titrations volumetric analysis procedures in which we measure the volume of reagent needed to react...

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Titrations Volumetric analysis

Procedures in which we measure the volume of reagent needed to react with an analyte

Titration Increments of reagent solution (titrant) are added to

analyte until reaction is complete.- Usually using a buret

Calculate quantity of analyte from the amount of titrant added.

Requires large equilibrium constant (Thermodynamic)

Requires rapid reaction (kinetic)

aA + tT → productsA: analyteT: titrant

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Titrations

Buret Evolution

Descroizilles (1806)Pour out liquid

Gay-Lussac (1824)Blow out liquid

Henry (1846) Copper stopcock

Mohr (1855) Compression clip

Used for 100 years

Mohr (1855) Glass stopcock

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Type of Titrations based on Measuring Techniquesi) 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 Titrations based on Chemical Reactionsi) Acid-Base Titrations, example:

H+ + OH– → H2Oii) Precipitation Titrations, example:

Ag+(aq) + Cl–

(aq) → AgCl(s)

iii) Redox Titrations:5 H2O2 + 2 MnO4

– + H+ → 5 O2 + 2 Mn2+ + 8H2Oiv) Complexometric Titrations, example:

EDTA + Ca2+ → (Ca–EDTA)2+

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Type of Titration Curves

Type Example. y-axis x-axis

Acid-base HCl/NaOH pH V. NaOH

Precipitation Ag+/Cl– pAg+ V. Ag+

Complexation Ca2+/EDTA pCa2+ V. EDTA

Redox MnO4–/Fe2+ Potential V. Fe2+

Type Example y-axis x-axis

Spectro-photometric

apotransferrin/Fe3+

Absorbance V.Fe3+

Thermo-metric

H3BO4/NaOH

Temperature V.NaOH

V. = volume

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Expressing concentrationFormalityMolarity (V & W)MolalityNormality

%W/W %W/V %V/V

part per thousand (ppt) - X's 1000parts per million (ppm) - X's 106

parts per billion (ppb) - X's 109

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relationship between titrant and analyte

# Eqg titrant = # Eqg analyte

(V*N)titrant =(V*N)analyte

# Eqg titrant = (V*N)titrant

#molestitrant=(V*M)titrant

#molesanalyte=(V*M)analyte

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Standardization: The process by which the concentration of a reagent is determined by reaction with a known quantity of a second reagent

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 and reaction- High purity- Nonhygroscopic - Chemically stable both in solid and solution- High MW or FW

Secondary standard: A standard solution which is standardized against a primary standard.

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Standardization

Required when a non-primary titrant is used

- Prepare titrant with approximately the desired concentration- Use it to titrate a primary standard- Determine the concentration of the titrant-

titrant known concentration

analyte unknown concentration

titrant unknown concentration

analyte known concentration

Titration Standardization

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Standardization of 0.1 M NaOH

1-selection the PS (e.g. KHP)

2-wheing the PS

10*0.1=mg/204.1 213.8

3-making solution

4-addind suitable indicator

5-titration 9.1ml

6-calculation 9.1*n=213.8/204.1 n=0.115

10

EqW

mgNml

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N1V1=N2V2


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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Example: 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:

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

MAns

5C2O42–

(aq) + 2MnO4–

(aq) + 16H+(aq) → 10CO2(g) + Mn2+

(aq) + 8H2O(l)

Standardization

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Example: 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)MnO4

–(aq) + 5Fe2+ + 8H+

(aq) → Mn2+(aq) + 5Fe3+ + 4H2O(l)

Unknown Analysis with a Blank Correction

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)/( %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


Back Titration

1)Add excess of one standard reagent (known concentration)

2)Titrate excess standard reagent to determine how much is left

- Add Fe2+ to determine the amount of MnO4- that did not react with oxalic acid

- Differences is related to amount of analyte - Useful if better/easier to detect endpoint

14

MnO4–

(aq) + 5Fe2+ + 8H+(aq) → Mn2+

(aq) + 5Fe3+ + 4H2O(l)

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Back TitrationExample: 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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)

1

1

1000.07610

1

06222.0 00.40(

3

33

SCNmmol

Agmmol

SCNmL

SCNmmolmL SCN.

