Question

You have 625 mL of an 0.23 M acetic acid solution. What volume (V) of 2.10 M NaOH solution must you add in order to prepare an acetate buffer of pH = 5.03? (The pKa of acetic acid is 4.76.)

Answer 1

For buffer pH we use the Hendersen-Hasselbalck equation,

pH = pKa + log[A-]/[HA]

Given,

pH required of buffer = 5.03

pKa of CH3COOH = 4.76

Feed values,

5.03 = 4.76 + log[A-]/[HA]

[A-]/[HA] = 1.8621

[A-] = 1.8621[HA]

Now we have,

Molarity of acetic acid = 0.23 M

Volume of acetic acid = 625 ml = 0.625 L

Molarity of NaOH = 2.10 M

The reaction of acetic acid and NaOH is as follows,

CH3COOH + NaOH ------> CH3COONa + H2O

So 1 mole of acetic acid would be neutralized with 1 mole of NaOH to form 1 mole of salt

moles of CH3COOH = M x V = 0.23 M x 0.625 L = 0.14375 moles

let x moles of NaOH is added, then,

remaining [HA] = 0.14375 - x moles

[A-] formed = x moles

we have from above,

[A-] = 1.8621[HA]

or,

x = 1.8621(0.14375-x)

x = 0.268 - 1.8621x

x = 0.094 moles = [A-] = moles of NaOH

Now,

moles of NaOH = M x V

or,

V = moles/M = 0.094/2.10 = 0.0448 L = 44.8 ml

Thus, 44.8 ml of 2.10 M NaOH is required to prepare a buffer of 5.03 pH.