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A 0.10 M butanoic acid (butyric acid) solution has a pH of 5.2. To the nearest...

A 0.10 M butanoic acid (butyric acid) solution has a pH of 5.2. To the nearest hundredth of a unit, what fraction of the butanoic acid/butanoate molecules are in the basic form? (The pKa of butanoic acid is 4.82.)

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Answer #1

Henderson - Hasselbalch equation is

pH = pKa + log( [ Base ] /[ acid ])

pH = 5.20

pKa of butanoic acid = 4.82

Butanate is the basic form

Butanoic acid is the acid form

5.20 = 4.82 + log([Butanate]/[Butanoic acid ])

log([Butanate]/[Butanoic acid ]) = 0.38

[ Butanate ]/ [ Butanoic acid ] = 2.40

So , for 1mole of Butanoic acid 2.40 mole of Butanate

[ Butanate ] = 2.40 × [ Butanoic acid ]

[Butanoic acid] + [Butanate] = 0.10M

[ Butanoic acid ] + 2.40[ Butanoic acid ] =0.10M

3.40[ Butanioc acid ] = 0.10M

[ Butanoic acid ] = 0.0294M

[ Butanate ] = 0.10M - 0.0294M =0.0706M

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