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Topic: ph=pKa  (Read 6520 times)

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Offline fk378

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ph=pKa
« on: May 06, 2008, 03:33:05 PM »
For the Arrhenius equation ph=pKa + log ([base]/[acid])
why does ph=pKa halfway to the equivalence point?  If it is halfway, and not AT the equivalence point, why would the concentration of the base equal the concentration of the acid?

Offline Borek

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Re: ph=pKa
« Reply #1 on: May 06, 2008, 03:54:38 PM »
Write neutralization reaction equation for - say - acetic acid.
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Offline Astrokel

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Re: ph=pKa
« Reply #2 on: May 07, 2008, 04:24:34 AM »
Let's say in a conical flask, you have CH3COOH and in burette you have NaOH.

Within the conical flask,

Before equivalence point: CH3COOH and some CH3COO- salt
At equivalence point: CH3COO- only
After equivalence point: CH3COO- + NaOH

Hence, buffer region happens before equivalence point due to presence of CH3COOH and CH3COO-Na+.

However, in order for maximum buffer capacity occurs, [CH3COO-] = [CH3COOH] and this occurs half way before equivalence point. It makes sense to be half way as at equivalence point you get CH3COO- only, and at half way before equivalence, you get [salt] = [acid], resulting in pH = pKa + lg 1 = pKa.

However it is not always at halfway before equivalence. If you are titrating NaOH with CH3COOH, you get pH = pKa occurs at twice the volume of the equivalence point volume.


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