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Physical Science Biological Science

Chapter # 1

SCIENCEThe study of all the things of universe is called science.OrThe knowledge based on observations experiments and results is called science.At the beginning, science was named as Natural Philosophy and it had two branches.

Natural Philosophy ( science )

Note:As there are only two types of living things from the beginning ie Animals and plants, soBiological science got two main farther subjects ie Botany( from Greek word Botany -----Herbs study plants ).Zoology ( from Greek word Zoo --- Animal ).On the other hand with the passage of time more and more development took place, inphysical science, therefore it is divided into the following main branches .Physics =, chemistry, Math’s, computer Astronomy etc.Matter:

Anything which has mass and volume is called matterOr Anything having mass and volume is called matter e.g

Pen, Air, Water, etcStates of matter.Solid stateLiquid stateGaseous statePlasma state

Chemistry:Is that branch of science which deals with the properties, compositions and structure ofmatter, changes occurring in matter and the laws under which these changes occur.

Branches of Chemistry: With the passage of time, more and more development took place in chemistry, therefore itwas felt necessary to divide chemistry into various branches to make its study moreconvenient and systematic, chemistry has been divided into the following important branches.

Organic Chemistry:The carbon containing compounds ( except CO2, CO, CO3

-2, HCO3-1 & CN ) are called

organic compounds and their study is called organic chemistry.

Inorganic Chemistry:The study of all the elements and compounds of the elements other than carbon is calledinorganic chemistry .

Study of Non-living matterStudy of Non-living matter

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Pure matter (substance)

ElementCompound

Metal Metalloid Non-metal

Molecule Formula unit

impure matter (sub stance)

Mixture

Homogeneous Mixture HeterogeneousMixture

Physical Chemistry:The branches of chemistry dealing with the study of laws and principles which areresponsible for the separation and combination of atoms during chemical reaction calledphysical chemistry.

Analytical Chemistry:The study of qualitative and quantitative analysis of a substance is called analyticalchemistry.

Biochemistry:The study of chemical reactions place inside the body of living things is called biochemistry.

Industrial Chemistry:The study of chemical reactions occurring during the formation of various synthetic productslike content, glass, plastic etc in industries is called industrial chemistry./

Nuclear ChemistryThe study of changes occurring in the nuclei of radio active elements as a result ofspontaneous emission of radiations & their effects is called nuclear chemistry.

Environmental ChemistryThe study of chemical reactions taking place in the environment and their effects upon us iscalled environmental chemistry.

Lecture# 3 Matter

Atom:The smallest particle of matter which may or may not have free existence is called atom,For example hydrogen element is composed of H atoms which don’t have free existence.Similarly Helium element is composed of He atoms which have free existence.Note:The word Atom is derived from a Greek word, Atoms which means indivisible. This namewas given by Democritus / 460-370 B.C.Actually the Greek, philosophers thought that matter could be divided into smaller andsmaller particles to into smaller and smaller particles to reach a basic unit, that can not befurther sub-divided.According to the modern concepts, atom is further composed of sub-atomic particles whichare more than 100. however there are the fundamental ones they are;1. Porton 2. Electon & 3. NeutronAll the particles of an atom are inside the nucleus of it. Only electrons are present outside, thenucleus, revolving around it.A Swedish chemist. Named as J. Berzelius ( 1779-1848) determined the atomic masses ofelements.

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He also provided the system of giving symbols to elements.

Elements:Element is a pure substance, which is composed of chemically identical atoms. e.g.; hydrogenis an element which consists of it atoms which are all chemically identical y the termchemically.Substance:Any pure matter which a uniform composition is known as substance. e.g.; elements,compound, molecule etc.

Identical wn mena that all atoms of an element will have same number of protons and sameno. of electrons. However their number of neutrons mean be different.i. H - element H,H,H,H,H,H AtomsSo far more than 110 elements are known. Most of them are natural while some are synthetic,Types:- Elements are of the following three types.1. Metals 2. Non-metals & 3. Metalloids

Metals Non-metals1. They are hard solids excepts hg which is liquid 1. They may be solid, liquid orgases2. There have shine 2. They have no shine3. They are good conductors of heart & electricity 3. They are bad conductorsexceptgraphite4. Some metals are ductile i.e.; their wires can be format 4. Non-metals arte brittle, i.e.;their& some are malleable i.e.; their sheets can be formed wires & sheets cannot beformed.5. They have light m.p & b.p.. 5. They have low m.p. &s.p.

Examples ExamplesNa, K, Mg, Cu, Ag, Au, Me etc. i. Sold:- C, S, P etc

ii. Liquid:- Br.iii. Gas:- H2, U2, O2 etc.

Metalloids:The elements which have properties in between metals and non-metals.e.g.; Ga, Ge. Etc.

Compound:When atoms of two or more elements chemically combine in a fixed ratio, a new substance isformed. Which is known as compound.For example when two atoms of hydrogen element chemically combine with one atom ofoxygen element, a compound, water (H2O) is formed in which the atoms are combined in 2:1

Molecule:The smallest particle of a substance which is capable of free existence is known as molecule.e.g.; H2O, HU, H2, O2 etc.Classification of Molecules:Molecules can be classified in two ways as:

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1. On the basis of nature of elements:On the basis of nature of elements, there are two types of moleculea. Molecule of Elements:The molecules which are made of atoms of same elements.e.g.: H2, O2, O3, S8, P4, N2, U2 etc.b. Molecule of Compound:The molecules which are made of atoms of different elements.e.g.: H2O, HU, H2

SO4, C6H12 06 etc.2. On the basis of number of atoms.On the basis of no. of atoms there are three types of molecules, which are described asfollow.a. Mono-atomic molecules:The molecules which are made of only one atom are called mono-atomic molecules.e.g.; He, Ne, Xe etc.b. Diatomic molecules:The molecules which are made of only two atoms (same or different).e.g.; H2, N2, CO, HU, U2 etc.C. Poly atomic molecules:The molecules which are made of more than two atoms are called ply atomic atoms are calledply atomic molecules.e.g.; H2O, H2 SO4, C6 H12 O6, CH4, NH3 etc.

