Class 12 Chemistry Volumetric Analysis Notes

CHEMISTRY • CHAPTER 1

Volumetric Analysis

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Volumetric analysis

Equivalent Weight

Equivalent wt of an element is defined as the no. of parts by wt of that element which can combine or displaced directly or indirectly 1.008 parts by wt of hydrogen, 8 parts by wt of oxygen or 35.5 parts by wt of chlorine.

Example: CH4, MgO, AlCl3

1) CH4

1.008 × 4 parts by wt of hydrogen combined with 12 parts by wt of carbon.

1 parts by wt of hydrogen combine with 12 / (1.008 × 4) parts by wt of carbon.

1.008 parts by wt of hydrogen combines with:

12 × 1.0081.008 × 4 = 3 parts by wt of carbon

Eq wt of C in CH4 is 3.

2) MgO

16 parts by wt of oxygen combine with 24 parts by wt of magnesium.

1 parts by wt of oxygen combines with 24/16 parts by wt of magnesium.

8 parts by wt of oxygen combines with:

2416 × 8 = 12 parts by wt of magnesium

Eq wt of Mg in MgO is 12.

3) AlCl3

35.5 × 3 parts by wt of chlorine combine with 27 parts by wt of Al.

1 parts by wt of chlorine combine with 27 / (35.5 × 3) parts by wt of Al.

35.5 parts by wt of chlorine combine with:

2735.5 × 3 × 35.5 = 9 parts by wt of Al

For an element

Eq wt = At. wtValency
E = A/V

eg: C, Ca, Al, Mg

Element At. wt (A) Valency (V) E = A/V
C1243
Ca40220
Al2739
Mg24212

For compound

Eq wt = mol. wtZ

where, Z = basicity for acid. [Total no. of replaceable H+ ion]

Example: Acid

Acid mol. wt Z = Basicity Eq wt = mol wt / Z
HCl36.5136.5
H2SO498249
HNO363163
CH3COOH60160
H3PO498332.66
(COOH)2·2H2O126263

Example Base

Base mol. wt Z = Basicity Eq wt = mol wt / Z
NaOH40140
KOH56156
NH4OH35135
Ca(OH)274237
Mg(OH)258229
Al(OH)378326

Example: Salt

Salt mol. wt Z = total no. of +ve or −ve charge Eq wt = mol wt / Z
MgCl295247.5
NaCl58.5158.5
K2SO4174287
Na2CO3106253
AlCl3133.5344.5
KCl74.5174.5

For radical

eg CO32−

Eq wt = wt of radical / total no of charge

= 60 / 2

= 30

SO42−

Eq wt = wt of radical / total no of charge

= 96 / 2

= 48

PO43−

Eq wt = wt of radical / total no of charge

= 95 / 3

= 31.66

NH4+

Eq wt = wt of radical / total no of charge

= 18 / 1

= 18

NO3−

Eq wt = 62

= 62

Cl−

Eq wt = 35.5 / 1

= 35.5

Ca2+

Eq wt = 40 / 2

= 20

For oxidising Reducing agent

eg KMnO4 (potassium permanganate)

a) In acidic medium

KMnO4 + H2SO4 → K2SO4 + MnSO4 + H2O + [O]

change in ON of Mn = 5

Eq wt = mol wt / Z

= 158 / 5

= 31.6

eq wt in acidic medium = 31.6

b) In basic medium

KMnO4 + KOH → K2MnO4 + H2O + [O]

change in ON of Mn = 1

Eq wt = mol wt / Z

= 158 / 1

= 158

eq wt in basic medium = 158

c) In neutral medium

KMnO4 + H2O → MnO2 + KOH + [O]

change in ON of Mn = 3

Eq wt = mol wt / Z

= 158 / 3

= 52.6

KMnO4 Eq wtAcidicBasicNeutral
Value31.615852.6

Volumetric Analysis

Gram equivalent wt

When equivalent wt of a substance is expressed in gram which is known gram equivalent wt.

No. of gram eq wt = wt of substance in gram / Eq wt
= W / E

NOTE: No of gram equivalent wt of any substance = no of eq wt of that substance.

1g eq wt of Na2CO3 = Eq wt of Na2CO3 = 53g of Na2CO3.

