Stoichiometry - 1 - SelfStudys

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Stoichiometry - 1

Vidyamandir Classes Stoichiometry - 1

In this chapter, we are going to build the basics of solving the numerical problems in chemistry. The conceptshere involve, understanding of mole concept, gram equivalents and their applications in various chemicalprocesses.

Concept of Gram Atom :

One gram-atom of an element means collection of 6.023 × 1023 atoms. This concept applies only to theelements, which exist in poly-atomic states (e.g. O as O2, Cl as Cl2, S as S8, P as P4 etc.).

Atomic mass th 12Mass of 1 atom of an element

1 / 12 the weight of C

(in a.m.u. where 1 a.m.u. = Atomic Mass Unit = 1.66 × 1027 kg)

(a) The number of gmatoms in ‘g’ gm of an element whose atomic mass is A is :

ggm - atomsA

Illustrating the Concept :

gm-atoms in 142 grams of chlorine = g 142 4A 35.5

gm-atoms in 16 grams of Oxygen = g 16 1A 16

(b) The number of atoms in ‘g’ gms of an element is given by :

No. of atoms = 0g NA

(N0 = Avogadro number = 6.023 × 1023)

Note : The concept of gm-atom is useful in Radioactivity as it gives us number of nuclei i.e.

No. of nuclei = Number of atoms = gm-atoms N0 = 0g NA

Number of atoms in 56 grams of Nitrogen = 23 2456 6.023 10 2.4092 1014

Note : Atomic masses of some common elements are given at the end of this chapter.

Section 1 1

BASIC IDEAS Section - 1

Section 12

Vidyamandir ClassesStoichiometry - 1

Concept of Gram Mole :

The collection of 6.023 1023 molecules of an element or ions or a compound constitute 1 mole of thatelement or of ions or of compound. This magical number 6.023 1023 is known as Avogadro Number (N0).

The number of moles (n) in ‘g’ grams of a substance, whose molecular weight is M0 is given by :

Number of moles = 0

The number of molecules in n moles :

Number of molecules = 0 00

gnN NM

The number of millimoles :

Millimoles (m.moles) = 0

g 1000M

Number of moles in 46 grams of sodium (Na)

g 46moles 2M 23

(Atomic Mass of Na = 23 = molecular mass as Na is monoatomic)

Number of moles in 100 gm of Chlorine (Cl2)

g 100moles 1.4M 71

(Mol. Mass Cl2 = 2 35.5 = 71)

Number of moles in 54 grams of phosphorus (P4)

g 54moles 0.4355M 124

(Mol. Mass of P4 = 4 Atomic weight = 4 31 = 124)

Number of moles in 49 grams of sulphuric acid (H2SO4)

49 1moles 0.598 2

(Mol. Mass = 2 1 + 32 + 16 4 = 98)

Number of moles in 25 gm of CaCO3 (calcium carbonate)

25moles 0.25100

(Mol. Mass = 40 + 12 + 16 × 3 = 100)

Section 1 3

Concept of Gram Equivalents :

We can define gram-equivalent (gm eq) in ‘g’ gms of a substance whose equivalent weight is E as follows :

grams ggram equivalents (gm.eq)Equivalent weight E

To understand the concept of gram equivalent, one must know the meaning of equivalent weight (E) of anelement or a compound.

Theoretically, Equivalent weight (E) of an element or a compound is defined as the weight of an elementor a compound which would combine with or displace (by weight) 1 part of hydrogen or 8 parts ofoxygen or 35.5 parts of chlorine.

Analytically, Equivalent weight (E) is defined for elements/ions/compounds as :

Molecular WeightEx

where x is known as Valence factor or ‘n’factor and its value varies depending upon the compound beingconsidered.

(i) Equivalent Weight (E) of an Element :

Atomic weight of an elementEValency of element

E(Aluminium) = 27 93 E(magnesium) =

24 122

E(oxygen) = 16 82 E(chlorine) =

35.5 35.51

(ii) Equivalent Weight (E) of an Acid :

E = Molecular weight of acid

Basicity

Acid : A species capable of giving H+ ions (protons) in its aqueous solution is an acid. It is generally representedas HA. The number of H+ ions furnished by an acid determines the basicity of acid.

Basicity of HCl (Hydrochloric acid) = 1 (As it furnishes only 1 H+ ion) Basicity of H2SO4 (Sulphuric acid) = 2 Basicity of H3PO4 (Phosphoric acid) = 3 Basicity of H2C2O4 (Oxalic acid) = 2

Note : It is not necessary that basicity of an acid is equal to the number of H+ ions contained in its formula.It is the number of furnishable H+ ions which determines the basicity of an acid.

Section 14

Stoichiometry - 1

Basicity of CH3COOH (Acetic acid) = 1Basicity of H3PO3 (Phosphorus acid) = 2

This can be deduced from its structure shown ( ) :

As it can be seen that number of –OH groups are 2, the number offurnishable H+ ions will also be 2.

Note that bond between P and –OH will not break as it is a stronger bond compared to the bond strengthbetween H and O in –OH group. So, H3PO3 is not a base although it contains 2 –OH groups. The Hattached directly to P atom will not be able to furnish in the solution as this bond is quite strong.

On the similar grounds, try to calculate the basicity of H3PO2 (Hypophosphorus acid).Its structure is shown on right..

(Ans: Basicity of H3PO2 = 1)

Also, note the structure of H3PO4 (Phosphoric acid) and its basicity (= 3).

HCl36.5E 36.5

1 2 4H SO

98E 49.02

2 2 4(oxalic acid)

H C O90E 45.02

3 4H PO98E 32.63

3 3H PO82E 41.02

3CH COOH60E 601

(iii) Equivalent Weight (E) of a Base:

E = Molecular weight of base

Acidity

Base:A species capable of giving OH ions (hydroxyl) in its aqueous solution. It is generally represented asBOH. The number of OH ions furnished by a base determines the acidity of base.

Acidity of NaOH (Sodium hydroxide) = 1 Acidity of Mg(OH)2 (Magnesium hydroxide) = 2 Acidity of Al(OH)3 (Aluminium hydroxide) = 3

NaOH40E 40.01

2Mg(OH)58E 292

2Ca(OH)

74E 37.02

(iv) Equivalent Weight (E) of an Ion :

E of an ion is determined by the charge on an ion irrespective of anion and cation.

E of an ion = Molecular weight of ion

Magnitude of charge on ion

Vidyamandir Classes

Section 1 5

Cl35.5E 35.5

1 2O16E 82 3Al

27E 9.03

60E 30.02 3

4PO95E 31.663 2

2 4(oxalate)C O

88E 44.02

(v) Equivalent Weight (E) of a Compound :

Equivalent of a compound (E) depends upon how a given compound reacts in a particular reaction.It means that equivalent weight can be variable. However, there are certain compounds for whichequivalent weight remains constant. The equivalent weight of alkali metal salts (Na, K etc) and alkaline-earth metals salts (Mg, Ca etc) is constant. It depends upon the magnitude of total charge on cation oranion.

