approach to titration - university of texas at austinwebb.cm.utexas.edu/courses/titrations and...

Approach to titration •  What type of titration being conducted (strong A-

strong B? strong A- weak B? strong B – weak A? Never weak-weak!)

•  What information is requested? (pH after volume added? Volume to reach eq. pt.?, etc.) (eq. pt. on next slide)

•  Where is eq. pt.? •  Where are you working relative to this volume of

titrant? (start of titration? before eq. pt.? at eq. pt.? after eq. pt.? )

Remember: Moles react with moles! Don’t forget to correct for dilution, i.e., solution volume after titrant added.

Equivalence point Equivalence point is when… moles of acid (or base) added = moles of base (or

acid) present in flask

ml base added

pH

Mentally picture a titration curve

ml base added

pH

pH

ml acid added

acid being neutralized

excess titrant (base) present

original solution of acid

original solution of base

base being neutralized

base analyte exactly neutralized

excess titrant (acid) present

acid analyte exactly neutralized

Remember: HA + OH- à H2O + A- K is BIG (… also for titration using strong acid)

Preliminary scouting of the titration curve construction or interpretation

( )( )titrantHAinitial

HAinitalHAinitialA VV

VFF

+=−

At equivalence point:

Moles of strong acid completely neutralize equal number of moles of strong or weak base ! Moles of strong base completely neutralize equal number of moles of strong or weak acid ! Moles react with moles BUT you need to correct for dilution (except when using HH eq. since volumes cancel).

Also remember:

basebaseacidacid FVFV =

For example, to determine the formal amount of conj base after titrating weak acid to e.p. ...

OH- + HA à H2O + A-

Sometimes seen as VAMA=VBMB or in Oxtoby et al.: c0V0=ctVe

moles

volume

Titration of strong acid with strong base

0

2

4

6

8

10

12

14

0 10 20 30 40 50

ml base added

pH

eq. pt. pH=7.00

Fa=0.01 M Va=40 ml Fb=0.02 M At equiv. point: Vb=(Va*Ma)/Mb=20 ml

Strong acid / strong base

titration

remaining strong acid determines pH

excess strong base determines pOH and, hence, the pH

Formal amount of acid in initial solution

The Key to Calculations Relate moles titrant to moles analyte.

…this requires that you know the reaction and stoichiometry!!

Regions of Titration Curve: Start of titration

Before Equivalence Point (Analyte is in excess)

At Equivalence Point (Analyte nominally neutralized by titrant)

After Equivalence Point (Titrant is in excess)

Weak acid – strong base titration

Weak acid titrated with strong base

0

2

4

6

8

10

12

14

0 10 20 30 40ml base added

pH

weak acid (ICE diagram) buffer (HH eq)

eq. pt.: conj. base (ICE diagram) -- remember that you’re solving for OH- and need to convert to pH

excess strong base:[OH-] = [excess base]

Weak base titrated with strong acid

0

2

4

6

8

10

12

14

0 10 20 30 40ml acid added

pHWeak base – strong acid titration

weak base (ICE diagram) solve for OH-

buffer (HH eq); remember to use Ka

eq. pt.: conj. acid (ICE diagram)

excess strong acid :[H+] = [excess acid]

A few quick examples…

1.  21. 56 mL of 0.100 M NaOH are needed to titrate 20.00 mL HCl. What is the concentration of HCl?

OH- + H+ à H2O

L 0.100 mol OH-

1mol OH- 1mol H+ 0.02156L

0.02000L

example

= 0.107 M HCl

AA

BB FVFV

=

mL 0.100 mmol OH-

1mmol OH- 1mmol H+ 21.56 mL

20.00 mL = 0.107 M HCl

or staying with mL…

0.020 L 0.150 mol H+

L 1 mol H+

1 mol OH-

0.100 mol OH-

L

= 0.030 L

2. 20.00 mL of 0.150 M HCl are titrated with 0.100 M NaOH. What is the pH at 0 mL, 25 mL, 30 mL and 35 mL titrant added?

