1. Unit 19 Acid Base Equilibria: Titrations
CHM 1046: General Chemistry and Qualitative Analysis
Unit 19 Acid Base Equilibria: Titrations
Dr. Jorge L. Alonso
Miami-Dade College – Kendall Campus
Miami, FL
Textbook Reference:
Chapter 19 (sec. 5-8)
Module 9

2. Titration
Soln-Unknown Concentration (M): Acid
A volumetric technique in which one can determine the concentration of a solute in a solution of unknown concentration, by making it react with another solution of known concentration (standard).
Standard-of known Conc.(M)
Known Volume (V)
Measure Vol to reach end pt.
MolesB = M x V
MolesA = M x V
If: MolesA = MolesB
{*TitrationMovie}
Then: (M x V)A = (M x V)B

3. Determining the Concentration of Solutions by Titration
A known concentration of base (or acid) is slowly added to a solution of acid (or base) until neutralization occurs.
(Standard)
Example:
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
xa HA (aq) + xb MOH (aq)  MA (aq) + H2O (l)
H2SO4(aq) + 2NaOH(aq)  Na2SO4(aq) + H2O(l)
Neutralization: equivalence point
# mol1(acid) = # mol2(base)
1 mol1(acid) = 1 mol2(base) xa xb

4. Titration Calculations: Stoichiometry using Molarities
xA HN + xB MOH  MN + HOH
Neutralization: moles(acid) = moles(base)
xA xB
Since moles = MV = moles x Liter Liter ηA ηB
MA x VA MB x VB = = = = =
Where
= coefficients from balanced equations
* Equation Useful for determining Molarities and Volumes at the Equivalence Point of a Titration *

5. Solution Stoichiometry Problems: Molarity
Problem: A volume of 16.3 mL of a 0.30M NaOH solution was used to titrate mL H2SO4. What is the concentration of H2SO4 in the solution of unknown concentration?
H2SO4 + 2 NaOH  2HOH + Na2SO4
MA x VA = MB x VB 1 2
= M H2SO4
Titration of a Strong Acid with a Strong Base

6. Titration Graph: pH vs. Volume
Titration Data:
Excess base
mL of NaOH pH
1.2 10
1.4 20
1.6 30
1.7 40
1.9 50
7.0 60
12.0 70
12.2 80
12.3 SB
Phenolphthalein Indicator
Acid = Base
Methyl Red Indicator pH
meter
Excess acid SA
{Titration2}

7. Titrations: The Strength of Acids & Bases
Strong Base with Strong Acid SA
Weak Base with Strong Acid SA
Phenolphthalein
8-10 SB WB
Weak Base with Weak Acid
Strong Base with Weak Acid WA WA SB WB

8. (2) Titration of a WA with a SB
With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.

9. Selecting Appropriate Indicators
Select appropriate indicator for following:
Phenolph
Meth Red
Meth Red
Phenolph
?????
an indicator is chosen so that it will change color at a pH just beyond the equivalence point (mid point of the steep vertical portion of the graph). The first point at which the indicator permanently changes color marks the end of the titration and is called the indicator end-point. Dropping a perpendicular from the indicator end-point to the x-axis is a very close estimation of the equivalence point.

10. (1) Titration of a SA with a SB (or SB with a SA)

11. Acid-Base Neutralization Equations
(1) Strong Acids and Bases are represented in completely dissociated state: as H+ and OH-
(2) Weak Acids and Bases are represented in undissociated state: as HA and B (or MOH)

12. (H2CO3 + K+)
(H2CO3 + Ca2+ + C2H3O2- )
(H2C O3 + Zn2+)
(H2CO3 + Zn2+ + SO42-)

13. Equations and Tables used in solving A-B Titration Problems
(1) Acid Base Neutralization: when you reach the end-point using SA or SB
(2) Acid Base Neutralization: when not at the end-point or using WA or WB
(3) WA or WB Equilibrium problems
HA ↔ H+ + A-
HA + OH-  H2O + A-
Use: Mole ICEnd Table
Use: [ICE] Table
Mole HA
MOH → MA
+ H20
HA ↔ H+ A-
(.15M)(.025L) (.10M)(.030L) II
0.061 C -x +x E x x II η η
0 η C η η
End η