E = Molecular weight

Magnitudeof totalchargeon cation or anion

2 3Na CO106E 53

2 (Magnitude of total charge on cation or anion = 2)

2 3 3Al CO234E 39

6 (Magnitude of total charge on cation or anion = 6)

(vi) Equivalent weight (E) of an Oxidising and Reducing Agent :

For the compounds taking part in redox reactions, the calculation of E becomes rather complex andone has to be very cautious. First, one should make sure which compound is acting as oxidising agent(O.A) and which is acting as reducing agent (R.A). Then find the number of electrons transferred byone mole of O. A or R.A in the reaction.

E of compound = Molecular weight

Electron transfer per mole of O.A or R.A

Note : We will discuss this concept in the study of Redox Reactions in more detail in upcoming modules.

gm. eq. in 71 gms of Chlorine atoms :71gmeq. 2

35.5 Cl(E 35.5)

gm. eq in 16 gms. of Oxygen atoms :g 16gm.eq 2E 8

O(E = 8)

gm. eq in 500 gms of H3PO4 :g 500gm.eq 15.3E 32.67

M(E )basicity

Section 16

gm. eq in 150 gms of Mg(OH)2 :g 150gm.eq 5.17E 29

M(E )acidity

Note : In practical cases, a smaller unit of moles and gram equivalent is used and is given by :

The number of milli moles (m.moles) in g grams = 0

g 1000M

The number of milli equivalents (meq) in g grams = g 1000E

Relationship between Moles and Gram Equivalents :

Gram equivalents = 0 0

g g gxE M / x M = x × moles

Expressing Concentration of Solutions :

Solution is a homogenous mixture of two or more components in which intermingling particles are of atomicor molecular dimensions. A solution consists of a dissolved substance known as solute and the substance inwhich the solute is dissolved is known as solvent. The concentration of a solution means the quantity ofsolute dissolved per unit volume of solution, or per unit quantity of solvent.

Concentration of solute Amount of soluteAmount of solution (or solvent)

Note : While discussing various methods for expressing concentration, we have taken solute as B dissolved insolvent as A and gB as grams of solute and gA as grams of solvent.

1. Mass fraction is the fractional part of a component that is contributed by it to the total mass of solution.

mass fraction of B(W ) B = B

mass fraction A(W ) A = A

Note : B BW W 1

2. Mole fraction is the fractional part of the moles that is contributed by each component to the total numberof moles that comprises the solution. In a solution containing nA moles of solvent and nB moles of solute ;

mole fraction of B B = B

mole fraction of A A = A

Note : A B 1.

Section 1 7

A given solution consists of 230 gm of C2H5OH (ethyl alcohol) and 720 gm of H2O (water). How todetermine the mole fraction of each component.

First, calculate moles of ethyl alcohol and water.

moles of C2H5OH = 230 5.046

and moles of H2O = 720 40.018

Now let A = mole fraction of H2O and B = mole fraction of C2H5OH

A40 0.88

and B A1 0.12

3. Molality (m) is expressed as number of moles of solute dissolved in 1000 gms (1.0 Kg) of solvent. It isdenoted by m. The unit is mol/kg. It can also be denoted by m.

i.e. moles of solutemkg of solvent

If nB represent moles of solute and gA represent gms of solvent, then, m = B

n 1000g

How to determine the molality of solution containing 87.7 gm of NaCl (sodium chloride) dissolvedin 1500 gm of water ?

First, calculate moles of NaCl :g 87.7moles 1.5M 58.5

Now, molality (m) = B

n 1000 1.5 1000 1.0mg 1500

4. Molarity (M) is expressed as moles of solute contained in one litre of solution or it is also taken as millimolesof solute in 1 cc (ml) of solution. Its unit is mol/ t. It is also denoted by M.

Molarity (M) = moles of solute millimoles of solute=litres of solution millilitres of solution

If nB represents number of moles of solute and VL be volume of solution in litres.

M = B B B

n g MV V

Note : moles of solute = Molarity Volume in litres nB = MVL (M : molarity).

or m.moles of solute = Molarity Volume in cm3 (ml) = MVcc1 mL = 1 millilitre, 1 cc = 1 ml = 1 cubic centimetres

Section 18

How to determine the molarity of a solution containing 149 gram of KCl (potassium chloride)dissolved in sufficient water to make 1500mL of a solution.

moles of KCl gMolarity where moleslitres of solution M

149moles of KCl 2.074.5

2.0Molarity 1.33molar

1500 /1000 or written as 1.33 M

How to calculate grams of solute (solid NaOH) in a 500 cc of 0.25M solution of NaOH ?Moles = MVL M : molarity of solution;VL : volume of solution in litres.

Moles 5000.25 0.1251000

gram of NaOH = 0.125 40 = 5 gm (mass = nM0).

5. Normality (N) is expressed as the number of gram equivalents (gmeq) of solute contained in one litre ofsolution or it can also be taken as number of mill equivalents (meq) in 1cc (mL) of solution. It is denoted by N.

Normality of solution (N) =gmeq of solute meq of solute

litres of solution millilitres of solution

If gB represents grams of solute, E represents Equivalent weight of solute and VL be volume of solution inlitres,

g / EgmeqNormalityV V

[A] (a) gm.eq. of solute = normality volume in litres = NVL

or meq of solute = normality volume in cm3 (mL) = NVcc

(b) moles of solute = Molarity VL = MVLor m.mol of solute = Molarity VmL = MVmL

[B]L L

gm.eq molesNormality MolarityV V

For acidic solutions : N = x M [x : basicity of acid ; M : molarity of acid]For basic solutions : N = x M [x : acidity of base ; M : molarity of base]

For a mono acidic base (acidity =1) and mono basic acid (Basicity = 1) Normality = Molarity

[C] Relation between molality(m), molarity (M), density(d) of solution and molar mass of solute (M0) :Molarity (M) means M moles of the solute in 1 L of the solution. If density is in g/mL and M0 is molarmass in 1g mol , then

Important Concepts in Stoichiometric Calculations :

Section 1 9

mass of solute = MM0 grams ; mass of solution = 1000 × d grams

mass of solvent = 01000d MM

Molality (m) = 0

M 10001000d MM

[D] Relationship between molality (m) and mole fraction (B) of the solute :

n nm 1000 1000g n M

m n Mn1000

or A AB

m n M /1000n m n M /1000

m M1000 m M

B A A A

11000 1000m1 M M

Illustrating the Concept :A solution was prepared by adding sufficient water to 100g of NaOH to make 0.50 L of solution.Calculate molarity and normality of solution.

gmeq 100 / 40Normality 5 NV 0.5

0Mggmeq , EE acidity

Now using N = x M ;

5M 5M solution1

[x = 1 for NaOH (Mono-acidic base)]

A solution was prepared containing 14.80 gm of Ca(OH)2 in 3000 ml of solution. Calculate:(a) molarity of solution (b) normality of solution(c) moles in a 2.5 L of same solution (d) gmeq contained in 1.5L of solution.

moles of Ca(OH) 14.8 / 74Molarity 0.067MV 3000 /1000

(b) 02

Mgm eq of Ca(OH) 14.8 / E 14.8 / 37 74Normality 0.133N EV 3 3 acidity 2

(c) moles in 2.5 L of sample = MVL = 0.067 2.5= 0.167 moles (in 2.5 L sample).(d) gmeq in 1.5 L of sample = NVL = 0.133 1.5 = 0.2 gmeq (in 1.5 L).