1g eq wt of H2SO4 = Eq wt of H2SO4 = 49g of H2SO4.

Concentration / strength of soln

1. Gram per litre (g/L)

Gram per litre of a substance define as a weight of substance in gram present in 1 litre of solution. It is denoted by (g/L) and given by:

Gram per litre = wt of substance in gram / volm of soln in litre

g/L = W / V

g/L = W × 1000 / Vml

2. Normality (N)

Normality of a soln is define as the no of gram equivalent wt of substance present in 1 litre of solution. It is denoted by N and given by:

Normality (N) = No. of g eq wt of sub / volm of soln in litre

= wt of substance in g / Eq wt × 1 / volm of soln in litre

Normality = g/L × 1 / Eq wt

N × E = g/L

Again:

g/L = W × 1000 / Vml   …(1)

g/L = N × E   …(2)

W × 1000 / Vml = N × E

W × 1000 = N × E × Vml

W = N × E × Vml / 1000

a) Normal solution (1N or N)

When 1g eq wt of any substance is present in 1 litre of solution then soln is called normal solution. It is denoted by 1N or N.

b) Semi-Normal solution (N/2 = 0.5N)

When half g eq wt of substance is present in 1 litre of solution then soln is called semi-solution. It is denoted by N/2 or 0.5N.

c) Deci-Normal soln (N/10 = 0.1N)

When 1/10 g eq wt of substance is present in 1 litre of solution then soln is called deci-normal soln. It is denoted by N/10 or 0.1N.

d) Centi-Normal soln (N/100 = 0.01N)

When 1/100 g eq wt of substance is present in 1 litre of solution then soln is called centi-normal soln. It is denoted by N/100 or 0.01N.

Molarity (M)

Molarity of a solution is define as the no of mole of substance present in 1 litre of solution. It is denoted by M and given by:

M = No. of mole of substance / Volm of soln in litre

= wt of substance in g / molecular wt × 1 / Volm of soln in litre

M = 1 / M.W × g/L

molarity × molecular wt = g/L

M × M.W = g/L

Again:

M × M.W = W × 1000 / Vml

W = M × M.W × Vml / 1000

a. Molar soln (1M = M)

When 1 mole of substance is present in 1 litre of solution then the solution is called molar soln. It is denoted by 1M or M.

b. Semi-molar soln (M/2 = 0.5M)

When half mole of substance is present in 1 litre of solution then the soln is called semi-molar soln. It is denoted by M/2 or 0.5M.

c. DECI-Normal solution (M/10 or 0.01M)

When 1/10 mole of substance is present in 1 litre of solution then the soln is called deci-normal solution.

d. Centi-molar solution (M/100 or 0.01M)

When 1/100 mole of substance is present in 1 litre of soln then the soln is called centi molar soln. It is denoted by M/100 or 0.01M.

Relation bet’n normality and molarity

We have:

g/L = N × E   …(1)

g/L = M × M.W   …(2)

from eqn (1) and (2)

N × E = M × M.W

N × M.W/Z = M × M.W

N = M × Z

For an acid:

N = M × Basicity

For a base:

N = M × Acidity

For a salt:

N = M × Total no of +ve or −ve charge

For an oxidising and reducing agent:

N = M × change in O.N.

Relation of % with normality and molarity

For normality:

Normality = % × 10 / Eq wt

If specific gravity (density) is given:

N = % × 10 × specific gravity / Eq wt

For molarity:

M = % × 10 / molecular wt

If specific gravity (density) is given:

M = % × 10 × specific gravity / molecular wt

4. Molality (m)

molality of a soln is define as the no of mole of substance present in 1 kg of solvent. It is denoted by m and given by:

m = No. of mole of substance / wt of solvent in kg

a) Molal soln

When 1 mole of substance is present in 1kg of solvent then the soln is called molal soln. It is denoted by 1m or m.

b) Semi molal soln (m/2 = 0.5m)

When 1/2 mole of substance is present in 1kg of solvent then the soln is called semimolal soln. It is denoted by m/2 or 0.5m.

c) Deci-molal soln (m/10 or 0.1m)

When 1/10 mole of substance is present in 1kg of solvent then the soln is called deci-molal soln. It is denoted by m/10 or 0.1m.

d) Centi-molal soln (m/100 or 0.01m)

When 1/100 mole of substance is present in 1kg of solvent then the soln is called centi-molal soln. It is denoted by m/100 or 0.01m.