Formula Unit:The simplest combination of atoms of a compd which possesses all the called formula unit.For example NaU is a formula unit. Sodium chloride is a cluster of NaT & U Ions and each Uion is surrounded by 6 NaT & U ions of sodium chloride there is no independent NAu Unit.Therefore, NaUis termed as formula unit rather than molecule.All ionic compounds are represented by formula units.Isotopes:Isotopes of an element may be defined as:"The atoms of an element having same atomic number but different atomic masses, are calledisotopes of that elements."OR "Atoms of an element having same no. of protons and same no. of electrons but differentno. of neutrons are called isotopes of that elements.OR " Atoms of an element having same chemical properties but different physical propertiesare called isotopes of that element.For example carbon element has three isotopes. Which are known as C-12, C-13, & C-14.All these isotopes of carbon have same no. of protons (e.g.; b) & same no. of electron (i.e.;6)but different no. of neutrons i.e.; 6,7 and 8 respectively in C-12, C-13, & C-14.

Lectures 4

ISOTOPESIsotopes of an element may be defined as:The atoms of an element having same atomic number but different atomic masses, are calledisotopes of that elements.OrAtoms of an element having same no of protons and same no, of electrons but different no ofneutrons are called isotopes of that element.OrAtoms of an element having same chemical properties but different physical properties arecalled isotopes of that element.For example carbon element has three isotopes, which are known as

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c- 12 c-13 c-14aAll these isotopes of carbon have same no, of protons ( ie, b ) same no of electrons ( ie, 6 )but different no of neutrons i-e 6, 7 and 8 respectively in c-12, c-13, c-14.

Other examples are, Hydrogen has 3 isotopes, oxygen has three,Nickle ( Ni ) has five, calcium has six, palladium has six, cadmium has nine and tin

has ii isotopes.Note:Round about 280 natural isotopes are known 40 of them are radio active isotopes.Beside these 280, about 300 unstable radio active artificial disintegration.Elements like Arsenic (As), fluorine (F), iodine ( I ) and gold ( Au ) etc, have only a singleisotope.It important to not e that elements of odd atomic number almost never possess more that twostable isotopes.On the other hand elements of even atomic number usually have larger number of isotopes.The isotopes whose mass number are multiple of 4, are particularly abundant out of 280isotopes that occur in nature, 154 have even mass number, and even atomic number.Ion:When an atom loses or gains one or more electrons have 8 electrons in their outermost(valence ) shell, they get a tire of –ive sign.

This positively or negatively changed species is k own as ion.e.g.

Electron configuration ofNa is K=2 , l=8, M=1

So sodium loses its single valence electron & thus forms sodium ion ( Na+ ). Such an ionhaving + ive charge is called caption

Na ----------------------- Na+ + Ie-

2, 8, 1 2, 8Similarly the electronic configuration on of 17CL is k=2, L=8, M=7 so it gains an electron toform chloride ion ( Cl -1 )i. e Cl + Ie- ---------------------------- Cl-1

2, 8, 7 2, 8, 8The ion having negative change is called “Anion ".It is important to note that properties of ions are totally different from their correspondingneutral atoms.The ions may be a group of atoms like So1

-2 , Co3-2, Hco3

-1 , no4-1 , Cr2O7

-2, PO4-3 etc

The + ive ion common e.g. NH4+ ion and some car bo captions in organic chemistry.

NOTE:Ion are always stable inside their solutionE.g.: Na+ & CL- ions are always stable inside water.

MOLECULAR ION:When a molecule loses or gains electron or electrons which is known as molecular ion.e. g

CH4+ , N2

+ , CO+ etcThe cationic molecular ions ( +ive ) are more abundant than the anionic ones.Molecular ions are formed by passing high speed electrons or x—particles through a

gas.

RADICAL:Any negatively or positively charged species or a specie with unpaired electron, which isunable to have free existence is known as radical.

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e. gH. + CL. ( Neutral radicals with unpaired electron )R- + CarbocationR- Carbocation. Etc

ATOMIC NUMBER:The no of protons or electrons present in an atom is called atomic number of that element.For example there are 11 protons and 11 electrons in an atom of sodium element, thereforeatomic mass of sodium is 11.

ATOMIC MASS OR MASS NO:Sum of the protons and neutrons present in the nucleus of an atom is called the atomic massof that atom.e. g

There are 11 protons ans 12 neutrons in the nucleus of Na atom, therefore the atomicmass of Na will be:

11 + 12 ======= 23 a. m. u

RELATIVE ATOMIC MASS OR TELATIVE MASS NO:The mass of one atom of an element compared with the mass of one atom of corbon – 12 iscalled relative atomic mass of that element.Carbon- 12 ( C-12 ) which is an isotopes of carbon element and it is considered as standard. Itmass is considered as exactly 12 a.m. uThe masses of all the other elements are compared with the mass of C- 12 which are knownas their relative atomic mass.

NOTE:The unit of atomic mass is a. m. u ie atomic mass unit.

ATOMIC MASS UNIT ( A. M. U )It may be defined as the 12th part of the mass of an atom of C-12OrThe mass of an atom of E-12 divided by 12 is known as one a. m. u.Let the mass of an atom of C-12 is Xg, then value of one a.m.u wil b:

1 a. m. u = x---------12

E.g. : the mass of an atom of hydrogen id 0.084 times the mass of C-12, so the relative atomicmass of Hydrogen is 0.084* 12

===1.008 a. m. u.We can find the mass of one atom of C-12 as follow.As atomic mass of C-12 = 12g which is equal to one mole of carbon – 12 containing6.023*1023 C-12 atomsi.e.12g of C= 6.023 * 1023 carbon atoms now as6.023* 1023 C atom have a mass = 12g mass of one C atom will be = 12* 1

----------------------6.023 * 1023

--- ---- ---- ----- ------ ==== 1.99 * 10-23gNow it is easy to find the value of 1 a. m. u

As1 a.m. u = mass of an atom f c-12/12

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So1 a.m. u = 1.99*10-23/12

1 a.m. u = 1.66*10-24

RELATIVE MOLECULAR MASS:Sum of the relative atomic masses of all the elements present in a molecule is called asrelative molecular mass of that molecule.For example

The relative atomic mass of H= 1.008 a. m. u---- ---- ---of Cl = 35-5 a. m. u

80 the relative molecular mass of HCL will be :1.008 + 35.5 === 36.508 a. m. u

AVERAGE ATOMIC MASS:The average of the atomic masses of all the isotopes of an element with respect t their relativeabundance is known as average atomic mass of that element for the determination of theaverage atomic mass of an element we multiply the masses of all the isotopes with relativeabundance and then add then up and finally divide them by 100.Fro example copper element ( Cu ) has two isotopes which are 63

29CU ( 69. 09% ) and 6529CU

( 30. 91 % ) so the average atomic mass of copper can be calculated asAverage Atomic == 63*69.69 + 65* 30.91Mass of CU -----------------------------

100Average Atomic mass of CU 63.55 a. m. u

RELATIVE ABUNDANCE:The no of atoms of an isotopes of an element present in 100 atoms of element is known asrelative abundance of that isotope.Eg if we take 100 atoms of Hydrogen element & let the no of protium atoms is 97 that ofdeuterium atoms is 2 and that of tritium, atoms is 1.Thus we can say that the relative abundance of protium is 97% , Deuterium is 2% and tritiumis 1 % .

