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

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

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

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.

J. Phys. Org. Chem. 2006; 19: 129–135

conformational change of the PAA from rod-like conformation to a random coil form,

proton release from PAA decreases withincrease in the degree of dissociation for the highest polymer concentration

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 OH-

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.

14w

10K1K


1.) Graph of How pH changes as Titrant is Added Remember, equivalence point is the

ideal goal

Actually measure End Point- Marked by a sudden physical

change: color, potential

Different Regions require different kinds of calculations- Illustrated examples

The “true” titration reaction is:

Titrant Analyte


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-

mL00.10V)M02000.0(mL00.50M1000.0)mL(V ee

mmol of HBrat equivalence point

mmol of OH-

being 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.


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

M0132.000.300.50

00.50M02000.000.10

00.300.10OH

][

Fraction of OH- Remaining

Initial concentration of OH-

Dilution Factor

Initial volume of OH-

Total volume

12.12pHM1058.70132.0100.1

OHKH 13

14

-w

][][

Calculate Remaining [OH-]:

Calculate [H+] and pH:


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 OH-

- pH determined by dissociation of water

- pH at the equivalence point for any strong acid with strong base is 7.00- Not true for weak acid-base titration

Kw Kw= 1x10-14

00.7pHM1000.1xxK 72w

x x


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

M1026.850.1000.50

50.0M1000.0H 4

][

Calculate excess [H+]:

Initial concentration

of H+

Dilution factor

Volume of excess H+

Total volume

Calculate volume of excess H+:

mL50.000.1050.10VV eequivalencadded

Calculate pH:

08.3)M1026.8log(H 4 ]-log[pH


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

With addition of strong base buffer- pH determined by Henderson Hasselbach equation

At equivalence point, all HA is converted into A-

- Weak base with pH determined by Kb

Ka

][][

HAAlogpKpH a

Kb


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

- Reaction goes to completion with addition of strong base

K

727.614awb

104.510/101

1K/K

1K

1K


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

mL00.10V)M02000.0(mL00.50M1000.0)mL(V ee

mmol of base mmol of HA


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

F - x x x

4a

221003.1HxK

x02000.0x

xFx

][

Ka Ka= 10-6.27Calculate [H+]:

Calculate pH:

99.3)M1003.1log(Hlog-pH 4 ][

103


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

Calculate [A-]/HA]:

Relative Initial quantities (HA≡1) 1 - -

Relative Final quantities - 103

107

Amount of added NaOH is 3 mL with equivalence point is 10 mL

90.510

710

3log27.6

HAAlogpKpH a

][][

Calculate pH:

Simply ratio of volumesSimply the differenceof initial quantities

103

21


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

21Relative Initial quantities (HA≡1) 1 - -

Relative Final quantities - 21

aaa pK2

12

1logpK

HAAlogpKpH

][][

21


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

The solution only contains A- weak base

Relative Initial quantities (HA≡1) 1 1 - -

Relative Final quantities - - 1 1

Kb

aw

b KKK

F - x x x

aw

b2

KKK

xFx


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

M0167.000.1000.50

00.50M02000.0'F


of HA

Dilution factor

Initial volume of HA

Total volume


5.) Region 3: At the Equivalence Point Titration of 50.00 mL of 0.02000 M MES with 0.1000 M NaOH

M1076.1OHx1086.1x0167.0

x

1086.110

101KKK

xFx

5--82

827.6

14

aw

b2

][

Calculate [OH-]:

Calculate pH:

25.90167.0101log

xKlogHlog-pH

14w

][

pH at equivalence point is not 7.00

pH will always be above 7.00 for titration of a weak acidbecause acid is converted into conjugate base at the equivalence point


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 A-

- pH determined by excess of OH-

M...