5. Formality (F)

Formality of a solution is defined as the no of gram of formula wt present in 1 litre of soln and it is denoted by (F) and given by:

Formality (F) = No. of gram of formula wt / volm of soln in litre

= wt of sub in gram / formula wt × 1 / volm of soln in litre

formality = 1 / formula wt × g/L

F × formula wt = g/L

Again:

F × formula wt = W × 1000 / Vml

F × formula wt × Vml = W × 1000

F × formula wt × Vml / 1000 = W

i) Formal soln (1F or F)

When 1g formula wt of substance is present in 1 litre of soln is called formal soln. It is denoted by 1F or F.

ii) Semi-formal soln (F/2 = 0.5F)

When 1/2 g formula wt of substance present in 1 litre soln then the soln is called semi-formal soln. It is denoted by F/2 or 0.5F.

iii) Deci-formal soln (F/10 or 0.1)

When 1/10 g formula wt of substance present in 1 litre of soln then the soln is called deci-formal soln. It is denoted by F/10 or 0.1.

iv) Centi-formal soln (F/100 or 0.01)

When 1/100 g formula wt of substance present in 1 litre of soln then the soln is called centi-formal soln. It is denoted by F/100 or 0.01.

6) Percentage concn

percentage concn of a soln is defined as the wt of substance in g or ml present in 100g or 100 ml of solution. It is of following type.

1) wt by wt percentage [(w/w) × %]

Wt by wt percentage of a soln is defined as the wt of substance in gram present in 100 gram of solution. It is denoted by (w/w)% and given by:

(w/w)% = wt of substance in g / wt of soln in g × 100

eg: 1% (w/w) NaOH means:

1g of NaOH is present in 100g of solution.

eg: 0.5% (w/w) KOH means:

0.5g of KOH is present in 100g of solution.

2. Wt by volume percentage (w/v) × %

Wt by volume percentage of a soln is defined as the wt of substance in gram present in 100 ml of soln. It is denoted by w/v and given by:

(w/v)% = wt of substance in g / volume of soln in ml × 100

Eg: 1% w/v sodium carbonate (Na2CO3) means:

1 gram of Na2CO3 is present in 100 ml of soln.

Eg: 0.1% w/v oxalic acid [(COOH)2] means:

0.1 gram of (COOH)2 is present in 100 ml of soln.

3. Volume by volume percentage (v/v)%

Volume by volume percentage of a soln is defined as the wt of substance in ml present in 100 ml of soln. It is denoted by v/v and given by:

(v/v)% = volume of sub in ml / volume of soln in ml × 100

eg 1% (v/v) H2SO4 means:

1 ml of H2SO4 is present in 100 ml of soln.

7) Parts per million (ppm)

parts per million soln is define as the wt of substance in gram present in 1 million parts by wt of soln and given by:

ppm = wt of substance (in g) / wt of soln × 106

8) Parts per billion (ppb)

parts per billion soln is define as the wt of substance in gram present in 1 billion parts by wt of soln. It is denoted by ppb and given by:

ppb = wt of substance (in g) / wt of soln × 109

9) Mole fraction

Mole fraction of soln is define as the ratio of no of mole of 1 component to the no of mole of all the component (solute and solvent).

let w consider n1 be the no of mole of solute and n2 be the no of mole of solvent then total no of mole of soln is (n1 + n2).

Mole fraction of solute = n1 / (n1 + n2)

mole fraction of solvent = n2 / (n1 + n2)

Mole fraction of solute + mole fraction of solvent:

n1/(n1+n2) + n2/(n1+n2)

= (n1 + n2) / (n1 + n2)

= 1

Imp Primary and secondary standard substances

The exact wt of some substance cannot be taken because they may be volatile, hygroscopic, deliquescent, efflorescent, [Unreadable in source] and composition is different in soln and solid state. When they are dissolved in fixed volume of soln then its exact concn is not known is called secondary substances and corresponding solution are called secondary standard solutions. eg sodium hydroxide, hydrochloric acid, sulphuric acid, nitric acid, potassium hydroxide, potassium permanganate (KMnO4).