DETEMINATION OF RELATIVE ATOMIC MASSES OF ISOPTPES BY MASSSPECTROMETRY:Mass spectrometry is a technique by means of which the relative abundance and atomic massof the isotopes of an element can be determined.The instrument used in this technique is called MASS SpectrometerAlthough there are several methods for the determination of atomic mass and molecularmass, but mass spectrometry is the most direct and accurate method.In this component is called vaporizer. Here the sample (element or molecule ) is vapourised.This vaporized sample is then passed into another component known as filament. Here highspeed electrons are passed through the vaporized sample. These high speed electrons collidewith the electrons of the atoms of sample and get them out of these atoms. Thus tiredlycharged ions.These tiredly change ions of mass " m" and change "e" are passed through tiredly chargedplates. Thus the +tive ions are accelerated.Then the +tive ions are padded through the poles of magnet (magnetic field) which forcesthem into a circular path.The radius of the path depend upon the charge to mass ration ( e/m) of the tire ions. The ionswith larger e/ m have smaller radius and the ions (+ ive ) with smaller e/m have larger radius.The ions with different masses and same charges are separated.

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+ ive ions with different e/m are thus separated and the detector receive them ofdifferent points producing different signals.Thus we easily come to know about the no of isotopes, their atomic masses and their relativeabundance.One of the first application of mass spectrometer was determined by F. W. Aston. It wasabout Neon ( Ne ). It was told or explained that the naturally occurring neo has B isotopes ieNe-20 ( 90.92 %)Ne- 21 ( 0.26 %) and Ne-22 ( 8.82% ).All this information is got from the graph as shown in the following diagram.

CHEMICAL FORMULA:The symbolic representation of a compound is known as its chemical formula.OrThe representation of a compound in terms of symbols of its elements is known as chemicalformula of that compound.

ADVANTAGES OR CHARACTERIOTICS FO CHEM. FORM:The chemical formula of a compound has the following advantages i.e. it provides us thefollowing information about the compound.It tells us about the elements present in the compound.it tells us about the no of atoms of each element.it tells us about the chemical composition of compound i.e. tells us the ratio b/w the atoms ofelements of the compound.It shows us about the no of moles of each element present in one mole of the compound.For example the chemical formula of water is H2O. Thus H2O provides us the followinginformation about water .There are H and O element in water.There are 2 atoms of H element.The ratio b/w the atoms of H & O is 2:1There are 2 moles of H and One mole of a elements in one mole of H20

TYPES OF CHEMICAL FORMULA:There are two types of chemical formula.Molecular formula 2. Empirical or simplest formulaMOLECULAR FORMULA:The formula of a compound which shows the actual no of atoms of its element is called asmolecular formula of that compound.EM[EROCAL OR SIMPLEST FORMULA:The formula of a compound which shows the simplest whole ratio b/w the atoms of itselements is known as empirical formula of that compound.

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For example:MOLECULE MOLECULAR FORMULA EPERICAL FORMULABENZENE C6H6 CHGLUCOSE C6H12O6 CH2OWATER H2O H2O

RELATIONSHIP B/W M.F.& E.FThe molecular and empirical formula of a compound ( molecule ) have the followingrelationship.There are some molecules for which both molecular and empirical formula are same.E.g. : water has H2O as molecular as well as empirical formulaM. F is an integral multiple of the E. Fi.e. M. F = N X E. FWhere "N" is an integer and its value may be 1, 2,3……………The value of "N" is determined as

N = Molecular Mass--------------------------------------------

Formula Mass Examples: The E.F of Acetic Acid is CH2O. Its Molecular Mass is 60g/mol. What is its M,f

DataE.g. Acetic Acid = CH2OThere fore Mass of Acetic Acid = 12+ 1+2+1630gMolecular Mass of Acetic acid = 60 g/molM. F = ?

M. F = n E.F= Mole mass/ For Mass (CH2O)= 60/30 (CH2O)= 2 (CH2O)

M.F = C2H4O (CH3 CooH)

PECENTAGE COMPOSITION:The components ( Part ) of each element ( by mass ) is known as percentage composition ofthe compound.The percentage of element of a compound may be determined in two cases.If the chemical formula of the compd is known, then the percentage of any element isdetermined as:Element = Mass of that element

----------------------------------------------* 100Molecular mass of compd

For example : what is the % age of C & O in CO2

If we don’t know about the chemical formula of the compound. Whose % age composition isto be determined, then first of all, we determine, the chemical composition of compound by aprocess known as chemical analysis and then find out the % age of each element.

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CHEMICAL ANALYSIS:The process by which the chemical composition of a compound is determined is calledchemical analysis:

Let we have an organic compound containing carbon, hydrogen and oxygen elements.During chemical this compound a weighted quantity ( known mass ) of this sample compd isplaced in the combustion tube which is fitted with a furnace. Oxygen is supplied at one end ofthe combustion tube .

At the other end of the combustion tube two consecutive jars are placed onecontaining a water absorbing substance like 50% KOH and the other jar containing a CO2

absorber like. Magnesium per chlorate [ Mg ( ClO4 )2 ]. These two jars are pre- weight.As a result of combustion of the organic compd, all the hydrogen converts into H2O

and carbon into CO2.These two gases ( H2O (g) & CO2 ) pass into the jars where they are absorbed and as a resultthe weights of both the jars increased. From the increase in the weight of jars, we can easily,determine the masses of H2O and CO2 formed from a known mass of the organic compd

Now the percentage of each element is determined as follow:

% C = Mass of Co2 / Mass of org. compd * Mass of C (12)/ M. Mss of Co2 * 100% H = Mass of H2o / Mass of org. compd * Mass of H (2)/ M. Mss of H2O (18) * 100

The %age of oxygen element is determined indirectly as;%O = 100 – ( %C + %H)

Example : During combustion analysis of an organic compd containing C,H and O, 1.039 gof CO2 and 0.636 g of H2O are produced from 0.5439 g of the org, compd.Determined the %age composition of compd.