6. Strength of a solution : It is expressed as grams of solute in 1 L of a solution.

grams of soluteStrengthV

We know, LgNV gmeqE

Section 110

gStrength NEV

Similarly, 0L

gStrength MMV

How to calculate normality when Strength is given ?(i) HNO3 containing 6.3 g/L of acid(ii) KOH solution containing 11.2 g/L of base

(i)3HNO

63E 631

6.3N 0.1N63

strengthNE

(ii) KOH56E 561

11.2N 0.2N56

How to calculate E when Strength is given ?

(i) 10N HCl acid solution containing 3.65 g/10 ml(ii) N/3.5 H2SO4 acid solution containing 14 g/L.

StrengthEq. wtNormality

(i) HClE ? Strength = 3.65 g /10 ml = 365 g /L

HCl365E 36.510

(ii) Similarly, 2 4H SO

14E 491/ 3.5

1. If molarity of a solute Ax By in a solution is ‘M’, assuming the solute to have complete dissociation (100%)Ax By x AAy+ + y Bx–

Then, molarity of Ay+ ions = x M ; molarity of Bx– ions = y M

1mole A B gives

moles of A and

moles of B

But, same is not applicable to normality.

If normality of solute Ax By is ‘N’, assuming the solute to have complete dissociation (100%) :Ax By x AAy+ + y Bx–

Then, normality of Ay+ ions = N ; normality of Bx– ions = Ngmeq.of A B gmeq.of A

gmeq.of B

(Correlate the above results, with the definitions of molarity and normality respectively).

Section 1 11

Illustrating the Concept :0.1 M H2SO4 0.2 M H+ 0.1 M 2

4SO 0.1 N H2SO4 0.1 N H+ 0.1 N 24SO

2. When Solute is Solid :

(a) x% by mass of solute (B) in the solution (A): It means x gm of solute (B) are present in 100 gmof solution (A)

(b) x% by volume of solute (B) in the solution (A): It means x gm of solute (B) are present in100 cm3 of solution (A).

3. When Solute is also a Liquid :(a) x% by mass of solute (B) in the solution (A): It means x gm of solute (B) are present in 100 gm

of solution (A)(b) x% by volume of solute (B) in the solution (A): It means x cm3 of solute are present in 100 cm3

of solution (A).

The molarity and normality of solution of H2SO4 containing x% H2SO4 by weight.(dsolution = d gm/cc.)

First, you must understand meaning of x% H2SO4 by weight.

This means x gms of actual amount of H2SO4 in 100 gms of solution or if W (in gms) be the total

mass of solution, then the mass of solute (H2SO4) contained = W gms100

Now coming back to original problem ;

Let us consider 1 L or 1000 cc of solution

Mass of solution in 1 L = 1000 dmassd

volume

Mass of actual H2SO4 in 1 L 1000d 10 d100

moles of H2SO4 in 1 L = 0

gmolesM

10 dMolarity MM

For normality, M0 is replaced by E

10 dnormality NEx

Section 112

Illustration -1 (a) Concentrated acid H2SO4 has a density of 1.8 g/ml and contains 49% acid by weight.Compute molarity of the solution. Also calculate the number of gmeq of H2SO4 contained in 1 L solution.

(b) What is the normality of a solution which is prepared by dissolving 100 ml of conc. H2SO4 inpart (a) in sufficient water to make 500 ml of solution?

(c) If we take 50 ml sample of above solution [in part (b)], find number of milli moles and milliequivalents in the sample.

Solution : (a) We have :0

10 dMolarityM

10 49 1.8molarity(M) 9.0M98

(x = 49, d = 1.8 g/mL = 1.8 g/cc, M0 = 98)

To calculate gmeq, it is better to calculate normality first.N = x M = 2 9 = 18 N x : Basicity of H2SO4 = 2gm.eq. = NVcc = 1 × 18 = 18 gmeq are contained in 1 L of given solution.

(b) N of H2SO4 = 18 [calculated above in part (a)]

gmeq in 100 cc of conc. H2SO4 18.0 100 1.81000

So, 1.8 gmeq of concentrated acid are to be added in water to make 500mL solution.

Normality of required solution = L

gmeq 1.8N 3.6V 500 /1000

(c) Milli equivalents = Normality x Volume (in ml) = NVCC = 3.6 × 50 = 180 meq.

Millimoles = Molarity × Volume (in ml) = MVCC

= 3.6 50 902 m.moles [N = x M and x = 2 for H2SO4, a dibasic acid]

Note : If a naturally occurring atom exists in the form of isotopes (say of molecular masses A1, A2, . . . . . etc) withtheir percentages as (x1 %, x2 % . . . . . . . etc), then average atomic mass, avgA of the atom is given as :

avg1 2

1 2A A A ............100 100x x

The mass spectrum of carbon shows that 98.892% of carbon atom are C-12 with a mass of 12.000amu and 1.108% are C-13 with a mass of 13.000335 amu. Calculate the atomic weight of naturallyoccurring carbon.

Atomic weight of carbon = 98.892 1.10812.000 13.00335

100 100 = 12.011 amu

Section 1 13

If isotopic distribution of isotopes (relative abundance) C-12 and C-14 is 98% and 2% respectively then the number of C-14 atoms in 12g of carbon is :

(A) 221.20 × 10 (B) 223.01 × 10 (C) 235.55 × 10 (D) 236.023 × 10

Average atomic mass = 98 212 14 12

100 100

Total No. of C atoms in 12g of 23 2312C 6.023 10 6.023 1012

And Total No. of C-14 atoms in 12g of C = 232 6.023 10100

= 21 2212.046 10 1.20 10 (A)

There are two isotopes of an element with atomic mass z. Heavier one has atomic mass z + 2 andlighter one has z – 1, then abundance of lighter one is :(A) 66.6% (B) 96.7% (C) 6.67% (D) 33.3%

Let x% be abundance of lighter one. Then (100 x) x(z 2) (z 1) z So, x=66.6 100 100

1. Dilution : Whenever a given solution of known concentration i.e. normality and molarity (known as standardsolution) is diluted (adding solvent), the number of millimoles (or milli equivalents) of solute remain unchanged.The concentration of solution however changes.In such a case if :

N1 = normality of original solution ; V1 = volume of original solution

and N2 = normality of diluted solution ; V2 = total volume of diluted solution

Since the number of milli equivalents remains same,

N1V1 = N2V2 (This also called equation of normality)

Also, M1V1 = M2V2 (This also called equation of molarity)

2. Whenever a small sample (volume) is taken from a standard solution, the concentration of sample is same asthat of standard solution. However number of millimoles (or milli equivalents) in sample differs from that ofstandard solution.From a 3.5 L of 0.5 N H2SO4 solution, a sample of 500mL is taken, then normality of a 500 mL solution isalso 0.5N.But meq in standard solution = 0.5 3500 = 1750 meqand meq in small sample = 0.5 500 = 250 meq

Section 114

Illustration - 2

Illustration - 4

Illustration - 3

What volume of water must be added to a 0.5 litre of 10 N acid solution to make itexactly 0.5 N solution?

Solution : Let Vcc of water is added to given solution of acid.

Apply equation of normality (for dilution) i.e. N1V1 = N2V2

V vol.of originalsolution 0.5L 500cc ; N normalityof solutionV final volumeof solution ; N normality of finalsolution

10 500 = 0.5 (500 + Vcc) Vcc = 9500 cc = 9.5 L.

