Question

Calculate the change in pH when 64.00 mL of a 0.575 M solution of NaOH is added to 1.00 L of a solution that is 1.00 M in sodium acetate and 1.00 M in acetic acid.

Answer 1

Given that volume of buffer = 1.0 L

molarity of acetic acid = 0.1 M

molarity of sodium acetate = 0.1 M

[acetic acid] = molarity x volume = 0.1 x 1.0 L = 0.1 mol

    [sodium acetate] = molarity x volume = 0.1 x 1.0 L = 0.1 mol

[NaOH] = molarity x volume = 0.575 x 0.064 L = 0.0368 mol

Original pH = pKa + log [sodium acetate]/[acetic acid] ----------(1)

pH after addition of NaOH:

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

Hence,

new pH = pKa + log {[CH3COONa] + [NaOH]/[CH3COOH]-[NaOH]} --------(2)

Since NaOH is a strong base, pH will be increased compared to original pH.

To get change in pH, subtract (1) from (2).

Then,

change in pH =  pKa + log {[CH3COONa] + [NaOH]/[CH3COOH]-[NaOH]} - {pKa+log [CH3COONa]/[CH3COOH]}

=   log {[CH3COONa] + [NaOH]/[CH3COOH]-[NaOH]} - { +log [CH3COONa]/[CH3COOH] }

= log { [0.1 + 0.0368]/ [0.1-0.0368]} - log [0.1/0.1]}

= 0.33

Therefore, change in pH = 0.33