AgNOmL

AgNOmmolAgNOmL

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(w/w) %612.4%100 010.1

132OAs

sampleg

1

1

3

1

43

43

AsOHmmol

Asmmol

mmolAg

AsOHmmol

32

32

32

3232

1

84.197

1000

1

2

1

OAsmol

OAsg

OAsmmol

OAsmol

Asmmol

OAsmmol


In a back titration analysis of HCO3-, 25 mL of a bicarbonate solution is reacted with 25.00 mL of 0.100 M NaOH. The excess NaOH was titrated with 0.100 M HCl. This required 14.82 mL. What is the concentration of bicarbonate in solution?

NaOH + HCO3- → Na+ + CO3

- + H2O

NaOH + HCl → NaCl + H2O

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Back Titration


Equivalence point

Quantity of added titrant is the exact amount necessary for stoichiometric reaction with the analyte- Ideal theoretical result

AnalyteOxalic acid(colorless)

Titrant(purple)

(colorless) (colorless)

Equivalence point occurs when 2 moles of MnO4- is added to 5 moles of Oxalic acid

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End point Occurs from the addition of a slight excess of titrant

- Marked by a sudden change in the physical property of the solution

- Change in color, pH, voltage, current, absorbance of light.

- End point ≈ equivalence point

AnalyteOxalic acid(colorless)

Titrant(purple)

(colorless) (colorless)

After equivalence point occurs, excess MnO4- turns solution purple Endpoint

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Titration Error

- Difference between endpoint and equivalence point

Corrected by a blank titration

1) repeat procedure without analyte2) Determine amount of titrant needed to observe change3) subtract blank volume from titration

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Calculation of ascorbic acid in Vitamin C tablet:

(i) Starch is used as an indicator: (ii) starch + I3

- starch-I3- complex

ascorbic acid was oxidized with I3-:

1 mole ascorbic acid 1 mole I3-

21

Standardization: Suppose 29.41 mL of I3- solution is required to react with 0.1970 g of pure

ascorbic acid, what is the molarity of the I3- solution?

Analysis of Unknown: A vitamin C tablet containing ascorbic acid plus an inert binder was ground to a powder, and 0.4242g was titrated by 31.63 mL of I3

-. Find the weight percent of ascorbic acid in the tablet.

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Titration of a MixtureExample: 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

mass moles

Total mixture 1.372

Na2CO3 x x/105.99 g/mol

NaHCO3 1.372 - x 1.372 – x/84.01 g/mol

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33

3232

33

33

3

3232

3232

32

648.0 )372.1(

724.0

01.84 372.1

99.105 2

02138.0

01.84

372.1 1

1 01.84

1 )372.1(

99.105 2

1 2

99.105 1

02138.0 1

7344.0 000 1

1 11.29

NaHCOgNaHCOgx

CONagCONagx

xx

Hmolx

NaHCOmolHmol

NaHCOgNaHCOmol

NaHCOgx

NaHCObyconsumedmmol H

Hmolx

CONamolHmol

CONagCONamol

CONagx

CONabyconsumedmmol H

HmolHClL

HmolmL

LHClmL

addedmmol HTotal

AnsAns

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Direct and back (indirect) titration of Aspirin

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A titration in which the reaction between the analyte and titrant involves a precipitation.

Ag+(aq) + Cl–

(aq) AgCl(s)

AgCl(s) Ag+(aq) + Cl–

(aq) Ksp = 1.8×10–10

s = [Ag+]=[Cl–]

[Ag+]=[Cl–]=1.35x10–5

pAg = 4.89 pCl = 4.89

Precipitation Titrations

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Titration curve of 50.0 mL of 0.0500 M Cl– with 0.100 M Ag+

pCl

pAg

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Example: 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

Solution:

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50*0.05=0.1*V2In eq point, 25 mlN1V1=N2V2

(a) 0 mL Ag+ added (At beginning)

[Ag+] = 0, pAg can not be calculated.

