analytic chemistry

8/10/2019 Analytic Chemistry

1/115


2/115


3/115

Solutions


4/115

Solution Compositions


5/115

Units of Concentration

*Formalityidentical to molarity that is used for

solutions of ionic salts that do not exist as

molecule in solid or in solution.


6/115



7/115


for salts: net charge of an ion

for redox reactants: no. of electrons lost or gain

Normality = fx Molarity

f:

for acids: no. of replaceable H+

for bases: no. of replaceable OH-


8/115


Mole fraction

Mole percent


9/115



10/115


1 ppm = 1 mg/L1 ppb = 1 g/L

1 ppt = 1 ng/L


11/115

Example 1

What is the molarity of water at 40C?

Ans. 55.5 M


12/115

Example 2

The concentration of glucose in normal

spinal fluid is 75 mg/100 g. What is

the molal concentration?

Ans. 4.2 x 10-3M


13/115

Example 3

A 34.00%-by-mass solution of H3PO4in

water has a density of 1.209 g/cm3at

200

C. What is the molarity andmolality of this solution?

Ans: 4.19 M, 5.26 m


14/115

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 CaCO3per liter of water.

What is the molar concentration of Ca2+

ions in a water sample with hardness

count of 175?

Ans. 1.75 x 10-3M.


15/115

Acids and Bases

Arrhenius Theory

Bronsted-Lowry Theory

Lewis Theory


16/115

Acids and Bases

Arrhenius Theory of Acid and Bases:

An acid produces H+and

a base produces OH-

in aqueous solutions.


17/115

Acids and Bases

Brnsted-Lowry Theory of Acid and Bases:

An acid is a proton donor, and

a base is a proton acceptor.


18/115

Acids and Bases


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

base, there is a conjugate acid.


19/115

Acids and Bases


Role of Solvent:

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

Amphotericrefer to a substance that can act as acid or base.


20/115

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.


21/115

Acids and Bases

Lewis Theory of Acids and Bases

A Lewis acidis a species (an atom, ion or

molecule) that is an electron-pair acceptor.

A Lewis baseis a species that is an electron-

pair donor.


22/115

Acids and Bases


Ligandsmolecules or ions that behave as

Lewis bases.


23/115

Acids and Bases



24/115

Strengths of Acids and Bases

STRONG WEAK

Acids Bases Acids Bases

HCl LiOH HF NH3

HI NaOH HCN Organic basesHBr KOH H3PO4 Amines

HClO4 RbOH H3BO3

HClO3 CsOH H2CO3

HNO3

Mg(OH)2

H2

SO3

HBrO3 Ca(OH)2 Carboxylic acids

*H2SO4 Ba(OH)2 Polyprotic acids


25/115

Strengths of Acids and Bases

*HClO4, HBr, H2SO4, HCl and HNO3are 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.


26/115

p-Functions

p-Functions are a method of expressing

concentrations, especially very large or very small

values.

pHcommonly known p-Function

- defined as negative logarithm of hydronium ionconcentration.

pH = - log[H3O+]


27/115

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


28/115

pH Calculations


29/115

pH Calculations

Other values of Kw...


30/115

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


31/115

pH Calculations

Kw = [H3O+] OH = 1 x 104

Kw = [MSA X] X = 1 x 104

X2 MSA X 1 x 104 = 0

Then,

pOH = -log [X]

pH = 14 + log [X]


32/115

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


33/115

pH Calculations


34/115

Example 5

What is the pH of 1 x 10-8HNO3solution?

Ans. 6.98


35/115

Example 6

What is the pH of pure water at 500C?

Ans. 6.63


36/115

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


37/115

Example 8

Calcium hydroxide solution has a

concentration of 0.05 M. Calculate its

pH.

Ans. 13


38/115

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


39/115

pH Calculations

For weak acids:

I: MWB 55.5 M 0 0

: - X -X + X + X

F: MWBX 55.5-X X X


40/115

pH Calculations

For weak acids:


41/115

pH Calculations

For weak bases:

I: MWB 55.5 M 0 0

: - X -X + X + X

F: MWB- X 55.5-X X X


42/115

pH Calculations

For weak bases:


43/115

pH Calculations

For weak acids and

acidic salt:

X2

K x M K = 0

* K = Kafor Weak

acid

K = Kw/K

bfor

Acidic salt

pH = -log [X]

For weak bases andbasic salt:

X2 K x M K = 0

* K = Kafor Weak base

K = Kw/K

afor Basic

salt

pH = 14 + log [X]


44/115

pH Calculations

Percent Ionization

equilibrium concentration of ionized acidinitial concentration of acid

X 100


45/115

Example 10

Calculate the pH of a 0.010 M solution

of iodic acid (HIO3, Ka= 0.17).

