1. NaOH  OH- + Na+ EXAM #1 M = mol #L 0.10 = mol 0.008 M = mol #L
EXAM #1a: Find moles of OH- in 8.00mL of 0.10M NaOH.
NaOH  OH- + Na+
M = mol #L
0.10 = mol
0.008
Mol (OH-) = 8 * 10-4
EXAM #1b: Find molarity of OH- AFTER of 8.00mL of 0.10M NaOH is titrated in. NOTE total volume = Vacid(50mL) + Vbase(8 mL) = 58 mL
M = mol #L
M = 8*10-4mol
.058L
M =

2. 0.005 mol 8.0*10-4mol -X X 0.005-X 8*10-4 M = mol #L 0.10 = mol 0.050L
EXAM #1c: Find the pH after 8.00 mL of 0.10 M OH- is added to the 50.0 mL acid sample.
FIRST FIND MOLES OF ACID ORIGINALLY PRESENT.
THEN EVALUATE NEUTRALIZATION STOICHIOMETRY; REMEMBER THE OH- IS LIMITING UP TO EQ POINT. Working in moles alleviates the need to keep track of dilutions!
M = mol #L
0.10 = mol
0.050L
Mol HA= 0.005 HA
OH-  A-
H2O
0.005 mol
8.0*10-4mol -X X
0.005-X
8*10-4
NOW USE THE FINAL MOLES
IN HENDERSON-HASSLEBACH
OH- IS LIMITING REAGENT AND IS CONSUMED TOTALLY

3. EXAM#1-c CONTINUED;
HA 
A- H+
0.005
8.0*10-4 -X X
0.005-X
8.0*10-4 +X
pH = pKa + LOG [A-]
[HA]
pH = LOG [8.0*10-4 ]
[0.0042]
Ok to work in moles for this equation.
pH = 8.49

4. EXAM #1d and e: AT THE pKa (HALFWAY) POINT:
Moles OH- added =:
mole HA neutralized
moles HA remaining
A- produced
*********AND mol OH- added = ½ HA INITIAL ******** HA
OH-  A-
H2O
0.005 mol
0.0025 -X X
0.005-X=0.0025
Neutralization stoichiometry: base added limits, OH- added = X and is completely consumed
#1e) Moles OH- at halfway point = ½ HA original ½ of =

5. pH = pKa + LOG [A-] [HA] pH = 9.21 + LOG [0.0025] [0.0025]
#1-f) AT THE pKa (HALFWAY POINT) point pH = pKa
pH = pKa + LOG [A-]
[HA]
pH = LOG [0.0025]
[0.0025]
pH = pKa = 9.21
#1-g) AT THE EQUIVALENCE POINT HA IS TOTALLY NEUTRALIZED TO 0.0 MOLE, AND A- IS THE ONLY ION REMINING.
OH- added total) = HA original = A- produced

6. 0.005 mol 0.0050 -X X 0.005 HA OH-  A- H2O M = mol #L
#1-h) AT THE EQUIVALENCE POINT HA IS TOTALLY NEUTRALIZED TO 0.0 MOLE, AND A- IS THE ONLY ION REMINING.
OH- added (total) = HA original = A- produced = mol HA
OH-  A-
H2O
0.005 mol
0.0050 -X X
0.005
M = mol #L
0.10 = mol
X L
vol OH- = .050 L = 50.0mL

7. #1-g) AT THE EQUIVALENCE POINT HA IS TOTALLY NEUTRALIZED TO 0
#1-g) AT THE EQUIVALENCE POINT HA IS TOTALLY NEUTRALIZED TO 0.0 MOLE, AND A- IS THE ONLY ION REMINING.
OH- added total) = HA original = A- produced, The Ph at equivalence is calculated by the hydrolysis equilibrium of A-.
THE EQUIVALENCE POINT STOICHIOMETRY GIVES THE A- MOLARITY FOR THE EQUIVALENCE
(HYDROLYSIS OF A-) EQUILIBRIUM. HA
OH-  A-
H2O
0.005 mol
0.0050 -X X
0.005
HYDROLYSIS EQUILIBRIUM OF A-
AT EQUIVALENCE PT. A-
HOH  HA
OH-
0.005 mol
IGNORE SOLVENT -X X
0.005-X

8. HYDROLYSIS EQUILIBRIUM OF A- AT EQUIVALENCE POINT A- HOH  HA OH-
0.005 mol
IGNORE SOLVENT -X X
0.005-X
Kb = Kw/Ka = [HA][OH-]
[A-]
Kb = 1.6 x 10-5 = (X)(X) ; X = 8.86 x 10-4 = [OH-]
(0.005/0.1L)
MUST CONVERT MOLES TO MOLARITY, TOTAL VOLUME IS 0.10 L
(0.05L ACID AND 0.05L BASE)
pOH- = -LOG(8.86 x 10-4) = 3.05
pH = at equivalence