0cddd9 ch07 let the titrations begin

8/13/2019 0cddd9 Ch07 Let the Titrations Begin

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ap er

Let the Titrations Begin

2007/10/31 QCA7e Chapter07 TMHsiung@2007 1/42

Contents in Chapter071. Overview of Titrimetry

2. Grades of Chemicals

3. Titration Calculations

4. Precipitation Titrations

1) Titration curve

2) Argentometric titration5. Establish Precipitation Titration Curves with a

Spreadsheet



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1. Overview of Titrimetry

1) Define Titration:Titrations (or titrimetric method) are based on

measuring the amount of a reagent of known

w u w .

A general equation can be expressed:

aA + tTproducts

A: analyte

T: titrant


2) 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.

u w

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 ex eriment.


Titration error: The determinate error in a titration

due to the difference between the end point and the

equivalence point.


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3) Type of Titrations based on Chemical Reactions

i) Acid-Base Titrations, example:H+ + OHH2O K= 1/K w

ii) Precipitation Titrations, example:

(aq) (aq) (s) = spii) Redox Titrations:

5 H2O2 + 2 MnO4+ H+5 O2 + 2 Mn

2+ + 8H2O

iv) Complexometric Titrations, example:

EDTA + Ca2+(CaEDTA)2+


4) 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

y w y .

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 titrimetr : Measurin total char e


(current x time) to complete the redox reaction, then

estimating analyte concentration by the moles of

electron transferred.


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5) Type of Titration Curves

Type Example. y-axis x-axis

Acid-base HCl/NaOH pH V. NaOH

+ + +.

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

2. Grades of Chemicals

1) Terms and Definitions

Reagent Grade: The reagents which meets or

surpasses the latest American Chemical Society

.

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 com osition


- High purity

- Nonhygroscopic

- Chemically stable both in solid and solution

- High MW or FW


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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

y y

consensus levels of the analyte concentration.

NIST Standard Reference Material (SRM): A CRM

issued by NIST that also meets additional NIST-

specific certification criteria and is issued with a

certificate. http://ts.nist.gov/MeasurementServices/ReferenceMaterials/USI


Standardization: The process by which the

concentration of a reagent is determined by reaction

with a known quantity of a second reagent

_ .c m

3. Titration Calculations

1) Terms and Definitions:

Blank Titration: Titration procedure is carried out

without analyte (e.g., a distilled water sample). It is

use o correc ra on error.

Back titration: A titration in which a (known)

excess reagent is added to a solution to react withthe analyte. The excess reagent remaining after its

reaction with the analyte, is determined by a


.


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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

KMnO to reach the titration end oint. What is the molarit M of

2) Standardization

MnO4stock solution? (FW Na2C2O4 134.0)

Solution:

422

242

422

422422

mL250

mL10

OCNamol1

OCmol1

OCNag134.0

OCNamol1

1

OCNag9.1129

5C2O42

(aq) + 2MnO4

(aq) + 16H+

(aq)10CO2(g) + Mn2+

(aq) + 8H2O(l)


42

42

4 MnO02250.0L1

mL1000

mL48.36

1

OCmol5

MnOmol2M Ans

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 ablank titration

re uire 0.00008 L of KMnO solution. What is the % Fe w/w in

3) Unknown Analysis with a Blank Correction

the ore? (AW Fe 55.847)

Solution:

MnO4

(aq) + 5Fe2+

+ 8H+

(aq)Mn2+

(aq) + 5Fe3+

+ 4H2O(l)

502250.002645.0

02645.000008.002653.0

24 FemolMnOmoltitrantL

LLvoltitrantNet


)/(%01.58%1002865.0

1

1

847.55

1titrantL11

2

4

wwFesamplegFemol

Feg

MnOmol

Ans


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4) 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):

)()( 33 aqaq AsOAgHAgAsOH

Solution:

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.

)()()( saqaq AgSCNSCNAg


mL x M = mmolTotal mmol Ag+ added

= (mmol Ag+ consumed by SCN) + (mmol Ag+ consumed by H3AsO4)

11000.0

4888.21

06222.000.40

3

33

AgmmolSCNmmol

SCNbyconsumedmmol Ag

mmolAgNOmL

AgNOmmolAgNOmL

addedmmol AgTotal

30760.14888.2

31

3

.1

4343

43

AgmmolxAgmmolAgmmol

AgmmolxAsOHmmol

AgmmolAsOmmol Hx

AsOHmmolxbyconsumedmmol Ag

SCNmmolSCNmL.


