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Analytical Chemistry

Scope

Solution Compositions

Acids and Bases

Gravimetric Analysis

Titrimetric Methods

Redox Reactions

Solutions

Solution Compositions

Units of Concentration

*Formality identical to molarity that is used for solutions of ionic salts that do not exist as

molecule in solid or in solution.


for salts: net charge of an ion

for redox reactants: no. of electrons lost or gain

Normality = f x Molarity

f:

for acids: no. of replaceable H+

for bases: no. of replaceable OH-


Mole fraction

Mole percent


1 ppm = 1 mg/L

1 ppb = 1 g/L

1 ppt = 1 ng/L

Example 1

What is the molarity of water at 40C?

Ans. 55.5 M

Example 2

The concentration of glucose in normal

spinal fluid is 75 mg/100 g. What is

the molal concentration?

Ans. 4.2 x 10-3 M

Example 3

A 34.00%-by-mass solution of H3PO4 in

water has a density of 1.209 g/cm3 at

200C. What is the molarity and

molality of this solution?

Ans: 4.19 M, 5.26 m

Example 4

The hardness of water (hardness count) is

usually expressed as parts per million

(by mass of CaCO3), which is equivalent

to milligrams of CaCO3 per liter of water.

What is the molar concentration of Ca2+

ions in a water sample with hardness

count of 175?

Ans. 1.75 x 10-3 M.

Acids and Bases

Arrhenius Theory

Bronsted-Lowry Theory

Lewis Theory

Acids and Bases

Arrhenius Theory of Acid and Bases:

An acid produces H+ and

a base produces OH-

in aqueous solutions.

Acids and Bases

Brnsted-Lowry Theory of Acid and Bases:

An acid is a proton donor, and

a base is a proton acceptor.

Acids and Bases


*For every acid, there is a conjugate base and for every

base, there is a conjugate acid.

Acids and Bases


Role of Solvent:

- Makes to act the substance to behave as an acid or base.

Amphoteric refer to a substance that can act as acid or base.

Acids and Bases


Classification of Solvents:

Solvent Example

1. Protophilic (proton seeking) H2O, NH32. Protegenic (proton generating) H2O,HOAc

3. Amphiprotic (both 1 & 2) H2O, EtOH

4. *Aprotic C6H6, CCl4

*independent of proton seeking and generating.

Acids and Bases

Lewis Theory of Acids and Bases

A Lewis acid is a species (an atom, ion or

molecule) that is an electron-pair acceptor.

A Lewis base is a species that is an electron-

pair donor.

Acids and Bases


Ligands molecules or ions that behave as

Lewis bases.

Acids and Bases


Strengths of Acids and Bases

STRONG WEAK

Acids Bases Acids Bases

HCl LiOH HF NH3

HI NaOH HCN Organic bases

HBr KOH H3PO4 Amines

HClO4 RbOH H3BO3

HClO3 CsOH H2CO3

HNO3 Mg(OH)2 H2SO3

HBrO3 Ca(OH)2 Carboxylic acids

*H2SO4 Ba(OH)2 Polyprotic acids

Strengths of Acids and Bases

*HClO4 , HBr, H2SO4, HCl and HNO3 are of

same strengths in water; but in glacial acetic

acid, their strengths are:

HClO4 > HBr > H2SO4 > HCl > HNO3

This phenomenon is known as levelling effect

of water.

p-Functions

p-Functions are a method of expressing

concentrations, especially very large or very small

values.

pH commonly known p-Function

- defined as negative logarithm of hydronium ion

concentration.

pH = - log[H3O+]

* pH can be less than 0 or greater than 14.

pH Calculations

pH Calculations

Other values of Kw...

pH Calculations

For strong acids, wherein water contributes to

H3O+ in solution;

H2O + H2O H3O+ + OH-

I: 55.5 M 55.5 M MSA 0

: - X -X + X + X

F: 55.5-X 55.5-X MSA+ X X

pH Calculations

Kw = [H3O+] OH = 1 x 1014

Kw = [MSA+ X] X = 1 x 1014

X2 + MSA X 1 x 1014 = 0

Then,

pOH = -log [X]

pH = 14 + log [X]

pH Calculations

For strong bases, wherein water contributes to

OH- in solution;

H2O + H2O H3O+ + OH-

I: 55.5 M 55.5 M 0 MSB

: - X -X + X + X

F: 55.5-X 55.5-X X MSB + X

pH Calculations

Example 5

What is the pH of 1 x 10-8 HNO3 solution?

