when I use henderson-hasselbach
Ka of HOP4^2- = 4.8 x10^-13
pKa= -lg(4.8 x10^-13)=12.3
substituting I get
7.44-12.3 = lg [base/acid]
e^-4.86=[base/acid}= 7.75 x10^-3
according to my textbook this answer is incorrect and should be 0.59
Please Help!
IF you are trying to work the problem you posted earlier, the 7.44 pH of a soution of H2PO4^- and HPO4^=, the numbers appear to me that you are using k3. You should be using k2.
In fact, if you look at my original post you will note that I specifically wrote pK2.
pH = pK2 + log (b/a)
And if you do it that way, I found
b/a = 1.7. The problem asked for a/b so the reciprocal of that is 0.59.
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