Question

You need to prepare an acetate buffer of 5.78 from a 0.626 M acetic acid solution and a 2.06 M KOH solution. If you have 825 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 5.78? The pKa of acetic acid is 4.76.

Answer 1

the buffer equation

pH = pKa + log(A-/HA)

initially

mmol of acid = MV = 0.626*825 = 516.45 mmol of acid

we need:

pH 5.78 solution

so find

A/HA ratio

pH = pKa + log(A-/HA)

5.78= 4.75 + log(A-/HA)

ratio = 10^(5.78-4.75) = 10.715

[A-] = 10.715*[HA]

so..

for that pH, we need to create conjguate, we can do this adding NaOH

so

mmol of base required = MV = 2.06 * Vbase

mmol of acid after addition of base = 516.45 - 2.06 * Vbase

mmol of conjugate formed after addition of base = 0 + 2.06 * Vbase

we know that ratio is

[A-] = 10.715*[HA]

so

2.06 * Vbase = 10.715*(516.45 - 2.06 * Vbase)

5533.76175 - 22.0729 *Vbase = 2.06*Vbase

Vbase ( 22.0729 +2.06) = 5533.76175

Vbase = 5533.76175 /   ( 22.0729 +2.06) = 229.3036 mL required