User:Marjan Dolatmoradi/Notebook/CHEM-671/2017/09/05

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AU NPs 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. Prepartion of 5mM NaOH Solution 100 mL NaOH × 5×10-3 mol/L NaOH× 1 mol/L NaOH stock solution = 500 μL NaOH stock Solution Prepartion of AU:AA:NaOH Solution at different ratios and calculating theoretical pH For the volumetric ratio of 1 mL:1 mL:0.5 mL : [HAuCl4] : 1×10-3 L × 2.5×10-3 mol/L = 2.5×10-6 mol HAuCl4 / 10×10-3 L = 0.25 mM [AA] : 1×10-3 L × 250×10-6 mol/L = 250×10-9 mol AA / 10×10-3 L = 2.5 μM [NaOH] : 500×10-6 L × 5×10-3 mol/L = 25×10-7 mol NaOH / 10×10-3 L = 0.25 mM pH: 2.5×10-6 mol HAuCl4 - 25×10-7 mol NaOH = 0 → pH = 7 For the volumetric ratio of 1 mL:1 mL:1 mL : [HAuCl4] = 0.25 mM [AA] = 2.5 μM [NaOH] : 1×10-3 L × 5×10-3 mol/L = 5×10-6 mol NaOH / 10×10-3 L = 0.5 mM pH: 5×10-6 mol NaOH - 2.5×10-6 mol HAuCl4 = 2.5×10-6 pOH = -log [OH] = -log [2.5×10-4] = 3.60 pH = 14 -3.6 = 10.4 For the volumetric ratio of 1 mL:1 mL:0.1 mL : [HAuCl4] = 0.25 mM [AA] = 2.5 μM [NaOH] : 0.1×10-3 L × 1 mol/L = 1×10-4 mol NaOH / 10×10-3 L = 10 mM pH: 1×10-4 mol NaOH - 2.5×10-6 mol HAuCl4 = 9.75×10-5 pOH = -log [OH] = -log [9.75×10-3] = 2.01 pH = 14 - 2.01 ≈ 12 For the volumetric ratio of 1 mL:1 mL:1 mL : [HAuCl4] = 0.25 mM [AA] = 2.5 μM [NaOH] : 1×10-3 L × 1 mol/L = 1×10-3 mol NaOH / 10×10-3 L = 0.1 M pH: 1×10-3 mol NaOH - 2.5×10-6 mol HAuCl4 = 9.975×10-4 pOH = -log [OH] = -log [9.75×10-2] = 1.001 pH = 14 - 1 = 13