Mass of Data / org. compd = 0.5439 gMass of Co2 = 1.039gMass of H2o = 0.6369 g

Solution:%C = 1.039/0.5439 * 12/44 * 100 = 52.108%%H = 0.6369/0.5439 * 2/18*100 = 13.11%%O = 100 – (52.108 + 13.11) = 34.77%

DETERMINATON OF EMPERICAL:FORMULA OF A COMPOUNDThe empirical formula is the simplest formula that gives information about the simple ratio ofatoms present in a compd.For the determination of Empirical formula, following steps are followed,Determination of the percentage of each element

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e. g C H Ox% y% z%

(2) The %age of each element is divided by its atomic (mass to get its no of moles)

E. gx/12 y/10008 z/16

Moles : C H O

(3) Divide the moles of all elements by the least one of them, to got the no, f atoms of eachelement

E. ga/e b/e c/c

Atoms : d e f

Note: if d, e, & f are whole numbers.Then they make the empirical formula of the compd.e : [CaHeOf]But if atoms of or more element are in fractions then all of the m are multiplied by a suitablenumber to convert them into whole numbers.e. g let d e f

1 1.33 1As here 1.33 is fractional figure, so: we will convert it into whole number by multiplying itwith 3i.e 3 ( 1 : 1.33: 1 )

3 : 3.99 : 3Or 3 : 4 : 3Now

E.F = C3H4O3

Example : Ascetic Acid ( vitamin C ) contains 40.92 % carbon, 4. 58% Hydrogen and 5. 1.5% oxygen.What is its empirical formula?Data.

%C = 40.92%H = 4.58%O = 54.5E.F = ?

Solution:1)Percentage : - 40.92% 4.58% 54.5%

2) Divide %age by Atomic Mass to get 40.92/12 4.58/1.008 54.5/16

3) Divide to by least get to atoms 3.41/12 4.54/3.4063.406/3.406

4)Multiplying by 3 to convert 1.33 into whole number to by least get to atoms

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3*1 : 3*1.33 : 3*13 : 4 : 3

SoEmp . Formula of Ascorbic Acid is = [C3H4O3]

DETERMINATION OF M.F. FROM E.F:The molecular formula of molecule can be determined from its empirical formula as followMol . Formula = n* Emp. formulaHere “n” is an integer and its value can be find as:N = Molecular MassFormula MassN= 1, 2, 3 ………………………Molecular Mass:It is the sum of the atomic masses of all the elements present in the molecular formula of amolecule.e.g :

The molecular formula of benzene is C6H6. Thus its molecular mass is12*6+1*6===== 78a. m. u or 78g/ mole.Formula Mass:

It is the sum of the atomic masses of all the elements present in the empiricalformula of a molecule or formula unit of a compound.e.gThe emp. Formula of benzene is CH, so its formula mass will be 12*1+1*1 ===13a. m. u or13g/mol

Example :The combustion analysis of an organic compd shows that it contains 65. 44%C, 5.50% H and29.06 % O. what is its emp. Formula.

If the molecular mass of the compound is 110.15 g/mol, then what will be themolecular formula of the compound Data.

Thermodynamics:The branch of science which deals with the study of transfer- motion of heat from a region ofhigh temp. to a region of low temp is known as thermodynamics.

1st law of thermodynamics (law of conservation of energy:this law states that “energy can nether be created nor destroyed, however it can be convertedfrom one from to another”.

CONCEPT OF MOLE & AVOGARO’S NO:MOLE:The atomic mass of an element, molecular mass of a molecule or formula mass of a formulaunit, expressed in grams, is called one mole of that element, molecule or formula unit,e.gGram At. Mass of O= 16 g = 1 mole oxygen element.

N= 14 g= 1 mole of N element,Na= 23g= 23g = 1 mole of Na elementH= 1g = 1 mole of H element

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Similarly : Gram Molecular mass of O =16 g = 1 mole oxygen element.--------------------------------------N = 14g = 1 mole of n element.----------------------------------- Na = 23 g = 1 mole of Na element.------------------------------------H = 1g = 1 mole of H element.Similarly:Gram Molecular Mass of H2O = 18 g = 1 mole of H2O molecules------------------------------HCL = 36.5 g = 1 mole of HCL molecules.----------------------------O2 = 32g = 1 mole of O2 moleculesSimilarly:Gram formula Mass of NACL = 58.5g = 1 mole of NACL formula units.----------------------------------- Mgo = 40 g = 1 mole of Mgo formula units.Note:No of moles of a substance ( element, molecule or formula unit) is represented by “ n” & canbe calculated by the following formulas.

Examples:What is the no of moles (or gram atoms of 0.1 of g on Na.

No of moles of Na (n Na) = ?Mass of Na (m Na) = 0.1 g

Solution:

n = Mass (g) of Na / At Mass of Na

n = 0.1/23n = 0.0043 moles or 4.3*10-3

AVOGADRO’S NO (NA):As one dozen of eggs , one dozen of bananas and one dozen of oranges have fixed no. ofitems (eggs, bananas & oranges) although they have different masses, different fast anddifferent nature , likewise it the case for one mole of a substance ( element, molecule orformula unit ).So we can say that : one mole of any substance contains a fixed no of particles. This fixed no,is called Avogadro’s No. it is represented by “Na” and its value is 6.023*1023.For example:1 mole ( 23g) of Na element = 6.023*1023 Na atoms.1 mole ( 1.008g ) of H element = 6.023*1023 H atoms.1 mole ( 16g ) of oxygen element = 6.023*23 O atoms.SIMILARLY1 mole ( 18g ) of H2O Molecule = 6.023* H2O Molecule.1 mole ( 32 ) of O2 Molecule = 60023*1023 O2 Molecule.1 mole ( 58.5g ) of Nacl formula unit.= 6.023*1023 Nacl formula units.----------------------------------------------= 6.023*1023 Na+ ions---------------------------------------------= 6.023*1023 Cl- ions.EXAMPLES:

What is the no of moles ( or gram atoms ) of 0.1g of Na.DATANo of moles of Na ( Nna ) = ?Mass of Na ( Mna ) = 0.1g

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Solv :As

Nng SoA = mass of mg SO4

---------------------------------Formula mass of mg SO4

10-3 = Mass pf MgSO4

--------------------120

Mass of Mg SO4 = 120 * 10-3

------------------- = 0.12 g3 . What is no. of moles of 12. 023*1026 molecules fo water ( H2O )DATAH2O = ?No. of H20 molecules = 12. 023 * 1026 moleculesSolve:As H2O = No of H2O molecules

------------------------------Avogadrp’s No

H2O = 12.023 * 1026

--------------------6.023 * 1023

H2O = 2 * 10 molecules

CHEMICAL REACTION:The process in which two or more substances combine in such a way that the old bondsbetween their atoms are broken down and new bonds are formed to form new substances isknown as chemical reaction.For example when two moles of Hydrogen gas ( H2 ) combine with one mole of oxygen gas (O2 ) the old bonds between the atoms of these two gas are broken down and these atoms formnew bonds in such a way t form two moles of H2O molecules.The substances which combine to undergo chemical reaction & whose bonds are broken arecalled the “Reactants” while the substances which are formed as a result of chemical reactionare known as products.

