Question

Find the pH of a solution at equilibrium when 23.76 mL of 0.200 F NaOH is added to 25.00 mL of 0.100 F H3Glu+ (fully protonated glutamic acid, using the same symbol here as in the lecture video “6a Polyprotic acid-base titrations”).

For H3Glu+: pKa1 = 2.19, pKa2 = 4.25, pKa3 = 9.67.

Answer 1

According to Henderson-Hassebalch equation,

pH = pKa + Log(n_salt/n_acid)

= 4.25 + Log(2.252 / 0.248)

= 4.25 + 0.958

= 5.208

(Note: 23.46 mL*0.2 M = 4.752 mmol of NaOH, 25 mL*0.1 M = 2.5 mmol of H3Glu+

2.5 mmol of H3Glu+ reacts with 2.5 mmol of NaOH to form 2.5 mmol of H2Glu.

The remaining mmol of NaOH = 4.752 - 2.5 = 2.252 mmol

2.252 mmol of NaOH reacts with 2.252 mmol of HGlu- (n_salt)

The remaining mmol of H2Glu = 2.5 - 2.252 = 0.248 (n_acid)

And for the deprotonation, H2Glu HGlu-, pKa2 = 4.25)