Question

You need to prepare an acetate buffer of pH 6.37 from a 0.667 M acetic acid solution and a 2.84 M KOH solution. If you have 675 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 6.37? The pKa of acetic acid is 4.76.

Answer 1

Let volume of KOH be V mL

mol of KOH added = 2.84*V mmol

Before adding KOH

Before Reaction:

mol of CH3COO- = 0 mmol

mol of CH3COOH = 0.667 M *675.0 mL

mol of CH3COOH = 450.225 mmol

2.84*V KOH will react with 2.84*V of CH3COOH to form extra 2.84*V of base

After adding KOH

mol of CH3COOH = 450.225-2.84*V mmol

mol of CH3COO- = 0+2.84*V mmol

we have below equation to be used:

This is Henderson–Hasselbalch equation

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

6.37 = 4.76+log {[CH3COO-]/[CH3COOH]}

log {[CH3COO-]/[CH3COOH]} = 1.61

[CH3COO-]/[CH3COOH] = 40.74

So,

(0+2.84*V)/(450.225-2.84*V) = 40.738

0+2.84*V = 18341.2786 - 115.696*V

(2.84+115.696)*V = 18341.2786-0

V = 155 mL

Answer: 155 mL