Ans. 6.98

Example 6

What is the pH of pure water at 500C?

Ans. 6.63

Example 7

How much water is to be added to 12 M HCl

in order to prepare 1600 mL solution of pH

= 1.50?

Ans. 1595.784 L

Example 8

Calcium hydroxide solution has a

concentration of 0.05 M. Calculate its

pH.

Ans. 13

Example 9

A solution is prepared by adding 125.0 mL of

0.025 M HNO3 to 150.0 mL of 0.020 M

HCl. Determine the concentration of pOH

of the resulting mixture.

Ans. 12.35

pH Calculations

For weak acids:

I: MWB 55.5 M 0 0

: - X -X + X + X

F: MWB X 55.5-X X X

pH Calculations

For weak acids:

pH Calculations

For weak bases:

I: MWB 55.5 M 0 0

: - X -X + X + X

F: MWB - X 55.5-X X X

pH Calculations

For weak bases:

pH Calculations

For weak acids and acidic salt:

X2 + Kx MK = 0

* K = Ka for Weak acid

K = Kw/Kb for Acidic salt

pH = -log [X]

For weak bases and basic salt:

X2 + Kx MK = 0

* K = Ka for Weak base

K = Kw/Kafor Basic salt

pH = 14 + log [X]

pH Calculations

Percent Ionization

equilibrium concentration of ionized acid

initial concentration of acidX 100

Example 10

Calculate the pH of a 0.010 M solution

of iodic acid (HIO3, Ka = 0.17).

Ans. 2.02

Example 11

What mass of benzoic acid, HC7H5O2 is

needed to dissolve in 350.0 mL of

water to produce a solution having a

pH of 2.85? Ka = 6.3 x 10-5.

Ans. 1.4 g

Example 12

Caproic acid, HC6H11O2, found in small

amounts in coconut and palm oils, is

used in making artificial flavors. A

saturated aqueous solution of the

acid contains 11 g/L and has pH =

2.94. Calculate the Ka for the acid.

Ans. 1.4 x 10-5

Example 13

What is the percent ionization of

propionic acid in a solution that is

0.45 M HC3H5O2? pKa = 4.89.

Ans. 0.53%

Example 14

What is the % ionization in 0.10 M NH3?

Ans. 1.33 %

Common Ion Effect

Common Ion Effect

Shift in equilibrium when one or more ions that

are part of the equilibrium are introduced from

an outside source.

reduction in the dissociation of the weak

electrolyte

Example 15

Calculate the [H3O+] in a 0.0045 M benzoic

acid (HC7H5O2) solution. Ka = 6.3 x 10-5.

Calculate the [H3O+] in a 0.0045 M benzoic

acid a solution which contains 0.001 M

NaC7H5O2.

Hydrolysis Reaction of Salts

Acidic Salt: NH4Cl

NH4+ + H2O H3O

+ + NH3KH = KW/KNH3

Basic Salt: NaCN

CN- + H2O HO- + HCN

KH = KW/KHCN

pH of Salts

Acidic Salt: pH = 7 log[Csalt/Kb]

when Csalt/KH >>> 1000

Basic Salt: pH = 7 + log[Csalt/Ka]

when Csalt/KH >>> 1000

pH of Salts

Example 16

What is the pH of an aqueous solution thatis 0.089 M NaOCl? pKa = 7.54

Ans. 10.24

Example 17

What weight (in grams) of NH4Cl is needed

to be dissolved in 200 mL of water to

provide a solution having a pH of 4.50?

Kb = 1.8 x 10-5

Ans. 19.2 g

Buffered Solutions

A buffered solution is one that resists a change in

pH when either hydroxide ions or protons are

added.

Buffered Solutions

Buffer Capacity

- refers to the amount of acid or base that a

buffer can neutralize before its pH changes.

Buffered Solutions

Buffered Solutions

pH of a Buffered Solution

or

Example 18

A buffered solution contains 0.25 M NH3and 0.40 M NH4Cl. Calculate the pH.