Ans. 2.02


46/115

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


47/115

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 Kafor the acid.Ans. 1.4 x 10-5


48/115

Example 13

What is the percent ionization of

propionic acid in a solution that is

0.45 M HC3H

5O

2? pKa = 4.89.

Ans. 0.53%


49/115

Example 14

What is the % ionization in 0.10 M NH3?

Ans. 1.33 %


50/115

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


51/115

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 benzoicacid a solution which contains 0.001 M

NaC7H5O2.


52/115

Hydrolysis Reaction of Salts

Acidic Salt: NH4Cl

NH4++ H2O H3O

++ NH3

KH= KW/KNH3

Basic Salt: NaCN

CN- + H2O HO- + HCN

KH= KW/KHCN


53/115

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


54/115

Example 16

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

Ans. 10.24


55/115

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


56/115

Buffered Solutions

A buffered solution is one that resists a change in

pHwhen either hydroxide ions or protons areadded.


57/115

Buffered Solutions

Buffer Capacity

- refers to the amount of acid or base that abuffer can neutralize before its pH changes.


58/115

Buffered Solutions


59/115

Buffered Solutions

pH of a Buffered Solution

or


60/115

Example 18

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

Ans. 9.05


61/115

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


62/115

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


63/115

Acid-Base Indicators

Acid-Base Indicatoris 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


64/115


HIn + H2O H3O+ + In-

acid color base color


65/115


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


66/115

Example 21

A particular indicator has a color red and acolor blue in its acid and base form

respectively. 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


67/115

Gravimetry


68/115

Gravimetry

Gravimetric Methods:

- are methods that depend upon measuring the

mass (i.e., gravity).


69/115

Gravimetry

Three types of gravimetric methods:

1) Volatilization Methods

2) Extraction Methods

3) Precipitation Methods


70/115

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.


71/115

Gravimetry


Extraction Method

The analyte to be determined is extracted w/ appropriate

solvent.

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


72/115

Gravimetry


Precipitation Method

A 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.


73/115

Gravimetric Methods

Basic calculations:Percentage of analyte in a sample is calculated

using a Gravimetric factor GF.


74/115

Gravimetric Methods

Basic calculations:


75/115

Example 22

A sample containing NaBr and KBr onlyweighs 312.54 grams. The samplewas dissolved in water and treated

with excess AgNO3. The precipitateformed was found to weigh 532.55grams. Calculate % KBr in the

sample.Ans. 49%

23


76/115

Example 23

A 0.1005 gram sample of an ioniccompound containing chloride ionsand an unknown metal is dissolved in

water and treated with an excess ofAgNO3. If 0.0445 g of AgClprecipitate forms, what is the % by

mass of Cl-

in the original compound?Ans. 10.95%

E l 24


77/115

Example 24

What weight of Mn ore should be taken

so that the percentage of MnO2in the

ore would be twice the mass of

Mn3O4 precipitate obtained in

milligram?

Ans. 57.0 mg


78/115


79/115

Titrimetric Methods

Tit i t i M th d


80/115

Titrimetric Methods

Titrimetric methods are analytical procedures in

which the amount of an analyte is determined

from the amount of a standard reagent requiredto react completley with the analyte.

Tit i t i M th d


81/115

Titrimetric Methods

Types:

1. Precipitimetry

2. Acidimetry/Alkalimetry3. Compleximetry

4. Reductimetry/Oxidimetry

Tit i t i M th d


82/115

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.

Tit i t i M th d


83/115

Titrimetric Methods

Standard Solution

Tit i t i M th d


84/115

Titrimetric Methods

Primary Standard

- a highly purifiedcompound that serves

as a reference materials.

Tit i t i M th d


85/115

Titrimetric Methods

Primary Standard

It must have the ff. characteristics:

a. High purity

b. Stable in airc. 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.

Tit i t i M th d


86/115

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)

Tit i t i M th d


87/115

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)

E l 26


88/115

Example 26

How many grams of KHP are neededto neutralize 167.33 mL of 0.99955 M

NaOH?

