Calculate the pH of a buffered solution that is 0.100M in C6H5COOH (benzoic acid, Ka = 6.4 x 10-5) and 0.100M in sodium benzoate (NaC6H5COO).
Use the HH equation.
I'm trying to do this right now too for a lab report and I don't get it at alllllll
The HH equation is the Henderson-Hasselbalch equation.
pH = pKa + log[(base)/(acid)]
The base is benzoate ion. The acid is benzoic acid. pKa is pKa for benzoic acid. Post your work if you get stuck.
But that's just .100 over .100 for the base over acid isn't it? then it would just be 6.4 X 10-5 because the log of 1 is 0?
yes, yes, and no.
Yes it is 0.1 over 0.1, and yes log 1 = 0, but no, pH isn't 6.4 x 10^-5.
pH IS, however, pKa since log 1 = 0.
How do you find pKa?
It's analogous to pH.
pH = -log(H^+).
pOH = -log(OH^-)
PKa = -log Ka
pKw = -log Kw.
so pKa for Ka of 6.4 x 10^-5 = 4.19. Plug 6.4 x 10^-5 into your calculator, hit the log button and it returns -4.19, then multiply by the - sign to obtain 4.19.