1. Simplify Titration Curves that Buffer With Visual Analysis
Amy Zitzelberger
Hazel Park High School
Hazel Park, MI
Jamie Benigna
The Roeper School
Birmingham, MI

2. Graph originally from: MIT Open ClassroomHA A- H+ HA HA
Point A. Before titration begins: Ka = x2 / [HA] where x = [H+]

3. HA
OH- A-
Na+
H2O
Early Buffering Region. Before ½ equivalence point when buffering occurs: pH = pKa + log [A-] / [HA] where [A-] = OH- used so far in moles, and [HA] is remaining moles.

4. HA A-
OH-
H2O A-
Na+
Na+
Early Buffering Region. Before ½ equivalence point when buffering occurs: pH = pKa + log [A-] / [HA] where [A-] = OH- used so far in moles, and [HA] is remaining moles.

5. Point C: ½ equivalence pt: pH = pKa [A-] = [HA], and log(1) = 0
OH- A-
Na+
OH-
Na+ A-
Na+
H2O
Point C: ½ equivalence pt: pH = pKa [A-] = [HA], and log(1) = 0

6. HA A-
OH- A-
Na+
OH-
Na+ A-
Na+
OH- A-
Na+
Later Buffering Region. After ½ equivalence point when buffering occurs: pH = pKa + log [A-] / [HA] where [A-] = OH- used so far in moles, and [HA] is remaining moles.

7. HA A-
OH- A-
Na+
OH-
Na+ A-
Na+ A-
OH-
OH- A-
Na+
Na+
Later Buffering Region. After ½ equivalence point when buffering occurs: pH = pKa + log [A-] / [HA] where [A-] = OH- used so far in moles, and [HA] is remaining moles.

8. HA
Na+ A-
OH-
OH- A-
H2O
OH- HA A-
Na+
OH-
Na+ A-
Na+ A-
OH-
OH- A-
Na+
Na+
Point D. Equivalence point: Kb = x2 / [A-] where x = [OH-], and [A-] = moles OH- used over total volume.

9. HA
Na+
OH-
Na+ A-
OH-
OH- A- A-
Na+
OH-
Na+ A-
Na+ A-
OH-
OH- A-
Na+
Na+
Beyond the Equivalence Pt. pOH = ̶ log[OH-] where the OH- is only the moles of excess OH- divided by the total volume.

10. HA
Na+
OH-
Na+ A-
OH-
OH- A- A-
Na+
OH-
Na+ A-
Na+ A-
OH-
OH-
OH- A-
Na+
Na+
Na+
Beyond the Equivalence Pt. pOH = ̶ log[OH-] where the OH- is only the moles of excess OH- divided by the total volume.

11. To Summarize E
A: Before the Equiv. Pt.
B: Buffering Region
C: ½ way to Equiv. Pt.
D: Equivalence Point D
E: After Equiv. Pt. B C A

12. A For example: 1) A solution of 0.20 M nitrous acid. To Summarize E
A: Before the Equiv. Pt.
B: Buffering Region
C: ½ way to Equiv. Pt.
D: Equivalence Point D
E: After Equiv. Pt. A B C A
For example:
1) A solution of M nitrous acid.

13. To Summarize E
A: Before the Equiv. Pt.
B: Buffering Region
C: ½ way to Equiv. Pt.
D: Equivalence Point D
E: After Equiv. Pt. E B C A
25.0 mL of 0.10 M nitrous acid with 26.5 mL of 0.10 M NaOH added to it.

14. D 25.0 mL of 0.30 M KOH is added to 25.0 mL of 0.30 M HF. To Summarize
A: Before the Equiv. Pt.
B: Buffering Region
C: ½ way to Equiv. Pt.
D: Equivalence Point D
E: After Equiv. Pt. D B C A
25.0 mL of 0.30 M KOH is added to 25.0 mL of 0.30 M HF.

15. To Summarize E
A: Before the Equiv. Pt.
B: Buffering Region
C: ½ way to Equiv. Pt.
D: Equivalence Point D
E: After Equiv. Pt. C B C A
A titration that has 25.0 mL of 1.00 M acetic acid with 25.0 mL of 0.50 M NaOH.

16. To Summarize E
A: Before the Equiv. Pt.
B: Buffering Region
C: ½ way to Equiv. Pt.
D: Equivalence Point D
E: After Equiv. Pt. B B C A
When 30.0 mL of 0.20M propanoic acid has been titrated with 20.0 mL of 0.20 M NaOH.