 g of sodium bicarbonate. | * A weight of .0112 g of luminol was added to 6 mL of water. The buffer composed of a direct addition of .0737 g of sodium carbonate and .4358 g of sodium bicarbonate. | ||
* Using a pH meter, the electrode detected the pH at 8.73. As suggested by Dr. Hartings, a solution of sodium carbonate was made to increase the pH of the solution. | * Using a pH meter, the electrode detected the pH at 8.73. As suggested by Dr. Hartings, a solution of sodium carbonate was made to increase the pH of the solution. | ||
* Several adjustments were made in increasing the pH. A total weight of 1.91 g of sodium carbonate dissolved in 15 mL of water was added to the 6 mL solution of luminol. | * Several adjustments were made in increasing the pH. A total weight of 1.91 g of sodium carbonate dissolved in 15 mL of water was added to the 6 mL solution of luminol. The final pH for luminol was 10.55. | ||
* The molarity of sodium carbonate (MW 105.9784 g/mol) added was calculated: | * The molarity of sodium carbonate (MW 105.9784 g/mol) added was calculated: | ||
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* Since there was limited amounts of the solid form of luminol, it was decided to take 6 mL of the 2.86 mM luminol stock solution and then add the appropriate amount of luminol to the 6 mL volume. The molarity of the 6 mL solution was very minute; the molarity was approximated as 2 mM. By making this assumption, .00106 g of luminol would make a 1 mM solution in 6 mL of water; this amount was multiplied by 8. The product was 0.00848 g of luminol was needed to be added into the 2 mM solution of luminol to increase the molarity to 10 mM. | * Since there was limited amounts of the solid form of luminol, it was decided to take 6 mL of the 2.86 mM luminol stock solution and then add the appropriate amount of luminol to the 6 mL volume. The molarity of the 6 mL solution was very minute; the molarity was approximated as 2 mM. By making this assumption, .00106 g of luminol would make a 1 mM solution in 6 mL of water; this amount was multiplied by 8. The product was 0.00848 g of luminol was needed to be added into the 2 mM solution of luminol to increase the molarity to 10 mM. The amount was weighed and added to the 6 mL 2.86 mM solution of luminol. | ||
* Due to time constraints, the chemiluminescence of the luminol at pH 10.55 was tested on a lab bench with the room lights turned off. The reaction produced a neon blue glow that lasted for more than 3 minutes. No photograph was taken since the apparatus of the camera was not suited for the fluorescence activity. | |||
Revision as of 23:53, 25 October 2012
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PCR mutation
0.46 mg = 0.46E6 ng
Continuation of Chemiluminescence
[math]\displaystyle{ \frac{1.91 g}{15 mL} }[/math] × [math]\displaystyle{ \frac{1 mol}{105.9784 g} }[/math] = [math]\displaystyle{ \frac{0.00120 mol}{mL} }[/math] × [math]\displaystyle{ \frac{1 mL}{1E(-3) L} }[/math] = [math]\displaystyle{ \frac{1.20 mol}{L} }[/math] = 1.20 M of sodium carbonate
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