Ans. 9.05

Example 19

How many mL of pure formic acid (s.g =

1.22) must be mixed to 325 mL of 0.0664

M NaOH solution to obtain a buffer

solution of pH 3.25? Ka = 1.7 x 10-4

Ans. 3.51 mL

Example 20

What is the pH of the resulting solution

made by mixing 5 mL of 0.2178 M HCl

and 15 mL of 0.1156 M NH3?

Kb = 1.8 x 10-5

Ans. 9.02

Acid-Base Indicators

Acid-Base Indicator is a substance whose color depends

on the pH of the solution to which it is added.

Two forms of acid-base indicators:

1. Weak acid (represented by HIn.

2. Conjugate base (represented by In-.

HIn + H2O H3O+ + In-

acid color base color


HIn + H2O H3O+ + In-

acid color base color


Name pH range pKa Color change

Thymolphthalein 1.70 1.2-2.8 R-Y

Methyl Orange 3.46 3.1-4.4 R-Y

Bromocresol Green 4.66 3.8-5.4 Y-B

Methyl Red 5.00 4.2-6.3 R-O

Bromothymol Blue 7.10 6.2-7.6 Y-B

M-cresol Purple 8.32 7.6-9.2 Y-Purple

Thymol Blue 8.96 8.0-9.6 Y-B

Phenolphthalein 9.00 8.3-10.0 C-Pink

Thymolphthalein 10.0 9.4-10.6 C-B

Example 21

A particular indicator has a color red and acolor blue in its acid and base formrespectively. If this indicator has a Ka = 3x 10-5 , by how much must the pHchange in order to change the indicatorfrom 75% red to 75% blue?

Ans. 0.95

Gravimetry

Gravimetry

Gravimetric Methods:

- are methods that depend upon measuring the

mass (i.e., gravity).

Gravimetry

Three types of gravimetric methods:

1) Volatilization Methods

2) Extraction Methods

3) Precipitation Methods

Gravimetry


Volatilization Method

The analyte are volatilized at suitable temperature.

The volatilized species are collected.

The collected samples are weighed directly or weighed by difference.

Gravimetry


Extraction Method

The analyte to be determined is extracted w/ appropriate solvent.

The mass of the purified extract is related to the amount of the analyte.

Gravimetry


Precipitation MethodA sample is dissolved in an appropriate solvent.

The analyte is precipitated by a reagent that yields a sparingly soluble product.

The precipitate is converted to a product of known composition by heat treatment.

Gravimetric Methods

Basic calculations:Percentage of analyte in a sample is calculated

using a Gravimetric factor GF.

Gravimetric Methods

Basic calculations:

Example 22

A sample containing NaBr and KBr onlyweighs 312.54 grams. The samplewas dissolved in water and treatedwith excess AgNO3. The precipitateformed was found to weigh 532.55grams. Calculate % KBr in thesample.

Ans. 49%

Example 23

A 0.1005 gram sample of an ioniccompound containing chloride ionsand an unknown metal is dissolved inwater and treated with an excess ofAgNO3. If 0.0445 g of AgClprecipitate forms, what is the % bymass of Cl- in the original compound?

Ans. 10.95%

Example 24

What weight of Mn ore should be taken

so that the percentage of MnO2 in the

ore would be twice the mass of

Mn3O4 precipitate obtained in

milligram?

Ans. 57.0 mg

Example 25

A 0.8715 gram sample containing chlorideand iodide ions gave a silver halideprecipitate which weighs 1.8561 grams.This was heated with Cl2 gas to convertsilver iodide precipitate to AgCl. Theresulting precipitate weighs 1.7223 gram.Calculate the percentage of chloride inthe sample.

Ans. 43%

Titrimetric Methods

Titrimetric Methods

Titrimetric methods are analytical procedures in

which the amount of an analyte is determined

from the amount of a standard reagent required

to react completley with the analyte.

Titrimetric Methods

Types:

1. Precipitimetry

2. Acidimetry/Alkalimetry

3. Compleximetry

4. Reductimetry/Oxidimetry

Titrimetric Methods

Standard Solution

- a reagent used to titrate the analyte.

It must:

1. Have a precisely known concentration.

2. Generally is added from a buret.

Titrimetric Methods

Standard Solution

Titrimetric Methods

Primary Standard

- a highly purified

compound that serves

as a reference materials.

