Asked by Delli
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.
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.
Answers
There are no human answers yet.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.