Question

You need to prepare an acetate buffer of pH 6.08 from a 0.812 M acetic acid...

You need to prepare an acetate buffer of pH 6.08 from a 0.812 M acetic acid solution and a 2.07 M KOH solution. If you have 925 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a bufer of pH 6.08? The pKa of acetic acid is 4.76.

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Homework Answers

Answer #1

apply buffer equation

pH = pKa + log(A-/HA)

substitute known values

6.08 = 4.76 + log(A-/HA)

A-/HA = 10^(6.08 -4.76) = 20.892

A-. = 20.892*HA

so...

initial mmol of acid = MV = 0.812*925 = 751.1 mmol of acid

after adding mmol of base:

mmol of base = MV = 2.07*Vbase

then

acid left = 751.1 - 2.07*Vbase

conjugate formed = 2.07*Vbase

so...

relate

A-. = 20.892*HA

2.07*Vbase = (20.892)*(751.1 - 2.07*Vbase)

2.07/20.892*Vbase = 751.1 - 2.07*Vbase

(2.07/20.892 + 2.07) * Vbase = 751.1

Vbas = 751.1 / (2.07/20.892 + 2.07) = 346.27 mL of NaOH required

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