Diprotic Acid Titrations

Concept #1: With the presence of 2 equivalence points we must now determine 2 equivalence volumes. 

Concept #2: Before the first equivalence point has been reached we have the formation of a buffer region. 

Concept #3: At the first equivalence point we now have an excess of the intermediate form for the diprotic acid. 

Concept #4: Before the second equivalence point has been reached we have the formation of another buffer region. 

Concept #5: At the second equivalence point we now have an excess of the basic form for the diprotic acid. 

Concept #6: After the second equivalence point there is an excess of strong base remaining. 

Example #1: Calculate the pH of 100 mL of a 0.25 M H₂CO₃ when 70.0 mL of 0.25 M NaOH are added. K_a1 = 4.3 x 10^-7 and Ka₂ = 5.6 x 10^-11. 

Example #2: Calculate the pH of 75.0 mL of a 0.10 M of phosphorous acid, H₃PO₃, when 80.0 mL of 0.15 M NaOH are added. K_a1 = 5.0 x 10^-2, Ka₂ = 2.0 x 10^-7. 

Example #3: Find the pH when 100.0 mL of a 0.1 M dibasic compound B (pK_b1 = 4.00; pK_b2 = 8.00) was titrated with 11 mL of a 1.00 M HCl. 

Example #4: Carbonic acid, H­₂CO₃, is a diprotic acid that dissociates, losing its two protons, to create bicarbonate, HCO₃^–, and carbonate, CO₃^2–, according to the following reactions given below:

H₂CO₃ (aq)  ⇌  HCO₃^– (aq)  +  H⁺ (aq)  ⇌  CO₃^2– (aq)  +  H⁺ (aq)

As a diprotic acid system, it has two dissociation constants that pK_a1 = 6.30 and pK_a2 = 10.30 for the two steps. In the reaction you titrate 50.0 mL solution of 0.50 M H­₂CO₃ with 1.00 M solution of NaOH. What would be the expected pH after the addition of 35 mL of the NaOH titrant?