Re: Acid dissolution + Caffeine. Please Help
Updated: 2010-07-31 19:06:40
We have a CH3COOH dissolution [0.1M], 10ml of dissolution.CH3COOH + H20 <=> CH3COO- H3O+ Ka= 1.8x10-5=[H3O+]x[CH3COO-]/[CH3COOH][H3O+]=[CH3COO-]= 0.01254 ph=-log[H30+]= 1.90 poh=14-1.9=12.1 [OH-]= 7.94x10-13Then, we add the caffeine 10mg. [Caffeine]=0.005Caffeine + H20 <=> CaffeineH+ H- kb= 4.0x10-11=[CaffeineH+]x[OH-]/[Caffeine] Which is the [OH-] to put in the formula? 7.94x10-13?