9.5 L of H2O will have to be added.

A commercial sample of oxalic acid is labelled as 22.5% H2C2O4 by weight (density =1.5 g/cc). Calculate (a) molarity (b) volume of acid having same amount of solute as in 1 L of0.2 M H2C2O4.Solution :

(a) Using standard formula : molarity = 0

10 22.5 1.5M 3.75 M90

(b) Let Vcc of acid is required.

m.moles of concentrated sample = m.moles of desired sample

Using M1V1 = M2V2 ; 3.75 × Vcc = 0.2 × 1000 Vcc = 53.34 cc

A 0.25 molar aqueous solution of NaOH is found to have a density of 1.26 g/cc.Determine its molality with respect to NaOH. Also determine its mole fraction.

Solution :0.25 M NaOH 0.25 moles/litre of NaOH

0.25 moles of NaOH in 1 L of solution

Use :moles of solutem 1000gms of solvent

First find the mass of water.

Mass of solution = Volume density

Mass of solution in 1.0 litre solution

= 1000 1.26 = 1260 gm

Now mass of NaOH = (moles M0)

= 0.25 40 = 10 gm

Mass of H2O = 1260 10 = 1250 gms

Hence molality (m) =0.25 10001250

= 0.2 m

For conversion of molality into mole fraction, use thefollowing relation.

Section 1 15

1000m1 M

B = mole fraction of solute

MA = molecular mass of solvent.

Note : TRY to derive the above relation yourself.

In given case : m = 0.202, MA = 2H OM 18

10000.21 18

Solve to get: NaOH = 0.00358

Illustration - 5 An aqueous solution of ethyl alcohol (C2H5OH) is found to be 10/9 molal. If the densityof the alcohol solution is 0.8 g/cc, find mole fraction and molarity of solution with respect to ethyl alcohol.

Illustration - 6

Solution :

Use m = B

10001 M

( = mole fraction of alcohol)

10 10009 1 18

= 0.0196

molal solution 109

mol of C2H5OH in

1000 gms of H2O.For molarity, find volume of solution and forvolume of solution, first calculate mass ofsolution.

Mass of solution= mass of solute (C2H5OH) + mass of water.

= (10/9 46) + 1000

= 51.1 + 1000 = 1051.1

Volume (in cc)mass 1051.1 1313.89cc 1.314L

density 0.8

= 1.314 L

moles 1019 10Molarity 0.845MV 1.382 11.825

One litre solution of 0.5N is heated. The volume of the solution is reduced to 750ccand 2.675 gm of HCl is lost. Calculate :(i) normality of the resultant solution.(ii) number of meq of HCl in 100 cc of the original solution.

Solution :

(i) First calculate grams of HCl in original solution

Use: LgNVE

g0.5 136.5 /1

g = 18.25 gm

Now wt. of HCl lost due to heating = 2.675 gm

wt. of HCl in resultant solution

= 18.25 2.675 = 15.575 gm

Now normality of new solution, N :

g / E 15.575 / 36.5N 0.57NV 750 /1000

(ii) meq in 100 cc of original sample

meq = NVcc = 0.5 100 50 meq

Section 216

STOICHIOMETRIC CALCULATIONS Section - 2

In this section, we will discuss the problems based on balanced chemical equations and application of moleconcept.

The analysis of a chemical reaction is generally carried out in the form of mass of reacting species taking partin a given reaction (gravimetric analysis) or in terms of concentrations of reacting species taking part in agiven reaction (volumetric analysis).

In Gravimetric Analysis, we generally analyse reactions such as : decomposition of compounds under heat toproduce a residue and a gas, or displacement reactions, action of acids on metals, or simple balancedchemical equations involving Weight (solid) Volume (gas) relationships.

In Volumetric Analysis, we generally analyse Neutralisation and Redox Titrations involving aqueous solutionsin general.

Neutralisation :A reaction in which an acid (or a base) completely reacts with a base (or an acid) to form salt and water iscalled as Neutralisation. If HA be the acid, BOH be the base and BA be the salt, then neutralisationreaction can be represented as follows :

HA + BOH BA + H2O

Redox Reactions :A reaction in which both oxidation and reduction takes place simultaneously is called as Redox reaction. Aredox reaction always involves a pair of oxidising agent and a reducing agent.

Note : We will deal with Redox Reactions in the next chapter.

Before we move on to study the balanced chemical equations, let us first analyse meaning of some terms.

Moles of NaCl : NaCl Na+ + Cl

1 mole 1 mole 1 mole 1 mole of NaCl contains 1 mole of Na+ (ion) & 1 mole of Cl ion

2 moles of NaCl contains 2 moles of Na+ ion & 2 moles of Cl ion

Moles of CaCl2 : CaCl2 Ca2+ + 2Cl

1 mole 1 mole 2 mole 4 moles 4 moles 8 moles

x-moles of H2SO4 : H2SO4 2H+ + SO42

1 mole 2 moles 1 molex moles 2x moles x moles

Application of Mole Concept on Balanced Chemical Equations :

Consider a general balanced chemical reaction or equation :

m A + n B p C + q D

where A and B are reactants; C and D are products; m, n, p, q are the stoichiometric coefficients.

The above balanced reaction is analysed as :

m moles of A react with n moles of B to produce p moles of C plus q moles of D.

This can be represented (written) as :

m moles of A n moles of B p moles of C q moles of D

Illustrating the mole concept :

What weight of KCl (Potassium Chloride) will be formed on heating 12.25 gm of KClO3 ?Also calculate weight of O2 will be liberated.

First write a balanced chemical equation for decomposition of KClO3.

3 2(2moles) (3 moles)(2 moles)

2KClO 2KCl 3O

From Stoichiometry, we have :

2 moles of KClO3 2 moles of KCl 3 moles of O2

moles of KClO3 = 12.25 0.1122.5

0(KCl)

M (KClO ) 122.5gmoles

MM 74.5

Now, 2 moles of KClO3 2 moles of KCl

0.1 moles of KClO3 0.1 moles of KCl

00.1 74.5 gmof KCl [ g moles M ]

7.45 gms of KCl are formed.

Similarly, 2 moles of KClO3 3 moles of O2

3 230.1 molesof KClO 0.1moles of O2

23 0.1 32 gms of O2

= 4.8 gm of O2.

