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What are the concentrations of HOAc and OAc in 0.6M acetate buffer, ph 6.0? The Ka...

What are the concentrations of HOAc and OAc in 0.6M acetate buffer, ph 6.0? The Ka for acetic acid is 1.70 x 10^-3 (pKa=4.77) Can you please show in detail how you got the answer?

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Homework Answers

Answer #1

Using Hasselbach-Handerson equation, ie

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

Where, base = acetate

acid = acetic acid

Putting the respective values, we get

6 = 4.77 + log [acetate] / [acetic acid]

1.23 = log [acetate] / [acetic acid]

10^1.23 = [acetate] / [acetic acid]

16.9 = [acetate] / [acetic acid]

16.9 x [acetic acid] = [acetate]

Now,

[acetate] + [acetic acid] = 0.6 M

16.9 x [acetic acid] + [ acetic acid] = 0.6 M

17.9 [acetic acid] = 0.6 M

[acetic acid] = 0.0335 M

Now, [acetate] = [acetic acid] x 16.9

[acetate] = 0.0335 x 16.9

[acetate] = 0.566 M

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