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Please decribe how you would make up 800 ml of .10 M acetate buffer pka 4.76...

Please decribe how you would make up 800 ml of .10 M acetate buffer pka 4.76 of Ph 4.2 using .1 M ch3cooh and .1 M ch3 coona
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Answer #1

For buffer we use Hendersen-Hasselbalck equation,

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

4.2 = 4.76 + log([acetate]/acetic acid])

[acetate] = 0.275[acetic acid]

we know,

[acetate] + [acetic acid] = 0.1 M x 0.8 L = 0.08 mols

0.275[acetic acid] + [acetic acid] = 0.08

[acetic acid] = 0.024 mols

[acetate] = 0.056 mols

Volume of 0.1 M CH3COOH needed for buffer preparation = 0.024/0.1 = 0.24 L

Volume of 0.1 M CH3COONa needed for buffer preparation = 0.056/0.1 = 0.56 L

So, we would mix 240 ml of 0.1 M CH3COOH and 560 ml of 0.1 M CH3COONa

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