Revised Ana Chem

Chemical and Physical Principles Analytical Chemistry

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ANALYTICAL CHEMISTRYAnalytical Chemistry

Branch of Chemistry that deals with the analysis, identification, separation andcomposition of matter

Involves methods used to identify the substances that are present in a sample(qualitative analysis) and the exact amount of the identified substances(quantitative analysis)

Quantitative Methods of AnalysisA. Classification of Methods of Analysis

1. Classical methodsa. Gravimetric method – measurement of the mass of a substance that is

chemically related to the analyteb. Volumetric method – measurement of the volume of solution necessary to react

completely with the analyte

2. Modern methodsa. Spectroscopic method – measurement of the electromagnetic radiation

produced by the analyte or its interactions with itb. Electroanalytic Method – involves measurement of the electrical properties of

the analyte such as current, potential or quantity of charge

3. Other methods – involves the measurement of the properties of the analyte suchas heat of reaction (calorimeter), index of refraction (refractometer), opticalactivity (polarimeter) or mass-to-charge ratio (mass spectrometer)

B. Typical Steps in Analysis

1. Selection of an appropriate method In the selection of method of analysis, it is necessary to consider the level of

accuracy, complexity and component of the sample, availability of equipmentand trained personnel and the time of analysis

Standard procedures are usually available from literature such as ChemicalAbstracts, Analytica Chimica Acta, Applied Spectroscopy, Journal of theAssociation of Analytical Chemists, etc.

2. Obtaining a representative sample The American Society for Testing and Materials (ASTM), National Bureau of

Standards (NBS) and Association of Official Analytical Chemists (AOAC) aresuch a few organizations that impose standard sampling procedures foranalysis of some samples

Three steps are generally followed in obtaining samples: obtaining a grosssample, obtaining a laboratory sample and obtaining an analysis sample

A gross sample is obtained from a bulk sample and obtained in such a mannerthat it is considered a representative of the bulk sample

A laboratory sample is a fraction of the gross sample weighing several gramswherein further reduction to few milligrams results into an analysis sample


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3. Measurement of the sample First step in preparing a sample involves measurement of either mass or

volume Solid samples are dried in an oven usually from 110C-120C for about 1-2

hours and cooled in a dessicator; analysis is said to be done in a dry basis.Samples that decompose upon heat treatment are analyzed on a wet basis oras-received basis

Replicate samples are taken for analysis to ensure accuracy of the method usedand quality of the results. Results obtained from these replicate samples aretreated using various statistical tests to establish reliability

Table 1. Classification of analysis based on sample sizeMethod Sample Mass Sample Volumemacro more than 100 mg more than 0.100 mL

semi-micro 10 mg – 100 mg 0.050 mL to 0.100 mLmicro 1 – 10 mg less than 0.050 mL

ultra-micro less than 1 mg –

4. Preparation of a solution of the sample Most methods are designed to process liquid samples specifically solutions,

since these samples are homogenous and are easy to handle The following solvents are commonly employed in preparation of the solutions

of the sample:a. Water. Samples of soluble salts readily dissolve in water at room

temperature and heating may be done to facilitate dissolution of the sample.b. Non-oxidizing acids. In many instances, some portion of the sample will

not dissolve in water and usually the addition of acids render the samplesoluble. Hydrochloric acid is a typical non-oxidizing acid along with dilutesulfuric and perchloric acid.

c. Oxidizing acids. For more stubborn samples, hot, concentrated sulfuricacid, nitric acid and aqua regia are used. Aqua regia is a mixture ofhydrochloric acid and nitric acid in 3:1 volume ratio. Hydrofluoric acid isalso used for dissolving silicate ores.

d. Fluxing agents. Samples which were not dissolved in aqueous solvents areusually fused with a molten solvent called flux. Fluxing agents may beclassified as acidic (K2S2O7, KHF2 and B2O3), basic (Na2CO3, K2CO3,NaOH or KOH) and oxidizing (Na2O2). Fusion is done by mixing a finelyground sample with the solid flux in an inert crucible and heated until theflux melts.

Reagents and chemicals used in the laboratory are classified as follows:a. Commercial or technical reagents. Reagents that undergo superficial

purification and not directly used for analysisb. United States Pharmacopoeia (USP Grade) or National Formulary

Reagents. Reagents used by pharmacists and unfit for analysisc. Chemical Pure (CP) reagents. Reagents that are more refined compared to

technical reagentsd. Reagent grade or analytical reagent (AR) or certified reagent. Reagents

analyzed by the manufacturer with the analysis found on the label of thecontainer

e. Primary standard grade. Chemicals with purity greater than 99.95%


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There are several ways of expressing concentration of a particular species insolution. The following are most commonly used to express concentrations ofsolutions:

a. Weight percent – usually used to express concentration of commercialaqueous reagents

100%solutionofweight

soluteofweight

wt

wt%

b. Volume percent – commonly used to specify the concentration of a pureliquid compound diluted with another liquid

100%solutionofvolume

soluteofvolume

vol

vol%

For alcoholic beverages, percentage of alcohol is usually expressed interms of proof as follows:

vol

vol%2proof

c. Weight/volume percent – used to indicate the concentration of a solidreagent in a dilute aqueous solution

100%(mL)solutionofvolume

(gram)soluteofweight

vol

wt%

d. Mole fraction (x) – commonly used in unit operations to expressconcentrations of solute present in a stream of gas or liquid

solventofmolesoluteofmole

soluteofmolex

e. Molality (m) – temperature-independent concentration term usedconveniently in physicochemical measurements of colligative properties ofsolutions

(kg)solventofkilogram

(mol)soluteofmolem

f. Molarity (M) – most commonly used in titration and denotes the amountof solute, in moles, dissolved in a solvent and diluting to a final volume of1L in a volumetric flask

(L)solutionofvolume

(mol)soluteofmoleM

Formality (F) – concentration term identical to molarity commonly usedfor solutions of ionic salts that do not exist as molecule in solid or insolution

g. Normality (N) – once popular unit of concentration still used by somechemists

(L)solutionofvolume

(eq)soluteofequivalentN


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An equivalent represents the mass of material providing Avogadro’snumber of reacting units. The number of equivalents is given by thenumber of moles multiplied by the number of reacting units per moleculeor atom.

fMW

m

EW

msoluteofequivalent

where m = mass of solute [g], EW = equivalent weight [g-equiv–1], MW =molar mass [g-mol–1] and f = molar equivalent [equiv-mol–1]

Therefore, normality (N) is related to molarity (M) according to thefollowing equation:

L

eq

L

mol

mol

eqMfN

Table 2. Molar equivalents of solutesNature of solute Molar equivalent

acid number of replaceable H+

base number of equivalent HO–

salt net charge of an ionoxidant gain of electron

reductant loss of electron

h. p-function - used to express concentrations at a magnitude of 10n where nis any integer less than zero defined as follows:

xxfp logwhere x = concentration of the species in molarity, M

5. Treatment of the sample Some samples has to be reduced or oxidized prior to analysis or sometimes

treated to become colored or converted to a form that it can be readilyvolatilized

More often, the accuracy of an analysis is affected by the presence ofunwanted components called interferences

Interferences can be eliminated by converting it into non-interfering form by aprocess called masking. Typical masking procedure may be done byconverting the interference into a stable complex ion that does not react withthe reagents added to the sample

Some separation processes are commonly employed to isolate the analyte fromthe interferences such as precipitation, electrodeposition, extraction, ionexchange, volatilization and chromatography

6. Measurement of the analyte Using classical methods of analysis, results can be accurate up to a few parts

per thousand or better, requires relatively large amount of sample and usuallyapplied to measurement of major constituents in a sample

Instrumental methods are generally more sensitive and selective. Analysis israpid, automated and capable of measuring more than one analyte at a time;however, these techniques are more expensive than classical methods.


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Table 3. Classification of constituents in a sampleComponent Relative Amount

major greater than 1.00%semi-micro 0.10% – 1.00%

micro 0.001% – 0.10%ultra-micro less than 0.001%

7. Calculation of results and reporting of data Results of analysis can be expressed depending on the nature of analyte

a. Solid Samples. Calculations on solid samples are based on mass. The mostcommon way of expressing results in a macro determination is by % massor % weight

100%samplewt.

analytewt.

wt

wt%

Table 4. Concentrations of analyte in solid samples in trace concentrationsUnit Definiton Unit

parts perthousand

310samplegram

analytegram

wt

wtpt

samplekg

analytegor

sampleg

analytemg

parts permillion

610samplegram

analytegram

wt

wtppm

samplekg

analytemgor

sampleg

analyteg

parts perbillion

910samplegram

analytegram

wt

wtppb

samplekg

analytegor

sampleg

analyteng

b. Liquid Samples. Similarly, concentrations of solid or liquid analytes inliquid samples obtained from a macro analysis is usually expressed as %weight by volume or % volume by volume defined as follows:

100%samplemL

analytegram

vol

wt%

or 100%samplevolume

analytevolume

vol

vol%

Table 5. Concentrations of analyte in liquid samples in trace concentrations(wt/vol and vol/vol)

parts permillion

610samplemL

analytegram

vol

wtppm

sampleL

analytemgor

samplemL

analyteg

parts perbillion

910samplemL

analytegram

vol

wtppb

sampleL

analytegor

samplemL

analyteng

parts pertrillion

1210samplemL

analytegram

vol

wtppb

sampleL

analytengor

samplemL

analytepg

parts permillion

610samplevol

analytevol

vol

volppm

sampleL

analyteLor

samplemL

analytenL

parts perbillion

910samplevol

analytevol

vol

volppm

sampleL

analytenLor

samplemL

analytepL


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Gravimetric Methods of Analysis