CHEMICAL EZUATION:The representation of a chemical reaction in terms of symbols and formulas of reactants andproducts is known as chemical equation.OrThe symbolic representation of a chemical reaction is known as chemical equation.e.g

2H2 + O2-------------------------- 2H2OIn a chemical equation, the reactants and the products are separated by means of an arrow.The reactants are at athe left side of the arrow while products are at the right side.The arrow is always dirceted from reactants to products.I .e

Reactants -----------------Products.

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STOICHOMETRY:The branch of chemistry which deals with the study of relationship between the quantities ofreactants and products is called stoichiometry. It is based on two laws.LAW OF CONSERVATION OF MASS:This law states that during a chemical reaction the total mass of reactants is always equal tothe total mass of the products.

LAW OF DEFINITE PROPORTIONThis law states that a particular compound consists of particular elements combined togetherin a fixed ratio, no matter from which source it is obtained.E . gWater is a molecule and it always consists of H and O elements in 2:1 no matter how it isobtained .Lets see how the stoichiometry depends upon these laws take the formation of H2O from H2

and O2

i.e.

H2 + O2 H2O2 g 32g 18 g 34g 18g

Non-Stoichimetich

H2 + O2 H2O2

2 g 32g 2g + 32 g 34g 34 gNon-Stoichimetich

Now2H2 + O2 2H2O

4g 32g 36 g 36g 36 g

Stoichimetich equation

From the above discussion it is clear that a stoichiomrtric equation must follow both theabove mentioned laws. If an equation follows one but not the other, then it will not bestoichiometric.From the knowledge of stoichiometry came to know that during balancing of a chemicalequation we should always put some digit as the co-efficient of any reactant a product , not inthe formula.

CHERACTERISTICS OF A STOICH. EQNA stoichiometric equation must have the following characteristics.It should be correct i . e we should write the correct symbols and formulas of the reactantsand productsIt should follow both the laws i. e law of conservation of mass & law of definite proportion.It should be balanced.

NOTE.We can study the following type of relationships from a stoicliometric (balanced) chemicalequation.

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Mass – Mass RelationshipFrom a stoichiometric equation we can easily find the mass of any reactant or product, if themass of any reactant or product is driven.For example.How much k2 SO4 will be formed from 14g of KOH according to the followingstoichiometric equation.2KOH + H2 SO4------------------ K2SO4 + 2H2O DATAMass of KOH = 14 gMass of K2SO4 = ?Solv

According to the stoichio. Equ.2KOH + H2SO4 --------- K2SO4 + 2H2O2 (39+16+1) 2*39+32+16*42*56 78+32+64112g/mol 174 g/molAs112f of KOH give = 174g K2SO4

So 14 g of KOH will give = 174 *14 K2SO4

---------------------112= 21.75g of K2SO4

Mole – Mole relationship:If no of moles of any reactant or product are given then the no of moles of any other reactantor product can easily be found from a stoichiometric equation.For example :

How many moles of oxygen are required to produce 8 moles of H2O accordingto the following equation.2H2 + O2----------------------------- 2H2ODATANO. of moles of O2 = ?No of moles of H2O = 8 moles.Solv :As according to St. equ:2H2 + O2-------------------------2H2O

1 mole 2 moles2 moles of H2O require = 1 mole of O2

8 moles of H2O will require = ½ *8= 4 moles fo O2

Mass – Mole relationship :If the mass of any reactany or product is given then the no of moles of any other reactant orproduct can be calculated from stoichiometric equFor example:

How many moles of water will be produced from 20 g of H2ODATA

H2O = ?Mass of H2 = 20 g

As2H2 + O2----------------- 2H2O4g 36g

4g of H2 give 36 g H2OSo 20g of H2 will give 36/4 * 20

17

--------------------- 180g H2ONow as H2O = mass of H2O/ m. mass of H2O = 180/18===== 10 molesMOLE – MASS RELATIONSHIPIF the no of moles of any reactant or product ae given then the mass of any other reactant orproduct can be calculated from the stoichiometric equation.For example:

How much nitrogen will be required to react with 16 moles H2 to produceammonia (NH3) according to the following stdoichiometric equation.N2 + 3H2-----------------------2NH3

DATAMass fo N2 = ?No of moles of H2= 15 molesSolv:As IN2 + 3H2-------------------2NH3

1 mole 3 moles3 moles of H2require = 1 mole N2

15 moles of H2 will require = i/3 * 15---------------------------------- = 5 moles N2

Now as N2 = Mass of N2/ mole. Mass of N2

5= Mass of N2/ 28Mass of N2 = 5 * 28 ====== 140 g

LIMITING REACTANT:The reactant which gives the leay no of moles of the product is called limiting reactant.OrThe reactant which is consumed earlier is called limiting reactant.OrThe reactant which stops a chemical reaction or which controls the formation of products iscalled delimiting reactant .For example it we have 3 molecules of a reactant > & 4 molecules of a reactant > to form theproduct then is the limiting reactant because it is consumedearlier or it gives the least no of moles of the produceI . e

Lets take another example consider the reaction between, H2and O2 as2H2(g)+ O2(g)------------------------------------2H2OLet we take 2 moles of Hydrogen ( 4g ) and react it with 2 moles of oxygen ( 64g ) then wewill get only 2 moles (36) of H2O. Here 2 moles (36g) of H2O are obtained because 2 molesof hydrogen (4g) reacy with only one mole (32g) of O2 according to the stoichiometricequation.Now as less H2 is O2, So H2 is the limiting reactant.Now let we react 4 moles ( 8g ) of H2 with 2 moles (64g) of O2 then we would obtained fourmoles (72g) of water.

IDENTIFICATTION OF LOMITING REACTANT:For the identification of the limiting reactant, the following three steps are performed.First of all find the no of moles of each reactant if their masses are given. If no of moles ofreactant are given directly, then this step is not required.Find out the no of moles of the product from all the reactants whose moles are givenThe reactant which gives or produces. The least no of moles of the products is selected as thelimiting reactant.