[Cl–] = 0.0500, pCl = 1.30

(b) 10 mL Ag+ added (Before Ve)

21

2211][

VV

VNVNCl

(d) 35 mL Ag+ added (After Ve)21

1122][

VV

VNVNAg

√Ksp=[Ag+]=[Cl-]=1.34*10-5

][][

Cl

KAg sp

][][

Ag

KCl sp


Diluting effect of the titration curves

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



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Ksp effect of the titration curves

25.00 mL 0.1000 M halide (X–) titrated with 0.05000 M Ag+

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(a) 40.00 mL of 0.0502 M KI + 0.0500 M KCl, titrated with 0.0845 M Ag+

(b) 20.00 mL of 0.1004 M KI titrated with 0.0845 M Ag+

Titration of a mixture (uncertainty concerned)

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Example: 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–?).Solution:(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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MN

N

VNVN

b

02073.02

2253333.055.15

2211

MN

N

VNVVN

c

03557.02

225)55.1523.42(3333.0

32)12(1


2) Argentometric Titration

Define Argentometric Titration: A precipitation titration in which Ag+ is the titrant.

Argentometric Titration classified by types of End-point detection:– Volhard method: A colored complex (back titration)– Fajans method: An adsorbed/colored indicator– Mohr method: A colored precipitate

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Mohr method

The Mohr method was first published in 1855 as a method for chloride analysis.

· In the precipitation of chloride by silver ion, chromate ion (CrO4

2) is used as an indicator in the formation of Ag2CrO4, a reddish-brown precipitate formed when excess Ag+ is present.

Ag+ + Cl AgCl(s)

white precipitate

2Ag+ + CrO42 Ag2CrO4(s)

red precipitate

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Ksp= 1.8 x 10-10 (S = 1.34 x 10-5 M)

CrO42

Ksp= 1.2 x 10-12 (s = 6.7 x 10-5M)


The titrations are performed only in neutral or slightly basic medium

to prevent silver hydroxide formation (at pH > 10).

2Ag+ + 2OH 2AgOH(s) Ag2O(s) + H2O

precipitate

to prevent chromic acid formation (at pH < 7).

CrO42 + H3O

+ HCrO4 + H2O

2 CrO42 + 2 H3O

+ H2CrO4 + H2O

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Volhard METHOD

Back titration for determination of Cl-.. First published in 1874.

Reactions:

Ag+ + Cl AgCl(s) Ksp = 1.82 x 10-10

(excess) white precipitate

SCN + Ag+ AgSCN(s) Ksp = 1.1 x 10-12

titrant white precipitate

SCN + Fe3+ FeSCN2+ Kf = 1.4 x 10+2

Indicator red complex

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H+, Fe3+


The titration is usually done in acidic pH medium

to prevent precipitation of iron hydroxides, Fe(OH)3.

Fe3+ +3(OH)- Fe(OH)⇄ 3 Ksp=1*10-39

If [Fe3+]=0.001 M pH=?????

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Since SAgSCN <SAgCl , equilibrium will shift to the right causing a negative error for the chloride analysis.

Ag+ + Cl AgCl(s) Ksp = 1.82 x 10-10

SCN + Ag+ AgSCN(s) Ksp = 1.1 x 10-12

SCN- + AgCl AgSCN + Cl-

To eliminate this error, AgCl must be filtered or add nitrobenzene before titrating with thiocyanate; nitrobenzene will form an oily layer on the surface of the AgCl precipitate, thus preventing its reaction with thiocyanate.

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Before Ve

(Cl– excess)Greenish yellow solution

After Ve (Ag+ excess)

Fajans Method: An adsorbed/colored indicator. Titrating Cl– and adding dichlorofluoroscein as indicator:

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5) Applications of argentometric titrations:

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titrations volumetric analysis procedures in which we measure the volume of reagent needed to react...

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