A. Types of Gravimetric Analysis

1. Extraction method. The analyte in the sample is obtained using as appropriatesolvent and the residue from the solution, after evaporation of the solvent, ischemically related to the analyte

2. Precipitation Method. The analyte is converted into a sparingly soluble solid,filtered, washed, dried or ignited and weighed

3. Volatilization Method. The sample is treated to yield a gas that is passed in anabsorbing medium; the analysis is based upon the change in mass of the medium

B. Gravimetric Factor and Precipitating Agents

1. Calculation in gravimetric analysisTo calculate the amount of analyte in the sample…

100%GFsampleofmass

formfinalofmassanalyte%

Gravimetric Factor (GF)

ratiomolarformfinalmol

analytemol

formfinalofmassmolar

analyteofmassmolarGF

y

x

2. Precipitating agent An ideal precipitating agent must give an insoluble product that has the

following properties: can be easily filtered and washed free from impurities or contaminants has very low solubility to avoid losses during filtration and washing inert towards components of the atmosphere has known composition after subjecting to appropriate heat treatment

Table 6. Precipitating agents used in precipitation gravimetrySpecies

PrecipitatedFinal Form Precipitant Species

PrecipitatedFinalForm

Precipitant

Cl AgClAgNO3

Al Al2O3

NH3Br AgBr Cr Cr2O3

I AgI Fe Fe2O3

SO4–2 BaSO4 BaCl2 Sn SnO2

As As2O3

H2S

Ba BaSO4 H2SO4

Bi Bi2S3 Cd CdSO4

Cd CdSO4 Sr SrSO4

Cu CuO Ca CaCO3

(NH4)2C2O4Sn SnO2 Mg MgCO3

Sb Sb2O3 Zn ZnCO3

Mg Mg2P2O7 (NH4)2HPO4K H2PtCl6 K2PtCl6

Zn Zn2P2O7 Hg HgS (NH4)2S

C. Theory of Precipitation

1. Properties of precipitates

a. Particle size. Solid particles formed from precipitation may vary accordingly: Colloidal – tiny particles with size ranging from 0.1 microns to 100

microns in diameter; these particles do not settle readily and cannot befiltered easily


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Crystalline – particles with size ranging from 100 microns or greater; theseparticles settle readily and easily filtered

Particle size is usually affected by temperature, concentration of reactants,solubility of precipitate and mixing rate

b. Appearance. Precipitates may appear to be colloidal (S) , curdy (AgCl), finecrystal (BaSO4), coarse crystal (PbCl2) or gelatinous (Al(OH)3)

c. Relative supersaturation (von Weimarn ratio)

S

SQationsupersaturrelative

where Q = concentration of the solute as precipitation begins and S = solubilityof the precipitate In order to obtain low relative supersaturation and form a crystalline

precipitate, Q must be minimized and S must be maximized. The followingmethods are done to accomplish such conditions: increase the temperature during precipitation (to maximize S) precipitate from dilute solution (to minimize Q) slow addition of precipitating agent with stirring (to minimize Q)

2. Mechanism of precipitationPrecipitation is assumed to occur in two ways:a. Nucleation

Prevails at high relative supersaturation Results in the formation of large number of small particles

b. Particle growth Prevails at low relative supersaturation Results in the formation of small number of large particles

3. Colloidal precipitatesa. Electrical nature of colloidal suspensions

Suspensions, which are stable since these particles are either positively ornegatively charged, hence repel each other

By heating, stirring and addition of electrolyte causes this suspension tocombine together and form a readily filterable solid

This process of converting a colloidal suspension into a readily filterablesolid is called coagulation or agglomeration

b. Adsorption of colloids Ions are attached directly to the

solid surface and comprise theprimary adsorption layer

The charge of this layer isdependent on the charge of theion present in excess

A charged primary adsorptionlayer attracts excess oppositelycharged ions to form a secondarylayer or counter-ion layer

These ions, held by electrostaticforces have higher mobility compared to the ions in the primary layer

The two layers, which constitute an electrical double layer, prevent otherparticles to come close thus inhibits the formation of larger aggregates

CounterIon

Layer

PrimaryAdsorption

Layer


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c. Factors affecting adsorption Common Ion Effect. Precipitates have a tendency to adsorb ions identical

to it more than any other ions Paneth-Fajans-Hahn Rule. In cases that there is more than one ion

adsorbed, the one having a lower solubility is adsorbed to a greater extent. Extent of Ionization of the Contaminant. The degree of adsorption

increases as the ionization of the contaminant decreases Effect of Concentration. Greater adsorption of contaminant ion increases as

its concentration in the liquid phase increases

Volumetric Methods of AnalysisA. Important Terminologies

1. Standard solution – solution of known concentration2. Standardization – process of determining the concentration of an unknown

solution3. Primary standard – a substance of high purity used for standardization4. Secondary standard – compound whose purity was established by a chemical

analysis and serves as reference material for volumetric analysis5. Equivalence point – point in titration where the amount of titrant added is

chemically equivalent to the analyte in the sample6. End point – an observable change in a titration process which estimates the

equivalence point7. Titration error – the difference between the actual volume of titrant required to

reach the end point and the theoretical volume of titrant required to reach theequivalence point

B. Conditions for a Volumetric Analysis1. The reaction must be rapid and can be represented by a simple balanced equation2. The reaction must be complete and no side reaction occurs3. An appropriate indicator must be available in order to detect the end point of the

reaction

C. Characteristics of a Good Primary Standard1. High purity and high equivalent weight2. Stable towards air, high temperature and humidity3. Soluble in water

D. Types of Titration

1. Direct Titration – type of titration where the analyte reacts with the standardsolution directly

2. Back Titration – type of titration where an excess standard solution is added andthe excess is determined by the addition of another standard solution

3. Replacement Titration – type of titration where the analyte is converted to aproduct chemically related to it and the product of such reaction is titrated with astandard solution


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E. Acid-Base Titration1. Theories of acids and bases

Theories Acid Base

Arrhenius theorySvante August Arrhenius (1859-1927)

substances thatdissociate in aqueous

solution to form aH3O+

substances thatdissociate in aqueoussolution to form bHO–

Brønsted-Lowry theoryJohannes Nicolaus Brønsted (1879-1947)Thomas Martin Lowry (1874-1936)

proton donor proton acceptor

Lewis theoryGilbert Newton Lewis (1875-1946)

species that acceptslone pair electrons

species that donateslone pair electrons

a hydronium ion, protonated water or solvated protonb hydroxide ion

2. Autoprotolysis or self-ionization reactions Involves spontaneous reaction of molecules producing a pair of ions Protic solvents have reactive H+ and undergo autoprotolysis

H2O + H2O H3O+ + HO– 14.0pK C25

auto

NH3 + NH3 NH4+ + NH2

– 8.92pK C25auto

CH3COOH + CH3COOH CH3COOH2+ + CH3COO– 5.14pK C25

auto

CH3OH + CH3OH CH3OH2+ + CH3O

– 7.16pK C25auto

CH3CH2OH + CH3CH2OH CH3CH2OH2+ + CH3CH2O

– 1.19pK C25auto

In these reactions, a molecule (or an ion) can act as an acid and as a base and istermed as amphoteric

Other similar terms are: Amphipatic compounds are those that possess both hydrophilic and

lipophilic properties Amphiprotic species are amphoteric molecules that can either accept or

donate a proton Ampholytes are amphoteric molecules that contain both acidic and basic

groups and commonly exist as zwitterions at a certain pH range

Table 7. Ion product constants for waterT,C KW1014 T,C KW1014 T,C KW1014

0 0.11 20 0.69 40 2.845 0.19 25 1.00 45 3.86

10 0.30 30 1.45 50 5.1815 0.46 35 2.05 100 49.87

Concentrations are expressed in molarity using density of water at each temperature. Source: W.L. Marshall andE.U. Franck, Ion Product of water Substance, 0-1000C, 1-10,000 Bars, J. Phys. Chem. Ref. Data 10(2), 1981, pp.295-304.