Ans. 34.157 grams

E l 27


89/115

Example 27

In standardizing a solution of NaOHagainst 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

A li ti f A id B Tit ti


90/115

Applications of Acid-Base Titration

1. Determination of Organic Nitrogen-Kjeldahl Method

2. Double indicator method for mixtureof basesWarder Titration

3. Acid Number

4. Saponification Number

Kjeldahl Method (Determination of Organic


91/115

j ( g

Nitrogen)

Step 1. Digestion The sample is oxidized in hot, concentrated

sulfuric acid, H2SO4and turns black. .

Step 2. Distillation

The oxidized solution is cooled and then

treated with NaOH to liberate ammoniagas:

NH4++ OH- NH3(g)+ H2O



92/115

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



93/115

Ammonia distilled is collected in a boricacid solution. . .

NH3+ H3BO3 NH4++ H2BO3

-1

Titrate the H3BO3-NH3solution with

standard acid. . .

H2BO3-1+ H3O

+ H3BO3+ H2O

Nitrogen)



94/115

Percentage Protein in the sample

%protein =%N * f

f = 5.70 (cereals)

= 6.25 (meat products)

= 6.38 (dairy products)

Nitrogen)

Example 28


95/115

Example 28

A 758-mg sample of full cream milk wasanalyzed by the Kjeldahl method; 38.61

mL of 0.1078 M HCl were required to

titrate the liberated ammonia. Calculatethe % N in the sample.

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

Example 29


96/115

Example 29

A 7.443-gram sample beef was analyzed for itsN 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 MNaOH 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


97/115

Example 30

A 3000 mg sample of flour was taken through aKjeldahl method. Upon digestion, the

ammonia liberated was collected into 200 mL

of 0.995 M H3BO3 solution. If this solutionrequired 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%

Example 31


98/115

A 500 mg sample of each mixture wasanalyzed 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 weightcompositions of each

mixture.

Acid Number


99/115

Acid Number

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

one gram of fat or oil

Acid no. = (VxM)KOHX MW KOH

gram fat or oil

Saponification no


100/115

Saponification no.

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

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

gram fat or oil

Molar mass of Fat or Oil. = 168,300

Sap. Value

Example 32


101/115

Example 32

The saponification no. of triglyceridesis 200. The average MW of the

triglycerides is:

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

Precipitation Titrations


102/115

Precipitation Titrations

Common Technique:

Argentometric Titrations:

1. Mohr method

2. Volhard method

3. Fajans method

Method Titrant Indicator


103/115

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


104/115

Example 33

A 1.500-gram sample of impure aluminumchloride was dissolved in water andtreated with 45.32 mL of 0.1000 M

AgNO3 using K2CrO4 as indicator.Express the analysis in % AlCl3

Ans. 13.43 %

Titration with EDTA


105/115

Titration with EDTA

EDTAEthylenediaminetetraacetic acid

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

Indicators:

1. Eriochrome Black T or Solochrome2. Calmagite

Example 34


106/115

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 required

35.57 mL of the EDTA solution. Theconcentration (ppm) in terms of CaCO3is

Ans. 315.95 ppm

Example 35


107/115

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 precipitatewas removed by filtration and the remaining

BaCl2consumed 6.79 mL of 0.2467 M EDTA

for titration to the Calmagite endpoint.

Calculate the purity of the sample.


108/115

Redox Reactions

Example 36


109/115

Example 36

Balance the reaction:

2342 MnONOMnONO

RULES:


110/115

U S 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 usually2, except in peroxides (2) and superoxides (1) Hydrogen is usually +1, except in hydrides (1)

Fluorine has1 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


111/115

Example 37

What is the molarity of a KMnO4 solutionstandardized 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


112/115

Example 38

A sample of pyrolusite weighing 0.2400gram was treated with excess KI. The

iodine liberated required 46.24 mL of

0.1105 M Na2S2O3solution. Calculate %MnO2(86.94) in the sample.

a. 46.27% b. 30.85% c. 92.54% d. 76.12%

Example 39


113/115

Example 39

A sample of iron ore weighing 385.6 mg wasdissolved in acid and passed through a

Jones reductor. The resulting solution 52.36

mL of 0.01436 M K2Cr2O7 for titration to thediphenylamine 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


114/115

Example 40

A 10.00 gram sample of cooked-ham waspured with 200 mL of water, filtered and

the resulting solution containing dissolved

potassium nitrite was acidified. This solutionwas treated with 25.00 mL of 0.00514 M

KMnO4was back titrated with 14.97 mL of

0.01678 M FeSO4. Calculate the amount of

nitrite (46.01) in ppm.

Ans. 90 ppm

Masking


115/115

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+2and 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

analytic chemistry

Documents