(w/w)%612.4%100010.1

1

1

84.197

1000

1

2

1

1

1

1

4709.0

4709.0

3232

32

32

3232

43

43

4343

OAssamplegOAsmol

OAsg

OAsmmol

OAsmol

Asmmol

OAsmmol

AsOHmmol

AsmmolAsOHmmol

AsOHmmolAsOHmmolx

Ans


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5) Kjeldahl Analysis ( ) for Total Nitrogen (TN)i) KD description:

Step 1: Kjeldahl digestion (decomposing and dissolving)

Or anic N NH CO H OH SO (K S O )Hg or Cu or Se

2 4 2 2 8

OHNHOH+NH 23(g)4

NH + H O NH H O+

Step 2: Neutralization by adding base

Step 3: Distillation NH3 into excess HCl standard


OH2OHOH 2+

3

Step 4: Titrating unreacted HCl withNaOH standard

PowerPoint [] []

:,,,

: (

,)" ",

,

,

"",

Protein + H2SO4CO2 + (NH4)2SO4 + SO2

(NH4)2SO4 + 2NaOHNa2SO4 + NH4OH


2NH4OH + H2SO4(NH4)2SO4 + 2H2O

Kjeldahl methodJohan Kjeldahl1800

:


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Example: A typical meat protein contains 16.2% (w/w) nitrogen.

A 0.500 mL aliquot of protein solution was digested, and theliberated NH3 was distilled into 10.00 mL of0.02140 M HCl.

The unreacted HCl required 3.26 mL of0.0198 M NaOH for

complete titration. Find the concentration of protein (mg

protein/mL) in the original sample.

Solution:


HmmolOHmmol

satndardNaOHbyconsumedmmol H

HmmolHClmL

HmmolHClmL

addedmmol HTotal

10198.0

2140.01

02140.000.10

HmmolxHmmolHmmol

HmmolxNdigestedmmol

HmmolNdigestedmmolx

Ndigestedmmolxbyconsumedmmol H

mmoOHmmolNaOHmL

am

0645.02140.0

1

1

.11

.


mLproteinmgsamplemL

Nmg

proteinmg

Nmmol

NmgNdigestedmmol

NigestemmoNigestemmox

/8.255.0

1

2.16

100

1

00674.141495.0

.

Ans


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6) 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

com lete titration:

Find the mass of each component of the mixture.

Solution:

22)(3

22)(32 22

COOHNaClHClNaHCO

COOHNaClHClCONa

aq

aq


32

32

221

02138.01

7344.0

0001

111.29

xHmolCONamol

CONabyconsumedmmol H

HmolHClL

Hmol

mL

LHClmL

addedmmol HTotal

33

33

3

323232

01.84

372.1

99.105

202138.0

01.84

372.11

101.84

1)372.1(

99.105199.105

xx

HmolxNaHCOmol

HmolNaHCOg

NaHCOmolNaHCOgx

NaHCObyconsumedmmol H

CONamolCONag


33

3232

648.0)372.1(

724.0

NaHCOgNaHCOgx

CONagCONagx

Ans

Ans


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4. Precipitation Titrations

A titration in which the reactionbetween the

analyte and titrant involves aprecipitation.

ra on curve

i) Guidance in precipitation titration calculation

Find Ve (volume of titrant at equivalence point)

Find y-axis values:

- At beginning

-


e- At Ve

- After Ve

Example: For the titration of 50.0 mL of 0.0500 M Clwith 0.100

M Ag+. The reaction is:

Ag+(aq) + Cl

(aq)AgCl(s) K = 1/Ksp = 1/(1.81010) = 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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(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)

5.1

0.600.100.50

50.1

1

1

1

100.00.10

1

0500.00.50

2

Clmmol

mLmLmLV

Clmmol

Agmmol

Clmmol

AgmL

AgmmolAgmL

ClmL

ClmmolClmL

ClmmoledprecipitatClmmoloriginalLeftClmmol

total


14.8

60.1

102.71050.2

108.1

][][

.0.60

9

2

10

pAg

pCl

MCl

KAg

mL

sp

(c) 25 mL Ag+ added (At Ve)

AgCl(s)Ag+(aq)+ Cl(aq) Ksp= 1.81010s = [Ag+]=[Cl]

Ksp= 1.81010 = s2

A + = Cl =1.35x105

pAg = 4.89

pCl = 4.89



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

001

1

100.00.25

1

100.00.35

Ammol

AgmL

AgmmolAgmL

AgmL

AgmmolAgmL

AgofwithVAgmmoledprecipitatAgmmoloriginalLeftAgmmole

82.7

1053.11018.1

108.1

][][

1018.10.85

00.1][

0.850.350.50

.

