# User:Klare Lazor/Notebook/Chem-496-001/2013/02/01

Biomaterials Design Lab Main project page
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## Objective

1. Make Fixant for cross linking of the films on Monday
2. Make film with no glutaraldehyde (control)
3. Make film with glutaraldehyde no isopropyl alcohol (control)
4. Make five films varying in concentration of isopropyl alcohol, determine amount of alcohol by calculating amount of OH groups on PVA
5. Let all seven films dry over the weekend

## Description

Films and components All films were heated on a hot plate until the PVOH dissolved in the water. For films 5,6, and 7, additional water was added to the gel to assist in the dissolution. Once dissolved, the film was removed from heat and isopropyl and glutaraldehyde were added. The isopropyl was added first and stirred for roughly 2 to 3 minutes. Glutaraldehyde was added last.

1. 0.5043 g PVOH + 6 mL water (this was the only small film prepared)
2. 0.5027 g PVOH + 6 mL water + 1 mL glutaraldehyde
3. 0.5029 g PVOH + 6 mL water + 87 μL isopropyl stock solution + 1 mL glutaraldehyde
4. 0.5029 g PVOH + 6 mL water + 87 μL 10% isopropyl solution (see notes) + 1 mL glutaraldehyde
5. 0.5060 g PVOH + 6 mL water + 87 μL 1% isopropyl solution + 1 mL glutaraldehyde
6. 0.5034 g PVOH + 6 mL water + 87 μL 0.1% isopropyl solution + 1 mL glutaraldehyde
7. 0.5050 g PVOH + 6 mL water

OH Group Calculations

• Was determined to be 6.86508xE21 OH groups in 0.5 g of PVOH
• PVOH MW = 146,000 to 186,000 calculated the average number of OH groups in a molecule of PVOH determined it was the most logical to do...
• PVOH repeating unit structure = CH2-CH-OH = molecular weight: 44g/mol
• molecular weight avg: 166,000/44 = 3772.73 units per molecule = 3772.73 OH groups per molecule
• 0.5g PVOH / 166,000 g/mol = 3.012E-6 moles PVOH in 0.5 grams of PVOH
• 3.012E-6 moles PVOH × 3773.73 OH groups per molecule = 0.0114 OH groups in 0.5 g PVOH
• 0.0114 OH groups in 0.5 g PVOH x 6.022 x 10E23 molecules/mol = 6.86508 x 10E21 OH groups in 0.5 g of PVOH

Isopropanol Solution Preparation

• We decided to start with 10% OH groups bound to isopropanol and decrease the concentration from there through serial dilution.
• [6.86508 x 10E21 OH groups x .1 molecule isopropanol/1 molecule OH] / 6.022 x 1023 molecules/mol = 0.00114 moles isopropanol
• 0.00114 moles isopropanol x 60.1 g/mol = 0.068514 g isopropanol
• 0.068514 g isopropanol / 0.786 g/mL = 0.0872 mL isopropanol
• Isopropanol solutions were then prepared by taking 10 mL of isopropanol. Serial dilutions with water were then carried out to create 10%, 1% and 0.1% solutions. 0.872 mL of each concentration of isopropyl was added to the specific film of study.

## Data

• Add data and results here...

## Notes

This area is for any observations or conclusions that you would like to note.

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