1. Equivalence point - point at which the titrant and the analyte are present in stoichometrically equivalent amounts

2. throughout the reaction1)
0.2 M HNO3
[H+] = 0.2 M
pH = 0.70 2)
= 1.00 x 10-3 mols NaOH
10 mL of 0.1 M NaOH
50 mL of 0.2 M HNO3
= 1.00 x 10-2 mols HNO3 *
i doesn’t change
throughout the reaction
HNO3 +NaOH → NaNO3 + H2O
HNO3
NaOH
NaNO3
H2O i
1.00 x 10-2 mols
1.00 x 10-3 mols
0 mols
0 mols c -y -y +y +y e
9.0 x 10-3 mols
0 mols
1.0 x 10-3 mols
1.0 x 10-3 mols

3. = 0.15 M HNO3 pH = 0.82 HNO3 + NaOH → NaNO3 + H2O 3)
9.0 x 10-3 mols HNO3
pH = 0.82
50 mL + 10 mL = 60 mL
0.06 L 3)
= 1.00 x 10-2 mols NaOH
100 mL of 0.1 M NaOH
50 mL of 0.2 M HNO3
= 1.00 x 10-2 mols HNO3
* duh
HNO3 + NaOH → NaNO3 + H2O
HNO3
NaOH
NaNO3
H2O i
1.00 x 10-2 mols
1.00 x 10-2 mols
0 mols
0 mols c -y -y +y +y e
0 mols
0 mols
1.0 x 10-2 mols
1.0 x 10-2 mols

4. pH = 7.00 HNO3 +NaOH → NaNO3 + H2O = 0.025 M NaOH [OH-] = 0.0254)
[H+] = [OH-] equivalence point
= 1.50 x 10-2 mols NaOH
150 mL of 0.1 M NaOH
50 mL of 0.2 M HNO3
* duh
= 1.00 x 10-2 mols HNO3
HNO3 +NaOH → NaNO3 + H2O
HNO3
NaOH
NaNO3
H2O i
1.00 x 10-2 mols
1.50 x 10-2 mols
0 mols
0 mols c -y -y +y +y e
= M NaOH
0 mols
5.0 x 10-3 mols
1.0 x 10-2 mols
1.0 x 10-2 mols
5.0 x 10-3 mols NaOH
[OH-] = 0.025
150 mL + 50 mL = 200 mL
pH = 12.4
0.2 L

5. Titration of a weak acid with a strong base ∙ before equivalence point use Ka equivalence point and beyond use Kb

6. HC2H3O2 + NaOH → NaC2H3O2 + H2O HC2H3O2 ⇆ H+ + C2H3O2-1)
HC2H3O2 + NaOH → NaC2H3O2 + H2O
Neutralization reaction
(this does not occur yet., no NaOH)
HC2H3O2 ⇆ H+ + C2H3O2-
Equilibrium dissociation
50 mL of 0.10 M HC2H3O2
HC2H3O2 H+
C2H3O2-
[i]
0.10 M
0 M
0 M c -y +y +y
[e] y y y
pH = 2.87
y = 1.34 x 10-3 M
[H+] = 1.34 x 10-3 M

7. throughout the reaction2)
HC2H3O2 + NaOH → NaC2H3O2 + H2O
= 1.00 x 10-3 mols NaOH
50 mL of 0.10 M HC2H3O2
= 5.00 x 10-3 mols HC2H3O2
10 mL of 0.10 M NaOH *
i doesn’t change
throughout the reaction
HC2H3O2
NaOH
NaC2H3O2
H2O i
5.00 x 10-3 mols
1.00 x 10-3 mols
0 mols
0 mols c -y -y +y +y e
4.00 x 10-3 mols
0 mols
1.0 x 10-3 mols
1.0 x 10-3 mols

8. Reaction to completion yields:
= M HC2H3O2
4.0 x 10-3 mols HC2H3O2
0.06 L
10 mL + 50 mL = 60 mL
= M C2H3O2-
1.0 x 10-3 mols C2H3O2-
0.06 L

9. HC2H3O2 ⇆ H+ + C2H3O2- pH = 4.14 [H+] = 7.22 x 10-5 M HC2H3O2 H+
[i]
0.067 M
0 M M c -y +y +y
[e] y y y
y = 7.22 x 10-5 M
[H+] = 7.22 x 10-5 M
pH = 4.14

10. @ equivalence point and beyond use Kb4)
HC2H3O2 + NaOH → NaC2H3O2 + H2O
= 5.00 x 10-3 mols NaOH
50 mL of 0.10 M HC2H3O2
= 5.00 x 10-3 mols HC2H3O2
50 mL of 0.10 M NaOH
HC2H3O2
NaOH
NaC2H3O2
H2O i
5.00 x 10-3 mols
5.00 x 10-3 mols
0 mols
0 mols c -y
at equivalence point!! -y +y +y e
0 mols
0 mols
5.0 x 10-3 mols
5.0 x 10-3 mols
@ equivalence point and beyond use Kb

