* HCl is a strong acid and will titrate first * Its equivalence point was at 35.00 mL of NaOH * NaOH concentration is 0.100 M * Moles of NaOH used = Volume x Concentration = 0.03500 L x 0.100 mol/L = 0.003500 mol * Moles of HCl = Moles of NaOH used = 0.003500 mol * H3PO4 is a weak acid and will titrate second * Its equivalence point was at 50.00 mL of NaOH * Additional NaOH used = 50.00 mL - 35.00 mL = 15.00 mL * Moles of additional Na

1. Acid Base Titration
Plot of pH vs. Volume of
Titrant (reagent)
Acid - Base Titrations
If know [titrant], Vequiv,
and stoich, you can
determine [analyte]
Use to determine the concentration of analytes (acid/base)

2. Ch. 11. Acid – Base (Neutralization) Titrations
Reagents:
Standard solutions (Known Concentration; one is
usually strong. Why?)
Acid: HCl, H2SO4, HClO4
Bases: Na2CO3, NaOH
11-7. Practical Notes

3. A2. Bases
Primary standards
Potassium hydrogen phthalate (KHP)
Benzoic Acid
Potassium hydrogen Iodate
Standardized Solution:
acid (i.e., HCl)
A1. Acids
Primary standards: Na2CO3,
tris(hydroxymethyl)aminomethane
Standardized soln: standardized base (i.e, NaOH)

6. Chapter 11: Acid - Base Titrations
Feasibility of a titration
concentration
reaction completeness (values of K)
choice of indicator or endpt detection
What acid when titrated with NaOH will
give rise to a larger change at the equivalence
point region
a. HCl
b. HAC (Ka = 10-5
)
c. HCN (Ka = 10-10
)

7. Effect of Acid Strength
Larger Ka, stronger the acid, larger the change

8. Effect of Base Strength
Larger Kb, stronger the base,
larger the changeFrom Skoog, West, Holler etal

9. Effect of Concentration
The larger the
concentrations,
the larger the change

10. Indicator is also important

11. Acid - Base indicators
evaluate end point
weak organic acid or base whose color depends of pH

15. Example Titrations
strong acid - strong base
weak acid - strong base
weak base - strong acid
Questions to ask
What is it?
What reacts?
write reaction
After reaction: what remains?
Identify it: Acid, Base, Buffer…
See helpful suggestions on blackboard

17. Strong Acid - Strong Base
Derive a curve for the titration of 50.00 mL of a 0.0500 M
HCl with 0.100 M NaOH
(1) Initial pH
strong acid - strong base
Four general regions
(2) Before the equivalence point
(3) At the equivalence point
(4) After the equivalence point

18. Strong Acid - Strong Base
Derive a curve for the titration of 50.00 mL of a 0.0500 M
HCl with 0.100 M NaOH
1. Initial pH
What is the pH of 0.050 M HCl ??
pH = -log [H+]
pH = -log [0.050] = 1.30

19. Strong Acid - Strong Base
Derive a curve for the titration of 50.00 mL of a 0.0500 M
HCl with 0.100 M NaOH
2. Before the equivalence point
add 10.00 mL of NaOH (0.100 M)
What happens
Acid reacts with base
pH determined by how much remains
Write the Reaction!!!

20. Strong Acid - Strong Base
add 10.00 mL of NaOH (0.100 M)
base reacts with acid
OH- + H+
 H2O
int. mmol
amt reacts
What’s left
Derive a curve for the titration of 50.0 mL
of a 0.050 M HCl with 0.10 M NaOH
1.0 2.5
-1.0 -1.0
0 1.5
What is concentration?
[H+] = 1.5 mmol/60.0 mL
pH = 1.6

21. Strong Acid - Strong Base
Derive a curve for the titration of 50.00 mL of a 0.0500 M
HCl with 0.100 M NaOH
Region 3
3. At the equiv. point (add 25.00 mL of NaOH)
Questions: what reacts
what is left
OH- + H+
 H2O
int. mmol
amt reacts
What is left
after rxn: just water
pH = 7.00
2.52.5
-2.5-2.5
0 0

22. Strong Acid - Strong Base
Derive a curve for the titration of 50.00 mL of a 0.0500 M
HCl with 0.100 M NaOH
Region 4
4. After the equiv. point (add 25.10 mL of NaOH)
OH- + H+
 H2O
int. mmol
amt reacts
What’s left
Calculate concentration
[OH-] = 0.010 mmol / 75.10 mL
2.510 2.500
-2.500-2.500
0.010 0
pOH = 3.88; pH = 10.12

