Transcript Chapter 11: Acid - Base Titrations

Acid-Base Titrations
Introduction
1.) Experimental Measurements of pKa

pKa of amino acids in an active-site of a protein are related to its function
-

Protein structure and environment significantly perturb pKa values
In medicinal chemistry, pKa and lipophilicity of a candidate drug predict how
easily it will cross a cell membrane
-
Higher charge  harder to cross membrane not a good drug
Acid-Base Titrations
Introduction
2.) Example:

impact of the Asp on the pKa of His in the His-Asp catalytic dyad.
-
Glucose 6-phosphate dehydrogenase (G6PD) catalyzes the oxidation of
glucose 6-phosphate using NAD+ or NADP+
His-240 is the general base that extracts a proton from the C1-OH of G6P
The pKa of His-240 in the G6PD apoenzyme is found to be 6.4, which corresponds to an
unidentified pKa value of 6.3 that was previously derived from the dependence of kcat on pH.
These results suggest that the pKa of His-240 is unperturbed by Asp.
Biochemistry, Vol. 41, No. 22, 2002 6945
Acid-Base Titrations
Introduction
3.) Overview

Titrations are Important tools in providing quantitative and qualitative data for
a sample.

To best understand titrations and the information they provide, it is necessary
to understand what gives rise to the shape of a typical titration curve.

To do this, acid-base equilibria are used to predict titration curve shapes.
proton release from PAA decreases with
increase in the degree of dissociation
for the highest polymer concentration
conformational change of the PAA from rodlike conformation to a random coil form,
J. Phys. Org. Chem. 2006; 19: 129–135
Acid-Base Titrations
Titration of Strong Base with Strong Acid
1.) Graph of How pH changes as Titrant is Added

Assume strong acid and base completely dissociate

Any amount of H+ added will consume a
stoichiometric amount of OHK  1 K  10 14
w

Reaction Assumed to go to completion

Three regions of the titration curve
-
Before the equivalence point, the pH is determined
by excess OH- in the solution
-
At the equivalence point, H+ is just sufficient to
react with all OH- to make H2O
-
After the equivalence point, pH is determined by
excess H+ in the solution.
Acid-Base Titrations
Titration of Strong Base with Strong Acid
1.) Graph of How pH changes as Titrant is
Added

Remember, equivalence point is the
ideal goal

Actually measure End Point
-

Different Regions require different kinds
of calculations
-

Marked by a sudden physical
change: color, potential
Illustrated examples
The “true” titration reaction is:
Titrant
Analyte
Acid-Base Titrations
Titration of Strong Base with Strong Acid
2.) Volume Needed to Reach the Equivalence Point

Titration curve for 50.00 mL of 0.02000 M KOH with 0.1000 M HBr

At equivalence point, amount of H+ added will equal initial amount of OH-
Ve ( mL )0.1000 M   50.00 mL ( 0.02000 M )  Ve  10.00 mL
mmol of HBr
at equivalence point
mmol of OHbeing titrated
When 10.00 mL of HBr has been added, the titration is complete.
Prior to this point, there is excess OH- present.
After this point there is excess H+ present.
Acid-Base Titrations
Titration of Strong Base with Strong Acid
3.) Before the Equivalence Point

Titration curve for 50.00 mL of 0.02000 M KOH with 0.1000 M HBr
Equivalence point (Ve) when 10.00 mL of HBr has been added
When 3.00 mL of HBr has been added, reaction is 3/10 complete
-
Initial volume of OHCalculate Remaining [OH-]:
50.00
 10.00  3.00 


[OH  ]  
0.02000 M 
  0.0132 M
10.00


 50.00  3.00 
Fraction of OH
Initial
concentration
Dilution Factor
Remaining
of OH
Total volume
Calculate [H+] and pH:
1.0  10 14
13
[H ] 


7
.
58

10
M  pH  12.12
0
.
0132
[OH ]

Kw
Acid-Base Titrations
Titration of Strong Base with Strong Acid
4.) At the Equivalence Point