Section 2 17

1. Some Important Anions used in Stoichiometry

X halides Cl :Chloride, Br : Bromide I : Iodide

2O Oxide 2S Sulphide3N Nitride 3P Phosphide

23CO Carbonate 3HCO Bicarbonate

3NO Nitrate 2NO Nitrite24SO Sulphate 4HSO Bisulphate

23SO Sulphite 3

4PO Phosphate

22 4C O Oxalate 2 4HC O Bioxalate

22 3S O Thiosulphate 2

4 6S O Tetrathionate

3ClO : Chlorate ; 3BrO : Bromate ; 3IO : Iodate

4MnO Permanganate 22 7Cr O Dichromate

2. Action of Heat on Some Important Compounds :Alkali Metals Alkaline Earth Metals

Carbonate Stable Unstable

2 3Na CO No Reaction 3 2CaCO CaO CO

Bicarbonates Unstable Unstable

3 2 3 2 22NaHCO Na CO CO H O 3 2 22Ca HCO CaO H O 2CO

Sulphates Stable Stable

2 4Na SO No Reaction

3. Some Basic Chemical Equations :

(i) 3 22 KClO 2 KCl 3 O (ii) 3 2 3 2 22 NaHCO Na CO CO H O

(iii) 3 2 21NaNO NaNO O2

(iv) 3 2 222 Pb NO 2 PbO 4 NO O

(v) 2 2 3 2 2 4 62 Na S O I Na S O 2 NaI (vi) 2 2 4 2 3Na C O Na CO CO

(vii) 4 2 3 2 32 FeSO Fe O SO SO (viii) 3 2 22 AgNO 2 Ag 2 NO O

2 Mg O 2 MgO

3 Mg N Mg N

Calcium behavessimilarly

(x) 4 3 2 2

4 2 2 2

NH NO N O 2H O

NH NO N 2H O

Section 218

Vidyamandir Classes

A flash bulb used for taking photograph in poor light contains 30 mL of O2 at 780 mmpressure at 27C. Suppose that metal wire flashed in the bulb is pure Aluminium (Al) and it is oxidised toAl2O3 in the process of flashing, calculate the minimum weight of Al-wire that is to be used for maximumefficiency.

Solution :Al is oxidised to Al2O3 as follows.

4 Al + 3 O2 2 Al2O3

First, calculate the moles (n) of O2.

Use PV = nRT

PV (780 / 760) (30 /1000)nRT 0.082 300

= 1.25 103

A mixture of NaCl and Na2CO3 is given. On heating 12 gm of the mixture with diluteHCl, 2.24 gm of CO2 is evolved at normal temperature. Calculate the amounts of two components in themixture.

Illustration - 7

Illustration - 8

PROBLEM SOLVING TECHNIQUE (Using Mole Concept)Follow the given sequence :1. First write a balanced chemical equation.2. Analyse the reactants and products according to their respective stoichiometric coefficients i.e.

i.e. m mole of A n moles of B p moles of C q moles of D3. According to data given, proceed as explained in the above illustration4. In stoichiometric problems involving gases, we will assume all the gases to be ideal and apply

Gas Equation : PV = nRT.(P : Pressure of gas in atm units, V : Volume of gas in L and T : Temperature in Kelvin and R : Universal gasconstants = 0.0821 atm L mol1 K1)

(a) At S.T.P., the volume occupied by 1 mole of a gaseous compound is equal to 22.4 L (or 22400 mL). Thisis also called as Molar volume.

(b) At room temperature, T = 298 K, P = 1.0 atm., volume of 1 mole of gas is 24.48 L.(c) N.T.P. (Normal temperature and pressure) condition is same as S.T.P. condition.

From stoichiometry of the reaction :

3 mole of O2 4 mole of Al

1.25 × 103 mole of O2 43 (1.25 × 103) mole of Al

gram of Al = 43 (1.25 × 103) 27 = 0.045 gm

Solution :Consider the effect of heat on the mixture.

NaCl + HCl nothing happens

Na2CO3 + 2 HCl 2 NaCl + CO2 + H2OLet x grams of Na2CO3 are there in the mixture.

moles of Na2CO3 = 0

gM 106

From stoichiometry, we have :

1 mole of Na2CO3 1 mole of CO2

moles of Na2CO3 = 106x

mol of CO2

106 44

xor x = 5.4 gm.

Hence grams of NaCl in the mixture= 12 x = 6.60 gm

Section 2

Section 2 19

Vidyamandir Classes

A mixture of FeO and Fe3O4 when heated in air to a constant weight, gains 5% of itsweight. Find the composition of the initial mixture.

Solution :In such type of problems, in order to simplify thecalculations, let us assume that the initial weight of themixture is 100 gm.

Then the final weight of the mixture after heating in air willbe 105 gm.Let x be the weight of FeO in the initial mixture, then theweight of Fe3O4 = 100 x

When the mixture is heated in air (O2):

4 FeO + O2 2 Fe2O3

4 Fe3O4 + O2 6 Fe2O3

4 moles FeO 2 moles of Fe2O3

moles 144x

moles of Fe2O3

Illustration - 9

4 moles of Fe3O4 6 moles of Fe2O3

232– x

moles 6 1004 232

– xmoles of Fe2O3

Total moles of Fe2O3 = 6 100

144 4 232

x – x

Weight of Fe2O3 =6 100

144 4 232

x – x160 = 105

Solving for x, we get ;

x = 20.25 gm = weight of FeO

weight of Fe3O4 = 100 – x = 79.75 gm.

% FeO = 20.25 and % Fe3O4 = 79.75

760 / 760 0.328n0.082 300

= 0.0133

The amount of Zn in the alloy

= 0.0133 65.4 = 0.87 g

or % Cu = 5.13 100 85.5%6.0

Illustration -10 Brass is an alloy of Cu-Zn. A sample of brass weighing 6.0 g, when treated with excessof dil. H2SO4 gives 328 mL of dry H2 at 27oC and 760 mm pressure. What is the percentage of Cu by weightin the alloy? (Atomic mass of Zn = 65.4 gm)

Solution :

Only Zn reacts with dil. H2SO4 and produces H2,whereas Cu does not.

Zn(s) + H2SO4 ZnSO4 + H2 (g)1 mol of Zn 1 mol of H2

moles of H2 evolved = PVnRT

Illustration -11 Calculate the weight of iron which will be converted into its oxide by the action of 18gof steam.Solution :

Remember that iron reacts with steam, itforms 3 4Fe O

2 3 4 23Fe 4H O Fe O 4H From stoichiometry of the reaction :

3 moles of Fe 4 moles 2H O

218Moles H O 118

3 3Moles Fe 14 4

3Weight of Fe 56 42g4

Section 220

Vidyamandir Classes

Important Concepts in Stoichiometric Calculations : 1. (a) Finding the Equivalent Weight of Basic Compounds :

Oxides, Carbonates and Bicarbonates of alkali metals and alkaline earth metal are basic in nature e.g.MgO, Na2CO3 and NaHCO3 are basic.The n-factor (acidity of carbonates, bicarbonates & oxides) is equal to the magnitude of total charge oncations or anions. It can also be confirmed by visualizing the neutralisation reaction as follows:

Na2CO3 + 2 HCl 2 NaCl + H2O + CO2

The above reaction shows that 1 mole of Na2CO3 requires two moles of H+ ions. Thus, its equivalent weight

is 106

2 i.e. 53 gm/mol. [Note: Acidity also refers to the number of H+ ions required to react with the base.]

(b) Finding the Equivalent Weight of Acidic Compounds :The oxides of non-metals are acidic in nature, such as SO2 & SO3 are popular acidic oxides used inneutralisation. The n-factor (basicity) can be determined as follows:

SO3 + 2 NaOH Na2SO4 + H2OThe above reaction shows that 1 mole of SO3 requires two moles of ions. Thus, its equivalent weight is80/2 = 40 gm/mol. Note : Basicity also refers to the number of OH- ions required to react with the acid.

2. Concept of Limiting Reagent :Whenever, the amount of two reacting species (reagents) is given, very rarely they react completely (unlesscalculated beforehand). In such cases, one of reagent reacts completely (as per Stoichiometry of the reaction)leaving behind the other in excess. The reagent which reacts completely is known as Limiting Reagent.Read the following example carefully to understand the concept.