8

2

10

2

pCl

MAg

KCl

MmL

AgmmolAg

mLmLmLV

sp

total


93.1pAg

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

iv) Diluting effect of the titration curves

25.00 mL 0.1000 M I

+ .

25.00 mL 0.01000 M I

titrated with 0.005000 M Ag+

25.00 mL 0.001000 M I




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v) Ksp effect of the titration curves

25.00 mL 0.1000 M halide (X)



(a) 40.00 mL of 0.0502

M KI + 0.0500 M

vi) Titration of a mixture (uncertainty concerned)

,

0.0845 M Ag+

(b) 20.00 mL of 0.1004

M KI titrated with0.0845 M Ag+



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Example: A 25.00 mL solution containing Brand Clwas

titrated with 0.03333 M AgNO3. Ksp(AgBr)=5x1013,Ksp(AgCl)=1.8x10

10.(a) Which analyte is precipitated first?

(b) The first end point was observed at 15.55 mL. Find the

concentrat on o t e rst t at prec p tate ror .

(c) The second end point was observed at 42.23 mL. Find the

concentration of the second that precipitated (Bror Cl?).

Solution:

(a)

Ag+(aq) + Br

(aq)AgBr(s) K = 1/Ksp(AgBr) = 2x1012


g (aq) +

(aq) g (s) = sp g = . xAns: AgBr precipitated first

(b)

BrM0.02073mL10001Brmol1

AgL1

Agmol0.3333

mLAg1000

AgL1

1

AgmL15.55

Ans

(c)

mL10001Clmol1

AgL1Agmol0.3333

mLAg1000AgL1

1AgmL15.55)-(42.23


.

L1mL25Agmol1Ans


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2) Argentometric Titration ( )1) General information:

Define Argentometric Titration: A precipitation

titration in which Ag+ is the titrant.

rgen ome r c ra on c ass e y ypes o n -

point detection:

Volhard method: A colored complex (back titration)

Fajans method: An adsorbed/colored indicator

Mohr method: A colored precipitate


1.(MohrMethod). 2.(FajansMethod). 3.(Volhard Method)

2) Volhard method (): A colored complex(back titration). Analysing Clfor example:

Step 1: Adding excess Ag+ into sample

Ag+ + ClAgCl(s) + left Ag+

(s)

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, SCNreacts 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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3) Fajans Method (): An adsorbed/coloredindicator. Titrating Cland adding

dichlorofluoroscein for example:

Before Ve(Cl excess)

Greenish

yellow solution

AgCl(s) Cl

1st

layer


After Ve

(Ag+ excess)

AgCl(s) Ag+ In

pink

1st

layer

4) Mohr Method (): A colored precipitateformed by Ag+ with anion, other than analyte, once

the Ve reached. Analysing Cl and adding CrO4

2

for example:

Precipitating Cl:

Ag

+

+ Cl

AgCl(s) Ksp = 1.8 x 10

10

End point determination by red colored precipitate,

Ag2CrO4(s):+ + 2 = 12


s sp .


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5) Applications of argentometric titrations:


5. Establish Precipitation Titration Curves with a

Spreadsheet

Example:

M+(aq) + X

(aq)MX(s) K = 1/Ksp

Xofvol.initial:V

Xofconc:C

Mofconc.:C

oX

oX

oM

T trant Ana yte


addedMofVwhenXofconc.:][X

addedMofVwhenMofconc.:][M

addedMofvol.:V

M

M

M


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(s)oXMM

oM MXmol)V](V[MVC

:MforbalanceMass

1) Derive the equation:

moles of M in

)][X][MC

][X][MC(VV

o

o

Xo

XM

(s)oXM

oX

oX MXmol)V](V[XVC

:ora anceass

precipitate

moles of X in

precipitate


Mass balance: The moles of an element in all

species in a mixture equal to the total moles of that

element delivered to the solution.

)][X][MC

][X][MC(VV

o

M

o

Xo

XM

2) Apply the equation:

ooo

K][X],[XCalculate4.

10][M],[MCalculate3.

pMInput2.

,,,.

sp

pM

sp


obtainedVequation,into][Xand][MSubstitute.5 M


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3) Example: Construct a titration curve for the titration

of 50.0 mL of 0.0500 M Clwith 0.100 M Ag+. (Kspof AgCl = 1.81010)



EXCEL

961_Ch07_Titration_X_with_Ag


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Homework: Problem 7-30/p.139, Due 2007/11/7

Exam les:

All

ExerciseA-E, G


End of Chapter07

Problems:1-7, 11-14, 31, 36

0cddd9 ch07 let the titrations begin

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