Titration curve for 50.00 mL of 0.02000 M KOH with 0.1000 M HBr
-
Just enough H+ has been added to consume OHpH determined by dissociation of water
Kw
x
x
Kw= 1x10-14
Kw  x 2  x  1.00  10 7 M  pH  7.00
-
pH at the equivalence point for any strong acid with strong base is 7.00
Not true for weak acid-base titration
Acid-Base Titrations
Titration of Strong Base with Strong Acid
5.) After the Equivalence Point

Titration curve for 50.00 mL of 0.02000 M KOH with 0.1000 M HBr
-
Adding excess HBr solution
When 10.50 mL of HBr is added
Calculate volume of excess H+:
Vadded  Vequivalence  10.50  10.00  0.50 mL
Calculate excess [H+]:
Volume of excess H+
0.50


4
[H  ]  0.1000 M 
  8.26  10 M
 50.00  10.50 
Initial
Total volume
concentration Dilution factor
of H+
Calculate pH:
pH  -log[ H  ]   log( 8.26  10 4 M )  3.08
Acid-Base Titrations
Titration of Strong Base with Strong Acid
6.) Titration Curve

Rapid Change in pH Near Equivalence Point
-

Equivalence point is where slope is greatest
Second derivative is 0
pH at equivalence point is 7.00, only for strong
acid-base
-
Not True if a weak base-acid is used
Acid-Base Titrations
Titration of Weak Acid with Strong Base
1.) Four Regions to Titration Curve

Before any added base, just weak acid (HA) in water
-
pH determined by Ka
Ka

With addition of strong base  buffer
-
pH determined by Henderson Hasselbach equation
 [ A ] 

pH  pK a  log 
 [HA] 



At equivalence point, all HA is converted into A-
Weak base with pH determined by Kb
Kb
Acid-Base Titrations
Titration of Weak Acid with Strong Base
1.) Four Regions to Titration Curve

Beyond equivalence point, excess strong base is added to A- solution
-
pH is determined by strong base
Similar to titration of strong acid with strong base
2.) Illustrated Example:

Titration of 50.00 mL of 0.02000 M MES with 0.1000 M NaOH
-
MES is a weak acid with pKa = 6.27
K
-
Reaction goes to completion with addition of strong base


7
K  1 K  1 K / K   1

5
.
4

10
b
w
a
1  10 14 / 10 6.27
Acid-Base Titrations
Titration of Weak Acid with Strong Base
3.)
Volume Needed to Reach the Equivalence Point

Titration of 50.00 mL of 0.02000 M MES with 0.1000 M NaOH
-
Reaction goes to completion with addition of strong base
Strong plus weak react completely
Ve ( mL )0.1000 M   50.00 mL ( 0.02000 M )  Ve  10.00 mL
mmol of base
mmol of HA
Acid-Base Titrations
Titration of Weak Acid with Strong Base
4.)
Region 1: Before Base is Added


Titration of 50.00 mL of 0.02000 M MES with 0.1000 M NaOH
Simply a weak-acid problem
Ka
Calculate [H+]:
F-x
Ka= 10-6.27
x
x
x2
x2

 K a  x  [H  ]  1.03  10  4
F  x 0.02000  x
Calculate pH:
pH  - log [H  ]   log( 1.03  10 4 M )  3.99
Acid-Base Titrations
Titration of Weak Acid with Strong Base
5.)
Region 2: Before the Equivalence Point



Titration of 50.00 mL of 0.02000 M MES with 0.1000 M NaOH
Adding OH- creates a mixture of HA and A-  Buffer
Calculate pH from [A-]/[HA] using Henderson-Hasselbach equation
Simply the difference
Simply
of volumes
of initialratio
quantities
Calculate [A-]/HA]:
Amount of added NaOH is 3 mL with equivalence point is 10 mL
Relative Initial quantities (HA≡1)
1
3
10
-
Relative Final quantities
7
10
-
3
10
Calculate pH:
3 
 [ A ] 
  6.27  log  10   5.90
pH  pK a  log 
 7 
 [HA] 


 10 
3
10
Acid-Base Titrations
Titration of Weak Acid with Strong Base
5.)
Region 2: Before the Equivalence Point

