# Chemistry

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).

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1. Use the HH equation.

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posted by DrBob222
2. I'm trying to do this right now too for a lab report and I don't get it at alllllll

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posted by Felicia
3. 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.

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posted by DrBob222
4. 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?

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posted by jack
5. 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.

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posted by DrBob222
6. How do you find pKa?

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posted by JJ
7. 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.

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posted by DrBob222

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