3. Strength of acids and basesStrong Weak

Acids Bases Acids BasesHClHBrHI

HNO3

HClO41H2SO4

LiOHNaOHKOH

RbOHCsOH

2R4NOH

carboxylic acidspolyprotic acids

metal cations

ammoniaamines

1 only the first ionization is complete; dissociation of the second proton has an equilibrium constant of 1.2 10–2

2 quaternary ammonium hydroxide; hydroxide salt of an ammonium cation


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4. Calculation of pH At 25C, the ion product constant for water, KW is equal to 1.00 10–14

At this temperature… 14pOHpHpK W For a dilute solution at 25C in which the contribution of water to the amount

of H3O+ and HO– in solution and the correction due to activity coefficients are

negligible, the following formulas can be used to calculate the pH:

a. Strong acids (SA) and strong bases (SB) HA + H2O H3O

+(aq) + A–

(aq) (SA): SAMlogpH MOHM+

(aq) + HO–(aq) (SB): SBMlog14pH

b. Weak acids (WA) and weak bases (WB) HA + H2O H3O

+(aq) + A–

(aq) (WA): WAa21 MKlogpH

B + H2O BH+(aq) + HO–

(aq) (WB): WBb21 MKlog14pH

c. Hydrolysis of salts As a general rule, salts coming weak acids or weak bases hydrolyze in water,

that is, only the strong conjugate hydrolyzes in water Acidic salt (AS) is formed from the reaction of a strong acid and weak base

HCl(aq) + NH3(aq) NH4+

(aq) + Cl–1(aq)

SA WB SCA SCB

Since only the strong conjugate hydrolyzes in water…NH4

+(aq) + H2O NH3(g) + H3O

+(aq)

][NH

]O][H[NH

K

KK

4

33

b

Wh

(AS):

b

AS21

K

Mlog7pH

Basic salt (BS) is formed from the reaction of a strong base and weak acidNaOH(aq) + HCN(aq) Na+

(aq) + CN–1(aq) + H2O

SB WA WCB SCA

Since only the strong conjugate hydrolyzes in water…CN–1

(aq) + H2O HCN(g) + HO– 1(aq)

][CN

][HCN][HO

K

KK

1

1

a

Wh

(BS):

a

BS21

K

Mlog7pH

Neutral salt (NS) is formed from the reaction of a strong base and strong acid Salts from weak acid and weak base (WAB) will have the following

hydrolytic equilibrium expressed by the equationNH4

+(aq) + CN–1

(aq) + H2O NH4OH + HCN

]O[H][NH][CN

]OH[HCN][NH

KK

KK

241

4

ba

Wh

(WAB):

b

aW21

K

KKlogpH

Amphoteric salts (HA–1 or HA–2)ionize as a weak acid and also a Brønstedbase that hydrolyzes

H3A + H2O H3O+ + H2A

–1 Ka1

H2A–1 + H2O H3O

+ + HA–2 Ka2

HA–2 + H2O H3O+ + A–3 Ka3

a2a121

12a1

12a2a1a1W

211

2 KKlog]A[HK

]A[HKKKKlogAHofpH

a3a221

2a2

2a3a2a2W

212 KKlog

][HAK

][HAKKKKlogHAofpH


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Table 8. Ionization constants of weak acids and basesWeak acids Formula Ka1 K a2 K a3

Acetic CH3COOH 1.75 10–5

Benzoic C6H5COOH 6.30 10–5

Hydrocyanic HCN 7.20 10–10

Hydrofluoric HF 6.70 10–4

Hydrogen sulfide H2S 9.10 10–8 1.20 10–15

Oxalic HOOCCOOH 6.50 10–2 6.10 10–5

Phosphoric H3PO4 1.10 10–2 7.50 10–8 4.80 10–13

Phosphorus H3PO3 5.00 10–2 2.60 10–7

Sulfuric H2SO4 1.20 10–2

Sulfurous H2SO3 1.30 10–2 5.00 10–6

Weak bases Formula Kb1 K b2

Ammonia NH3 1.75 10–5

Aniline C6H5NH2 4.00 10–10

Diethyl amine (C2H5)2NH 8.50 10–4

Dimethyl amine (CH3)2NH 5.90 10–4

Ethyl amine C2H5NH2 4.30 10–4

Methyl amine CH3NH2 4.80 10–4

THAM (CH2OH)3CN 1.20 10–6

Triethyl amine (C2H5)3N 5.30 10–4

Trimethyl amine (CH3)3N 6.30 10–5

Ethylenediamine H2NC2H4NH2 8.50 10–5 7.10 10–8

Zinc hydroxide Zn(OH)2 4.40 10–5

d. Buffer solutions Solution that has the ability to resist changes in hydrogen ion concentration

upon the addition of small amounts of acid or base (buffer action) Usually consists of a mixture of weak acid (HA) and its conjugate salt (A–1)

or of a weak base (B) and its conjugate salt (BH+) Henderson-Hasselbalch equation

]H[M

]H][M[HK

acidic

basicA

][M

][MlogpKpH

basic

acidicA

]HO[M

]HO][M[HOK

1basic

1acidic

1

B

][M

][MlogpKpKpH

basic

acidicBW

Buffer capacity or buffer intensity or buffer index is the number of moles ofstrong acid or strong base for a liter of solution to cause a unit change in pH

pHd

Cd

pHd

Cdβ BHA

where CHA and CB = number of moles per liter of strong base or strong acid,respectively to cause d[pH]. For a buffer solution containing weak acid andits conjugate salt with concentrations greater than 0.001 M, the buffercapacity is estimated as

1

1

AHA

AHA

CC

CC2.303β


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To account for the contribution of water to either H3O+ or HO–1 in solution, use the

following scheme:

5. Commercial concentrated acids and bases

Acids %wtSpecificGravity Molarity Bases %wt

SpecificGravity Molarity

HAc 99.7 1.05 17.4 NH3 29.0 0.90 15.3HF 49.0 1.17 28.9 KOH 45.0 1.46 11.7HCl 37.3 1.18 12.1 NaOH 51.0 1.48 18.9HBr 48.0 1.50 8.9HI 47.0 1.50 5.5

HNO3 70.0 1.42 15.8HClO4 70.5 1.67 11.7H2SO4 96.5 1.84 18.1H3PO4 85.0 1.70 14.7

6. Primary standards for acid-base titrationa. Acidic substances for standardizing basic solutions

Name Formula MolarMass

Molarequivalent

Benzoic acid C6H5COOH 122.125 1Potassium hydrogen bis(iodate) KH(IO3)2 389.915 1Potassium hydrogen o-phthalate C6H4(COOH)(COOK) 204.22 1Sulfamic Acid HSO3NH2 97.09 1

b. Basic substances for standardizing acidic solutions

Name Formula MolarMass

Molarequivalent

Sodium Carbonate Na2CO3 105.989 2Mercuric Oxide HgO 216.59 2Sodium tetraborate decahydrate Na2B4O710H2O 204.22 2Tris(hydroxymethyl)aminomethane (CH2OH)3CNH2 121.137 1

7. Indicators for acid-base titration

Common Name pKapH transition

Range Color change

methyl orange 3.46 3.1-4.4 r-ybromocresol green 4.66 3.8-5.4 y-b

methyl red 5.00 4.2-6.3 r-obromothymol blue 7.10 6.2-7.6 y-b

m-cresol purple 8.32 7.6-9.2 y-pphenolphthalein 9.00 8.3-10.0 c-rthymolphthalein 10.0 9.4-10.6 c-b

thymol blue1.70 1.2-2.8 r-y8.96 8.0-9.6 y-b

Indicator pH range is pKa±1 and the appropriate indicator for an acid-basetitration is one with pKa close to the equivalence point pH

0MKK2 xx

xlog14pH

xlogpH

weak base (WB)basic salt (BS)

weak acid (WA)acidic salt (AS)

BS

AS

WB

WA

/KK

/KK

K

K

K

AW

BW

B

A

0KM W2 xx

strong base (SB)

strong acid (SA)


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8. Applications of acid-base titrationa. Determination of nitrogen – Kjeldahl Method

Developed by Johan Gustav Christoffer Kjeldahl (1879-1900) in 1883Step 1: Digestion involves oxidation of the sample with hot and concentrated sulfuric acid to

convert the carbon and hydrogen to CO2 and H2O, respectively and thenitrogen (amides or amines) to NH4

+

for inorganic nitrates and nitrites, sample is reduced to NH4+ using either

Devarda alloy (50% Cu-45% Al-5%Zn) or Arnd’s alloy (60% Cu-40% Mg) HgO and H2SeO3 are added as catalyst while K2SO4 is added to increase

the boiling point of the solutionStep 2: Distillation The oxidized solution is treated with NaOH to liberate NH3 gas

NH4+

(aq) + HO–1(aq) NH3(g) + H2O(l)

Glass or Porcelain beads are added to avoid bumping In some modifications, hydrogen peroxide is added to decompose organic

matrix formed If mercuric oxide, HgO is used as a catalyst, it is necessary to add Na2S2O3

to precipitate mercuric sulfideHg2+

(aq) + S2O3-2

(aq) +2HO-1(aq) HgS(s) + SO4

-2(aq) + H2O(l)

Step 3: Titration Ammonia gas is collected in an known excess of standard HCl and the

excess is titrated with a standard solution of NaOH using methyl red orbromocresol green as indicator (back titration)

NH3(g) + HCl(aq) NH4Cl(aq)

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

The distillate can also be collected using excess of boric acid and theresulting solution is titrated using standard HCl (replacement titration)

NH3 (g) + H3BO3 (aq) NH4+

(aq) + H2BO3-2

(aq)

H2BO3-2

(aq) + HCl(aq) H3BO3(aq) + Cl–1(aq) + H2O(l)

For conversion of nitrogen content to protein content of selected foodproducts, the following factors apply:

Table 9. Jones factor for protein conversionAnimal origin Vegetable originFood Factor Food Factor Food Factor

Eggs 6.25 Barley 5.83 WheatMeat 6.25 Corn 6.25 Whole kernel 5.83Milk 6.38 Oats 5.86 Bran 6.31In general… Rice 5.95 BeansCereals 5.70 Rye 5.83 Castor 5.30Meat Products 6.25 Sorghums 6.25 Soybean 5.71Dairy Products 6.38 Peanuts 5.46 Velvet 6.25