Titrimetric Methods

Primary Standard

It must have the ff. characteristics:

a. High purity

b. Stable in air

c. Absence of hydrated water molecules

d. Moderate cost and easy availability

e. Solubility in the titration solutions

f. Large formula weight (molecular weight)

*Compounds that do not meet all these criteria are called secondary standards.

Titrimetric Methods

Some commonly used Primary Standards:

For Bases:

Benzoic Acid, C6H5COOH (f = 1)

Oxalic Acid, H2C2O42H2O (f = 2)

Potassium Biiodate, KH(IO3)2 (f =1)

Potassium Hydrogen Phthalate (KHP),

C6H4(COOH)(COOK) (f =1)

Sulfamic Acid (HSO3NH2) (f =1)

Titrimetric Methods

Some commonly used Primary Standards:

For Acids:

Calcium Carbonate, CaCO3 (f = 2)

Mercuric oxide, HgO (f = 2)

Sodium Carbonate, Na2CO3 (f = 2)

Tris-hydroxymethylaminomethane (THAM),

(CH2OH)3CNH2 (f =1)

Example 26

How many grams of KHP are needed

to neutralize 167.33 mL of 0.99955 M

NaOH?

Ans. 34.157 grams

Example 27

In standardizing a solution of NaOH

against 1.431 gram of KHP, the

analyst uses 35.50 mL of the alkali

and has to run back with 6.12 ml of

acid (1mL = 4.1 mg NaOH). What is

the molarity of the NaOH solution?

Ans. 0.2151 M

Applications of Acid-Base Titration

1. Determination of Organic Nitrogen-Kjeldahl Method

2. Double indicator method for mixture of bases Warder Titration

3. Acid Number

4. Saponification Number

Kjeldahl Method (Determination of Organic

Nitrogen)

Step 1. Digestion

The sample is oxidized in hot, concentratedsulfuric acid, H2SO4 and turns black. .

Step 2. Distillation

The oxidized solution is cooled and thentreated with NaOH to liberate ammoniagas:

NH4+ + OH- NH3(g) + H2O


Nitrogen)

Step 3. Titration

Using an excess amount of HCl. . .

NH3 + HCl NH4Cl

The excess HCl is determined using a standard

NaOH solution

HCl + NaOH NaCl + H2O

Ammonia distilled is collected in a boric

acid solution. . .

NH3 + H3BO3 NH4+ + H2BO3

-1

Titrate the H3BO3-NH3 solution with

standard acid. . .

H2BO3-1 + H3O

+ H3BO3 + H2O


Nitrogen)

Percentage Protein in the sample

%protein =%N * f

f = 5.70 (cereals)

= 6.25 (meat products)

= 6.38 (dairy products)


Nitrogen)

Example 28

A 758-mg sample of full cream milk was

analyzed by the Kjeldahl method; 38.61

mL of 0.1078 M HCl were required to

titrate the liberated ammonia. Calculate

the % N in the sample.

a.12.04% b. 7.69% c. 15.59% d. 10.93%

Example 29

A 7.443-gram sample beef was analyzed for its

N content and the liberated NH3 was

collected in a 43.25 mL of 0.4330 M HCl and

a 15.00 mL back titration with 0.0250 M

NaOH was required. Determine the % protein

in the sample using 6.25 as factor for meat

products.

a.12.44% b. 21.57% c. 32.54% d. 10.98%

Example 30

A 3000 mg sample of flour was taken through a

Kjeldahl method. Upon digestion, the

ammonia liberated was collected into 200 mL

of 0.995 M H3BO3 solution. If this solution

required 25.25 mL of 0.3315 M HCl for

titration to methyl red end point, what is the

percentage of protein in flour? Use 5.70 for

cereal products.

Ans. 22.27%

A 500 mg sample of each mixture was

analyzed for its alkaline content using 0.1025

M HCl via double indicator method.

Example 31

Mixture V0->Ph V0->MR

1 4.27 10.18

2 0.01 6.19

3 5.12 10.24

4 6.37 6.38

5 5.63 9.04

Determine the weight

compositions of each

mixture.

Acid Number

Acid number - mass (mg) of KOH that will neutralize the acid produced fromwater degradative reaction of

one gram of fat or oil

Acid no. = (VxM)KOH X MW KOH

gram fat or oil

Saponification no.