24. Note: large
change at equil.
point. Why??

25. Weak Acid - Strong Base
Derive a curve for the titration of 50.00 mL of a 0.100M
HAc with 0.100 M NaOH. Ka = 1.75 x 10-5
weak acid - strong base
four regions
1. Initial pH
present: weak acid
calculate the pH of 0.100 M HAc
HAc + H2O = H3O+
+ Ac-
[H3O+] = (1.75 x 10-5
x 0.100)1/2
pH = 2.88

26. Weak Acid - Strong Base
Derive a curve for the titration of 50.00 mL of a 0.100 M
HAc with 0.100 M NaOH. Ka = 1.75 x 10-5
2. After initial addition of OH-
(10.00 mL)
base reacts with acid
what do you have after rxn?
OH- + HAc  Ac- + H2O
int. mmol
reacts
What’s left
Buffer!
pH = pKa + log [B]/[A]
pH = pKa + log [1.00/4.00]
1.00 5.00
0.00
-1.00 -1.00
4.00
1.00
1.00
after reaction: HAc, Ac-
pH = 4.15

27. Weak Acid - Strong Base
OH-
+ HAc  Ac-
+ H2O
int. mmol
reacts
What’s left
Region 3
At equiv. point ( add 50.00 mL of OH-)
equal moles of acid and base
base reacts with acid
what do you have after rxn?
5.00 5.00
-5.00 -5.00 5.00
0.000.00 5.00
after rxn: [Ac-] = 5.00 mmol/100.0 mL
What is it ??
A. Acid
B. Base
C. Buffer

28. Calculate the pH of a 0.0500 M Ac-
Ac- + H2O = OH- + HAc
Kb = Kw/Ka = 5.56 x 10-10
[OH] = (5.56 x 10-10
* 0.050)1/2
pH = 8.72

29. Weak Acid - Strong Base
4) After initial addition excess OH-
(60.00 mL)
base reacts with acid
what do you have after rxn
OH-
+ HAc  Ac-
+ H2O
int. mmol
reacts
5.00
-5.00 -5.00
0.00
5.00
1.00
6.00
after
5.00
after rxn: excess OH- and Ac-
What are they
A. Acids
B. Bases
C. Acid and a Base

30. [OH] = 1.00 mmol / 110. mL
pOH = -log [ 0.00909] = 2.04
pH = 11.96

31. Weak acid-Strong base

32. Titration of a Weak Base with a Strong Acid
(see text for details)

33. Problem to work:
Calculate the pH of a solution prepared by mixing 10.00 mL
of 0.1200 M NaOH with 20.00 mL of 0.1100 M Acetic Acid.
The Ka for acetic acid is 1.75 x 10-5
.

35. Polyprotic Acids-Base Titrations

38. Titration of a weak base with a strong acid

39. clicker questions
Calculate the pH at the first equivalence point for the titration of
25.00 mL of 0.100 M Na2CO3 with 0.100 M HCl.
Ka1 = 4.45 x 10-7
; Ka2 = 4.69 x 10-11
Calculate the pH of a solution prepared by mixing 25.00 mL
of 0.100 M Na2CO3 with 30.0 mL of 0.100 M HCl.

44. Sometimes it can be hard
to determine the end point

45. Mixtures of acids and bases:
2. If you have two strong acids or two weak
acids with approximately the same Ka, you
will only see one equivalence point
1. If one acid is strong and the other weak (Ka < 10-5
),
it should be possible to titrate each separately.
Stronger acid will titrate first and will possibly
give a pH break at its equivalence point.
titration of the weaker acid will follow and give a
pH break at its equiv. point
Examples of Mixtures
HCl/HAc, H2SO4, HCl/H3PO4

46. Clicker questions
1. How many endpoints would be observed in
the titration of a mixture of HCl and Acetic Acid
(Ka = 10-5
)
a. 0
b. 1
c. 2
d. 3
1. How many endpoints would be observed in
the titration of H2SO4 (Ka = very large; Ka2 = 10-1
)
a. 0
b. 1
c. 2
d. 3

47. A mixture of hydrochloric acid and phosphoric acid
is titrated with 0.100 M sodium hydroxide. The first
endpoint required 35.00 mL of NaOH and the second
required 50.00 mL. Calculate the mmoles of each acid.
Example: