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2 of 2 What concentration of sodium acetate do you need to prepare a pH=4.500 buffer...

2 of 2 What concentration of sodium acetate do you need to prepare a pH=4.500 buffer from 0.137 M acetic acid (Ka=1.75E-5)

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Answer #1

Given Ka = 1.75*10-5

We know pKa = -log Ka = -log(1.75*10-5)= -log1.75 +5 log10

pKa = -0.24 + 5*1 = 4.76

We know from Henderson equaqution

pH = pKa + log{ [conjugate base]/[Acid]}

4.5 = 4.76 + log {[sodium acetate] / [acetic acid]}

4.5 - 4.76 = log {[sodium acetate] / [acetic acid]}

-0.24 = log {[sodium acetate] / [acetic acid]}

10^-0.24 = {[sodium acetate] / [acetic acid]}

0.575 = {[sodium acetate] / [acetic acid]}

[sodium acetate] = 0.575 * [ acetic acid] = 0.575 * 0.137 M =0.079 M

Concentration of sodium acetate = 0.079 M

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