Titration of strong acid with strong base

First, calculate Vep (Volume of titrant needed to reach equivalence point)

example

Vbase(ep) =

Remember… at equiv. point, moles of acid = moles of base basepebaseacidacid FVFV .).(=

H+ + OH- à H2O

What is the pH at 25 mL titrant added?

Moles analyte (H+) remaining = 0.0005 mol H+

(initial volume + volume added = 0.04500 L) = 0.0111 M H+

What is the pH at 0 mL titrant added? (Only HCl present)

pH = -log[H+] = -log(0.150) = 0.824

(volume analyte)(conc. Analyte) – (volume titrant added)(conc. titrant)

(0.0200 L H+)(0.150 mol H+/L) – (0.02500 L OH-)(0.100 mol OH-/L)

moles of acid initially moles of OH added = moles acid neutralized

Before Eq. Pt. - Analyte in excess:

pH = 1.96

Remember: H+ + OH- àH2O K = 1/Kw = 1014 ! ! !

What is the pH at 30 mL titrant added? (This is equiv. point !!!)

At Eq. Pt. - Neither in excess. Solution contains water, Na+ and Cl-.

Na+ and Cl- have neither acid or base character... solution is neutral.

Kw = [H+][OH-] = 10-14

[H+] = [OH-] = 10-7M pH = 7.00


After Eq. Pt. - Titrant in excess: excess moles base = (VB-Veq)FB

molarity of solution containing excess strong base:

96.1114041.2

][1009.92035100.0)3035()( 3

=−==

==+

−=

+

− −−

pOHpHpOH

OHxVVFVV

AB

BeqB

Mentally picture the titration curve

ml base added

pH


excess titrant (base) present

original solution of acid

acid analyte exactly neutralized

example Titration of weak acid with strong base

20.00 mL of 0.150 M HA are titrated with 0.100 M NaOH. Ka=2.4x10-4 What is the pH at 0 mL, 25 mL, 30 mL and 35 mL titrant added?

example

0.020 L 0.150 mol H+

L 1 mol H+

1 mol OH-

0.100 mol OH-

L

= 0.030 L

First, calculate Vep (Volume of titrant needed to reach equivalence point)

Vep =

ml base added

pH

30

Example titration problem

(Only HA present)

pH = 2.22

This is simply a solution containing a weak acid, HA, at 0.150 F

HA ⇄ H+ + A-

][]][[

HAAHKa−+

=

HA H+ A- i 0.15 0 0 c -x +x +x e 0.15-x x x

MxHOKassumptionMxHx

xx

xxKa

3

3

224

106][)(106][

150.0150.0104.2

−+

−+

−

=

==

≈−

==


(volume analyte)(conc. Analyte) – (volume titrant added)(conc. titrant)

(0.0200 L HA)(0.150 mol HA/L) – (0.02500 L OH-)(0.100 mol OH-/L)

moles of acid initially moles of OH added = moles acid neutralized

Before Eq. Pt. - Analyte in excess: HA + OH- à H2O + A- Neutralization reax: K=LARGE,

MHA

xxsolutionofvolHAmolesHA

0111.0][2520

)1.025()15.020(][

=

+

−== NOTE: we can keep

vol in mL if we are consistent

ml base added

pH


][][log

HAApKpH a

−

+=

Henderson-Hasselbach equation (Buffer region of curve!)


pH = 4.32

MA

xsolutionofvoladdedOHmolesA

0556.0][2520)1.025(][

=

+==

−

− NOTE: we can keep vol in mL if we are consistent... units will cancel !

0111.00556.0log62.3

][][log +=+=

−

HAApKpH a

(volume titrant added)(conc. titrant)

(0.02500 L OH-)(0.100 mol OH-/L)

moles of OH- added = moles acid neutralized = moles A- produced

HA + OH- à H2O + A-

Henderson-Hasselbach equation

ml base added

pH


What is the pH at 25 mL titrant added? (continued)

(This is equiv. point !!!)

At Eq. Pt. – All HA has nominally be converted to A-. Solution contains water, Na+ and A- (a weak base!)

