Question

You need to prepare an acetate buffer of pH 6.50 from a 0.644 M acetic acid solution and a 2.75 M KOH solution. If you have 575 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 6.50? The pKa of acetic acid is 4.76.

Answer 1

Let volume of KOH be V mL

mol of KOH added = 2.75*V mmol

Before adding KOH

Before Reaction:

mol of CH3COO- = 0 mmol

mol of CH3COOH = 0.644 M *575.0 mL

mol of CH3COOH = 370.3 mmol

2.75*V KOH will react with 2.75*V of CH3COOH to form extra 2.75*V of base

After adding KOH

mol of CH3COOH = 370.3-2.75*V mmol

mol of CH3COO- = 0+2.75*V mmol

we have below equation to be used:

This is Henderson–Hasselbalch equation

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

6.5 = 4.76+log {[CH3COO-]/[CH3COOH]}

log {[CH3COO-]/[CH3COOH]} = 1.74

[CH3COO-]/[CH3COOH] = 54.95

So,

(0+2.75*V)/(370.3-2.75*V) = 54.9541

0+2.75*V = 20349.4986 - 151.1237*V

(2.75+151.1237)*V = 20349.4986-0

V = 132 mL

Answer: 132 mL