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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)
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
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)
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
)
Effect of Acid Strength
Larger Ka, stronger the acid, larger the change
Effect of Base Strength
Larger Kb, stronger the base,
larger the changeFrom Skoog, West, Holler etal
Effect of Concentration
The larger the
concentrations,
the larger the change
Indicator is also important
Acid - Base indicators
evaluate end point
weak organic acid or base whose color depends of pH
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
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
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
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!!!
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
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
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
Note: large
change at equil.
point. Why??
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
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
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
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
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
[OH] = 1.00 mmol / 110. mL
pOH = -log [ 0.00909] = 2.04
pH = 11.96
Weak acid-Strong base
Titration of a Weak Base with a Strong Acid
(see text for details)
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
.
Polyprotic Acids-Base Titrations
Titration of a weak base with a strong acid
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.
Sometimes it can be hard
to determine the end point
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
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
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:

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Neutralizationtitrations 140617143109-phpapp01

  • 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)
  • 4.
  • 5.
  • 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
  • 12.
  • 13.
  • 14.
  • 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
  • 16.
  • 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
  • 23.
  • 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
  • 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 .
  • 34.
  • 36.
  • 37.
  • 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.
  • 40.
  • 41.
  • 42.
  • 43.
  • 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: