Asked by Rachel
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).
Answers
Answered by
DrBob222
Use the HH equation.
Answered by
Felicia
I'm trying to do this right now too for a lab report and I don't get it at alllllll
Answered by
DrBob222
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.
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.
Answered by
jack
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?
Answered by
DrBob222
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.
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.
Answered by
JJ
How do you find pKa?
Answered by
DrBob222
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.
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.