1. See summary: top of p.778 in textbook

2. Half-Equivalence Point
Volume of titrant added is half the volume of the equivalence point
pH = pKa
Why?
pH = pKa + log (CB / A)
At the half equivalence point, there is the same amount of original acid and conjugate base (formed from neutralization).

3. Half-Equivalence Point
Volume of titrant added is half the volume of the equivalence point
Moles of strong base added is half that of original weak acid
HA + OH-  A H2O
pH = pKa + log
.25
pH = pKa
@ half-equivalence point

4. Base Titrated with Strong Acid
                                                                                                

5. Weak Base Titrated with Strong Acid
Initial pH:
use Kb for weak base, ICE table
Between initial and equiv pt:
NH3 + H3O+  NH4+ + H2O
Complete reaction stoichiometry.
Use in H-H equation.

6. Weak Base Titrated with Strong Acid
3. At equiv pt:
All NH3 reacted. NH4+ has been formed.
NH H2O  NH3 + H3O+
Use Ka and ICE table
4. After equiv pt:
Any additional H+ added has nothing to react with. Use excess H+ to calc pH

8. Acid-Base Titrations
Add solution (either the acid or the base) from buret to solution in flask
The concentration of one solution is known. One concentration is unknown.
Often, the goal is to find the unknown concentration.

9. What is the concentration of NaOH when 30
What is the concentration of NaOH when 30.0mL of the base neutralizes 26mL of 0.50M HCl?
When 25 mL of HCl is titrated with 6.00 M Ba(OH)2, it takes 39.3 mL of base to reach the end point. What is the pH of the acid solution?

10. What is the concentration of NaOH when 30
What is the concentration of NaOH when 30.0mL of the base neutralizes 26mL of 0.50M HCl?

11. When 25 mL of HCl is titrated with 6. 00 M Ba(OH)2, it takes 39
When 25 mL of HCl is titrated with 6.00 M Ba(OH)2, it takes 39.3 mL of base to reach the end point. What is the pH of the acid solution?