# Difference between revisions of "IGEM:American University/2009/Notebook/Advanced Experimental Chemistry: Fun Times in Beeghly/2013/03/08"

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# Today's Procedure

Same procedure as 2/15: Preparing PVOH and 3-Chloro-2-Hydroxypropyltrimethylammonium chloride reacted crystals.

4 sets of 2g PVOH were left to dissolve (as much as possible on a stir plate) in 5mL 0.2M NaOH for 30 minutes. 0.855g of 3-Chloro-2-Hydroxypropyltrimethylammonium chloride was left to dissolve in 5mL 0.2M NaOH for 30 minutes.

0.01%: x/0.855g=0.01%/10%, x=0.000855g. 0.000855g/0.171(g/mL)= 0.005mL of the "stock" 3-Chloro-2-Hydroxypropyltrimethylammonium chloride solution. Thus to make our 0.01% 3-Chloro-2-Hydroxypropyltrimethylammonium chloride:PVOH, would require the 2gPVOH dissolved in 5mL NaOH 0.2M + 0.005mL of the "stock" 3-Chloro-2-Hydroxypropyltrimethylammonium chloride filled to 10mL NaOH 0.2M [4.995mL extra NaOH 0.2M]. Similarly...

0.1%: Required the 2gPVOH dissolved in 5mL NaOH 0.2M + 0.05mL of the "stock" 3-Chloro-2-Hydroxypropyltrimethylammonium chloride filled to 10mL NaOH 0.2M [4.95mL extra NaOH].

1%: Require the 2gPVOH dissolved in 5mL NaOH 0.2M + 0.5mL of the "stock" 3-Chloro-2-Hydroxypropyltrimethylammonium chloride filled to 10mL NaOH 0.2M [4.5mL extra NaOH].

To make the "10%" solution, we used the 4.445mL left over "10% stock" 3-Chloro-2-Hydroxypropyltrimethylammonium chloride solution. To solve for the actual percentage this makes... (4.445mL)(0.171g/mL 3-Chloro-2-Hydroxypropyltrimethylammonium chloride) = 0.760095g 3-Chloro-2-Hydroxypropyltrimethylammonium chloride (0.760095g)/(0.855g)=x%/10% x=8.89% so, the solution is actually 8.89% of PVOH OH groups reacting w 3-Chloro-2-Hydroxypropyltrimethylammonium chloride.

Once dissolved, acetone was added to the solutions until, with shaking the vials, they filled with white precipitate. The vials were then left to continue precipitating in the hood until next lab.

Liz and Klare: They gathered the DSC and FTIR data while I set up the new Quaternary Ammonium Films