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You have a 250.-mL sample of 1.22 M acetic acid (Ka = 1.8 × 10–5). Calculate...

You have a 250.-mL sample of 1.22 M acetic acid (Ka = 1.8 × 10–5). Calculate the pH of the best buffer.

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Answer #1

Best buffer is the one that shows zero pH change ,when small amounts of strong acid or base is added to it.This is so because the best buffer has equal concentrations of weak acid and its conjugate base.

So that any small amount of strong acid or base (say dX) when added to it changes the ratio of the concentration of the conjugate base and acid negligibly,

[conjugate base]/[weak acid]1

So, when strong acid is added, it neutralizes equal amount of conjugate base and produces equal amount of weak acid

([conjugate base]-dX)/([weak acid]+dX ) which is approximately equal to 1

According to henderson hasselbach equation,

pHpka +log [conjugate base]/[weak acid]

So, for the given acetic acid sample,

pHpka +log [acetate ions]/[acetic acid]

for best buffer, [acetate ions][acetic acid]

So pHpka + log 1pka +0

pHpka-log ka-log (1.8*10^-5)4.74

pH4.74

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