Source: Food and Agriculture Organization of the United Nations - http://www.fao.org/docrep/006/y5022e/y5022e03.htm

b. Double indicator method for mixture of bases – Warder Titration The presence of hydroxide, carbonate and bicarbonate in water is also

referred to as alkalinity which is a measure of the acid-neutralizing capacityof water

One method requires titration of the mixture to reach the phenolphthaleinendpoint with the volume recorded as V0-Ph. On the same solution, methylred is then added and an additional volume is required to reach the end pointrecorded as VPh-MR


127

Another method requires two duplicate samples. One sample is treated withphenolphthalein and the other with methyl red. Volumes of titrant required toreach the phenolphthalein and methyl red endpoints are recorded as V0-Ph andV0-MR, respectively.

m-Cresol purple can also be used to detect phenolphthalein alkalinity (P)while bromocresol green or methyl orange for the total alkalinity (T)

Table 10. Alkalinity relationshipsResult from titration NaOH Na2CO3 NaHCO3

V0-Ph > VPh-MR V0-Ph – VPh-MR 2VPh-MR 0V0-Ph < VPh-MR 0 2V0-Ph VPh-MR – V0-Ph

V0-Ph = VPh-MR 0 2V0-Ph = 2VPh-MR 0V0-Ph 0 ; VPh-MR > 0 0 0 VPh-MR

V0-Ph > 0 ; VPh-MR 0 V0-Ph 0 0Note: If analysis involves measurement of volume due to total alkalinity (V0-MR), use theconversion: VPh-MR = V0-MR – V0-Ph

The following relationship can be summarized using the following diagram:

c. Acid number or acid value Defined as the mass (mg) of KOH that will neutralize the acid produced from

water degradative reaction of one gram of fat or oil

oilorfatofgram

)(56.10))(M(VnumberAcid KOH

mLKOH

d. Saponification number or Koettstorfer number

+ 3 KOH +

OHH2C

CH

H2C OH

HO

R1COOK

R2COOK

R3COOK

H2C

HC

H2C OOCR3

OOCR2

OOCR1

Defined as the mass (mg) of KOH required to saponify one gram of fat or oil Can be used to determine the approximate molar mass of fat or oil The sample is refluxed with ethanolic KOH and the resulting solution is

titrated with standard HCl

oilorfatofgram

)(56.10))(MV(VValueSap HCl

mLsamplelwith

mLblank

ValueSap

8,30061OilorFatofMassMolar


128

F. Precipitation Titration1. Important terminologies

Saturated solution – solution that contains the maximum amount of solutedissolved in a given amount solvent at a specific temperature

Solubility – the maximum amount of solute dissolved in a given solvent at aspecific temperature

Solutions that contain dissolved solute less than the maximum are calledunsaturated while those that contain dissolved solute more than the maximumare called supersaturated solutions

2. Solubility rules for ionic compounds in water at 25CSoluble compounds Insoluble compounds

All nitrates, bicarbonates, chloratesand compounds containing alkalimetal ions and ammonium ion.

All carbonates, phosphates, chromatesand sulfides except that of alkali metal

ions and ammonium ionAll halides except that of Ag+, Hg2

2+

and Pb2+ All hydroxides except that of alkalimetal ions and Ba++All sulfates except that of Ag+, Ca++,

Sr++, Ba++ and Pb++

3. Solubility product constant (KSP) Consider an aqueous saturated solution of a sparingly soluble salt represented by

the equation:AxBy (s) x A+y

(aq) + y B–x(aq)

The equilibrium constant for this reaction would be:

]BA[

][B][AK

y(s)x

(aq)x

(aq)y

eq

yx

However the concentration of the solid AxBy in the solution will be constant (theratio of the number moles of AxBy and the volume of the solid is constant). Thus,

yx ][B][A]BA[KK (aq)x

(aq)y

y(s)xeqSP

Table 11. Solubility product constants at 25CCOMPOUND KSP COMPOUND KSP

AgCl 1.82 10–10 Mg(OH)2 7.10 10–12

AgBr 5.00 10–13 Ca(OH)2 6.50 10–6

AgI 8.30 10–7 PbCl2 1.70 10–5

BaCO3 5.00 10–9 PbI2 7.90 10–9

Ba(IO3)2 1.57 10–9 Hg2Cl2 1.20 10–18

Al(OH)3 3.00 10–34 Hg2Br2 5.60 10–23

CaCO3 (calcite) 4.50 10–9 Hg2I2 4.70 10–29

4. Argentometric titrations One of the oldest analytical techniques that started in the mid–1800’s Silver nitrate (AgNO3) is commonly used as titrant and the end point are

observed as follows:a. Formation of a secondary colored precipitate – Mohr method

Developed by Karl Friedrich Mohr (1806-1879) in 1865

Titrant: AgNO3

Titration reaction: Ag+(aq) + Cl–1

(aq) AgCl(s)white


129

Indicator: K2CrO4

Indicator reaction: 2Ag+(aq) + CrO4

–2(aq) Ag2CrO4(s)

yellow red

In practice, the indicator concentration is kept between 0.002 M to 0.005 M Titration is done at a pH of 8 to avoid precipitation of silver as hydroxide

(above pH of 10) and eliminate formation of HCrO4–1 (below pH of 6) which

results to consumption of more titrant Usually a low concentration of chromate is desired to detect the end point

clearly since a chromate ion imparts an intense yellow color

b. Formation of a colored complex – Volhard method Developed by Jacob Volhard (1834-1910) in 1874

Titrant: KSCNBack titration: Ag+

(aq) + Cl–1(aq) AgCl(s)

excess whiteAg+

(aq) + SCN–1(aq) AgSCN(s)

white

Indicator: ferric alum, NH4Fe(SO4)212H2OIndicator reaction: Fe+3

(aq) + SCN–1(aq) Fe(SCN)+2

(aq)red

Titration is done in acidic medium using HNO3 with indicator concentrationof about 0.01 M

For the titration of chloride, the resulting precipitate is filtered off before theback titration since it reacts with the titrant and is more soluble than AgSCN

For the titration of iodide, the indicator is not added until all iodide isprecipitated since the dissolved iodide is oxidized by the ferric ion

c. Formation of a colored adsorption complex – Fajans Method Developed by Kazimierz Fajans (1887-1975) in 1926

Titrant: AgNO3

Titration reaction: Ag+(aq) + Cl–1

(aq) AgCl(s):Cl–1. . . . DCF–1

excess white greenish-yellowIndicator: fluorescein, dichlorofluorescein or eosinIndicator reaction: Ag+

(aq) + Cl–1(aq) AgCl(s):Ag+1:DCF–1

excess white pink

Before the equivalence point, chloride anion adsorbs to the precipitate in theprimary adsorption layer and drives the adsorption dye anion away byelectrostatic repulsion and the dye imparts a greenish-yellow color insolution

As soon as the equivalence point is just exceeded with the presence of excesssilver ion, this ion now adsorbs to the precipitate in the primary adsorptionlayer where the oppositely-charged adsorption dye anion adsorbs to thecounter-ion layer and imparts a pink color in solution

For titration of chlorides, fluorescein may be used at an optimum pH rangebetween 7-10 while dichlorofluorescein is used in acidic solution of pHgreater than 4.4

For bromides, iodides and thiocyanates, eosin is used for titration in acidicmedium of pH between 1-2 imparting magenta color at the end point

Dextrin is added to prevent excessive coagulation of the AgCl precipitate


130

G. Complexation Titration1. Important terminologies

Ligand – molecule or ion which possesses at least one unshared pair of electroncapable of forming coordinate covalent bond with an ion

Coordination complex or metal complex – formed when a metal ion is bonded tomonodentate ligands

Chelate complex – formed when a metal ion is bonded to a polydentate ligand Chelants – chemicals that form soluble complex molecules with a metal ion

which results to inactivation of the ion’s ability to react with other elements toproduce precipitates

2. Titration methods involving complexesa. Determination of cyanide – Liebig method

The titration is carried by the dropwise addition of AgNO3 in a solution of acyanide forming a soluble cyanide complex of silver:

2CN–1 + Ag+ Ag(CN)2–1

The endpoint of the titration is the formation of a permanent faint turbidity:Ag(CN)2

–1 + Ag+ Ag [Ag(CN)2](s)

b. Determination of nickel An ammoniacal solution of nickel is treated with a measured excess of

standard cyanide solution:Ni(NH3)6

+3 + 4CN–1 + 6H2O Ni(CN)4–1 + 6NH4OH

The excess cyanide is determined according to the Liebig method

c. Titration with ethylenediaminetetraacetic acid (EDTA) The structure suggests six potential sites (hexadentate) for metal bonding –

the four carboxyl groups and two amino groups Reagents for EDTA titration Free acid, H4Y – can be used as a primary standard when dried for

several hours from 130C to 145C and dissolved completely with smallamount of base

Disodium EDTA dihydrate, Na2H2Y2H2O – analytical reagent grade iscommercially available and usually dried at 80C for 24 hours

NH3-NH4Cl buffer solution (pH = 10) – prepared from 17.5 grams ofNH4Cl and 142 mL of concentrated NH3 and

Indicators for EDTA titration Eriochrome Black T or Solochrome – used for titrations with pH more

than 6.5 since it polymerizes in strongly acidic solutions; color changesbetween pH range of 7-11 from royal blue to wine red;

