Chemistry Help Please

Question

Calculate the volume of 0.2M NaOH required to bring the pH of 20 mL of 5% acetic acid solution to 4.74, which is the pKa of acetic acid.

Answer 1

20 ml x 5 g acetic acid / 100 ml = 1 g acetic acid

moles acetic acid = 1 g x 1 mol / 60.1 g = 0.0166 mols acetic acid (CH3COOH)

The reaction taking place to form the buffer is as follows:

CH3COOH + NaOH ==> CH3COONa + H2O

Using the Henderson Hasselbalch equation:

pH = pKa + log [salt] / [acid]

4.74 = 4.74 + log [salt]/[acid]

For the pH to equal the pKa, the [salt] must equal the [acid]

To accomplish this, we can add 1/2 the moles of the CH3COOH thus converting it to CH3COONa

CH3COOH + NaOH ==> CH3COONa + H2O

0.0166..........0.0083.............0...............Initial

0.0166-0.0083...-0.0083.......+0.0083...Change

0.0083.............0...................0.0083......Equilibrium

So, we need to add 00083 mols of NaOH

Volume of 2 M needed to do this is...

0.0083 mol x 1 L / 0.2 mol = 0.0415 L = 41.5 mls