Posted by Petty on Thursday, September 22, 2011 at 7:33am.
Start with the acid dissociation equation:
HAc -> H+ + Ac-
Ka is then
[H+][Ac-]/[HAc] {given in the question}
at the start
HAc -> H+ + Ac-
and we start from A mole/L and B mol/L for the acid and buffer
A M ___0__ B M
at the end
HAc -> H+ + Ac-
A-x____x___B+x
so Ka is
Ka=(x)(B+x)/(A-x)
Assume that x is small wrt A and wrt B so that the equation reduces to
Ka=x(B)/(A)
which we can rearrange to
x=Ka(A)/(B)
take logs of each side
log(x)=log(Ka(A)/(B))
which is
log(x)=log(Ka) + log((A)/(B))
multiply through by -1
-log(x)=-log(Ka) - log((A)/(B))
which is
-log(x)=-log(Ka) + log((B)/(A))
or
pH = pKa + log((B)/(A))
or
pH = pKa + log [base]\[acid]
As an added note (perhaps picky but I think needed) is that / is used as a divide sign while \ is used for other purposes.
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