Ch.10 - Acid-Base TitrationsWorksheetSee all chapters
All Chapters
Ch.1 - Chemical Measurements
Ch.2 - Tools of the Trade
Ch.3 - Experimental Error
Ch.4 + 5 - Statistics, Quality Assurance and Calibration Methods
Ch.6 - Chemical Equilibrium
Ch.7 - Activity and the Systematic Treatment of Equilibrium
Ch.8 - Monoprotic Acid-Base Equilibria
Ch.9 - Polyprotic Acid-Base Equilibria
Ch.10 - Acid-Base Titrations
Ch.11 - EDTA Titrations
Ch.12 - Advanced Topics in Equilibrium
Ch.13 - Fundamentals of Electrochemistry
Ch.14 - Electrodes and Potentiometry
Ch.15 - Redox Titrations
Ch.16 - Electroanalytical Techniques
Ch.17 - Fundamentals of Spectrophotometry
BONUS: Chemical Kinetics

In a Strong Acid-Strong Base Titration we have the formation of a sigmodial shaped graph. 

Strong Base-Strong Acid Titration

Concept #1: In these series of titrations the strong base is the starting solution or analyte and the strong acid is the titrant. 

Concept #2: At all points before the equivalence point we calculate the amount of strong base remaining through an ICF Chart.

Example #1: At the equivalence point for two strong species the pH will equal 7.00

Concept #3: After the equivalence point there will remain an excess of strong acid. 

Example #2: Calculate the pH of the solution resulting from the titration of 150.0 mL of 0.20 M NaOH with 80.0 mL of 0.15 M HBr.

Example #3: Calculate the pH of the solution resulting from the titration of 100.0 mL of 0.30 M LiH with 150.0 mL of 0.40 M Hl.

Strong Acid-Strong Base Titration

Concept #4: In these series of titrations the strong acid is the starting solution or analyte and the strong base is the titrant. 

Concept #5: At all points before the equivalence point we calculate the amount of strong acid remaining through an ICF Chart.

Concept #6: At the equivalence point for two strong species the pH will equal 7.00

Concept #7: After the equivalence point there will remain an excess of strong base. 

Example #4: Calculate the pH of the solution resulting from the titration of 50.0 mL of 0.10 M HI with 20.0 mL of 0.30 M NaOH.

Example #5: Calculate the pH of the solution resulting from the titration of 90.0 mL of 0.40 M HClO3 with 50.0 mL of 0.50 M KOH.