Question

You need to prepare an acetate buffer of pH 6.36 from a 0.714 M acetic acid solution and a 2.69 M KOH solution. If you have 475 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 6.36? The pKa of acetic acid is 4.76.

Answer 1

Let volume of KOH be V mL

mol of KOH added = 2.69*V mmol

Before adding KOH

Before Reaction:

mol of CH3COO- = 0 mmol

mol of CH3COOH = 0.714 M *475.0 mL

mol of CH3COOH = 339.15 mmol

2.69*V KOH will react with 2.69*V of CH3COOH to form extra 2.69*V of base

After adding KOH

mol of CH3COOH = 339.15-2.69*V mmol

mol of CH3COO- = 0+2.69*V mmol

we have below equation to be used:

This is Henderson–Hasselbalch equation

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

6.36 = 4.76+log {[CH3COO-]/[CH3COOH]}

log {[CH3COO-]/[CH3COOH]} = 1.6

[CH3COO-]/[CH3COOH] = 39.81

So,

(0+2.69*V)/(339.15-2.69*V) = 39.8107

0+2.69*V = 13501.8047 - 107.0908*V

(2.69+107.0908)*V = 13501.8047-0

V = 123 mL

Answer: 123 mL