CLASSICAL METHODS IN TECHNIQUES OF ANALYTICAL
CHEMISTRY: TITRIMETRIC METHODS OF ANALYSIS
ERT 207 ANALYTICAL CHEMISTRY
SEMESTER 1, ACADEMIC SESSION 2015/16
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Overview
OVERVIEW OF TITRIMETRY GRADES OF CHEMICALS TITRATION CALCULATIONS PRECIPITATION TITRATIONS- TITRATION CURVE
ARGENTOMETRIC TITRATION
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OVERVIEW OF TITRIMETRY
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Defination: Titrations (or titrimetric method) are
based on measuring the amount of a reagent of known concentration that reacts with the unknown.
It is the process of determining unknown concentration by adding the small increments of standard solution until the reaction is just complete.
A general equation can be expressed:aA + tT → productswhere, A: analyte; T: titrant
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OVERVIEW OF TITRIMETRY
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Revisiting Keywords of Titration Equivalence point: The point (e.g., volume of
titrant) in a titration where (theoretically) stoichiometrically equivalent amounts of analyte and titrant react.
Indicator: A colored compound whose change in color signals the (experimental) end point of a titration.
End point: The point (e.g., volume of titrant) in a (experimental) titration where we stop adding titrant in an experiment.
Titration error: The determinate error in a titration due to the difference between the end point and the equivalence point.
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OVERVIEW OF TITRIMETRY
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Type of Titrations based on Chemical Reactions:
i. Acid-Base Titrations:H+ + OH– → H2O K= 1/Kw
ii. Precipitation Titrations:Ag+
(aq) + Cl–(aq) → AgCl(s) K=1/Ksp
iii. Redox Titrations:5 H2O2 + 2 MnO4
– + H+ → 5 O2 + 2 Mn2+ + 8H2O
iv.Complexometric Titrations:EDTA + Ca2+ → (Ca–EDTA)2+
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OVERVIEW OF TITRIMETRY
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Type of Titrations based on Measuring Techniques:i. Volumetric titrimetry: Measuring the
volume of a solution of a known concentration (e.g., mol/L) that is needed to react completely with the analyte.
ii. Gravimetric (weight) titrimetry: Measuring the mass of a solution of a known concentration (e.g., mol/kg) that is needed to react completely with the analyte.
iii. Coulometric titrimetry: Measuring total charge (current x time) to complete the redox reaction, then estimating analyte concentration by the moles of electron transferred.
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OVERVIEW OF TITRIMETRY
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Type of Titration Curves:
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GRADES OF CHEMICALS
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Terms and Definitions:i. Reagent Grade:
The reagents which meets or surpasses the latest American Chemical Society specifications.
ii. Primary standard: The reagent which is ready to be
weighted and used prepare a solution with known concentration (standard).
Requirements of primary reagent are:
- Known stoichiometric composition- High purity - Nonhygroscopic - Chemically stable both in solid & solution- High MW or FW
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GRADES OF CHEMICALS
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Secondary standard: A standard which is standardized
against a primary standard. Certified reference materials (CRM):
A reference material, accompanied by a certificate, which has been analysed by different laboratories to determine consensus levels of the analyte concentration.
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GRADES OF CHEMICALS
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NIST Standard Reference Material® (SRM): A CRM issued by NIST that also meets
additional NIST-specific certification criteria and is issued with a certificate.
Standardization: The process by which the
concentration of a reagent is determined by reaction with a known quantity of a second reagent.
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TITRATION CALCULATIONS
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Terms and Definitions: Blank Titration: Titration procedure is
carried out without analyte (e.g., a distilled water sample). It is used to correct titration error.
Back titration: A titration in which a (known) excess reagent is added to a solution to react with the analyte. The excess reagent remaining after its reaction with the analyte, is determined by a titration.
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TITRATION CALCULATIONS
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(i) Standardization To standardizing a KMnO4 stock solution,
the primary standard of 9.1129 g Na2C2O4 is dissolved in 250.0 mL volumetric flask.