11. Reaction to completion yields:
pH = 8.72
= M C2H3O2-
5.0 x 10-3 mols C2H3O2-
C2H3O2- ⇆ HC2H3O2 + OH-
0.10 L
C2H3O2-
HC2H3O2
OH-
[i]
0.050 M
0 M
0 M c -y +y +y
[e] y y y
pOH = 5.28
y = 5.3 x 10-6 M
[OH-] = 5.3 x 10-6 M

12. @ equivalence point and beyond use Kb5)
HC2H3O2 + NaOH → NaC2H3O2 + H2O
= 6.00 x 10-3 mols NaOH
50 mL of 0.10 M HC2H3O2
= 5.00 x 10-3 mols HC2H3O2
60 mL of 0.10 M NaOH
HC2H3O2
NaOH
NaC2H3O2
H2O i
5.00 x 10-3 mols
6.00 x 10-3 mols
0 mols
0 mols c -y -y +y +y e
0 mols
1.0 x 10-3 mols
5.0 x 10-3 mols
5.0 x 10-3 mols
@ equivalence point and beyond use Kb

13. Reaction to completion yields:
pH = 11.96
= M C2H3O2-
5.0 x 10-3 mols C2H3O2-
= M OH-
0.11 L
C2H3O2- ⇆ HC2H3O2 + OH-
1.0 x 10-3 mols OH-
0.11 L
C2H3O2-
HC2H3O2
OH-
[i]
0.045 M
0 M
0.009 M c -y +y +y
[e] y y y
pOH = 2.04
y = 3.11 x 10-9 M
[OH-] = 9.0 x 10-3 M

14. Titration of a weak base with a strong acid ∙ before equivalence point use Kb equivalence point and beyond use Ka

15. HCl + NH3 → NH4Cl NH3 ⇆ NH4+ + OH- 100 mL of 0.05 M NH3 pH = 11.0 1)
Neutralization reaction
(this does not occur yet., no HCl)
NH3 ⇆ NH4+ + OH-
Equilibrium dissociation
100 mL of 0.05 M NH3
NH3
NH4+
OH-
[i]
0.05 M
0 M
0 M c -y +y +y
[e] y y y
pH = 11.0
y = 9.5 x 10-4 M
[OH-] = 9.5 x 10-4 M

16. HCl + NH3 → NH4Cl 15 mL of 0.10 M HCl 100 mL of 0.05 M NH3 2)
= 1.5 x 10-3 mols HCl
100 mL of 0.05 M NH3
= 5.00 x 10-3 mols NH3
15 mL of 0.10 M HCl
HCl
NH3
NH4Cl i
1.5 x 10-3 mols
5.00 x 10-3 mols
0 mols c -y -y +y e
0 mols
3.5 x 10-3 mols
1.5 x 10-3 mols

17. Reaction to completion yields:
= M NH4+
1.5 x 10-3 mols NH4+
0.115 L
15 mL mL = 115 mL
= M NH3
3.5 x 10-3 mols NH3
0.115 L

18. NH3 ⇆ NH4+ + OH- pH = 9.62 [OH-] = 4.2 x 10-5 M NH3 NH4+ OH- [i] c [e]-y +y +y
[e] y y y
y = 4.2 x 10-5 M
[OH-] = 4.2 x 10-5 M
pH = 9.62

19. @ equivalence point and beyond use Ka3)
HCl + NH3 → NH4Cl
= 5.00 x 10-3 mols HCl
100 mL of 0.05 M NH3
= 5.00 x 10-3 mols NH3
50 mL of 0.10 M HCl
HCl
NH3
NH4Cl i
5.0 x 10-3 mols
5.00 x 10-3 mols
0 mols c -y -y +y e
at equivalence point!!
0 mols
0 mols
5.00 x 10-3 mols
@ equivalence point and beyond use Ka

20. Reaction to completion yields:
= M NH4+
5.0 x 10-3 mols NH4+
NH4+ ⇆ NH3 + H+
0.15 L
NH4+
NH3 H+
[i]
0.033 M
0 M
0 M c -y +y +y
[e] y y y
pH = 5.37
y = 4.3 x 10-6 M
[H+] = 4.3 x 10-6 M

21. @ equivalence point and beyond use Ka3)
HCl + NH3 → NH4Cl
= 7.50 x 10-3 mols HCl
100 mL of 0.05 M NH3
= 5.00 x 10-3 mols NH3
75 mL of 0.10 M HCl
HCl
NH3
NH4Cl i
7.5 x 10-3 mols
5.00 x 10-3 mols
0 mols c -y -y +y e
2.5 x 10-3 mols
0 mols
5.00 x 10-3 mols
@ equivalence point and beyond use Ka

22. Who cares about the equilibrium??
Reaction to completion yields:
= M HCl
2.5 x 10-3 mols HCl
Who cares about the equilibrium??
0.175 L
Not me
This will be y
Which will be
[H+] =
pH = 1.84