Why Didn't They Use the H-H Equation in the End?

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Discussion Overview

The discussion revolves around the choice between using the Henderson-Hasselbalch (H-H) equation and the equilibrium expression for the acid dissociation constant (Ka) in solving a chemistry problem related to pH calculation. Participants explore the implications of using one equation over the other, particularly in the context of understanding and accuracy.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • Some participants note that both the H-H equation and the Ka expression use the same inputs and can be used to find pH, raising the question of how to choose between them.
  • One participant suggests that using the Ka expression may lead to a clearer understanding of the problem compared to the H-H equation, which they describe as a "forgettable" formula.
  • Another participant claims to have obtained a different answer using the H-H equation compared to the Ka expression, expressing frustration over the discrepancy.
  • There is a suggestion that the H-H equation may require careful manipulation, and participants discuss the potential for mixing up terms, which could lead to errors in calculations.
  • One participant emphasizes that the H-H equation is just another form of Ka, implying that they should yield the same results if applied correctly.

Areas of Agreement / Disagreement

Participants express disagreement regarding the equivalence of the H-H equation and the Ka expression, with some asserting they yield different results in practice. The discussion remains unresolved as participants have not reached a consensus on which equation is preferable or under what circumstances.

Contextual Notes

Participants highlight potential pitfalls in using the H-H equation, such as misplacing terms or misunderstanding the logarithmic relationships involved, which may affect the accuracy of results.

Hidemons
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Here's a solved chemistry problem:

http://answers.yahoo.com/question/index?qid=20100731143735AAjKe10

Why did the answerer not use the H-H equation at the very end?
Please Help.


The H-H equation and the Ka= (H)(A)/(HA) equation (what was used in the problem) both use the same inputs and both are used to find pH. How do I pick one over the other?



(the template provided)
 
Last edited:
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Hidemons said:
Here's a solved chemistry problem:

http://answers.yahoo.com/question/index?qid=20100731143735AAjKe10

Why did the answerer not use the H-H equation at the very end?
Please Help.The H-H equation and the Ka= (H)(A)/(HA) equation (what was used in the problem) both use the same inputs and both are used to find pH. How do I pick one over the other?
(the template provided)
First he found [H+] then he took logs and found pH.

Ka= (H)(A)/(HA) is equivalent practically to the HH equation.

How to pick one over the other? Use Ka= (H)(A)/(HA) because, at the stage you're at you are going to understand clearly what you're doing better than if you just apply a (forgettable) formula; this will be needed for more difficult problems you might get next.
 
epenguin said:
Ka= (H)(A)/(HA) is equivalent practically to the HH equation.

Practically or completely? I worked out the linked problem using H-H and it gives you a different answer. very frustrating
 
Practically - you have to switch things about maybe.

the HH eq. is usually given in a form like

well see here: http://en.wikipedia.org/wiki/Henderson–Hasselbalch_equation

I don't know what you've done but someone like you :-p easily gets pH and pK on the wrong side and mixes + and - log and (equivalently) gets the fraction upside down etc. That's why I recommend not using it, but sticking with something you understand easier.
 
Hidemons said:
Practically or completely? I worked out the linked problem using H-H and it gives you a different answer. very frustrating

Show what you did, answers should be identical. If they are different you are probably doing something wrong.

Henderson-Hasselbalch equation is just another form of Ka (see derivation on the page).
 

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