Titration of 50.00 mL of 0.02000 M MES with 0.1000 M NaOH

pH = pKa when the volume of titrant equals ½Ve
Relative Initial quantities (HA≡1)
Relative Final quantities
1
1
2
1
2
-
-
1
2
1 
 [ A ] 
  pK  log  2   pK
pH  pK a  log 
a
a
1 
 [HA] 


 2
1
2
Acid-Base Titrations
Titration of Weak Acid with Strong Base
5.)
Region 3: At the Equivalence Point


Titration of 50.00 mL of 0.02000 M MES with 0.1000 M NaOH
Exactly enough NaOH to consume HA
Relative Initial quantities (HA≡1)
1
1
-
-
Relative Final quantities
-
-
1
1

The solution only contains A-  weak base
Kb
F-x
Kb 
x
K
x2
 Kb  w
Fx
Ka
x
Kw
Ka
Acid-Base Titrations
Titration of Weak Acid with Strong Base
5.)
Region 3: At the Equivalence Point

Titration of 50.00 mL of 0.02000 M MES with 0.1000 M NaOH
Calculate Formal concentration of [A-]:
A- is no longer 0.02000 M, diluted by the addition of NaOH
Initial volume of HA
50.00


F'  0.02000 M 
  0.0167 M
50
.
00

10
.
00


Initial
Total volume
concentration Dilution factor
of HA
Acid-Base Titrations
Titration of Weak Acid with Strong Base
5.)
Region 3: At the Equivalence Point

Titration of 50.00 mL of 0.02000 M MES with 0.1000 M NaOH
Calculate [OH-]:
Kw 1  10 14
x2
8
 Kb 


1
.
86

10
Fx
Ka
10 6.27
x2
 1.86  10  8  x  [OH - ]  1.76  10 -5 M
0.0167  x
Calculate pH:
 1  10 14

   log 
 0.0167



  9.25


pH at equivalence
point is not 7.00
pH will always be above 7.00 for titration of a weak acid
because acid is converted into conjugate base at the equivalence point
K
pH  - log[H ]   log  w
 x

Acid-Base Titrations
Titration of Weak Acid with Strong Base
5.)
Region 4: After the Equivalence Point


Titration of 50.00 mL of 0.02000 M MES with 0.1000 M NaOH
Adding NaOH to a solution of A-
NaOH is a much stronger base than ApH determined by excess of OH-
Calculate volume of excess OH-:
Amount of added NaOH is 10.10 mL with equivalence point is 10 mL
Vadded  Vequivalence  10.10  10.00  0.10 mL
Calculate excess
[OH-]:
Volume of excess OH-
0.10


[OH- ]  0.1000 M 
  1.66  10  4 M
 50.00  10.10 
Initial
Total volume
Dilution
factor
concentration
of OHCalculate pH:
 1.00  10 14 
 Kw 


  10.22
pH  - log[H ]   log 


log

 1.66  10  4 
 [OH  ] 


Acid-Base Titrations
Titration of Weak Acid with Strong Base
5.)
Titration Curve


Titration of 50.00 mL of 0.02000 M MES with 0.1000 M NaOH
Two Important Features of the Titration Curve
Equivalence point: [OH-] = [HA]
Steepest part of curve
Maximum slope
Maximum Buffer
Capacity
pH=pKa
Vb = ½Ve
Minimum slope
Acid-Base Titrations
Titration of Weak Acid with Strong Base
5.)
Titration Curve


Depends on pKa or acid strength
Inflection point or maximum slope decreases with weaker acid
-
Equivalence point becomes more difficult to identify
weak acid  small slope change in
titration curve
Difficult to detect equivalence point
Strong acid  large slope change in
titration curve
Easy to detect equivalence point
Acid-Base Titrations
Titration of Weak Acid with Strong Base
5.)
Titration Curve


Depends on acid concentration
Inflection point or maximum slope decreases with
lower acid concentration
-
Equivalence point becomes more difficult to
identify
Eventually can not titrate acid at very low
concentrations
High
Lowconcentration
concentrationlarge
smallslope
slopechange
changeinin
titration
titrationcurve
curve
Easy
to detect
equivalence
point
Difficult
to detect
equivalence
point
At low enough concentration, can not detect change
Acid-Base Titrations
Titration of Weak Base with Strong Acid
1.)
Simply the Reverse of the Titration of a Weak Acid with a Strong Acid