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Lets explain all this process by means of a numerical example.Example.

NH3 gas can be prepared by heating together two solids i.e NH4CL and Ca(COH). If amixture containing 100 g of each solid is heated then:Calculate the mumber of grams of NH3producedCalculate the excess amount of reagent left unreacted.Equation: 2NH4CL + CA( OH )2 --------CACL + 2NH3 + 2H2oSolution: DATAMass of NH4CL = 100 gMass of Ca(OH) = 100 gMass of NH3= ? For finding the limiting reactant,Finding no of moles of reactants.NH4CL= 100/53.5 = 1.87 molesCa(OH)2= 100/74 = 1.35 molesNow finding th no of moles of producty ( NH3 ) from the both the reactants as from equationit is clear that2NH4CL ---------------------------2NH3

2 moles of NH4CL give = 2 moles NH3

moles of NH4CL will give = 2/2 * 1.87=1.87 moles NH3

similarlyCaCOH2=============2NH3

1 moles of Ca(OH)2 gives = 2 moles NH3

So1.35 moles of Ca(OH)2 will give = 2/1* 1.35 = 2.70 moles NH3

C= As NH4CL gives theleast no of moles of the product ( NH3), so it is the limiting reactantAnd thus the no of moles of NH3 produced are = 1.87 molesNow as

NH3 = mass (g) of NH3 /M. Mass of NH3

1.87 = mass (g) of NH3

17Mass of NH3 = 1.87* 17Mass of NH3 = 31.179g

B . Amolunt of the reagent present in excess ?As

2NH4CL + CA(OH)2-------------------- CaCl2 + 2NH3+ 2H2OAs

2 moles pf NH4CL require = 1 mole CA(OH)2

So1.87 moles of NH4CL will require = ½ * 1.87= 0.935 moles Ca(OH)2

Now as total no of moles of Ca(OH)2 = 1.35 molesThe reacted ono of moles of Ca(OH)2 = 0.935 molesSoThe moles of Ca(OH)2 left unreacted = 1.35- 0.935= 0.45 molesHenceMass of excess Ca(OH)2 left unreacted = n*x m. moles = 0.415* 74

==========30.71g

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Lecture # 10YIELD:The amount of products obtained is known as yield.

Types:There are two types of yield.

Theoretical yieldActual Yield

Theoretical yield:The amount of products calculated from a stoichiometric equation is known as theoreticalyield.E .g

2H2 + O2 --------------------2H2O4g + 32g ------------------ 36g theoretical yield.

Actual yield:The amount of products obtained during chemical reaction or experiment is called actualyield.E .g

When we combine 4g of H2 and 32gO2 in the presence of electric spar, then water isobtained whose, weight is known as actual yield.NOTE:

The actual yield of a reaction is usually less then the theoretical yield.It is because of the following reasons,Due to the formation of by- produce is that product which unwanted whether it is indicated ornot is the chemical equation.Due to reversible dreactopm e . gN2 + 3H2----------------2NH3

Here some of the NH3 converts back into reactants before it is collected.Due to mechanical loss during the process of weighing, drying, filtration etc.Usually yield is indicated as percent yield to note the efficiency of the process. The percentyield is calculated as.

% Y = A.Y/T.Y * 100

CHAPTER 1

BASIC CONCEPTS EXERCISEQ2.A . SEA WATER: it is a mixture are dissolved it.B . Helium gas: it is an element because Helium lags consists of He atoms whose valenceshall is complete and don’t make bonds.C . Sodium Chloride : it is a compound because it is made of atoms of Na and Cl elements.D . Botle Of Soft drink: it is a mixture because CO2 gas So dissolved in it .e. Air : it is a mixture because it contains various non- reacting gases like N2(78%), O2,(21%) etcf. Concrete: it is a mixture because it contains various non-reacting substances like sandcement and pebbles etc.Q. 3 a. 11 X/B A (PS+NS)

5 B Z(Ps or es)No of protons = No of electrons = 5 No of Neutrons = A-Z

20

--------------------=11-5--------------------=6

Q 4 . ans.Almost everuy element has isotopes lhaving diferent atomic masses & different

relativ eabundance. Thus the atomic mass of an element is actually the average of the atomicmasses of its isotopes with respect to their relative abundance carbon has isotopes C-12 & C-13 with their relative abundance as 98.89% and 1.11% respectively. Thus the average atomicmass of carbon will be:Average At.Mass of C = 12*98.89 + 13*1.11/100------------------------- = 12.01 a. m .u.So it is clear that the average of atomic masses of the isotopes of C is 12.01, that is whyatomic mass of crbon is taken as 12.01 a. m. u. rather than 12.a. m. u.

DATA:Q.5

Atomic Mass of C-35 = 35a. m. u.Relative abundance of C-35 = 75.53%

Similarly:Atomic mass of C-37 = 37 a. m. u.Relative Abundance of C-37 = 24.47%Average Atomic Mass of cl= ?

Soln:Average At. Mass of cl = 35*75.53+37*24.47/100

= 35.489 a. m. uQ 6. DATANo. of person =5 billionsNo of particles = 6.023 *1023 particlesNo of particles counted per head per second= 2 particlesThusTotal no of padrticles counted per second= 2*5-------------------------------------------- = 10 billions------------------------------------------- = 1*1010 particlesTime ( in year ) taken for counting 6. 023*1023 particles =?Soln : 1st of all finding the time in seconds required for counting 6.02*1023 particles.As :

1*1010 particles are counted in = 1 second.So

6.023*1023 particles will be counted in = 1/1*1010 * 6.023*1023

SoTime taken for counting 6.023*1023 particles = 6.023*1023/1*1010

---------------------------------------------------------= 6.023*1013 secpndsNow converting this time into days & then into years.As

86400 seconds wil make = 1/86400 * 6. 023*1013

----------------------------------= 0.000070 * 1013 days--------------------------------- = 7. 0 *108 days

Now as365 days make = 1 year.

So7.0*108 days will make = 1/365 *7.0* 108 years.