14. (1) Titration of a SA with a SB
Example: 25 mL of 0.15M HCl with 0.10M NaOH.
(1) What volume of NaOH is required to reach the equivalence point?
(2) What is the pH of the initial strong acid? (strong acid problem)
In strong acid [HA]=[H+], so pH=-log [H+]
= -log (0.15) = 0.82
(3) What is the pH prior to the equivalence point? Let’s say after 30. mL of NaOH. (excess SA problem)
*what is used for neutralization Rx?*
Mole HX
MOH → MX
H20
(.15M)(.025L) (.10M)(.030L) II η η C η η
End η
pH = -log (1.4 x 10-2) = 1.9
Salt of SA & SB: will not Hydrolyze

15. (1) Titration of a SA with a SB
Example: 25 mL of 0.15M HCl with 0.1M NaOH.
(4) What is the pH at the equivalence point?
Only salt + water present and salt will not hydrolyze water since it is derived from SA & SB. So pH = 7
(5) What is the pH after the equivalence point? Lets say after 40. mL of NaOH. (excess SB problem)
*what is used for neutralization Rx?*
mole HX
MOH → MX
H20
(.15M)(.025L) (.10M)(.040L) II η η
0 η C η η
End
0 η η
Salt of SA & SB: will not Hydrolyze
pOH = -log(3.1x10-3) = 2.5
pH + pOH = pKw pH = pKw - pOH
pH = 14 – 2.5 = 11.5

16. (2) Titration of a WA with a SB
SB (OH-)
HA ↔ H+ + A-
HA + OH-  H2O + A-
*what is used for neutralization Rx?*
Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed.
The pH >7 at the equivalence point.
*what is used for equilibrium Rx?*

17. (2) Titration of a WA with a SB
Example: 25 mL of 0.15M HC2H3O2 (Ka= 1.8 X10-5) with 0.10M NaOH.
(1) What volume of NaOH is required to reach the equivalence point?
(2) What is the initial pH of the acetic acid? (Before titration, WA Equilibrium problem)
HA ↔ H+ A- II
0.15 C -x +x E x x

18. (2) Titration of a WA with a SB
Example: 25 mL of 0.15M HC2H3O2 (Ka= 1.8 X10-5) with 0.10M NaOH.
(3) What is the pH prior to the equivalence point? Let’s say after 30. mL of NaOH. (WA Buffer problem)
mole HA
MOH → MA
H20
(.15M)(.025L) (.10M)(.030L) II η η
0 η C η η
End η
Salt of WA & SB: WILL Hydrolyze H2O

19. (2) Titration of a WA with a SB
Example: 25 mL of 0.15M HC2H3O2 (Ka= 1.8 X10-5) with 0.1M NaOH.
(4) What is the pH at the equivalence point? This 37.5 mL (Hydrolysis of Salt derived from a WA & SB)
mole HA
MOH → MA
H20
(.15M)(.025L) (.10M)(.038L) II η
0 η C η η
End
0 η
Salt of WA & : will Hydrolyze water
Salt is NaC2H3O2
Na+ = derived from SB (NaOH), will not hydrolyze.
C2H3O2- = derived from WA (acetic acid) it WILL hydrolyze water.

20. (2) Titration of a WA with a SB
Example: 25 mL of 0.15M HC2H3O2 (Ka= 1.8 X10-5) with 0.1M NaOH.
(4) What is the pH at the equivalence point? This 37.5 mL (Hydrolysis of Salt derived from a WA & SB)
mole HA
MOH → MA
H20
(.15M)(.025L) (.10M)(.038L) II η
0 η C η η
End
0 η
[C2H3O2-] =.
Salt is NaC2H3O2
C2H3O2- + H2O
↔ HC2H3O
OH- II
0.061 C -x +x E x x
pH + pOH = pKw pH = pKw - pOH
pH = 14 – 5.2 = 8.8

21. 2005B Q1

22. Mole ICEnd Table:

23. (3) Titration of a WB with a SA
SA (H3O+)
MOH ↔ M+ + OH-
H3O+ + MOH  2H2O + M+
The pH at the equivalence point in these titrations is < 7.
Methyl red is the indicator of choice.