.M. 410661

1010005010010000

][OH-

Calculate excess [OH-]:


of OH-

Dilution factor

Volume of excess OH-

Total volume

Calculate volume of excess OH-:

mL10.000.1010.10VV eequivalencadded

Calculate pH:

22.101066.1

1000.1logOHKlogHlog 4

14w

][][-pH

Amount of added NaOH is 10.10 mL with equivalence point is 10 mL


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

pH=pKa

Vb = ½Ve

Minimum slope

Maximum Buffer Capacity


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

Strong acid large slope change in titration curve

Easy to detect equivalence point

weak acid small slope change in titration curve

Difficult to detect equivalence point


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 concentration large slope change in titration curve

Easy to detect equivalence point

Low concentration small slope change in titration curve

Difficult to detect equivalence pointAt 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

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

Kb

][][

HAAlogpKpH a

F - x x x

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

Acid-Base Titrations

Kb2

Two equivalence points Kb1

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)

Volume at Second Equivalence Point Must Be 2Ve

- Second reaction requires the same number of moles of HCl

mL00.10V)M1000.0(mL00.10M100.0)mL(V ee

mmol of HCl mmol of B


2.) A Typical Case Point A

- Before Acid Added- Weak base problem

Kb1

0.100 - x x x

49.11pHx

KH

1011.3x1000.1x100.0

x

w

342

][


2.) A Typical Case Point between A &B

- Before First Equivalence Point- Buffer problem

104

14

1bw

2a 101101101

KKK

Point (1.5 mL) is before first equivalence point (10 mL)

6755158

51510010

...

ml.ml.ml.

BHB

][][

751067500102 ..log.BHB

logpKpH a

][][


2.) A Typical Case Point B

- Before First Equivalence Point- Buffer problem

00.101log00.10BH

BlogpKpH 2a

][

][

104

14

1bw

2a 101101101

KKK

Point B (5 mL) is halfway to first equivalence point (10 mL)

155

ml5ml5ml00.10

BHB

][][

pH = pKa2=10.00


2.) A Typical Case Point C

- First Equivalence Point- Intermediate form of the Diprotic

acid

FKKKFKKH

1w112

][

Account for dilution for formal concentration (F) of BH+

M0500.000.1000.10

00.10M1000.0'F


of B

Dilution factor

Initial volume of B

Total volume50710163

101630500010

1010050001010

8

85

145105

.)x.log(pH

x..

))(().)()((H

][

Solve for pH using intermediate form equation


2.) A Typical Case Point D

- Before Second Equivalence Point

- Buffer Problem

00.51log00.5BHBHlogpKpH 2

22a

][][

59

14

1bw

2a 101101101

KKK

Point D (15 mL) is halfway to second equivalence point (2x10 mL). First, subtract Ve (10 mL)

155

ml5ml00.5ml00.10

BHBH

22

][][

pH = pKa1=5.00


2.) A Typical Case Point E

- Second Equivalence Point- Weak acid problem

Account for dilution for formal concentration (F) of BH2

+2

M0333.000.1000.20

00.10M1000.0'F


of B

Dilution factor

Initial volume of B

Total volume

pH determined by acid dissociation of BH2+2

Kb2


2.) A Typical Case Point E

- Second Equivalence Point- Weak acid problem

0.0333 - x x x

24.3pH1072.5x

100.1x0333.0

xxF

x

4

522

2bw

1a KKK

Ka1


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:

M1043.100.1000.25

00.5M1000.0H 2

][

mL00.500.2000.25VV eequivalencadded Excess acid:

Concentration of H+:

pH:

85.1)1043.1log(Hlog 2 ][-pH


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


4.) Using Derivatives to Find End Point Useful when End points overlap End Point of titration curve is where slope is

greatest - dpH/dV is large- pH change in pH between consecutive points- V average of pair of volumes- Second derivative is similar difference using

first derivative values ph = 4.400-4.245=0.155

End point: 1st derivative is maximumEnd point: 2nd derivative is zero


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

beaA

HApHb VVK10V

where: Vb = volume of strong base addedVe = volume of base needed to reach equivalence point

A-, HA = activity coefficients ≈ 1


5.) Using Gran Plot to Find End Point Plot is a straight line

- If ratio of activity coefficients is constant- Slope = -KaHA/a-

- X-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

Never Goes to Zero, approximation that every mole of OH- generates one mole of A- is not true as Vb approaches Ve

Slope Gives Ka

x-intercept gives 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


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


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 green will change colorSignificantly past the equivalence point resulting in an error.

Bromocresol purple color change brackets the equivalence point and is a good indicator choice


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


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?

Documents

Acid-Base Titrations