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