10.00 mL of the Na2C2O4 solution require 48.36 mL of KMnO4 to reach the titration end point. What is the molarity (M) of MnO4
– stock solution? (FW Na2C2O4 134.0) Solution:5C2O4
2–(aq) + 2MnO4
–(aq) + 16H+
(aq) → 10CO2(g) + Mn2+(aq) +
8H2O(l)
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TITRATION CALCULATIONS
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(ii) Unknown Analysis with a Blank Correction A 0.2865 g sample of an iron ore is dissolved in
acid, and the iron is converted entirely to Fe2+. To titrate the resulting solution, 0.02653 L of
0.02250 M KMnO4 is required. Also a blank titration require 0.00008 L of KMnO4 solution. What is the % Fe (w/w) in the ore? (AW Fe 55.847)
4242
4
422
242
422
422422
MnO 02250.0L 1mL 1000
mL 48.361
OC mol 5
MnO mol 2
mL 250mL 10
OCNa mol 1OC mol 1
OCNa g 134.0OCNa mol 1
1OCNa g 9.1129
M
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TITRATION CALCULATIONS
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(iii) Back Titration The arsenic in 1.010 g sample was
pretreated to H3AsO4(aq) by suitable treatment. The 40.00 mL of 0.06222 M AgNO3 was added to the sample solution forming Ag3AsO4(s): )(43)()()(43 33 saqaqaq AsOAgHAgAsOH
Solution:MnO4
–(aq) + 5Fe2+ + 8H+
(aq) → Mn2+(aq) + 5Fe3+ +
4H2O(l)
)/( %01.58%100 2865.0
1
1
847.55
1
5 titrantL 1
02250.01
02645.0
02645.0 00008.0 02653.0
2
4
24
wwFesamplegFemol
Feg
MnOmol
FemolMnOmoltitrantL
LLvoltitrantNet
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TITRATION CALCULATIONS
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(iii) Back Titration The arsenic in 1.010 g sample was
pretreated to H3AsO4(aq) by suitable treatment. The 40.00 mL of 0.06222 M AgNO3 was added to the sample solution forming Ag3AsO4(s):
The excess Ag+ was titrated with 10.76 mL of 0.1000 M KSCN. The reaction was:
Calculate the percent (w/w) As2O3(s) (fw 197.84 g/mol) in the sample.
)(43)()()(43 33 saqaqaq AsOAgHAgAsOH
)()()( saqaq AgSCNSCNAg
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TITRATION CALCULATIONS
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(w/w) %612.4%100 010.11
1 84.197
1000 1
2 1
1
11
4709.0
4709.0
3 0760.1 4888.2
3 1
3
0760.11
1
1000.07610
4888.2 1
06222.0 00.40
3232
32
32
3232
43
43
4343
4343
43
3
33
OAssamplegOAsmol
OAsg
OAsmmolOAsmol
AsmmolOAsmmol
AsOHmmolAsmmolAsOHmmol
AsOHmmolAsOHmmolx
AgmmolxAgmmolAgmmol
AgmmolxAsOHmmol
AgmmolAsOmmol Hx
AsOHmmolxby consumedmmol Ag
AgmmolSCNmmol
Agmmol
SCNmL
SCNmmolmL SCN.
SCNby consumedmmol Ag
mmolAgNOmL
AgNOmmolAgNOmL
addedmmol AgTotal
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TITRATION CALCULATIONS
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(iv) Kjeldahl Analysis for Total Nitrogen (TN)
(a) KD description:
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TITRATION CALCULATIONS
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Example 1: A typical meat protein contains 16.2%
(w/w) nitrogen. A 0.500 mL aliquot of protein solution
was digested, and the liberated NH3 was distilled into 10.00 mL of 0.02140 M HCl. The unreacted HCl required 3.26 mL of 0.0198 M NaOH for complete titration.
Find the concentration of protein (mg protein/mL) in the original sample.
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TITRATION CALCULATIONS
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Example 2: A 0.7121g sample of wheat flour was
analyzed by the Kjeldahl method. The ammonia formed by addition of
concentrated base after digestion with H2SO4 was distilled into 25.00 ml of 0.04977 M HCl.
The excess HCl was then back-titrated with 3.97 mL of 0.04012 M NaOH.
Calculate the percent protein in the flour using the 5.70 factor for cereal.
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TITRATION CALCULATIONS
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(v) Titration of a Mixture A solid mixture weighing 1.372 g containing
only sodium carbonate (Na2CO3, FW 105.99) and sodium bicarbonate (NaHCO3, FW 84.01) require 29.11 mL of 0.7344 M HCl for complete titration:
Find the mass of each component of the mixture.
22)(3
22)(32 22
COOHNaClHClNaHCO
COOHNaClHClCONa
aq
aq
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TITRATION CALCULATIONS
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PRECIPITATION TITRATIONS
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A titration in which the reaction between the analyte and titrant involves a precipitation.
(I) Titration curve: Guidance in precipitation titration
calculation Find Ve (volume of titrant at equivalence point) Find y-axis values:
- At beginning- Before Ve
- At Ve
- After Ve
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PRECIPITATION TITRATIONS
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EXAMPLE 3: For the titration of 50.0 mL of 0.0500 M Cl–
with 0.100 M Ag+. The reaction is: Ag+
(aq) + Cl–(aq) AgCl(s)
K = 1/Ksp = 1/(1.8×10–10) = 5.6 x 109
Find pAg and pCl of Ag+ solution added(a) 0 mL (b) 10.0 mL (c) 25.0 mL (d) 35.0 mL
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PRECIPITATION TITRATIONS
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(ii) Construct a titration curve: Example: Titration of 50.0 mL of 0.0500
M Cl– with 0.100 M Ag+
pCl
pAg
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PRECIPITATION TITRATIONS
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(iii) End point determination
dy/dx
d2y/dx2
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PRECIPITATION TITRATIONS
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(iv) Diluting effect of the titration curves
25.00 mL 0.1000 M I– titrated with 0.05000 M Ag+
25.00 mL 0.01000 M I– titrated with 0.005000 M Ag+
25.00 mL 0.001000 M I– titrated with 0.0005000 M Ag+
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PRECIPITATION TITRATIONS
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EXAMPLE 6 A 25.00 mL solution containing Br– and
Cl– was titrated with 0.03333 M AgNO3. Ksp(AgBr)=5x10–13, Ksp(AgCl)=1.8x10–10.