Illustrating the Concept:

A closed vessel is found to certain 192 gm of Mg and 96 gm of 2O . This mixture is burnt.(a) Which is the limiting reagent ?(b) Find the weight of reagent in excess (the other one) ?

22Mg O 2MgO

2 moles of Mg 1 mole of 2O 2 moles of MgO

Initial moles of 192Mg 824

Initial moles of 296O 332

1 mole of 2O 2moles of Mg

223moles of O 3 6 moles of Mg1

Mg is in excess.Hence O2 is the limiting reagent as it is consumed fully.

Excess of Mg = 8 – 6 = 2 moles Grams of excess Mg = 2 24 = 48 gmmoles of MgO formed

= moles of Mg consumed = 6

Grams of MgO formed = 6 40 =240 gm

Read the following Illustrations carefully to understand the concept of limiting reagent.

Section 2 21

Vidyamandir Classes

Illustration -12 Consider the reaction : 2A + 3B 4C + 5DIn the above reaction A and B are reactants and C and D are products. If one mole of each of A and Bare reacted. Then:I. 2.25 mole of of D is formed II. 1.6 mole of D is formedIII. 0.33 mole of A are left after complete reaction IV. 1.33 mole of C is formedThe correct choice is :(A) I, II (B) II, III, IV (C) I, III, IV (D) I, II, IV

Solution : (B)

2A 3B 4C 5D 2 moles of A 3 moles of B 4 moles of C5 moles of D [From Stoichiometry]Initial A Bn n 1. [For 1 mole of A,1.5 moles of B are needed. But only 1 mole of B is given. Hence, B is the limiting reagent]

A B C D2 1 4 5n 1 0.33 ; n 1 1 0 ; n 1.33 ; n 1.663 3 3 3

In the following reaction : 2 2 2 2MnO + 4HCl MnCl + 2H O + Cl .When 2 moles

of 2MnO reacted with 4 moles of HCl, 11.2 L 2Cl was collected at STP. Find the percent yield of 2Cl .(A) 25% (B) 50% (C) 100% (D) 75%

Solution : (B)

2 2 2 2MnO + 4HCl MnCl + 2H O + ClFrom Stoichiometry 1 mole 4 moles 1 mole 2 moles 1 mole

Initial (Given) 2 moles 4 moles [For 2 moles of 2MnO , 8 moles of HCl are needed,but only 4 moles of HCl are given. Hence, HCl is the limiting reagent]Finally 1 mole 0 mole 1 mole 2 moles 1 mole = 22.4 L at STP (Theoretical yield)

% yield of 211.2Cl 100% 50%22.4

What is the number of moles of 3Fe OH s that can be produced by allowing 1 mole

of ,2 3Fe S 2 moles of 2H O and 3 moles of 2O to react as : 2 3 2 2 32Fe S + 6H O + 3O 4Fe OH + 6S ?(A) 1 mol (B) 1.84 mol (C) 1.34 mol (D) 1.29 mol

Solution : (C)

2 3 2 2 32Fe S 6H O 3O 4Fe OH 6S

From Stoichiometry 2 moles 6 moles 3 moles 4 moles 6 molesInitial 1 moles 2 moles 3 moles[For 1 mole of 2 3Fe S , 3 moles of 2H O are needed & 1.5 mole of 2O . But only 2 moles of 2H O are

given and 3 moles of 2O are given. Hence, 2H O is the limiting reagent.]Finally 1/3 mole 0 moles 2 moles 4/3 moles 2 moles

Illustration -13

Illustration -14

Section 222

NEUTRALISATION Section - 3

As we know that an acid may be monobasic (HCl, HNO3 etc.), dibasic (H2SO4, H2C2O4 etc.) or tribasic(H3PO4 etc.) and similarly a base may be monoacidic (NaOH, NaHCO3 etc.), diacidic [Ca(OH)2, Na2CO3etc.] or triacidic [Al(OH)3 etc.], so it is better to define the neutralisation reaction in the following manner:

A reaction in which 1 gram equivalent (or 1 meq) of an acid (or a base) completely reacts with 1 gramequivalent (or 1 meq) of a base (or an acid) to form 1 gram equivalent (or 1 meq) of corresponding salt.

A stage at which the process of neutralisation is complete is known as end point or neutral point and theprocess carried out to study neutralisation of acids and base is called as Titration.At equivalence point :

gm eq. (or meq) of acid = gm eq. (or meq) of base

Na Va = Nb Vb (N : normality, V : volume in mL or L, a: acid, b : base)

This relation can be used in two ways (modified form) :

1. If ga grams of acid are neutralised by a base whose normality is Nb and volume required for neutralisation ofgiven acid is Vb (in cc), then

g1000 N V

basicity

2. If gb grams of a base are neutralised by an acid of normality Na and volume Va (in cc), thenb

g1000 N V

acidity

How many grams of borax (Na2B4O7.10H2O) are required to neutralize 25 ml of 0.2 M of HCl andH2SO4 separately.

Borax in water gives : 2B O 7H O 4H BO 2OH4 7 2 3 3 (Remember it as fact)

1 mol of borax 2 mol OH acidity = 2

(i) meq of borax = meq of HCl

g 1000 0.2 1 25382 / 2

[ HCl is monobasic acid]

g = 0.955 gm = mass of Borax(ii) meq of borax = meq of H2SO4

g 1000 0.2 2 25382 / 2

[ H2SO4 is diabasic acid] g = 1.91 gm = mass of Borax

Section 3 23

How many mL of 0.5 N HCl acid will be required to completely neutralise 500 ml of a 0.1 N NaOHsolution?Let VmL of acid is required for neutralisation. Using equation of neutralisation we have : NaVa = NbVb

0.5 Vcc = 0.1 500 Vcc = 100 cc = 100 mL

0.5 gm of fuming H2SO4 (oleum) is dilute with water. This solution is completely neutralised by30.0 mL of 0.4 N NaOH. Find the % age of free SO3 in the sample.

1. Salts formed by the reactions between a strong acid (like HCl, H2SO4, HNO3, H3PO4 etc. . . ) and astrong base like NaOH, KOH, Mg(OH)2, Ca(OH)2) when dissolved in water does not effect the acidityor basicity of the solution. For example : A solution having Na2SO4 (formed by the reaction betweenNaOH and H2SO4) as its solute is neutral as Na2SO4 is a salt of strong acid and strong base.

2. Back Titration : This concept comes into picture while analyzing the Neutralization in case any of the acidor base is found to be in excess (Over stepping of the end point).Both of the above concepts can be clearly understood on careful examination of the followingillustrations :

Oleum is (H2SO4 + SO3 = H2S2O7)Note that SO3 is acidic oxide and hence reactswith NaOH.

2 NaOH + SO3 Na2SO4 + H2O ;

i.e., it acts as a dibasic acid.

Let gms. of SO3 in 0.5 gm = x ;Then, gms of H2SO4 = 0.5 xAt neutralisation stage : meq of sample

= meq of NaOH

3 2 4SO H SO

0.5 1000E E

x x= 0.4 30

0.5 1000

80 / 2 98 / 2

x x= 0.4 30

x 0.4 gm = mass of SO3

% age of SO3 = 0.4 100 80%0.5

What volume of 3 4M H PO6 solution will completely react with 200 ml of a solution of

M2 aluminium carbonate ?

