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Author Topic: ph=pKa  (Read 3485 times)

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fk378

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ph=pKa
« on: May 06, 2008, 09:33:05 AM »

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?
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Borek

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Re: ph=pKa
« Reply #1 on: May 06, 2008, 09:54:38 AM »

Write neutralization reaction equation for - say - acetic acid.
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Astrokel

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Re: ph=pKa
« Reply #2 on: May 06, 2008, 10:24:34 PM »

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.


;D
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