Again, Titration Reaction Goes to Completion:

Again, Four Distinct Regions to Titration Curve

Before acid is added  just weak base reaction
-
pH determined from Kb
Kb
F-x

x
x
Before equivalence point,  buffer
-
pH determined from Henderson Hasselbach equation
-
Va=½Ve then pH = pKa (for BH+)
pKa and pKb can be determined from titration curve
 [ A ] 

pH  pK a  log 
 [HA] 


Acid-Base Titrations
Titration in Diprotic Systems
1.)
Principals for Monoprotic Systems Apply to Diprotic Systems


Multiple equivalence points and buffer regions
Multiple Inflection Points in Titration Curve
Two equivalence points
Kb1
Kb2
Acid-Base Titrations
Titration in Diprotic Systems
2.)
A Typical Case

Titration of 10.0 mL of 0.100 M base (B) with 0.100 M HCl
-

pKb1 = 4.00 and pKb2 = 9.00
Volume at First Equivalence Point (Ve)
Ve ( mL )0.100 M   10.00 mL ( 0.1000 M )  Ve  10.00 mL
mmol of HCl

mmol of B
Volume at Second Equivalence Point Must Be 2Ve
-
Second reaction requires the same number of moles of HCl
Acid-Base Titrations
Titration in Diprotic Systems
2.)
A Typical Case

Point A
-
Before Acid Added
Weak base problem
Kb1
0.100 - x
x
x
x2
 1.00  10  4  x  3.11  10  3
0.100  x
[H  ] 
Kw
 pH  11.49
x
Acid-Base Titrations
Titration in Diprotic Systems
2.)
A Typical Case
Point between A &B

-
Before First Equivalence Point
Buffer problem
Kw 1  10 14
10
K a2 


1

10
K b1 1  10  4
Point (1.5 mL) is before first
equivalence point (10 mL)
[ B]
[ BH  ]

10.00 ml  1.5 ml 8.5

 5.67
1.5 ml
1.5
 [ B] 
  10.00  log 5.67   10.75
pH  pKa 2  log 
 [ BH  ] 


Acid-Base Titrations
Titration in Diprotic Systems
2.)
A Typical Case
Point B

-
Before First Equivalence Point
Buffer problem
Kw 1  10 14
10
K a2 


1

10
K b1 1  10  4
Point B (5 mL) is halfway to first
equivalence point (10 mL)
[B]


[BH ]
10.00 ml  5 ml 5
 1
5 ml
5
 [B] 
  10.00  log 1  10.00
pH  pK a2  log 
 [BH  ] 


pH = pKa2=10.00
Acid-Base Titrations
Titration in Diprotic Systems
2.)
A Typical Case

Point C
-
[H  ] 
First Equivalence Point
Intermediate form of the Diprotic
acid
K 2 K1F  K1Kw
K1  F
Account for dilution for formal
Solve
for pH using
concentration
(F) of BH+
intermediate form equation Initial volume of B
10.00


5
10
5

F'  0(.10
1000
M
0

)( 10
)( 0.0500 )  ( 10 )( 1014
) .0500 M
[H ] 
 3.16 x10 8
1010
5 .00  10.00 
 0.0500

8
Initial
pH   log(
3.16 x10 )  7.50
Total volume
concentration Dilution factor
of B
Acid-Base Titrations
Titration in Diprotic Systems
2.)
A Typical Case
Point D

-
Before Second Equivalence Point
Buffer Problem
Kw 1  10 14
5
K a2 


1

10
K b1 1  10  9
Point D (15 mL) is halfway to second
equivalence point (2x10 mL). First,
subtract Ve (10 mL)
[BH  ]

2
[BH2 ]
10.00 ml  5.00 ml 5
 1
5 ml
5
 [BH  ] 
  5.00  log 1  5.00
pH  pK a2  log 
 [BH 2  ] 
2


pH = pKa1=5.00
Acid-Base Titrations
Titration in Diprotic Systems
2.)
A Typical Case

Point E
-
Second Equivalence Point
Weak acid problem
Account for dilution for formal
Initial volume of B
concentration (F) of BH2+2
10.00