Calmagite – similar to EBT but color change is sharper and its aqueoussolution is stable; used for titration at pH = 10 using NH3-NH4Cl buffer

Solutions of EDTA combines with any metal ions in a 1:1 ratio

N

H2C C

O

N

CH2

C

O

M

H2C

H2C

H2C

H2C

C

O

C

O

O

O

O

O

HO

C CH2

N

H2C

C OH

O

CH2

H2C N

CH2

C

OH

O

H2C C

OH

O

O

+ M+n


131

Types of EDTA titration Direct titration – solution containing the metal cation is buffered to the

desired pH and titrated directly with EDTA using auxiliary complexingagent such as citrate, tartrate or triethanolamine to avoid precipitation ofthe metal as hydroxide

Back titration – in the absence of metal indicator, the solution is treatedwith excess EDTA, buffered to the desired pH and the excess isdetermined using standard solution of either sulfates or chlorides of zincor magnesium

Replacement titration – applied for cations that do not react with themetal indicator like Ca+2; in the determination of Ca+2, small amount ofmagnesium chloride is added to EDTA where Ca+2 initially displacesMg+2 in the EDTA complex and displaced Mg+2 combines with EBTproducing a red complex; when all the calcium is titrated, the liberatedMg+2 is released, combines with EDTA and the endpoint is observedwith the formation of blue uncomplexed indicator

H. Oxidation-Reduction Titration

1. Important terminologies Oxidation – process which involves increase in oxidation state as a result of loss

of electron Reduction – process which involves decrease in oxidation state as a result of

gain of electron Disproportionation – process in which an element in an intermediate oxidation

state yields products in both lower and higher oxidation states Oxidant or oxidizing agent – substance that accepts electron and undergoes

reduction Reductant or reducing agent – substance that donates electron and undergoes

oxidation

2. Oxidation numbers and balancing oxidation-reduction reactions

a. Rules in assigning oxidation numbers An atom in its free or elemental form has oxidation equal to zero For monoatomic ions, the oxidation number is equal to its charge Metals have positive oxidation number such as alkali metals (+1), alkaline

earth metals (+2), aluminum (+3), zinc (+2) and silver (+1) Nonmetals usually have negative oxidation numbers: Oxygen is usually –2, except in peroxides (–2) and superoxides (–1) Hydrogen is usually +1, except in hydrides (–1) Fluorine has –1 oxidation state; other halogens are usually in the –1

oxidation state, except when combined with oxygen, they are positive;when different halogens are bound to each other, –1 is assigned to themore electronegative halogen

The sum of oxidation number of elements in a compound is equal to zero The sum of oxidation number of elements in a polyatomic ion is equal to the

charge of the ion

b. Balancing oxidation-reduction reactions A reaction is balanced when the number of atoms of each element and the

net charge on both sides are equal


132

Step 1: Assign oxidation numbers to each of the species in the reactionNO2

–1 + MnO4–1 NO3

–1 + MnO2+3 –2 +7 –2 +5 –2 +4

Step 2: Identify oxidation and reduction reactions and indicate the number ofelectrons lost or gained, respectively

Oxidation: NO2–1 NO3

–1 + 2e–+3 +5

Reduction: MnO4–1 + 3e– MnO2

+7 +4

Step 3: Balance the reaction by multiplying a factor on both sides of thereaction so that the numbers of electrons on both reactions are the same

Oxidation 3: 3 NO2–1 3 NO3

–1 + 6e–+3 +5

Reduction 2: 2 MnO4–1 + 6e– 2 MnO2

+7 +4

3 NO2–1 + 2 MnO4

–1 3 NO3–1 + 2 MnO2

Step 4: Balance the charges (by adding H+ or HO–) and number of hydrogenand/or oxygen atoms (by adding H2O) on both sides of the equation

In acidic medium, add H2O to the oxygen-deficient side and supply H+ tobalance the hydrogen

3 NO2–1 + 2 MnO4

–1 + 2 H+ 3 NO3–1 + 2 MnO2 + H2O

In basic medium, balance assuming reaction was in acidic medium.Neutralize H+ by adding HO–1 on both sides of the reaction and simplify3 NO2

–1 + 2 MnO4–1 + 2 H+ + 2 HO–1 3 NO3

–1 + 2 MnO2 + H2O + 2 HO–1

3 NO2–1 + 2 MnO4

–1 + 2 H2O 3 NO3–1 + 2 MnO2 + H2O + 2 HO–1

3 NO2–1 + 2 MnO4

–1 + H2O 3 NO3–1 + 2 MnO2 + 2 HO–1

3. Standard electrode potential The potential of a half-cell reaction with the standard hydrogen electrode (SHE)

used as anode when the activities of all reactant and products are taken as unity,that is, 1M concentration and 1 atm partial pressure

Usually listed as standard reduction potential (red) where a positive valueimplies that the electrode was used as a cathode and the SHE as anode

High value of a reduction potential indicates that the electrode is a goodoxidizing agent

Nernst equation Formulated by Walther Hermann Nernst (1864-1941) Accounts for the effect of concentration on electrode potentials

For a half-cell reduction reaction…Ox Red + ne–

Ox

RedRedRed a

aln

nF

RTεε

where Red = actual cell potential [V], Red = standard reduction potential [V],R = 8.314 J-mol–1-K–1, T = temperature [K], n = number of electrons thatappear in the half-cell reaction [mol], a = activity [ ] and F, Faraday’sconstant = 96485.3399 coul-(mol e–)–1

At 25C and for a given cell…

Qlogn

0.05916εε cellcell

where Q = reaction quotient [ ]


133

The equilibrium constant and the standard electrode potential are related asfollows:

V0.05916

nεKlog

(T = 298.15 K, = 0 and Q = K)

4. Oxidation-reduction titration methods

a. Auxiliary oxidants and reductants Pre-reductants Jones reductor - consists of zinc metal treated with 2% solution of HgCl2

(amalgamated zinc) and used to reduce Fe+3 (Fe+2), Cu+2 (Cu), TiO+2

(Ti+2), UO2+2 (U+3 orU+4) and Cr+3(Cr+2)

Walden reductor - consists of a column filled with silver metal or aninsoluble salt of silver and does not reduce Fe+3 and TiO+2

Na2SO3, NaHSO3, or SnCl2

Pre-oxidants NaBiO3, (NH4)2S2O8, K2S2O8, Br2, Cl2, Na2O2 or H2O2

b. Permanganate titration Titration is carried out in acidic medium using sulfuric acid In the presence of HCl, titrant is consumed to oxidize Cl–1

Acidic and basic solutions of KMnO4 are less stable than neutral ones andkept in dark-colored bottles to avoid decomposition

Titrant: KMnO4

Half-cell reactions: acidic medium MnO4

–1 + 8H+ + 5e–Mn+2 + 4H2O basic medium MnO4

–1 + 2H2O + 3e–MnO2 + 4HO–

Primary standards: As2O3 H3AsO3 + H2O H3AsO4 + 2H+ + 2e–

Na2C2O4 C2O4–2 2CO2 + 2e–

Fe metal Fe Fe+2 + 2e–

FeSO4(en)SO44H2O Fe+2 Fe+3 + e–

Indicator: self-indicatingEndpoint: pale pink color that persists for 30 s

c. Dichromate titration Titration is carried out in acidic medium only Titrant is stable towards light and less easily reduced in the presence of

organic matter compared to permanganate

Titrant: K2Cr2O7

Half-cell reaction: Cr2O7–2 + 14H+ + 6e– 2Cr+3 + 7H2O

Primary standards: Fe metal Fe(s) Fe+2 + 2e–

FeSO4(en)2SO44H2O Fe+2 Fe+3 + e–

Indicator: sodium diphenylamine sulfonateN-phenylanthranilic acid

Endpoint: first appearance of blue-violet


134

d. Cerium (IV) titration Titration is carried out in acidic medium using sulfuric acid at concentrations

0.5 M or higher In the presence of HCl, titrant is consumed to oxidize Cl–1

Titrant: Ce(SO4)2 and (NH4)4[Ce(SO4)4]2H2OHalf-cell reaction: Ce+4 + e– Ce+3

Primary standard: As2O3

Half-cell reaction: H3AsO3 + H2O H3AsO4 + 2H+ + 2e–

Indicator: ferroin / N-phenylanthranilic acidEnd point: orange-red to pale blue / yellowish-green to purple

e. Iodimetry: Direct titration with iodine Titration is carried out in neutral, weak alkaline or weak acidic solutions

Titrant: I2 dissolved in concentrated solution of KIHalf-cell reaction: I3

–1 + 2e– 3I–1

Primary standard: As2O3

Half-cell reaction: H3AsO3 + H2O H3AsO4 + 2H+ + 2e–

Indicator: Starch solutionEndpoint: Formation of intensely blue-colored complex

f. Iodometry: Indirect titration with iodine The analyte is an oxidizing agent which reacts with I–1 added to the solution

in excess to liberate I2 equivalent to the amount of analyte present

Titrant: Na2S2O3

Half-cell reaction: 2S2O3–2 S4O6

–2 + 2e–

Primary standard: KIO3 or K2Cr2O7

Half-cell reaction: 2IO3–1 + 12H+ + 10e– I2 + 6H2O

Cr2O7–2 + 14H+ + 6e– 2Cr+3 + 7H2O

Indicator: Starch solutionEndpoint: Color change from blue to colorless

g. Summary of oxidants and reductants used in titration

Oxidants Half-cell reaction (V)Combining

ratioKMnO4 (acidic) MnO4

–1 + 8H+ + 5e–Mn+2 + 4H2O 1.51 5KMnO4 (basic) MnO4

–1 + 2H2O + 3e–MnO2 + 4HO– 1.695 3MnO2 MnO2 + 4H+ + 2e–Mn+2 + 2H2O 1.23 2

K2Cr2O7 Cr2O7–2 + 14H+ + 6e– 2Cr+3 + 7H2O 1.33 6

Ce(SO4)2 Ce+4 + e– Ce+3 1.61 1I2 in KI I3

–1 + 2e– 3I–1 0.5355 2I2 (satd) I2 + 2e– 2I–1 0.5345 1I2(aq) I2 + 2e– 2I–1 0.6197 1KIO3 2IO3

