Crosslinking PVOH (MW:146,000-186,000) Films With 0.7M HCl and neutralizing .6 Molar sodium bicarbonate
- To high of a molecular weight will remake with with a lower molecular weight
- MW: 22,000 will be used instead for this....higher molecular weight is to thick for this type of reaction with crosslinking with glutaraldehyde
- 20ml of 100% H2So4 in 1 L of H20 was use for the fixant will determine amount of H+ ions in solution to determine concentartion of HCL
- 20ml* 1.84g/ml ÷ 98.079g/mol = 0.375moles of H2SO4 ÷ 1 Liter= 0.369 Molar H2SO4
- assume complete dissociation
- 0.369mols of H2SO4/L * 2mols HCl/1mol of H2SO4 = 0.7M HCl, which is equal to 6ml of 12 M HCl in 94 ml of H2O
- To neutralize a NaHCO3 solution will be made, 0.7 M NaHCO3
Switching to lower molecular weight: PVOH (MW: 22,000) + phenol + glutaraldehyde + 30 ml of water
- 22,000 MW PVOH → 44g/mol
- 1 gram PVOH/ 22,000g/mole = 4.57E-5 m oles PVOH
- 22,000 g/mol PVOH / 44 g/mol = 500 OH groups per a molecule of PVOH
- 4.54E-5moles PVOH * 5000 oh groups/molecule of PVOH * 6.002 E23 molecules/mol = 1.369 E 22 oh groups in 1 gram of PVOH 22000 mw
- we want one OH in Phenol to react with 10%, 1%,.1% and 0.01% of oh groups in PVOH
- used the 10% stock left over from last time and reduce by a factor of ten for each percentage
- 1%= 1ml of stock
- .1%=0.1ml of stock
- .01%= 0.01 ml of stock
- 10%= whatever ml left over will end up close to 10% will calculate after measuring how much stock is left over
- Add data and results here...
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