(A) 600 mL (B) 900 mL (C) 1200 mL (D) 1800 mL

Solution : (C)

Applying equation of neutralisation :

meq.of H3PO4 (acid) = meq of Al2(CO3)3 (base)

n-factor of Al2 (CO3)3 as base = 6

[Total charge on all cations or on all anions]

mL1 13 V 6 2006 2

[Meq. = n-factor mmoles]VmL = 1200 mL

Illustration -15

Section 324

Illustration -16

Illustration -17

Illustration -18

100 mL of mixture of NaOH and Na2SO4 is neutralised by 10 mL of 0.5 M H2SO4.Hence, and mass of NaOH in 100 mL solution is :(A) 0.2 g (B) 0.4 g (C) 0.6 g (D) None of these

Solution : (B)As Na2SO4 , neutral being a salt of strong acid and strong base, only NaOH will be neutralised by H2SO4.Applying equation of neutralisation : meq. of H2SO4 (acid) = meq of NaOH (base)

3NaOH2 0.5 10 10 1 n n(moles) 0.01

Mass of NaOH = 0.01 40 = 0.4 g1.0 g of the carbonate of a metal was dissolved in 25 mL of 1.0 N HCl. The resulting

liquid required 5 mL of 1.0N NaOH for neutralization. The equivalent weight of the metal carbonate is :(A) 50 (B) 30 (C) 20 (D) None of these

Solution : (A)Meq. of acid (HCl) used = 25 1 25 [Meq = n-factor mmoles]

Excess of meq. of acid (HCl) = Meq. of NaOH = 5 1 5 [Back titration]Meq. of acid used for metal carbonate = 25 – 5 = 20 = Meq. of carbonate

Meq of carbonate = 120 1000 E 50E

gMeq. 1000E

5.3g of M2CO3 is dissolved in 150 ml of 1 N HCl. Unused acid required 100 ml of 0.5 NNaOH. Hence equivalent weight of M is :(A) 53 (B) 46 (C) 2 (D) 23

SOLUTION : (D)gmeq of Acid (HCl) = gmeq of M2CO3 + gmeq of NaOH gmeq of M2CO3 = gmeq of Acid gmeq of NaOH = (150 1 10–3) – (100 0.5 10–3)

2 3M COn 2 0.1 ; 2 3M COn 0.05

M 2 3M CO

5.3M 2M 600.05

2M + 60 = 106 M = 23

100 ml solution of 0.1N HCl was titrated with 0.2 M NaOH solution. The titration wasdiscontinued after adding 30 ml of NaOH solution. The remaining titration was completed by adding 0.5M KOH solution. The volume of KOH required for completing the titration is:(A) 8ml (B) 16ml (C) 32ml (D) 64ml

SOLUTION : (A)

meq of HCl 0 1 100 10 .meq of NaOH soln 30 0 2 6 .meq of HCl remaining 10 6 4

For excess of HCl, now KOH is used.meq of KOH=meq of HCl remaining

0 5 4 . mLV 8 ( )mLV KOH mL

Illustration -19

Section 3 25

Vidyamandir Classes

5 mL of 8 N 3HNO , 4.8 mL of 5 N HCl and a certain volume of 17 M 2 4H SO are mixedtogether and made upto 2 L. 30 mL of the acid mixture exactly neutralises 42.9 mL of 2 3Na CO solutioncontaining 0.1 gm of 2 3 2Na CO .10H O in 10 mL of water. Calculate : (a) the volume of 2 4H SO added tothe mixture (b) the amount (in gm) of the sulphate ions in the solution.Solution :

meq. of acid mixture = meq of 3HNO + meq of HCl + meq of 2 4H SO

Let N be the normality of the acid mixture and CCV be the volume of 2 4H SO added.

CCN 2000 8 5 5 4.8 2 17 V ......(I)

Now find N of carbonate as follows : strengthNE

Strength = 0.1 g/10 ml 10 g/L ; E= 0M /2 = 286/2 = 143 ;( 0M 106 180 ; adding the mass of 210 H O)10N143

Now meq of acid mixture = meq of 2 3Na CO solution10N 30 42.9 N 0.1 normality of acid mixture143

Substituting in equation (I), we get : CC0.1 2000 40 24 34 V CCV 4cc For gms of sulphate ions :

meq of 2 4 CC CCH SO 2 17 V 136 V 4

Now, meq of 24 2 4SO meq of H SO

g g1000 136 1000 136 g grams of SO ions 6.53gmE 96 / 2

A sample of chalk 3CaCO is contaminated with calcium sulphate 4CaSO ; 1 gm ofthe solid mixture is dissolved in 230 mL of N/10 HCl ; 40 mL of N/10 NaOH is required to neutralise theexcess of acid. Find the percentage of chalk in the mixture.Solution :Here, note that 4CaSO does not react with HCl as it gives a netural solution in aqueous medium (a salt of strong

acid and strong base) whereas, 3CaCO being basic reacts with HCl.

Illustration -20

Illustration -21

So now it is a simple problem of neutralisation,Note that volume of HCl used against 3CaCO is notgiven. To calculate the volume of HCl for 3CaCO ,proceed as follows :

Excess of HCl is neutralised by NaOHmeq of excess of HCl = meq. of NaOH

First find the meq of NaOH used for excessof HCl.

meq of NaOH 1/10 40 4 meq of excess of HCl =

4 (def. of neutralisation)Now Find the initial meq of HCl taken.

Section 326

Vidyamandir Classes

So meq of 2 3Na CO (pure) in 50 mL = 3.4

meq of pure 2 3Na CO is 250 mL

250 g3.4 17 or 1000 1750 E

17 106 / 2g 0.901gm

2 3Na CO is a diacidic base E = 106/2

So mass of pure

2 30.901Na CO 100 90.1%

Illustration -22

Initial meq of HCl = (1/10)230 = 23 meq ofHCl used for neutralisation of carbonate

= 23 – 4 = 19

3meq of CaCO 19

1 gm of impure 2 3Na CO is dissolved in water and the solution is made upto 250 mL. Too50 mL of this solution, 50 mL of 0.1N HCl is added and the mixture after shaking well, required 10 mL of0.16 N NaOH solution for complete neutralization. Calculate % purity of the sample of 2 3Na CO .

Solution :1.0 gm of (impure)

2 3 2Na CO H O 250mL

2 350mL Na CO 50mLof 0.1N HCl

10mL of 0.16 N NaOH

In this question, HCl is in excess. excess meq of HCl = meq of NaOH

0.16 10 1.6

meq of HCl added to 2 3Na CO 0.1 50 5

meq of HCl used to neutralised 2 3Na CO

5 1.6 3.4

Yield of a Chemical Reaction:

In general, in any chemical reaction, the amount of product formed is always less than the calculatedamount due to reversibilities in the chemical reaction. Therefore, yield of a chemical reaction (Y) comesinto picture and is given by :

Actual yield of the productY 100 Actual yield % yield Theoretical valueTheoretical yield of the product

g 1000 19E

g 0.95 gm 95%

Section 3 27

Vidyamandir Classes

Illustration -233NH is formed in the following steps :

I. 2Ca 2C CaC 50 % yield

II. 2 2 2CaC N CaCN C 100% yield

III. 2 2 3 3CaCN 3H O 2NH CaCO 50% yield

Find the moles of calcium needed to produce 2 moles of ammonia

Solution :Actual yield = % yield (Theoretical value) ; Let x mole of Ca are needed.

