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henderson-hassalbach equation
Started by globy321
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No...you also use it when finding the equilibrium pH in acid-base reactions.
Especially helpful in finding the amount of ionized acid/or base with WEAK acids and bases (because strong acids and bases dissociate completely).
H-H is just a shortcut formula based on the regular law of mass action formula used to calculate the pH of a buffer.
If you are using the H-H equation, you already know that information. I suppose you could use it to find the concentration of the acid or base form if you already know the concentration of the complement and the pH of the solution, but if you already have all that information, it is unlikely that you lack the concentration of both members of the complementary acid-base pair.
You use regular equilibrium calculations if you lack that information.
If you are using the H-H equation, you already know that information. I suppose you could use it to find the concentration of the acid or base form if you already know the concentration of the complement and the pH of the solution, but if you already have all that information, it is unlikely that you lack the concentration of both members of the complementary acid-base pair.
You use regular equilibrium calculations if you lack that information.
This works for titrations where a buffer is produced, but if you go to equivalence, you end up with an imaginary number in the equation because the -log(base/acid) becomes -log(base/0) or -log(0/acid) both of which are imaginary. At that point you need to go back to the law of mass action type calculations from which H-H is derived.
