CL1825_Week 3_Acid-Base & Complexation Titrations (1)

8/14/2019 CL1825_Week 3_Acid-Base & Complexation Titrations (1)

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Outlinesof

Lecture

1. AcidBaseTitration

2. TitrationCurveofaStrongAcidwithStrongBase

3. ChoiceofIndicators

4. Complexation Titration5. Complexation TitrationwithEDTA


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References

1. Fundamentals of Analytical Chemistry, 8th

ed.,Douglas A. Skoog, Donald M. West, F. James Holler,

Stanley R. Crouch

Brooks/Cole

2. Instant Notes Analytical Chemistry, 1st ed.,

D. Kealey, P. J. Haines, BIOS Scientific Publishers Ltd.


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AcidBaseTitrations

Acid-base titrations or neutralisation titrations arewidely employed to determine the amounts of acids

and bases.

The titration involves the neutralisation of an analyte

of unknown concentration and a standard solution of

known concentration.

The standard solutions are strong acids (eg. HCl,

HClO4, H2SO4) or strong bases (eg. NaOH, KOH,

Ba(OH)2).


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AcidBaseTitrations

The strong acids / bases react more completely withthe analytes compared to weak acids / bases.

Therefore, they give sharper end points.

Standard solutions of acids are prepared by diluting

concentrated acids.

Nitric acid is seldom used because its oxidising

properties can cause undesirable side reactions that

may affect the titration results.

Standard solutions of bases are prepared by

disssolving solid NaOH, KOH in water.


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TitrationCurve

of

A

Strong

Acid

withAStrongBase

pH change in a titration can be calculated by applying

the formulae below:

pH = - log [H3O

+

] pH + pOH = pKw = 14


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Question

For a titration of 50.00 mL of 0.05 M HCl with 0.10 MNaOH, calculate the pH of the solution:

(a) before any base is added

(b) when 10.00 mL of base is added (c) when 25.00 mL of base is added

(d) when 25.10 mL of base is added


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(a) HCl(aq)

H+

(aq) + Cl-

(aq)

[H+] = [HCl] = 0.05 M

pH = - log [H+] = -log (0.05) = 1.30


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(b) No. of mol NaOH = 0.1 0.010 = 1 10-3 mol

No. of mol HCl = 0.05 0.05 = 2.5 x 10-3 mol

No. of mol HCl left = (0.05 0.05) - 1 10-3

= 1.5 10-3 mol

Total volume of solution = 50 + 10 = 60mL

Concentration of H+ = Concentration of HCl

= 1.5 10-3 / 0.06

= 0.025 M

pH = - log [H+] = -log (0.025) = 1.60


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(c)After the addition of 50.00-mL HCl, the equivalence point

is reached. Neither HCl nor NaOH is in excess, therefore

the concentrations of H+ and OH- are the same.

[H+][OH-] = Kw = 1.00 10-14

[H+][H+] = 1.00 10-14

[H+] = (1.00 10-14)0.5 = 1.00 10-7

pH = -log [H+] = -log (1.00 10-7) = 7.0


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(d) No. of mol NaOH added = 0.1 0.0251

= 2.51 10-3 mol

No. of mol HCl = 0.05 0.050 = 2.5 10-3

molNo. of mol NaOH left = 2.51 10-3 2.5 10-3

= 0.01 10-3 mol

Total volume of solution = 50 + 25.1 = 75.1mLConcentration of OH- = 0.01 10-3 / 0.0751

= 1.332 x 10-4 M

pOH = - log [OH

-

] = -log (1.332 x 10

-4

) = 3.876pH = pKw pOH

= 14 3.876

= 10.12


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Learning Check Calculate the pH of a solution if 25.00mL of 0.200M

HBr is mixed with 12.50mL of 0.100 M Ba(OH)2 isadded.


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Titration

Curve

of

A

Strong

Acid

withAStrongBase

The value of pH is plotted against the volume oftitrant added to form a pH titration curve.

The titration curve makes it possible to identify the

equivalence point of a titration, the point at which the

stoichiometric amount of acid and base have been

mixed together.

The end point of a titration is based on the

experimental observation, for example change of

colour for an indicator.


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Titration

Curve

of

A

Strong

Acid

withAStrongBase

Features:(i) A gradual increase of

pH before and after the

equivalence point.

(ii) pH rises sharply near

the equivalence point.

(iii) At the equivalence

point, pH = 7


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ChoiceofIndicators

During acid-base titration, an indicator is added. Indicators are weak organic acids or bases whose

undissociated form differs in colour from its

conjugate base or conjugate acid form.

H In + H2O In- + H3O

+

During a titration, an indicator which changes colour

at a pH range on the steep vertical portion of the

titration curve is chosen.