From I : Yield of 2CaC 50% / 2 x x

From II : Yield of 2CaCN 100% / 2 / 2 x x

From III : Yield of 3NH 50% 2 / 2 / 2 x x

Given : Moles of ammonia 2 / 2 4 moles x x

USE OF DOUBLE INDICATORS IN NEUTRALISATION Section - 4

For studying the titrations of alkali mixtures such as NaOH and Na2CO3 ; NaOH and NaHCO3 ;Na2CO3 and NaHCO3, two indicators phenolphthalein and methyl orange are used.

Phenolphthalein is a weak orange acid and gives end point between pH range of 8-10, while methylorange, a weak base gives end point between pH range of 3-4.4.

When methyl orange is used as an indicator for studying the neutralisation titrations for abovemixtures, it indicates complete neutralisation for these, i.e. at the end point (colour change forindicator) the above mixtures are fully neutralised.

When phenolphthalein is used as an indicator for the above mixtures:

(a) it indicates complete neutralisation of NaOH (or KOH, i.e. strong alkali).

(b) it indicates half neutralisation of Na2CO3 (at the end point NaHCO3 is formed).

Note: It fails to indicate the neutralisation of NaHCO3 at all. Why? Because CO2 is the product of final neutralisation of NaHCO3.CO2 is acidic oxide and in acidic medium phenolphthalein fails.

So, in an analysis of such alkali mixtures, both the indicators are used:

one after the other in the same volumetric mixture

two indicators are used seperately in two different titrations.

Section 4

Vidyamandir Classes

FOR EXAMPLE :1. In the Neutralisation mixture of NaOH and Na2CO3 :(a) Use of methyl orange:

NaOH and Na2CO3, both are fully neutralised. i.e.milliequivalent (meq or gmeq) of acid used = meq of NaOH + meq of Na2CO3

(b) Use of Phenolphthalein :NaOH is fully neutralised and Na2CO3 is half neutralised

milliequivalent (meq. or gm.eq) of acid used = meq of NaOH + 2l

meq of Na2CO3

2. In the Neutralisation mixture of NaHCO3 and Na2CO3 :(a) Use of Methyl Orange :

NaHCO3 and Na2CO3, both are fully neutralised, i.e.milliequivalent (meq or gmeq) of acid used = meq of NaHCO3 + meq of Na2CO3

(b) Use of Phenolphthalein :Na2CO3 is half neutralised and NaHCO3 is not neutralised at all.

milliequivalent (meq or gmeq) of acid used = 12 meq of Na2CO3

A solution of NaOH and Na2CO3 is prepared. 25 ml of this solution required forneutralisation:(a) 25.0 mL of 0.10 N HCl when phenolphthalein is used as indicator.(b) 35.0 mL of 0.10 N HCl when methyl orange is used as indicator.Find the strength of NaOH and Na2CO3.Solution :

25mL, 0.1 N HCl

(Phenolphthalein) Experiment - I

35mL, 0.1 N HCl

(Methyl Orange) Experiment - II

(25 mL)

Illustration -24

Let m moles of NaOH be x and that of Na2CO3 be y in 25 ml 3.5 = x + 2y ...(ii) Experiment - I : Meq. of Acid = Meq. of Base (For complete neutralisation)

125 0.1 1 22

[ meq = n-factor m moles]

[Note: Na2CO3 is half neutralised with phenolphtha-lein as indicator]

25 x y .....(i)

Experiment - II:Meq of Acids = Meq. of Base (For completeneutralisation)

35 0.1 1 2 x y

Section 4

Solving.....(i) and ....(ii), we get: y = 1 and x = 1.5

NaOH( )

1.5 40g 0.061000

in 25 ml

2 3Na CO( )

1 106g 0.1061000

in 25 ml

Thus, strength of NaOH 10000.06 2.4g / L

and strength of Na2CO3 10000.106 4.24g / L

8 gm of a mixture of anhydrous Na2CO3 and NaHCO3 was dissolved in water andmade upto 1000 mL, 25 mL of this solution required for neutralisation:(a) 32.0 mL of N/10 HCl using methyl orange and(b) 12.0 mL of N/10 HCl using phenolphthalein.Find the strength of NaHCO3 and Na2CO3.Solution :

Let m moles of Na2CO3 be x and that of NaHCO3 be y in 25 ml sample.Experiment - I : Meq. of Acid = Meq. of Base (For complete Neutralisation)

1 112 2 1 010 2

x y 1.2 = x + 0 x = 1.2 ....(i)

Experiment - II : Meq. of Acids = Meq of Base (For complete neutralization)132 2 1 1 1

10 x y 3.2 = 2x + y

Solving (i) and (ii), we get : x = 1.2 and y = 0.8

Thus,2 3

gx 1000

M 2 3Na CO

(in 25 ml)

1.2 106g 0.127 gm1000

gy 1000

M 3NaHCO

(in 25ml)

0.8 84g 0.0672gm1000

Illustration -25

phenolphthalein12 mL, N/10 HCl

Experiment I25 mLsample

Methyl Orange

32 mL, N /10 HClExperiment II

28 Section 4

Thus, strength of Na2CO3 in the sample 10000.127 5.28gm / L

and strength of NaHCO3 in the sample 10000.0672 2.69 gm / L

THINGS TO REMEMBER

1. Atomic Weights of Most Commonly used Elements :

ELEMENT SYMBOL ATOMIC WEIGHT ELEMENT SYMBOL ATOMIC WEIGHT

Hydrogen H 1.0 Helium He 4.0Carbon C 12.0 Nitrogen N 14.0Oxygen O 16.0 Fluorine F 19.0Sodium Na 23.0 Magnesium Mg 24.0Aluminium Al 27.0 Phosphorus P 31.0Sulphur S 32.0 Chlorine Cl 35.5Potassium K 39.0 Calcium Ca 40.0Chromium Cr 52.0 Manganese Mn 55.0Iron Fe 56.0 Copper Cu 63.5Zinc Zn 65.4 Arsenic As 75.0Bromine Br 80.0 Silver Ag 108.0Tin Sn 119.7 Iodine I 127.0Barium Ba 137.4 Gold Au 197.0Mercury Hg 200.0 Lead Pb 207.0

2. Recapitulating and Summarizing:

gmoles MVM

gmmoles 1000 MVM

ggmeq NVEgmeq 1000 NVE

Thing to Remmeber 29

gmeq molesmeq m.molesN M

xx x = n - factor

n 1000 1000molality(m)g M

m1000mM

10 dMM gStrength = = N E = M M

litres of solution10 dNE

Note : (i) In experiments involving neutralisation, we should use concept of gmeq (or meq) i.e.,

gmeq of acid = gmeq of base

(ii) In experiments (usually gravimetric analysis) we should use mole concept i.e. first balance the chemicalequation and then correlate the reactants and products as per their stoichiometric coefficient.

Thing to Remmeber

Stoichiometry - 1 - SelfStudys

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