[migar]

We have a CH3COOH dissolution [0.1M], 10ml of dissolution. We add 10mg of caffeine C8H10N4O2.

CH3COOH + H20 <=> CH3COO-   +  H3O+       Ka= 1.8x10-5
Caffeine + H20 <=> CaffeineH+   +   OH-        kb= 4.0x10-11

Which is the [CaffeineH+] and the ph?

Thank you

[discodermolide]

ka = conc of product1 x conc of prod 2 x conc of product …. (in moles) divided by the conc of reactant1 x conc of reactant 2 x reactant of product …. (in moles) ,
in your case

1.8x10-5 = molar conc of CH3COO- x molar conc of H3O+ / molar conc of CH3CO2H x molar conc of water

rearrange it and you have the pH which is the molar con of H3O+

[migar]

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-]= 7.94x10-13

Then, we add the caffeine 10mg. [Caffeine]=0.005

Caffeine + H20 <=> CaffeineH+   +   OH-       

kb= 4.0x10-11=[CaffeineH+]x[OH-]/[Caffeine] 

Which is the [OH-] to put in the formula? 7.94x10-13?