= 0.01917 * 108 years= 1.917 * 106 years

21

So1.917*1.106 years are taken for counting 6.023*1023 particles ( Avogadro’s No of

particles ).Q . 7 ans:As

Mass of 1 a. m. u = 1.66*10-24g.ThusMass of 13.2a. m. u will be = 1.66*10-24 * 13.2---------------------------------- = 21.912 * 10-24g-------------------------------- = 2.1912 * 10-23

Q. 8 DATANO OF S atoms = ?No. of moles of S = 5.10 molesSoln ans

N = NO. OF S Atoms/NA

So5.10 = No of s atoms /6.023*1023

SoNo. of S atoms = 5.10 * 6.023*1023

----------------------------= 30.717 * 1023 S atoms------------------- = 30.0717*1024

Q.9DATA

NO.of moles 0f Co= ?No. of Co atoms = 6*109 atoms.Soln: As

Nco = No. of Co atoms /NA= 6*109/6.023*1023 = 0.996*1009-23

= 9.96 * 10-17 moles.Q. 10

Ethene (C2H4) and propenc (C3H6) are the molceules which have same empericalformula but different molecular formula.I . eEthene PropeneM . F C2H4 C3H6 differentE . F CH2 CH2 sameQ. 11

Ans. The expression P4 s signifies or shows that it is a molecule of phosphorouselement containing four phophorous atoms bonded together by means of covalent bonds.The difference between P4 & 4p is that P4 means one mole of P4 molecule.While 4p means 4 moles of phosphorous element.Q.12A I H2 ii O2

B I HCL ii COC I P4 ii S8D I H2O ii C6 H12 O6

Question . 13 solve by yourself.Q. 14 : formula of chlorophyll = C55 H72 MgN4O5 Molecular Mass= ?Soln:

AS Atomic mass of :C =12 H =1 MG = 24 N = 14 & O = 16The Molar Mass of chlorophyll (Css H/2 Mg N4O5) is

= 12.55 11*72 + 24* 1 + 14*4 + 16 *5

22

Molar Mass of chlorophyll = 660+72+24+56+80= 892 g/ mole

Q. 15 DATAFormula of hydrated Aluminium sulphate = Al2(So4) * H2O

% Al = 8.20 %No . of molecule (X) = ? in Al2 (SO4)2*H2OSoln: Mass of Al = 27 *2 = 54g

As Mass of S = 32*2 = 64g% Al = Mass of Al/Molar Mass of *100 Mass of O = 16*8 =1289

Al2(SO4)2 *H2O Mass of H2O = 18 *X = 18xg8.20 = 54/264x * 100 So molecular mass of Al2(SO4)2 x H2OOr = 54 +64 +128 +18x8.20 * 264x + 54 * 100 *1 = 264 xg/ moleX = 54*100/8.20 *264X = 2.49Q . 17A . Balance

C3H8 + O2---------------------CO2 + H2OBalanced equation is :

2C3H8 + 100-----------------6CO2 +8H2OB . DATA

Mass of CO2 = ?No. of moles of C3H8 = 3.65 moles.Stoichiometric equation is

2C3H8 + 100 ------------ 6CO2 + 8H2O2 moles 6 moles

As 2moles of C3H8 give = 6 moles of CO2

So 3.65 mole of C3H8 will give = 6/2 * 3.65-------------------------------------------- = -10.95 moles CO2

Now asMass of CO2 = nCO2

= 10.95 * 44= 481.8g CO2

So481.8g of CO2 are formed from = 3.65 moles of C3H8

Q . 18 DATANO . OF moles of Mno2 = 0.86 molesMoles of HCL = 48.2gSo

No .of modes of HCL = 48.2/36.5 ===== 1.32 molesI . which reagent is used first ( limiting reactant )= ?Ii . Mass of CL2 = ?Reaction :

Mno2 + 4HCL --------Mncl2 + Cl2 + 2H2OI . Limiting reactant = ?Mno2=? HCL = ?As As1 Mno2-------------1CL2 4 moles of HCL give = 1 mole Cl2

1 moles Mno2 give = .1 mole Cl2 1.32 moles will give = 1/4 * 1.32So = 0.33 moles Cl20.86 moles will give = 0.86 moles CL2

As HCL gives the least no. of moles of CL2 ( product ) thus HCL so the limiting reactant.And thus

23

No. of moles of CL2 = 0.33 moles.Therefore,

Mass of Cl2 = ncl * Mol. Mass of Cl2= 0.33 *71

--------------------= 23.43g CL2

Q . 25Wrong or Ambiguous statementsA . 1 moles of hudrogen b . the molecular moles of Nacl is 58.5 a. m. uCorrect statements.A . 1 mole of hydrogen element or hydrogen molecule.B . Formula mass of Nacl is 58.5 a. m. u becaise Naccl is a formula unit hot a molecule andthe mass of a formula unit is called as formula mass.Q . 26A . SO2 compd and MoleculeB . S8 molecule but not compdC . C5 element.D . N2O5 compd & MolecleE . O elementF . O2 Molecule but not compdG . O3 Molecule but not compdH . CH4 Molecule & compdI . KBR compd but not molecule ( ionic bond )J . S elementK . P4 Molecule but cont compdL . Lif compd but molecule because of ionic bond.Q . 27.A 2.5 mole CH4

No .of atoms = ?First of all, we will determine the no. of CH4 molecule in 2.5 molesThus nCH4 = no. of CH4 Molecules/ avogadro’s no= No . of CH4 molecules /6.023 * 1023

OrNo. of CH4 molecules = 2.5*6.023*1023

-------------------------= 15.05* 1023 CH4 moleculesNow as Molecule of CH4 has = 5 atomsThus 15.05 * 1023molecules of CH4 will have = 5/1 * 15.15*1023

SoNo. of atoms in 15.05 * 1023molecules of CH4 = 75.28 * 1023

BNO .Of moles of He = 10 molesNo. of atoms = ?AsNO. OF He atoms = n * avosadro’s No-----------------------= 10*6.023 * 1023

-----------------------= 60.23 *1023 He atomsCNo. of moles of SO2 = 4 molesNo. of atoms = ?As

No. of moilecules = n * avogadro’s No---------------------------------= 4* 6.023 *1023

----------------------------------= 24.092 * 1023 SO2 molecules

24

Now as1 moleclules of So2 has = 3 atoms

So 24.092 * 1023 molecules of So2 will have = 3/1 * 24.092*1023

= 72.27* 1023

DNO .Of moles of So8 = 1.8 molesNo. of atoms = ?As

No. of molecules = n * NA----------------------= 1.8 *6.023 *1023

---------------------= 10.84* 1023 So8 moleculesNow as1 molecules of SO8 contains = 9 atomsSo10.84* 1023 So8 molecules will have = 9/2 * 10.84* 1023

= 97.57 * 1023 atomsENo .of moles of NH3= 3 molesNo. of atoms = ?As no .of NH3 moleclules = 3* 6.023* 1023 NH3 moleculesNow as1 molecules of NH3 has = 4 atomsSo

18.069 * 1023 molecules will have = 4/1 * 18.069 * 1023

------------------------------------------= 72.77 * 1023 atoms

So it is clear that 1.8 moles of So8 has the largest number of atoms.Application of Buffer solutions :Imporatnce of B. SolnsBuffer solutions have tramendous applications. Some important applications of B. sons are,The Human vlood is a buffer whose PH is 7.4. if this PH changes slightly then seriouslyproblems may happen. Inspite of having so many acidic and alicalins substances the PHdoesn’t changes.Buffer solutions arte mostly used in Analytical laboratories for carrying out particularsreactions at particular PH.Buffer solutions are used in various industries.Buffer solutions are used in Biological labartories.

