# Chemistry

I have the answer to this question but the book does not explain WHY this is a good buffer solution? It uses pH=pKa - log(base/acid) and shows 7.2=7.2 - log(1).
I don't understand why they can only use pKa7.2 and ignore the pKa 12.7.

You are instructed to prepare 100ml of a 0.02 M sodium phosphate buffer, pH 7.2, by mixing 50ml of 0.02 M Na2HPO4 and 50ml of 0.02 M NaH2PO4. Explain why this procedure provides an effective buffer at the desired pH and concentration.

Provided Table info:
H2PO4(-): Ka(M) 6.23x10^-8 pKa 7.2
HPO4(2-): Ka(M) 2.20x10^-13 pKa 12.7

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1. First, a correction. The Henderson-Hasselbalch equation is pH = pKa + log (base)/(acid). That negative sign is either a typo in the book or by you in the post. What you are asking isn't exactly clear. I think you want to know why
a. is the solution a buffer
b. why it is effective
c. why use pk2 and not pK3?

a. The solution is a buffer BECAUSE you have an acid and its conjugate base.
b. Since 7.2 = 7.2 + log 1 (log 1 is zero) is 7.2, the most effective buffer is when pKa = pH. To get that you make (base) = (acid) and that ALWAYS makes (base)/acid) = 1 and that ALWAYS makes log 1 = 0 and that ALWAYS makes pH = pKa. Since base = acid, adding acid OR base affects the pH the same way so it is effective against BOTH acids and bases equally.
c. Why use pk2 and not pK3?
Look at k2 expression here.
H2PO4^- ==> H^+ + HPO4^2-
then
k2 = (H^+)(HPO4^2-)/(H2PO4^-)

Here is the buffer equation if a base is added.
H2PO4^- + OH^- ==> HPO4^- + H2O and here is the secret.
Note that k2 is the ONLY k expression (write out k1 and k3 expressions to see this if you will) that includes H2PO4^- and HPO4^2 and nothing else-. k1 for H3PO4 = (H^+)(H2PO4^-)/(H3PO4) and you don't have H3PO4 anywhere in that buffer. Also k1 = (H^+)(PO4^3-)/(HPO4^-) and you don't have PO4^3- anywhere in that buffer. You have only the salts NaH2PO4 and Na2HPO4 so k1 and k3 are not applicable. Get it?

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