Question

How many mL of 1M NaOH do you need to add to 300 mL of 1 M acetic acid to make a buffer with a pH of 5.46? Water will be added to bring the total volume of the buffer to 1 L. The pKa of acetic acid is 4.76.

Answer 1

Let us say that Z mL of 1 M NaOH need to be added.

Z mL of 1 M NaOH = Z mL x 1 M = Z mmol of NaOH

300 mL of 1 M acetic acid = 300 mL x 1 M = 300 mmol of acetic acid

Now, Z mmol of NaOH reacts with Z mmol of acetic acid to form Z mmol of sodium acetate.

Remaining mmol of acetic acid = (300 - Z) mmol

Total volume of the buffer = 1 L = 1000 mL       

Hence, the concentration of acetate (CH₃COO^-) in the solution = Z mmol / 1000 mL = Z/1000 M

New concentration of acetic acid (CH₃COOH) = (300 - Z) mmol / 1000 mL = (300 - Z)/1000 M

Now,

[CH₃COO^-]/[CH₃COOH] = Z/(300 - Z)

From Henderson-Hasselbalch equation,

pH = pK_a + log[CH₃COO^-]/[CH₃COOH]

or, 5.46 = 4.76 + log Z/(300 - Z)

or, log Z/(300 - Z) = 0.7

or, Z/(300 - Z) = 10^0.7

or, Z/(300 - Z) = 5.01

or, Z = 1503 - 5.01Z

or, 6.01Z = 1503

or, Z = 250 mL

Hence, the volume of NaOH that need to be added = 250 mL