24. (3) Titration of a WB with a SA:
Calculation of pH
(1) Acid Base Neutralization
(2) WB Equilibrium problem
B+ H2O ↔ BH+ + OH-
B + H+  H2O + A-
Weak base and strong acid
Use: Mole ICEnd Table
Use: [ICE] Table
Mole B
H → A-
+ H20
B H2O ↔
BH+
OH-
(.15M)(.025L) (.10M)(.030L) II
0.061 C -x +x E x x II η η
0 η C η η
End η

25. 2007A Q1
Titration: weak acid with strong base

26. Use: Mole ICEnd Table (e) What’s pH? For F-: For HF: = 0.15 M F-
OH → A-
+ H20
(.40 M)(.025L) (.40M)(.015L) II
0.010 η η
0 η C η η
End η
(e) What’s pH?
For F-:
= M F-
For HF:
= M HF

27. Use: [ICE] Table II 0.10 0.15 C -x +x E 0.10 - x x 0.15 + x 0.15 M F-
0.10 M HF
Use: [ICE] Table HA H+
↔ A- II
0.10
0.15 C -x +x E x x x

28. Titrations of Polyprotic Acids
Ka3
In these cases there is an equivalence point for each dissociation.
Ka2
Ka1

29. Titration: weak acid strong base
2005B Q1

30. 2005B Q1

34. Titration: weak base strong acid
2000 QA

36. Titration: weak acid strong base
2001 Q3

37. Answers Q3

38. Titration: weak acid strong base
2002A Q1

40. 2003A Q1
Titration: weak base strong acid

43. 2006B Q1
Titration: weak acid strong base

46. Expressing Concentrations of Solutions: Molarity (& Normality*)
* For MDC students only!

47. Molarity (M) xA HN + xB MOH  MN + HOH mol mol =
moles of solute
Liters of solution =
xA HN + xB MOH  MN + HOH
mol mol A = B
(mol/L)A x LA (mol/LB) x LB =
MA x VA = MB x VB
Where
= coefficients for the acid (A) and the base (B) from the balanced neutralization equations

48. (mol/L)A x LA (mol/LB) x LB =
mol mol A = B
(mol/L)A x LA (mol/LB) x LB =
MA x VA MB x VB =
molesA MB x VB =

49. xA HN + xB MOH  MN + HOH
For titrations:
Since
MB x VB =

50. Molarity (M) vs. Normality (N)
Lesson for MDC students only:
Molarity (M) vs. Normality (N)
equiv of solute
L of solution
N =
mol of solute
L of solution
M =
Where:
M = N
When n = 1
That is when using HCl, KHP NaOH
But not when using H2SO4, Ca(OH)2
A/B = # H+ or #OH-
Redox = #e- involved in balanced
redox equation.

51. Molarity (M) vs. Normality (N)
Acid g-MM (g/)
+ H20 to
HCl 36 g L =
H2SO4 98 g L =
H3PO4 98 g L =
Molarity (/L)
Normality (eq/L)
g-EW
36/1
=36
98/2
=49
98/3
=32.7
Eq(g/gEW)
36/36
98/49
98/33 1M = 1N 2N 3N

52. N = M or M = N NA x VA = NB x VB 2H3PO4 + 3 Ca(OH)2  6 HOH + Na3PO4
2M H3PO4
3M Ca(OH)2
Using Molarity
1N H3PO4
1N Ca(OH)2
Using Normality
N = M or M = N
NA x VA = NB x VB
Using Normality for titrations:

53. Titration of a WA with a SB
Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed.
The pH at the equivalence point will be >7.
Phenolphthalein is commonly used as an indicator in these titrations.

54. Titration: measuring the Equivalence Point (H+ = OH-)
Methyl red in base (range R4-6Y)
Phenolphthalein in base (range C 8-10 F)
A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base. The end-point of a titration is when indicator changes color.

55. Titration of a SA with a SB
xA HN + xB MOH  MN(aq) + HOH
1. From the start of the titration the pH goes up slowly. Just before the equivalence point, the pH increases rapidly.
Excess base
2. At the equivalence point, moles H+ = moles OH-, and the solution contains only water and the salt from the cation of the base and the anion of the acid.
H+ = OH-
3. Just after the equivalence point, the pH increases rapidly. As more base is added, the increase in pH again levels off.
Excess acid

56. Titration xa xb Solution of known concentration (M2 x V2 = #mol2)
Solution of unknown concentration (M1 x V1 = #mol1)
Neutralization: mola = 1 molb (equivalence point) xa xb
{*TitrationMovie}

57. Stoichiometric/Volumetric Calculations
xA HN + xB MOH  MN + HOH
ACID
ACID
ACID xA xB
BASE
BASE
BASE
MA x VA MB x VB = = =

58. 2006 (B)

61. 2007 (A)