F'  0.1000 M 
  0.0333 M
20
.
00

10
.
00


Initial
Total volume
concentration Dilution factor
of B
pH determined by acid dissociation of BH2+2
Kb2
Acid-Base Titrations
Titration in Diprotic Systems
2.)
A Typical Case

0.0333 - x
Point E
-
Second Equivalence Point
Weak acid problem
Ka1
K a1 
x
x
x2
x2

 1.0  10  5
F  x 0.0333  x
 x  5.72  10  4  pH  3.24
Kw
K b2
Acid-Base Titrations
Titration in Diprotic Systems
2.)
A Typical Case

Beyond Point E
-
Past Second Equivalence Point
Strong acid problem
pH from volume of strong acid added.
Addition of 25.00 mL:
Excess acid:
Vadded  Vequivalence  25.00  20.00  5.00 mL
Concentration of H+:
5.00


2
[H  ]  0.1000 M 
  1.43  10 M
 25.00  10.00 
pH:
pH  - log [H  ]   log( 1.43  10 2 )  1.85
Acid-Base Titrations
Titration in Diprotic Systems
3.)
Blurred End Points

Two or More Distinct Equivalence
Points May Not be Observed in
Practice
-
Depends on relative difference in
Kas or Kbs
-
Depends on Relative strength of
Kas or Kbs
Only one Equivalence point is clearly evident
Second Ka is too strong and is not
a weak acid relative to titrant
Acid-Base Titrations
Titration in Diprotic Systems
nd derivative is zero
End
point:
End
point:
1st2derivative
is maximum
4.)
Using Derivatives to Find End Point


Useful when End points overlap
End Point of titration curve is where slope is
greatest
-
Dph = 4.400-4.245=0.155
dpH/dV is large
DpH change in pH between consecutive points
DV average of pair of volumes
Second derivative is similar difference using
first derivative values
Acid-Base Titrations
Titration in Diprotic Systems
5.)
Using Gran Plot to Find End Point


Method of Plotting Titration Data to Give a Linear
Relationship
A graph of Vb10-pH versus Vb is called a Gran plot

Vb 10  pH  HA K a Ve  Vb 
 A
where:
Vb = volume of strong base added
Ve = volume of base needed to reach equivalence point
A-, HA = activity coefficients ≈ 1
Acid-Base Titrations
Titration in Diprotic Systems
5.)
Using Gran Plot to Find End Point

Plot is a straight line
-


If ratio of activity coefficients is constant
Slope = -KaHA/aX-intercept = Ve (must be extrapolated)
Measure End Point with data Before Reach End Point
Only use linear region of Gran Plot
-
Changing ionic strength changes activity coefficients
added salt to maintain constant ionic strength
Slope Gives Ka
x-intercept gives Ve
Never Goes to Zero, approximation that every
mole of OH- generates one mole of A- is not
true as Vb approaches Ve
Acid-Base Titrations
End Point Determination
1.)
Indicators: compound added in an acid-base titration to allow end
point detection


Common indicators are weak acids or bases
Different protonated species have different colors
Acid-Base Titrations
End Point Determination
1.)
Indicators: compound added in an acid-base titration to allow end
point detection

Color Change of Thymol Blue between pH 1 and 11
pK = 1.7
pK = 8.9
Acid-Base Titrations
End Point Determination
2.)
Choosing an Indicator


Want Indicator that changes color in the vicinity of the equivalence point
and corresponding pH
The closer the two match, the more accurate determining the end point will
be
Bromocresol purple color change
brackets the equivalence point and
is a good indicator choice
Bromocresol green will change color
Significantly past the equivalence
point resulting in an error.
Acid-Base Titrations
End Point Determination
2.)
Choosing an Indicator
The difference between the end point
(point of detected color change) and the
true equivalence point is the indicator
error
Amount of indicator added should be
negligible
Indicators cover a range of pHs
Acid-Base Titrations
End Point Determination
3.)
Example:
a) What is the pH at the equivalence point when 0.100 M hydroxyacetic acid
is titrated with 0.0500 M KOH?
b) What indicator would be a good choice to monitor the endpoint?