A- + H2O ⇄ HA + OH-

][]][[

−

−

=AHAOHKb

A- OH- HA i 0.06 0 0 c -x x x e 0.06-x x x

MxOHOKassumptionMxOHx

xx

xxKb

6

6

2211

1058.1][)(1058.1][

06.006.01017.4

−−

−−

−

=

==

≈−

==

MmlmlMxA initial 06.0)3020(2015.0][ =

+=−

pH =14-pOH= 8.20


After Eq. Pt. - Titrant in excess: excess moles base = (VB-Veq)MB

molarity of solution containing excess strong base:

96.1114041.2

][1009.92035100.0)3035()( 3

=−==

==+

−=

+

− −−

pOHpHpOH

OHxVVFVV

AB

BeqB

Note: this is exactly the same calculation and conclusion that we reached when titrating the strong acid in the earlier example.


Recap of weak acid titrated by

strong base

Weak acid titrated with strong base

0

2

4

6

8

10

12

14

0 10 20 30 40ml base added

pH

weak acid (ICE diagram) buffer (HH eq)

eq. pt.: conj. base (ICE diagram) -- remember that you’re solving for OH- and need to convert to pH

excess strong base:[OH-] = [excess base]

Weak base titrated with strong acid

0

2

4

6

8

10

12

14

0 10 20 30 40ml acid added

pH

Recap of weak base titrated

by strong acid

weak base (ICE diagram) solve for OH-

buffer (HH eq); remember to use Ka

eq. pt.: conj. acid (ICE diagram)

excess strong acid :[H+] = [excess acid]

Acid-base indicators are weak acids (bases)!

HIn ⇋ H+ + In-

1110log

1101log

][][log

+=+=

−=+=

+=−

aa

aa

a

pKpKpH

pKpKpH

HInInpKpH

In solution when [HIn]=10 [In], then…

In solution when [In]=10 [HIn], then…

BLUE RED

BLUE

RED

Color change: pH = pKa + 1

8.6

(Henderson-Hasselbalch eq.)

Why do try to use the minimum amount of indicator?

pH

mL of base added

Approximate pKa of indicator

Use of correct indicator

Use of incorrect indicator

pH

mL of base added

Approximate pKa of indicator

Some Acid-Base Indicators

http://www.precisionnutrition.com/ie-how-ph-strips-work

Acid bas indicators in nature Hydrangeas can change color with soil acidity

Add dolomitic lime (CaMg(CO3)2 ) several times a year. This will help to raise the pH.

Al must be present. To make the Al available to the plant, the pH of the soil should be low (5.2-5.5). Adding aluminum sulfate will supply Al and tend to lower the pH of the soil.

Al3+ + H2O ⇋ AlOH2+ + H+ CO3

2- + H2O ⇋HCO3- + OH-

Titration of polyprotic acids

0

2

4

6

8

10

12

14

0 5 10 15 20 25 30

ml strong base added

pH

Analyzing titration curve of polyprotic acid (H3A, H2A-, etc.)

Original solution was mixture (50% each) of H3A and H2A-

All converted (nominally) to HA2-

All converted (nominally) to A3-

All converted (nominally) to H2A-

0

2

4

6

8

10

12

14

0 5 10 15 20 25 30


pH


Original solution is mixture (50% each) of H3A and H2A-

Treat as soluiton of weak base, A3- (ICE diagram)

pH=1/2{pK2+pK3}

pH=1/2{pK1+pK2}

pH=pK3

pH=pK2

pH=pK1 **

** Unique in this case since we start with 50/50 mix of H3A and H2A

0

2

4

6

8

10

12

14

0 5 10 15 20 25 30


pH


Original solution was mixture (50% each) of H3A and H2A-

BUFFER REGIONS (use HH eq.)

Polybasic titration (e.g., initial solution of Na2HPO4 and Na3PO4)

0

2

4

6

8

10

12

14

0 5 10 15 20 25 30

ml strong acid added

pH

approach to titration - university of texas at austinwebb.cm.utexas.edu/courses/titrations and...

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