classical methods in techniques of analytical chemistry: titrimetric methods of analysis ert 207...

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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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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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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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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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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(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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(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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(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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(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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(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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(iv) Kjeldahl Analysis for Total Nitrogen (TN)

(a) KD description:

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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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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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(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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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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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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(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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(iii) End point determination

dy/dx

d2y/dx2

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(iv) Diluting effect of the titration curves

25.00 mL 0.1000 M I– titrated with 0.05000 M Ag+



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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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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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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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Fajans Method: An adsorbed/colored indicator.

Titrating Cl– and adding dichlorofluoroscein:

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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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classical methods in techniques of analytical chemistry: titrimetric methods of analysis ert 207...

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