–1 + 12H+ + 10e– I2 + 6H2O 1.20 5

Reductants Half-cell reaction (V)Combining

ratioAs2O3 H3AsO3 + H2O H3AsO4 + 2H+ + 2e– –0.559 4Na2C2O4 C2O4

–2 2CO2 + 2e– 0.49 2Fe metal Fe Fe+2 + 2e– 0.440 2FeSO4 Fe+2 Fe+3 + e– –0.771 1Na2S2O3 2S2O3

–2 S4O6–2 + 2e– –0.08 1


135

C C C C

I Br

5. Application of oxidation-reduction titrationa. Iodine number of oils and fats

Measure of the degree of unsaturation of fats or oils Expressed as the number of centigrams of iodine absorbed by 100 grams of

fat or oil Sample is dispersed in chloroform, treated with solution of iodine

monochloride in glacial acetic acid (Wij’s solution) and allowed to react inthe dark for 30 min

+ IBr (excess) + IBr (unreacted)

KI is added to liberate the unreacted iodine and titrated with standardNa2S2O3 solution

IBr (unreacted) + KI I2 + KBr

(g)samplemass

MVVNumberIodine

mg10cg1

Immol1Img53.802

OSNa2I1

OSNablankmL,

OSNasamplemL,

OSNa 2

2

322

2

322322322

b. Peroxide value of oils and fats Measure of the extent of oxidative rancidity of fats and oils during storage Expressed as the number of milliequivalent or millimole of peroxide per

kilogram of sample Sample is dissolved in a mixture of chloroform and acetic acid (2:3), bubbled

with nitrogen gas to remove remaining oxygen and treated with excess KI toliberate iodine

2 I–1 + RO2H + H2O ROH + 2 HO–1 + I2

(kg)samplemass

MVVValuePeroxide 322

322

322322322 OSNammol1OSNameq1

OSNablankmL,

OSNasamplemL,

OSNa

c. Dissolved oxygen (DO) – Winkler Method Measure of the amount of oxygen dissolved or carried in a given medium Measure of the ability to oxidize organic impurities in a body of water Sample is treated with manganese (II) hydroxide which is converted to a

brown precipitate of manganese (III) hydroxide in the presence of oxygen4 Mn(OH)2 + O2 + 2 H2O 4 Mn(OH)3

Alkaline iodide-azide solution is added and the precipitate is then dissolvedin concentrated H3PO4 to release the iodine

Mn(OH)3 + I–1 + 3H+ Mn+2 + ½ I2 + 3 H2O

(L)samplevolume

MVoxygenDissolved 2

2

322

2

322322 Ommol1Omg32

OSNammol4Ommol1

OSNamL

OSNa

d. Chemical oxygen demand (COD) Measure of the amount of oxygen necessary to oxidize all the organic

material in a water sample Expressed as milligrams of oxygen required for oxidation per liter of sample Sample is refluxed in the presence of HgSO4, Ag2SO4/H2SO4 solution and a

known excess amount of standard K2Cr2O7 solution and back titrated withstandard (NH4)2Fe(SO4)2 solution


136

(L)samplevolume

MVVCOD 2

2

722

22

722222 Ommol1

Omg32OCrKmmol4

Ommol6

Femmol6

OCrKmmol1

FesamplemL,

FeblankmL,

Fe

Molecular Absorption SpectrometryA. Absorption Process and Beer-Lambert Law

If a beam of light passes through a glass container filled with liquid, the emergentradiation is always less powerful than that entering

Consider a block of absorbing matter where a beam of monochromatic radiation ofradiant power, P0 strikes the surface perpendicularly and passes through the lengthof the material, b

The emergent or transmitted radiation willalways have less radiant power, P than theentering or incident radiation

The fraction of incident radiation transmitted bythe solution, P/P0 is called transmittance andrelated to absorbance according to the equation:

0P

PlogTlogA

where A = absorbance, T = transmittance, P0 =incident radiant power [W], and P = transmittedradiant power [W]

Beer-Lambert’s law states that the absorbanceis directly proportional to the concentration of the absorbing species and to thepath length

εbcP

PlogTlogA

0

where = molar absorptivity [L-mol–1-cm–1], b = path length [cm] and c =concentration [mol-L–1]

B. Quantitative Analysis

1. Standard addition method Involves addition of several increments of a standard solution to aliquots of

samples of the same size and the resulting solution upon adding the colordevelopment reagent is then diluted to a fixed volume (VT) and measured for itsabsorbance

Assume several identical aliquots of the unknown solution of volume Vx weretreated with several increments of standard solution of volume Vs of knownconcentration Cs and diluted to a fixed final volume VT.

If each of these solutions were assumed to obey Beer’s law, the absorbance (AS)of each solution is described by:

bm SXXSST

XX

T

SSs CCkVCkV

V

CεbVV

CεbVA

where k = b/VT, m = kVS and b = kCXVX

P0 P

b

dx

Po P


137

2. Analysis of mixtures The total absorbance of a solution at a given wavelength is equal to the sum of the

absorbances of the components in the mixtureAt 1: yyxx1 bCεbCεA

11

At 2: yyxx2 bCεbCεA22

3. Photometric titration Plot of absorbance versus volume of titrant where the curves consist of two

straight line regions with different slopes The end point is the intersection of the extrapolated linear regions

REVIEW QUESTIONS AND PROBLEMS

1. All of the following is used as a hygroscopic material in desiccators excepta. CaSO4 b. Mg(ClO4)2 c. P2O5 d. H2SO4

2. Analytical methods classified as micro analysis uses sample mass ranging froma. < 1 mg b. 1-10 mg c. 10-100 mg d. > 100 mg

3. Chemical which are tested by the manufacturers showing the actual percentages ofimpurities and labeled on the containers are called __________.

a. reagent grade chemicalsb. analytical reagent

c. certified reagentd. all of these

4. Vessels intended to contain definite volumes of liquid at a certain temperatureusually 20C are labeled as follows except

a. TC b. C c. In d. TD

5. What proportion by weight of Na2C2O4 (134) to KHC2O4H2C2O4 (218.2) that mustbe mixed in a solution so that the normality of the resulting solution as a reducingagent is three times the normality as an acid?

a. 0.33 b. 0.65 c. 1.54 d. 3.07


138

6. Platinum crucibles can be used for the following processes without significant lossexcept

a. Fusion with sodium carbonate, borax or alkali bifluoridesb. Evaporation with hydrofluoric acidc. Ignition of oxides of Ca and Srd. Heating with sulfides

7. What volume of water must be added to concentrated HCl solution to prepare 100mL 0.955 M HCl solution?

a. 7.9 mL b. 15.8 mL c. 46.0 mL d. 92.1 mL

8. What grade of water as defined by the British Standard 3978 is suitable for thedetermination of trace quantities which can be prepared by the distillation of de-ionized water?

a. Grade 1 b. Grade 2 c. Grade 3 d. Grade 4

For numbers 9 to 11…A 20% wt/wt aqueous solution of NaCl (58.45) at 25C has a density of 1.145 g-mL–1.Express the amount of solute in this solution as follows:

9. % wt/vola. 17 b. 19 c.21 d. 23

10. molarity (M)a. 0.98 b. 1.96 c. 3.92 d. 5.88

11. molality (m)a. 3.42 b. 5.13 c. 6.84 d. 8.56

12. The following describes colloidal suspensions formed during precipitation excepta. These particles are almost invisible to the naked eyeb. They settle readily from a given solutionc. They are not easily filteredd. none of the these

13. A 1.5176 g sample of a CaCO3 was dissolved in an acidic solution. The calcium wasprecipitated as CaC2O4H2O (146.11) and the ignited precipitate at 230C was foundto weigh 0.8249 g. What is the percentage of CaO (56.08) in the sample?

a. 20.9% b. 23.8% c. 41.8% d. 47.6%

For numbers 14 to 16…To a solution containing a precipitate of PbS, an excess of H2S was added.

