Practice: Calculate the pH of the solution resulting from the titration of 100.0 mL of 0.08 M LiH with 150.0 mL of 0.10 M Hl.

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

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 a dilution factor equation.

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 #1: Calculate the pH of the solution resulting from the titration of 115.0 mL of 0.32 M KOH with 120.0 mL of 0.150 M HCl.

Practice: Calculate the pH of the solution resulting from the titration of 100.0 mL of 0.08 M LiH with 150.0 mL of 0.10 M Hl.

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 a dilution factor equation.

Concept #6: After the equivalence point there will remain an excess of strong base.

Example #2: Calculate the pH of the solution resulting from the titration of 250.0 mL of 0.370 M HBrO_{4} with 150.0 mL of 0.200 M NaH.

Example #3: Initially 7.1 g of HCl is titrated with 9.5 g of NaOH thereby making a solution that has a volume of 12.85 L. What is the pH of the solution?

a) 1.82

b) 2.48

c) 7.00

d) 11.52

e) 12.27