COMMPM ION EFFECT:The decreases in solubility of an electrolyte in its soln by introduction of another electrolytesuch that both the electrolytes produce common ions is known as common ion effect.

It is important to note that the common ion may be caption ( + ive ) or anion (- ive ).Explanation :As we know that ionization is never 100% in case of strong electrolytes, most of the conc. Ofthe electrolyte goes into ionized form while small conc., remains an-ionized and thus anequilibrium us established. In case of work electrolyte gets ionized while most of it remainsunionized and thus an equilibrium is established.It we dissolve on electrolyte in a solvent ( like water ) to form al soln, then an equilibrium isestablished among the unionized amount of electrolyte and its ions produced.Which produces a common ion to the already existing electrolyte, here an equilibrium isestablished among the unionized amount of the 2nd electrolyte and its ions.

25

One of the two electrolytes will be more stronger. So the reverse reaction of the weakelectrolyte speeds up due to common ion and thus its solubility decreases.

Application :Purification of commercials saltCommercial salt ( Nacl ) contains impurities like dust etc. To purify it , it is dissolved inwater thus all impurities suspend in water and pure Nacl dissolves as

Nacl====== Na+(aq) + Cl-

(aq)

When another electralyte is introduced into the soln which produces a common ion with thealreay existing electrolyte. Lets consider Acl. Then

HCL =====H+(aq) + CL-

(aq)

As HCL is more stronger than NACL therefore the reverse reaction of Nacl increases andthus its solubility decreases thus pure Nacl precipitates out, the water containing theimpurities is poured out and thus pure Nacl is left out which is washed with fresh water andthen dried up.Of salt AB

SIMILARLYAgel ( silver chloride ) is a sparingly soluble salt: then its solubility product expression will

be,KSP = [AG+] [CL-]

b/cAfcl =====H2O=============== Ag+

(aq) + Cl-(aq)

Kc (Agcl ) = [ Ag+] [ CL-]KSP = [ Ag+] [ CL-]It is important to note that KSP depends rpon temperature.

BUFFER SOLUTIONS:The solution which resists change in its Ph, even with addition of small amount of a strongacid or a strong base: is known as buffer solution.A buffer solution is made as.When a weak acid ( like CH3COOH ) is mixed with its salt made by it with a strong base ( i.eCH3COONa ). They are mixed in a particular ration. This mixture is known as buffer and thesoln. of this mixture is known as buffer solution. Such a buffer is known as Acidic buffer.I . e CH3COOH / CH3COONa bufferWhen a weak base ( like NH4OH ) is mixed with its salt made by it with a strong acid ( I .eNH4CL ) in a fixed ratio and then dissolved in water. We get a buffer soln. such a buffer isknown as basic buffer.I . e NH4OH / NH4CL bufferThe buffer soln, has a particular PH range and this range remanins xonstant even for longtime or even after the addition of small amount of strong add or strong base.

BUFFER ACTION:The ability of a buffer to resist change in its PH, even after the addition of small amount ofstrong acid or base : is known as buffer action.Now the question arises that how a buffer soln resists change in its ph even with the additionof small amont of a strong acid or base.It can be explained as, consider the CH3COOH / CH3COONa soln:If we add small amount of a strong acid like HCL the PH doesn’t change. It is b/c of the factthat in CH3COOH/ CH3COONa: CH3COONa ionizes to large extent: so there are

26

CH3COOO- in large excess which come from CH3COONa. Here common ion bothproduce H+. As a result the solubility of CH3COOH is furthur decreased.change Thus most of H+ ion of HCL ( which are responsible for change of PH ) are convertedinto a weak acid, CH3COOH which is already in excess in buffer soln.Thus the PH of buffer remains constant b/c H+ ions of HCL don’t remain in soln.Strong electrotypeCH3COONA================= CH3OO-

(aq)

Weak electrotypeHCL ========= H+

(aq) + Cl-(aq)

Then:H+

(aq) + CH3COO-(aq) ------------- CH3COOH

From HCLFrom (CH3COONa) weak (acid)

11. if we add, small amount of a strong base like NaoH. The OH- ions of NAOH which areresponsible for change of PH of buffer combine with H+ ions of CH3COOH producing aneutral compd H2O.Thus the PH of buffer soln doesn’t.I . e

CH3COONa =========== CH3COO-(aq) + Na+

(aq)

CH3COOH============CH3COO-(aq) + H+

(aq)

NAOH==================Na+(aq) + OH-

(aq)

ThenH+

(aq) + OH-(aq) ------------- H2O ( Nuetral )

Amount of strong add or strong base.

Buffer capacity:The extent till which a buffer can resist change in its PH is known as buffer capacity.A buffer can resist change in its PH if we add particular amount of strong acid or base. If weexceed the amount of strong acid or base, then the PH of PH will certainly change.So we cabin say that there is a certain limit till which a buffer can resist change in its PH.This limit is known as buffer capacity.CALCULATIOHN OF PH BUFFER SOLUTION:The ph of a buffer solution is calculated by an equation known as “Hander so n Horsebackequation” . This equation is

PH = Pka + log base/ acidThe hander son Hesselback equation can be derived asConsider the ionization of a weak. Acid CH3COOH.CH3COOH CH3COOH- + H+

The ionization constant expression will be

Ka = [H+] [CH3COOH-] / [CH3COOH]

Solving the above equation for [H+]

[H+] [CH3COOH-] = Ka [CH3COOH]

[H+] = Ka [CH3COOH] / [CH3COOH]

Taking -10g on both sides;

-10g [H+] = -10g (Ka [CH3COOH] / [CH3COOH])

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Ph = (-10g Ka) +(-10g [CH3COOH] / [CH3COOH]

PH = -10g Ka - 10g [Acid]/[Base]

But – 10g Ka = pkaPH = Pka – 10g (Acid)/(Base)

OrPH = Pka + 10g [Acid]/[Base]