Saponification no./Koettstorfer no. = mass of KOH reqdto saponify 1 gram fat or oil

Sap. no. = (Vblank, mL Vsample, mL ) (MHCl)(56.10)

gram fat or oil

Molar mass of Fat or Oil. = 168,300

Sap. Value

Example 32

The saponification no. of triglycerides

is 200. The average MW of the

triglycerides is:

A. 200 B. 280 C. 600 D. 840

Precipitation Titrations

Common Technique:

Argentometric Titrations:

1. Mohr method

2. Volhard method

3. Fajans method

Method Titrant Indicator

Mohr

AgNO3 K2CrO4

Volhard

KSCN ferric alum, NH4Fe(SO4)212H2O

Fajans

AgNO3

Fluorescein

Dicholro-fluorescein

End point: greenish-yellow to

pink

)()(: saq AgClClAgRxn )(42)(24)(2 saqaq CrOAgCrOAg

)()()(

)()(

:

saqaq

saq

AgSCNSCNAg

AgClClAg

ionBacktitrat

)(

2)(

1)(

3 )( aqaqaq SCNFeSCNFe

)()( saq AgClClAg

Example 33

A 1.500-gram sample of impure aluminumchloride was dissolved in water andtreated with 45.32 mL of 0.1000 MAgNO3 using K2CrO4 as indicator.Express the analysis in % AlCl3

Ans. 13.43 %

Titration with EDTA

EDTA Ethylenediaminetetraacetic acid

combines w/ any metal ion in ratio of 1:1

Indicators:

1. Eriochrome Black T or Solochrome

2. Calmagite

Example 34

An EDTA solution was prepared by dissolvingthe disodium salt in 1 L of water. It wasstandardized using 0.5063 gram of primarystandard CaCO3 and consumed 28.50 mLof the solution. The standard solution wasused to determine the hardness of a 2 Lsample of mineral water, which required35.57 mL of the EDTA solution. Theconcentration (ppm) in terms of CaCO3 is

Ans. 315.95 ppm

Example 35

The 300 mg sample of impure Na2SO4 (142.04)

was dissolved in sufficient water and the

sulfate was precipitated by the addition of

35.00 mL of 0.1022 M BaCl2. The precipitate

was removed by filtration and the remaining

BaCl2 consumed 6.79 mL of 0.2467 M EDTA

for titration to the Calmagite endpoint.

Calculate the purity of the sample.

Redox Reactions

Example 36

Balance the reaction:

2342 MnONOMnONO

RULES: 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 the more electronegative halogen

The sum of oxidation number of elements in a compound is zero.

The sum of oxidation number of elements in a polyatomic ion is the

charge of the ion

Example 37

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

Example 38

A sample of pyrolusite weighing 0.2400

gram was treated with excess KI. The

iodine liberated 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%

Example 39

A sample of iron ore weighing 385.6 mg was

dissolved in acid and passed through a

Jones reductor. The resulting solution 52.36

mL of 0.01436 M K2Cr2O7 for titration to the

diphenylamine sulfonic acid endpoint.

Calculate % Fe3O4 (231.55 g/mol) in the ore

sample.

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

Example 40

A 10.00 gram sample of cooked-ham was

pured with 200 mL of water, filtered and

the resulting solution containing dissolved

potassium nitrite was acidified. This solution

was treated with 25.00 mL of 0.00514 M

KMnO4 was back titrated with 14.97 mL of

0.01678 M FeSO4. Calculate the amount of

nitrite (46.01) in ppm.

Ans. 90 ppm

Masking

Example 41: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 in nitric acid and diluted to 500 mL. A 10mL aliquot was treated with KI to mask the Hg and the resultingsolution required 7.06 mL of 0.0348 M EDTA solution. A second 25mL aliquot was treated with ascorbic acid and the pH was adjustedto 2.00 to reduce Hg+2 and the metallic Hg was removed from thesolution. To this solution, thiourea was then added to mask the Cuand the resulting solution required 8.58 mL for titration. The leadion was titrated in a 250 mL in the presence of NaCN to mask Cu,Zn and Hg and required 3.11 mL for titration. Calculate thepercentage of Cu and Hg in the sample of alloy.

Ans. 47.08% Cu, 3.48% Hg

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