14. What is the ion present in the primary adsorption layer?a. Pb+2 b. H+ c. S–2 d. HO–

15. The charge of this layer is __________.a. negative b. positive c. neutral d. any of these

16. The ion present in the counter ion layer is __________.a. Pb+2 b. H+ c. S–2 d. HO–

17. A mixture containing FeCl3 (162.20) and AlCl3 (133.33) only weighs 750.8 mg. Thechlorides were precipitated using ammonia and ignited to Fe2O3 (159.69) and Al2O3

(101.96), respectively. The oxide mixture weighs 351.3 mg. Calculate the percentageof Al (26.98) in the sample.

a. 15.5% b. 41.2% c. 43.3% d. 58.8%


139

18. Which of the following does not describe the correct way to wash precipitates?a. Minimum volume of washing liquid must be used to wash the precipitateb. Wash with small portions of washing liquidc. For very soluble precipitates, ionic salts containing common ion must be

added to the washing liquidd. Gelatinous precipitates requires more washing than crystalline precipitates

19. What weight of Mn ore should be taken so that the percentage of MnO2 (86.94) inthe ore would be twice the mass of Mn3O4 (228.82) precipitate obtained inmilligram?

a. 19.0 mg b. 38.0 mg c. 57.0 mg d. 76.0 mg

20. Process by which an agglomerated colloid return to it dispersed state during washingdue to leaching of electrolyte responsible for its coagulation

a. nucleation b. coagulation c. agglomeration d. peptization

21. A 5.488-gram sample containing MgCl2 (95.21) and NaCl (58.45) was dissolved insufficient to give 1L solution. Analysis of chloride content of a 250 mL aliquotresulted in the formation of 2.462 gram AgCl. The magnesium in a second 100 mLaliquot resulted in the formation of 0.2610 gram of Mg2P2O7 (222.53) after treatmentwith (NH4)2HPO4. Determine the percentage of NaCl in the sample.

a. 12.7% b. 23.2% c. 36.1% d. 40.7%

22. It is the expressed as the volume of a solution chemically equivalent to a mass of asolid reagent

a. titer b. aliquot c. molarity d. ppm

23. What is the extent of ionization of the second proton of 0.5 M H2SO4 at 25C?a. 1% b. 2% c. 4% d. 8%

24. Which of the following aqueous solutions of the same concentration will have thelargest pH value at 25C?

a. NH3 b. KCl c. NH4Cl d. NaF

25. Which of the following solutions at 25C will have the lowest pH value?a. 0.15 M Na2SO4 (KA2 of H2SO4 = 1.2 10–2)b. 5.2 10–8 M HClc. 0.05 M NaCN (KA of HCN = 6.2 10–10)d. 0.01 M NH4Cl (KA of NH4

+ = 5.6 10–10)

26. Which of the following acid-base pairs will result in the formation of a buffersolution when titration is done before the equivalence point?

a. NaOH and HClb. KOH – HNO3

c. NH3 – HBrd. all of these

27. What volume of 0.1025 M HCl must be added to 15.64 mL of 0.0956 M NH3 toproduce a solution of pH = 9.00?

a. 5.3 mL b. 7.3 mL c. 9.3 mL d. 11.3 mL

28. Which of the following statements is not correct?a. The buffer capacity is always a positive numberb. The larger the buffer capacity, the more resistant the solution is to pH changec. In general, alkaline buffering capacity is maximum over a pH range of pKb±1d. The acid buffering capacity is maximum at pH equal to pKa


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29. What concentration of acetic acid is needed to prepare a buffer solution with pH =4.50 and a buffer capacity of 0.10 M-pH–1?

a. 0.067 M b. 0.122 M c. 0.174 M d. 0.225 M

30. Approximately how many grams of NH4Cl (53.45) should be dissolved in a liter of0.125 F NH3 to reduce the concentration of hydroxide ions to one-thousandth of itsoriginal value?

a. 79.1 g b. 62.5 g c. 11.7 g d. 7.91 g

31. The conjugate base of H2PO4–1 is __________

a. HPO4–2 b. PO4

–3 c. H3PO4 d. P2O5

For numbers 32 to 35…Calculate the pH of the resulting solution when 25.0 mL of 0.100 M H2C2O4 was treatedwith the following volumes of 0.100 M NaOH:

32. 0 mLa. 1.26 b. 2.84 c. 4.07 d. 5.12

33. 15 mLa. 1.57 b. 1.71 c. 1.86 d. 2.04

34. 25 mLa. 1.44 b. 2.88 c. 5.76 d. 6.64

35. 49.9 mLa. 4.39 b. 7.39 c. 8.37 d. 12.18

For numbers 36 to 38…The acid dissociation constant of acetic acid in methanol is 3.02 10–10. Calculate the pHof the following solutions in acetic acid:

36. 0.035 M CH3COOHa. 3.10 b. 5.49 c. 6.90 d. 8.51

37. 0.035 M CH3COOH + 0.070 M CH3COONaa. 4.44 b. 5.05 c. 9.22 d. 9.82

38. 0.035 M CH3COONaa. 2.97 b. 4.32 c. 11.03 d. 12.38

39. In the standardization of an acid solution with primary standard sodium carbonate,why is it necessary to boil the solution before completing the titration?

a. to eliminate the reaction product, carbon dioxide and carbonic acidb. to destroy the buffering action of the resulting solution due to the presence of

carbonic acid and unreacted hydrogen carbonatec. to achieve a sharper endpoint with methyl red indicator due to the large

decrease in pHd. all of the these

40. Calculate the molarity of NaOH solution if 12.25 mL was used to titrate 0.2615 gramof primary standard KHP.

a. 0.1045 b. 0.1354 c. 0.2509 d. 0.1697


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41. What is the best basis for choosing the right indicator for a given acid – base titrationfrom among the following?

a. type of acidb. type of base

c. pH at equivalence pointd. molarity of the acid or base

42. In standardizing a solution of NaOH against 1.431 grams of KHP, the analyst uses35.50 mL of the alkali and has to run back with 8.25 mL of acid (1mL = 10.75 mgNaOH). What is the molarity of the NaOH solution?

a. 0.2118 M b. 0.2044 M c. 0.7831 M d. 0.2598 M

43. In the titration of a weak acid with a strong base, which of the following is the bestindicator to be used?

a. bromocresol greenb. methyl orange

c. methyl redd. phenolphthalein

44. A 0.2055-g sample of calcite (impure CaCO3) is treated with 27.18 ml of 0.0712 NHCl and the excess is found to require 5.44 ml of 0.0869 N NaOH for back titration.Calculate the percentage purity of calcite in terms of % wt/wt CaCO3 (100).

a. 17.8% b. 35.6% c. 53.4% d. 71.2%

45. Process of determining the nitrogen content of organic materials by mixing thesample with powdered copper (II) oxide and ignited to a combustion tube givingCO2, H2O, N2 and small amounts of nitrogen oxides.

a. Kjeldahl Methodb. Dumas Method

c. Winkler Methodd. Wij’s Method

46. A 640 mg sample of P2O5 (141.94) contains some H3PO4 (97.995) impurity. Thesample is reacted with water and the resulting solution is titrated with 0.867 MNaOH requiring 20.7 mL to the thymolphthalein end point. Calculate the percentageof impurity in the sample.

a. 0.90% b. 1.80% c. 2.70% d. 3.60%

47. In the titration of phosphoric acid, which of the following statements is true?a. Titration curve contains three inflection points since it is a triprotic acidb. Phenolphthalein indicator is used to detect the second end point.c. It can be treated as a monoprotic or diprotic acid during titrationd. all of these

For numbers 48 to 50…A 500-mg sample of each mixture was analyzed for its alkaline content using 0.1025 MHCl via double indicator method.

Mixture 1 2 3 4 5V0-Ph (mL) 4.27 0.01 5.12 6.37 5.63V0-MR (mL) 10.18 6.19 10.24 6.38 9.04

48. Which of the following mixtures contains NaHCO3?a. Mixtures 2 and 4b. Mixtures 4 and 5

c. Mixtures 3 and 4d. Mixtures 1 and 2

49. Calculate the purity of the sample containing NaHCO3 only.a. 1.82% b. 5.22% c. 10.64% d. 11.13%

50. What is the % wt NaOH for the sample containing a mixture of NaOH-Na2CO3?a. 1.82% b. 5.22% c. 10.64% d. 11.13%


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51. In the analysis of nitrogen using Kjeldahl Method, which of the following is added todecompose organic matrices present in the sample?

a. H2SO4 b. HgO c. K2SO4 d. H2O2

52. A 7.279-gram sample of meat was analyzed for its nitrogen content using KjeldahlMethod. Upon digestion, the ammonia liberated was collected in 250 mL of 0.855 MH3BO3. The resulting solution was titrated with 37.25 mL of 0.3122 M HCl usingmixed indicator. Determine the % protein in the sample using 6.25 as factor for meatproducts.

a. 13.98% b. 2.24% c. 19.69% d. 3.14%

53. Which of the following is NOT used as primary standard for the standardization ofalkali solutions?

a. HgO b. H2C2O4 c. HSO3NH2 d. C6H5COOH

For number 54 and 55…A 1.05 gram sample of butter is refluxed with ethanolic KOH and required 10.4 mL of0.1875 M HCl to reach the phenolphthalein end point. If blank determination required27.7 mL of the same standard acid,

54. Calculate the saponification valuea. 45 b. 87 c. 125 d. 173

55. Assuming butter comprised mainly of fat, what is its molar mass?a. 475 b. 970 c. 1445 d. 1940

56. Which of the following statements is true about precipitation titration?a. In Volhard method, AgCl is more soluble than AgSCN thus requiring filtration

of AgCl prior to back titrationb. Titration involving adsorption indicators is slow but has wide range of

applicationc. In Mohr titration, the concentration of the chromate ion in more basic

solutions is too low to produce the precipitate near the equivalence pointd. all of these

