Greg S. answered • 12/04/20
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Here, recognize that LiOH (a group 1 metal hydroxide) is a strong base, and therefore fully ionized in solution. Then recall that pH =-log([OH-] and you will have the answer.
Greg S.
Absolutely correct, JRS. Thank you. pH will be 14 - pOH, and the pOH is -log([OH-]). Quite right. Did you happen to notice my answer to the question you addressed with the Henderson Hasselbach equation? Unless I'm (again) mistaken, you can't use the small-dissociation approximation there...
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12/04/20
J.R. S.
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I had not noticed your answer until just now, and I agree with you because of the strength of the "weak" acid H3PO4. The Ka is considerably greater than most other weak acids. This, of course, accounts for the significant dissociation, and you are quite correct that one should not use the approximation that I used. Thanks for the follow-up.
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12/04/20
J.R. S.
12/04/20