Question

The titration of weak diprotic acid H₂A with strong base is conducted. What is the pH of the solution at the following volumes, 0,10,25,50,60,75,100,125,and 150mL if K^_a1=1.0x10^-4 and K_a²=3.5x10^-4 and the concentrations for both HA^- and OH^- are 0.1M?

Answer 1

The titration of weak diprotic acid 50ml of H2A with strong base is conducted.
What is the pH of the solution at the following volumes, 0,10,25,50,60,75,125,and 150mL
if Ka1=1.0x10^-4 and Ka2 =3.5x10^-7and
the concentrations for both HA- and OH- are 0.1M?

[HA-] + H2O = [H2A] + [OH-]; Kb2 = Kw/Ka1 = [H2A]*[OH-]/[HA-]
[H2A] = [HA-]/[OH-]* Kw/Ka1 = [HA-]*[H3O+]/Ka1;                              

[HA-] + H2O = [A-2] + [H3O+]; Ka2 = [A-2]*[H3O+]/[HA-]
[A-2]=Ka2*[HA-]/[H3O+]

For titration with NaOH
Mass balance equation:        [NaHA]= [H2A] + [HA-] + [A-2]
Charge balance equation:     [NaHA] + [H3O+] = [HA-] + 2[A-2] + [OH-]
From above two equations, [H3O+] = [A2-] + [OH-] - [H2A]

Solving above equations, [H3O+] = [(Ka2*[HA-]+Kw)/(1+[HA-]/Ka1)]^0.5
Ka1=1.0x10^-4 and Ka2 =3.5x10^-7
Ka1/Ka2 = Ka1=(1.0x10^-4)/3.5x10^-7 = 0.286E3

[HA-] = 0.1M; [OH-] = 0.1M

[H3O+] = [(3.5E-7*0.1+1.0E-14)/(1+0.1/1.0E-4)]^0.5
= [(3.5E-8+1.0E-14)/(1+ 1.0E3)]^0.5 = 5.91E-6
pH = -Log [H3O+] = 5.23