1. PART A. 10 M Acetic Acid (5. 00 mL of 1. 0 M Acetic Acid diluted to 50
PART A 0.10 M Acetic Acid (5.00 mL of 1.0 M Acetic Acid diluted to 50.0 mL)
EX6-1 (of 14)

2. 1 Theoretical pH of 0.10 M Acetic Acid
HC2H3O2 (aq) H2O (l) ⇆ H3O+ (aq) C2H3O2- (aq)
Initial M’s
Change in M’s
Equilibrium M’s
0.10
- x
+ x
+ x x x x
Ka = [H3O+][C2H3O2-]
____________________
[HC2H3O2]
1.8 x 10-5 = x2
___________
(0.10 – x)
1.34 x = x
2.87 = pH
EX6-2 (of 14)

3. 2 Theoretical pH with the Addition of HCl (1.00 mL of 1.0 M HCl)
1.0 mol HC2H3O2 x L
_____________________ L
= mol HC2H3O2
1.0 mol HCl x L
_______________ L
= mol HCl
= mol H3O+
The strong acid does not have a base to react with
The pH comes from the molarity of unreacted strong acid in the solution
mol H3O+
_______________________ L
= M H3O+
pH = -log( M) =
EX6-3 (of 14)

4. the weak acid neutralized the weak conjugate base
3 Theoretical pH with the Addition of NaOH (1.00 mL of 1.0 M NaOH)
1.0 mol HC2H3O2 x L
_____________________ L
= mol HC2H3O2
1.0 mol NaOH x L
__________________ L
= mol NaOH
= M OH-
Strong bases react completely with acids
OH- (aq) HC2H3O2 (aq) → H2O (l) C2H3O2- (aq)
the weak acid neutralized
by the strong base
the weak conjugate base
Is produced
EX6-4 (of 14)

5. 3 Theoretical pH with the Addition of NaOH (1.00 mL of 1.0 M NaOH)
OH- (aq) HC2H3O2 (aq) → H2O (l) C2H3O2- (aq)
Initial mol’s
Change in mol’s
Final mol’s – –
pH = pKa + log nC2H3O ___________
nHC2H3O2
= log ( mol) _________________
( mol) =
EX6-5 (of 14)

6. PART B. 1 M Sodium Acetate (5. 00 mL of 1
PART B 0.1 M Sodium Acetate (5.00 mL of 1.0 M Sodium Acetate Diluted to 50.0 mL)
EX6-6 (of 14)

7. 4 Theoretical pH of 0.10 M Sodium Acetate
0.10 M NaC2H3O2
 M Na+ and 0.10 M C2H3O2-
weak base
KaKb = Kw
Kb = Kw
____ Ka
= x 10-14
______________
1.8 x 10-5
= x 10-10
EX6-7 (of 14)

8. 4 Theoretical pH of 0.10 M Sodium Acetate
C2H3O2- (aq) H2O (l) ⇆ HC2H3O2 (aq) OH- (aq)
Initial M’s
Change in M’s
Equilibrium M’s
0.10
- x
+ x
+ x x x x
Kb = [HC2H3O2] [OH-]
____________________
[C2H3O2-]
5.56 x = x2
___________
(0.10 – x)
7.46 x 10-6 = x
5.13 = pOH
8.87 = pH
EX6-7 (of 14)

9. the weak conjugate base neutralized by the strong acid
5 Theoretical pH with the Addition of HCl (1.00 mL of 1.0 M HCl)
1.0 mol C2H3O2- x L
____________________ L
= mol C2H3O2-
1.0 mol HCl x L
_______________ L
= mol HCl
= mol H3O+
Strong acids react completely with bases
H3O+ (aq) C2H3O2 - (aq) → H2O (l) HC2H3O2 (aq)
the weak conjugate base neutralized by the strong acid
the weak acid
is produced
EX6-8 (of 14)

10. 5 Theoretical pH with the Addition of HCl (1.00 mL of 1.0 M HCl)
H3O+ (aq) C2H3O2 - (aq) → H2O (l) HC2H3O2 (aq)
Initial mol’s
Change in mol’s
Final mol’s – –
pH = pKa + log nC2H3O ___________
nHC2H3O2
= log ( mol) __________________
( mol) =
EX6-9 (of 14)

11. 6 Theoretical pH with the Addition of NaOH (1.00 mL of 1.0 M NaOH)
1.0 mol C2H3O2- x L
____________________ L
= mol C2H3O2-
1.0 mol NaOH x L
__________________ L
= mol NaOH
= mol OH-
The strong base does not have an acid to react with
The pH comes from the amount of unreacted strong base in the solution
pOH = -log( M) =
pH =
EX6-10 (of 14)

12. PART C 0.1 M Acetic Acid / 0.1 M Sodium Acetate
(5.00 mL 1.0 M Acetic Acid and 5.00 mL 1.0 M Sodium Acetate diluted to 50.0 mL)
EX6-11 (of 14)

13. 7 Theoretical pH of 0.10 M Acetic Acid / 0.10 M Sodium Acetate
pH = pKa + log [C2H3O2-] _____________
[HC2H3O2]
= log (0.10 M) ___________
(0.10 M) =
EX6-12 (of 14)

14. 8 Theoretical pH with the Addition of HCl (1.00 mL of 1.0 M HCl)
9 Theoretical pH with the Addition of NaOH (1.00 mL of 1.0 M NaOH)
EX6-13 (of 14)

15. EX6-14 (of 14)