User:Raquel Lara/Notebook/project 1/2017/09/05

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Au:AA:NaOH Preparation AuNPs solutions made with AA last week did not form any colloids or fibers so remaking solutions with low Au:AA ratios over different pH range Total of 12 solutions made ( 4 volumetric ratios with differing NaOH concentrations for each of the 3 amino acids- arginine, phenylalanine, and Isoleucine) Concentration of HAuCl4= 2.5 mM (1 mL) (1/1000 mL) (2.5 x 10^-3 mol/L) = 2.5 x 10^-6 mol HAuCl4 Volumetric ratios of Au:AA:NaOH Using 5 mM NaOH solution to prepare the following volumetric ratios: (1) 1:1:0.5 (0.5 mL) (1L/1000 mL) (5 x 10^-3 mol/L) = 2.5 x 10^-6 mol NaOH (2) 1:1:1 (1 mL) (1L/1000 mL) (5 x10^-3 mol/L) = 5 x 10^-6 mol NaOH Switching to 1 M NaOH solution and prepare mixtures using the following volumetric ratios: (3) 1:1:0.1 (0.1 mL) (1L/1000 mL) (1 mol/1 L) = 1 x 10^-4 mol NaOH (4) 1:1:1 (1 mL) (1 L/1000 mL) (1 mol/L) = 1 x 10^-3 mol NaOH Theoretical pH Calculations Moles of OH= Moles NaOH - Moles HAuCl4 (1) (2.5 x 10^-6 mol NaOH) - ( 2.5 x 10^-6 mol HAuCl4) = 0 mol OH [OH]= (0 mol OH)/(10 x 10^-3 L total solution) = 0 M OH pH = 7 (2) (5 x 10^-6 mol NaOH) - ( 2.5 x 10^-6 mol HAuCl4) = 2.5 x 10^-6 M OH [OH] = (2.5 x 10^-6 mol OH)/(10 x 10^-3 L total solution) = 2.5 x 10^-4 M OH pOH = -log [OH] = -log [ 2.5 x 10^-4 M] = 3.602059991 pH = 14 - pOH = 14 - 3.602059991= 10.39794001 (3) (1 x 10^-4 mol NaOH) - ( 2.5 x 10^-6 mol HAuCl4) = 9.75 x 10^-5 mol OH [OH] = (9.75 x 10^-5 mol OH)/(10 x 10^-3 L total solution) = 0.00975 M OH pOH = -log [OH] = -log [ 0.00975 M] = 2.010995384 pH = 14 - pOH = 14 - 2.010995384 = 11.98900462 (4) (1 x 10^-3 mol NaOH) - ( 2.5 x 10^-6 mol HAuCl4) = 9.975 x 10^-4 [OH] = ( 9.975 x 10^-4 mol OH)/(10 x 10^-3 L total solution) = 0.09975 M OH pOH = -log [OH] = -log [ 0.09975 M] = 1.001087096 pH = 14 - pOH = 14 - 1.001087096 = 12.9989129 AuNP Formation with BSA -The Au:BSA molar ratio for the right test tube is equal to 80:1 which led to the formation of Au nanoparticles. -The Au:BSA molar ratio for the left test tube is equal to 160:1 which led to the formation of gold-protein fibers.