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ChoiceofIndicators

Indicator pH Range ofColour Change Colour Change

Methyl orange 3.1-4.4 Red to orange

Methyl red 4.2-6.3 Red to yellow

Bromothymol

blue

6.2-7.6 Yellow to blue

Phenol red 6.8-8.4 Yellow to red

Phenolphthalein 8.3-10.0 Colourless to

pink


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For weak acid strongbase titration (bluecurve), the pH at the

equivalence point (a) is8.72. Phenolphthalein issuitable because itchanges colour at the pH

range of 8.2-9.8, which isat the vertical portion ofthe titration curve.

Methyl red is unsuitable

because it changescolour at pH 4.2-6.0,which is not at the verticalportion of the curve.


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For strong acid strong

base titration (red curve),

the pH at the equivalence

point (b) is 7. Bothphenolphthalein and

methyl red are suitable

because they changecolour at the pH ranges

which are within the

vertical portion of the

titration curve.


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Complexation Titration Complexation titration (also known as chelatometry) is a

volumetric analytical technique in which the end point of

a titration is indicated by the formation of a coloured

complex.

This technique has been widely used to determine theconcentrations of different metal cations in solution.

Usually, an organic dye or indicator that can produce a

distinctive colour change is added. A complexing reagent that can form a stable complex

with the metal cation is added from the burette to the

solution.


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Complexation Titrationwith

EDTA

EDTA (ethylenediaminetetraacetic acid) is the most

widely used complexing agent in the titration.

EDTA has 4 carboxyl groups and 2 amine groups

which can act as electron pair donors. It is a

hexadentate ligand as it has the potential to form 6

coordinate bonds with the metal cation.


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Complexation TitrationwithEDTA

Titrations with EDTA are typically carried out at pH >12 because at this pH, all the carboxyl groups are

deprotonated. The structure is shown below:


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Complexation TitrationwithEDTAAt pH > 12, EDTA is acting as a hexadentate ligand

and is able to form 6 coordinate bonds with the metal

cation. The shape of the metal complex is

octahedral.


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EDTA

EDTA will bind to the metal cations in 1:1 ratioregardless of the charges of the metal cations.

Sodium EDTA dehydrate salt is usually used forpreparing the EDTA solution.

The formation constants KMY for EDTA complexes canbe calculated using the equation below:

Y4- refers to EDTA with all 4 carboxyl groups fully

deprotonated.


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EDTA

For examples,

The bigger the value of KMY, the more stable is the

EDTA-metal complex. EDTA is used extensively in the standardisation of

metal cation solution because the formation

constants KMY is very high, meaning that theequilibrium for the reaction lies far to the right to form

the metal-EDTA complexes.


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Metalcations andKMYvalues

Cation KMY Cation KMY

Ag+ 2.1 107 Cu2+ 6.3 1018

Mg2+ 4.9 108 Zn2+ 3.2 1016

Ca2+ 5.0 1010 Cd2+ 2.9 1016

Co2+ 2.0 1016 Al3+ 1.3 1016

Ni2+ 4.2 1018 Fe3+ 1.3 1025


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Complexation TitrationwithEDTA During the titration, an organic dye is added to

indicate when the end point has been reached.

Common organic dyes are Fast Sulphon Black,

Eriochrome Black T and Eriochrome Red B.

These dyes bind to the metal cations to form

coloured complexes.

When EDTA is added, EDTA will displace the organic

dye and bind strongly to the metal cations.


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EDTA

When all the dyes have been displaced from the

metal cations, there is a colour change which

indicates the end point has been reached.

Examples:

Eriochrome Black T forms a red complex with Ca2+

and Mg2+. The free form of Eriochrome Black T is blue.

Fast Sulphon Black F forms a purple complex with

Cu2+. The free form of Fast Sulphon Black F is green.

(red) (colorless) (colorless)(blue)

End Point indicated by a color

change from red to blue


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EDTATitrations

Titration of Mg2+ by EDTA

- Eriochrome Black T Indicator

Addition of EDTA

Before Near After

Equivalence point


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QuestionCalculate the volume of 0.05M EDTA solution that is

needed to titrate 25.0 mL of 0.075 M Mg(NO3)2

No. of mol Mg2+ = No. of mol Mg(NO3)2

= 0.075 0.025

= 1.875 10-3 mol

No. of mol EDTA = 1.875 10-3 mol

Volume of EDTA = 1.875 10-3

/ 0.05= 0.0375 L

= 37.5 mL


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Learning CheckA 25.00 mL sample though to contain 0.0125M Ca2+

is titrated at pH 10.0 using a 0.0105M solution of EDTA as the titrant. Write the titration reaction for this

analysis and determine the volume of titrant that

required to reach the end point.


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Learning CheckCa2+ + EDTA4- [Ca(EDTA)]2-

No. of mol Ca2+ = 0.0125 0.025

= 3.125 10-4 mol

No. of mol EDTA = 3.125 10-4 mol

Volume of EDTA = 3.125 10-4 / 0.0105

= 0.02976 L= 29.76 mL

CL1825_Week 3_Acid-Base & Complexation Titrations (1)

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