57. In the determination of chloride using Mohr method, what should be the theoreticalmolar concentration of the chromate ion indicator given that KSP of Ag2CrO4 is 1.1 10–12?

a. 0.002 b. 0.004 c. 0.006 d. 0.008

58. Which of the following is not a correct analytical method–titrant pair?a. Mohr–AgNO3

b. Fajans – AgNO3

c. Volhard – AgNO3

d. Liebig – AgNO3

59. A 1.500-gram sample of impure aluminum chloride was dissolved in water andtreated with 45.32 mL of 0.1000 M AgNO3 using K2CrO4 as indicator. Express theanalysis in %AlCl3 (133.33).

a. 40.28% b. 13.43% c. 4.48% d. 27.36%

60. In Volhard Method, why is it necessary to carry out titration in acidic solution?a. To prevent precipitation of iron as hydrated as hydrated oxideb. To prevent formation of AgSCN precipitatec. To prevent reduction of halided. To prevent precipitation of silver as hydrated as hydrated oxide


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61. A mixture of LiBr (86.845) and BaBr2 (297.22) weighing 800 mg is treated 50.00mL of 0.1879 M AgNO3 and the excess is found to require 8.76 mL of 0.3179 MKSCN for back titration, using ferric alum as indicator. What is the percentage ofBaBr2 in the sample?

a. 67.95% b. 32.05% c. 35.62% d. 64.38%

62. In Mohr titration, which of the following statement is CORRECT?a. The indicator is usually kept at a concentration of 0.2-0.5 M so as not to

obscure the red precipitate colorb. At low pH, part of the indicator is present as HCrO4

–1 and less Ag+ arerequired to reach the endpoint

c. At high pH, silver is precipitated as silver hydroxide thus produces error in theamount of titrant added

d. all of these

63. A 750.25-gram alloy of nickel was dissolved and treated to remove the impurities. Itsammoniacal solution was treated with 50 mL of 0.1075 M KCN and the excesscyanide required 2.25 mL of 0.00925 M AgNO3. Determine % Ni (58.69) in thealloy.

a. 20.86% b. 37.69% c. 10.53% d. 41.72%

64. Which of the following is true about Liebig method for determination of cyanide?a. The titration process requires an indicator to signal the end of titrationb. The addition of excess amount of AgNO3 produces an insoluble compoundc. A red complex of silver with cyanide is formed which signals the end of the

titration processd. The ratio of silver to cyanide is 2:1 with a permanent faint turbidity as the

endpoint

65. A 500-mg sample containing NaCN required 23.50 mL of 0.1255 M AgNO3 toobtain a permanent faint turbidity. Express the result of this analysis as % CN–.

a. 15.34% b. 23.01% c. 17.25% d. 30.67%

66. Which of the following ions is best titrated with EDTA at minimum pH less than 7?a. Ca+2 b. Sr+2 c. Mg+2 d. Fe+3

67. Which of the following affects the stability of metal complexes?a. geometrical factorsb. macrocyclic effect

c. chelate effectd. all of these

68. An EDTA solution was prepared by dissolving the disodium salt in 1L of water. Itwas standardized using 0.5063 gram of primary standard CaCO3 and consumed28.50 mL of the solution. The standard solution was used to determine the hardnessof a 2L sample of mineral water, which required 35.57 mL of the EDTA solution.Express the analysis in terms of ppm CaCO3.

a. 89 ppm b. 316 ppm c. 158 ppm d. 269 ppm

69. Which of the following statements is true?a. Multidentate chelating agents form weaker complexes with metal ionsb. All metals can be determined with high precision and accuracy using

chelometric titrationc. Eriochrome black T gives a sharp endpoint for the titration of calciumd. Ca-EDTA complex is more stable than Mg-EDTA complex


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70. The 300 mg sample of impure Na2SO4 (142.04) was dissolved in sufficient water andthe sulfate was precipitated by the addition of 35.00 mL of 0.1022 M BaCl2. Theprecipitate was removed by filtration and the remaining BaCl2 consumed 6.79 mL of0.2467 M EDTA for titration to the Calmagite endpoint. Calculate the purity of thesample.

a. 80% b. 85% c. 90% d. 95%

71. A 0.8521 gram sample of an alloy was found to contain Cu (63.55) and Zn (65.41)with small amounts of Pb (207.2) and Hg (200.59). The sample was dissolved innitric acid and diluted to 500 mL. A 10 mL aliquot was treated with KI to mask theHg and the resulting solution required 7.06 mL of 0.0348 M EDTA solution. Asecond 25 mL aliquot was treated with ascorbic acid and the pH was adjusted to 2.00to reduce Hg+2 and the metallic Hg was removed from the solution. To this solution,thiourea was then added to mask the Cu and the resulting solution required 8.58 mLfor titration. The lead ion was titrated in a 250 mL in the presence of NaCN to maskCu, Zn and Hg and required 3.11 mL for titration. Calculate the percentage of Cuand Hg in the sample of alloy.

a. 47% Cu and 3% Hgb. 44% Cu and 5% Hg

c. 53% Cu and 7% Hgd. 56% Cu and 5% Hg

72. Commonly, the analyte in a sample is present in two different oxidation states. Pre-reduction is then necessary before titration. One of the metallic reductors is zincsoaked in a dilute solution of mercuric chloride. This reductor is known as__________.

a. Walden reductorb. Devarda Alloy

c. Lindlars catalystd. Jones reductor

73. At pH = 7 and a pressure of 1 bar, the potential for the half reaction, 2H+(aq) + 2e–

H2 (g) is __________.a. 0 V b. –0.414 V c. –0.828 V d. –1.255 V

74. Which of the following is false about iodine as an oxidizing agent in titration?a. Standard iodine solutions have low smaller electrode potentialb. Sensitive and reversible indicators are readily availablec. Iodine is very soluble in water and losses are minimald. The solution lacks stability and requires regular standardization

75. What is the molarity of a KMnO4 solution standardized against 1.356 gram Na2C2O4

(134 g/mol) requiring 25.1 mL of the solution in acidic medium?a. 0.161 M b. 0.403 M c. 1.008 M d. 0.856 M

76. All of the following is used as oxidant in redox titrations excepta. KMnO4 b. Cerium (IV) c. K2Cr2O7 d. Iodide

77. A sample of iron ore weighing 385.6 mg was dissolved in acid and passed through aJones reductor. The resulting solution 52.36 mL of 0.01436 M K2Cr2O7 for titrationto the diphenylamine sulfonic acid endpoint. Calculate % Fe3O4 (231.55 g/mol) in theore sample.

a. 15.05% b. 45.15% c. 90.30% d. 67.98%

78. A sample of pyrolusite weighing 0.2400 gram was treated with excess KI. The iodineliberated required 46.24 mL of 0.1105 M Na2S2O3 solution. Calculate % MnO2

(86.94) in the sample.a. 46.27% b. 30.85% c. 92.54% d. 76.12%


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79. Which of the following metal is not reduced by Walden reductor?a. Cr b. Fe c. Cu d. V

80. A 10.00 gram sample of cooked-ham was pureed with 200 mL of water, filtered andthe resulting solution containing dissolved potassium nitrite was acidified. Thissolution was treated with 25.00 mL of 0.00514 M KMnO4 was back titrated with14.97 mL of 0.01678 M FeSO4. Calculate the amount of nitrite (46.01) in ppm.

a. 450 b. 900 c. 1350 d. 1800

81. Which of the following is used as a pre-oxidant?a. sodium bismuthateb. ammonium peroxydisulfate

c. sodium peroxided. all of these

82. Which of the following statements is true on the determination of chemical oxygendemand?

a. For solutions with high chloride content, addition of sulfamic acid is necessaryto eliminate the chloride interference

b. Chemical oxygen demand accounts for the oxygen demand of bothbiodegradable and non-biodegradable oxidizable pollutants in a body of water

c. The interference caused by the nitrite ion is eliminated by the addition ofHgSO4 which oxidizes this ion to its nitrate form

d. all of these

83. A 100 mL water sample was analyzed by Winkler Method. If 7.52 mL of 0.0124 MNa2S2O3 was used for titration, determine ppm O2 in the water sample.

a. 2.8 b. 3.5 c. 6.4 d. 7.5

84. The following volumes of standard solutions containing 20 ppm Fe+2 were added to10-mL aliquot samples of waste water in 500-mL volumetric flasks. Excessthiocyanates ion was added and the resulting solutions were diluted up to the markand the following absorbance were obtained in a 1.00-cm cell:

Volume of standards, ppm 0 5 10 15 20 25Absorbance 0.095 0.281 0.433 0.628 0.806 0.972

Calculate the concentration of Fe+2 in the waste water.a. 5.5 ppm b. 8.3 ppm c. 11.0 ppm d. 16.5 ppm

85. A 500 mg sample of a solution containing cobalt (58.93) and nickel (58.69) wasdissolved and diluted to 50 mL. A 25 mL aliquot was treated with a complexingagent to produce a colored complex and the volume was adjusted to 50 mL. Thefollowing data were obtained for the simultaneous determination in a 1.00-cm cell:

Wavelength(nm)

Molar absorptivity, [M–1-cm–1] Absorbance ofsolution, ACo Ni

510 36400 5520 0.815656 1240 17500 0.314

Calculate the concentration of Co in the solution in ppma. 1172 ppm b. 1940 ppm c. 2142 ppm d. 2343 ppm

Revised Ana Chem

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