Posted by Jude on Thursday, February 12, 2009 at 7:46pm.
Let's call propionic acid HP, then it will ionize as
HP ==> H^+ + P^-
First convert pH = 2.50 to (H^+) by
pH = -log(H^+).
Then set up the expression for Ka.
Ka = (H^+)(P^-)/(HP).
You know (H^+) now from the pH calculation. You know (P^-) because it equals the same as (H^+). (HP) = y-(H^+).
Solve for y which = (HP).
Then M x molar mass propanoic acid = grams propanoic acid.
Post your work if you get stuck.
When you do the -log of 2.50 you get a negative number, so I am not really sure what you do with that, do you not use the pKa for anything?
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