(a)Which analyte is precipitated first?(b)The first end point was observed at
15.55 mL. Find the concentration of the first that precipitated (Br– or Cl–?).
(c)The second end point was observed at 42.23 mL. Find the concentration of the second that precipitated (Br– or Cl–?).
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ARGENTOMETRIC TITRATION
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General information: Define Argentometric Titration: A
precipitation titration in which Ag+ is the titrant.
Argentometric Titration classified by types of End-point detection:
i. Volhard method: A colored complex (back titration)
ii. Fajans method: An adsorbed/colored indicator
iii. Mohr method: A colored precipitate
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ARGENTOMETRIC TITRATION
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Volhard method: A colored complex (back titration). Analysing Cl– for example:Step 1: Adding excess Ag+ into sample
Ag+ + Cl– → AgCl(s) + left Ag+
Step 2: Removing AgCl(s) by filtration/washing
Step 3: Adding Fe3+ into filtrate (i.e., the left Ag+)
Step 4: Titrating the left Ag+ by SCN–:Ag+ + SCN– → AgSCN(s)
Step 5: End point determination by red colored Fe(SCN)2+ complex. (when all Ag+ has been consumed, SCN– reacts with Fe3+)SCN– + Fe3+ → Fe(SCN)2+
(aq)
Total mol Ag+ = (mol Ag+ consumed by Cl–) + (mol Ag+ consumed by SCN–)
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ARGENTOMETRIC TITRATION
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Fajans Method: An adsorbed/colored indicator.
Titrating Cl– and adding dichlorofluoroscein:
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ARGENTOMETRIC TITRATION
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Mohr Method: A colored precipitate formed by Ag+ with anion, other than analyte, once the Ve reached.
Analysing Cl– and adding CrO42–:
Precipitating Cl–:Ag+ + Cl– → AgCl(s) Ksp = 1.8 x 10–10
End point determination by red colored precipitate, Ag2CrO4(s):
2Ag+ + CrO42– → Ag2CrO4(s)
Ksp = 1.2 x 10–12
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EXAMPLE 1
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EXAMPLE 2
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Amount HCl =
= 1.2443 mmol
Amount NaOH =
= 0.1593 mmol
Amount N =Amount HCl – Amount NaOH = (1.2443 – 0.1593) mmol = 1.0850 mmol
mLHCl
mmol.HClxmL. 0497700025
mLNaOH
mmol.NaOHxmL. 040120973
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EXAMPLE 2
% N =
= 2.1342
% protein =
= 12.16
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%xsampleg.
Nmmol
Ng.Nxmmol.
10071210
014007008501
N%
protein%.xN%.
70513422
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EXAMPLE 3
(a)0 mL Ag+ added (At beginning)[Ag+] = 0, pAg can not be calculated.[Cl–] = 0.0500, pCl = 1.30
bblee@unimap
AgmL.
Agmmol.
AgmL
Clmmol
Agmmol
ClmL
Clmmol.ClmL.Ve
025
1000
1
1
1
1
05000 050
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EXAMPLE 3(b) 10 mL Ag+ added (Before Ve)
bblee@unimap
M.mL.
Clmmol.]Cl[
mL.mL.mL.V
Clmmol.
Agmmol
Clmmol
AgmL
Agmmol.AgmL.
ClmL
Clmmol.ClmL.
ClmmoledprecipitatClmmoloriginalLeftClmmol
total
10502 060
51
060 010 050
501
1
1
1
1000 010
1
05000 050
2
148
601
102710502
1081 92
10
.pAg
.pCl
M..
.
]Cl[
K]Ag[ sp
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EXAMPLE 3
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(c)25 mL Ag+ added (At Ve)
AgCl(s) Ag+(aq) + Cl–(aq)
Ksp = 1.8×10–10
s = [Ag+]=[Cl–] Ksp = 1.8×10–10 = s2
[Ag+]=[Cl–]=1.35x10–5
pAg = 4.89 pCl = 4.89
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EXAMPLE 3
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(d) 35 mL Ag+ added (At Ve)
93.1
82.7
1053.11018.1
108.1
][][
1018.1 0.85
00.1][
0.85 0.35 0.50
00.1
1
100.0 0.25
1
100.0 0.35
82
10
2
pAg
pCl
MAg
KCl
MmL
AgmmolAg
mLmLmLV
Agmmol
AgmL
AgmmolAgmL
AgmL
AgmmolAgmL
AgofwithVAgmmoledprecipitatAgmmoloriginalLeftAgmmol
sp
total
e
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EXAMPLE 4(a) Ag+
(aq) + Br–(aq) AgBr(s)
K = 1/Ksp(AgBr) = 2x1012
Ag+(aq) + Cl–(aq) AgCl(s)
K = 1/Ksp(AgCl) = 5.6x109
Ans: AgBr precipitated first
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EXAMPLE 4
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