When exact wt of some substances can be taken and its concn is known by dissolving in fixed volume of soln is called primary standard substances and corresponding soln are called primary standard soln.

eg: oxalic acid, sodium carbonate (Na2CO3), copper sulphate, sodium chloride (NaCl), potassium chloride, potassium dichromate (K2Cr2O7), mohr’s salt [FeSO4·(NH4)2SO4·6H2O].

Condition to be of primary standard substances

  • The substance should be pure, dry and soluble in water.
  • The substance should have high molecular wt and equivalent wt.
  • The substance should not be volatile, hygroscopic, deliquescent, [Unreadable in source] & efflorescent.
  • The substance should be stable (i.e) its composition should not change in soln or in solid for long time.

Titration

The process of determining the concn of unknown soln by gradual addition of one soln in the other solution in presence of third chemical substance i.e indicator is called titration. It is of following types.

1) Acid base titration

The process in which the concn of an acid is determine with the help of standard base (alkali soln) in presence of indicator or vice versa is called acid base titration.

Eg: titration betn H2SO4 and Na2CO3 – NaOH / (COOH)2.

i) Acidimetry

The process of determining concn of an acid by finding the exact volume of it by neutralizing with known volume of standard base (alkali) in presence of suitable indicator is called acidimetry.

ii) Alkalimetry

The process of determining concn of a base by finding the exact volume of it by neutralizing with known volume of standard acid in presence of suitable indicator is called alkalimetry.

2) Redox titration

The titration in which the concn of oxidising agent is determine with the help of primary standard reducing agent or vice versa is called redox titration.

Eg: titration bet’n KMnO4 and oxalic acid soln in presence of sulphuric acid (H2SO4).

Imp: KMnO4 is an oxidizing agent and oxalic acid is reducing agent whereas KMnO4 act as self indicator.

Different terms used in titration

i) Titrant

The volm of soln which is taken in a burette of known concn is called titrant.

ii) Titrand

The volume of soln which is taken in a conical flask of unknown concentration is called titrand.

iii) End point

The point at which rxn complete is indicated by change in colour of indicator is called end point. It is a practical point.

iv) Equivalence point

The point at which equal amount of titrant is neutralized by titrand is called equivalence point. It is the theoretical end point.

The difference betn equivalence point or end point is called titration error.

v) Indicator

The chemical substance which indicate the completion of rxn by change in its own colour is called indicator. eg: methyl orange, methyl red, phenolphthalein.

Selection of indicator in acid-base titration

Nature of titration – Acid Base pH range of soln at equivalence point Indicator pH range of Indicator
StrongStrong3–11Any indicator—
StrongWeak3–8Methyl orange(3.1–4.4)
WeakStrong(6–11)Phenolphthalein(8.2–10)
WeakWeak6–8——

Principle of volumetric analysis

1) Equal volm of an acid is neutralized equal volume of base if their concn are same.

eg: 10ml of 0.1N HCl is neutralized 10ml of 0.1N NaOH.

100ml of 1N H2SO4 is neutralized 100ml of 1N KOH.

2) Normality and volm are reciprocal for the same soln.

eg: 1000ml of 1N Na2CO3 = 500ml 2N Na2CO3.

1000 × 5ml of 1N Na2CO3 = 5000ml of 0.2N Na2CO3.

1000/20 ml of 20N Na2CO3 = 50ml of 20N Na2CO3.

3) If an acid is neutralize by a base then N1V1 = N2V2.

NaVa = NbVb
S1V1 = S2V2
or SaVa = SbVb
These are acidity eqn.

Law of Equivalence (Normality equation)

let us consider an acid react with base.

Acid + Base neutralization rxn → Salt + H2O

At equivalence point:

No. of eq wt of an acid = No. of g eq wt of a base   …(1)

we know:

Normality = No. of eq wt of substance / volm of soln in litre

Normality × volume of soln in litre = no of eq wt of substance.

For acid:

No. of g eq wt of an acid = Normality of acid × volume of acid soln in litre   …(2)

For base:

No. of g eq wt of base = Normality of base × volume of base soln in litre   …(3)

Putting value of eqn (2) and (3) in eqn (1):

Normality of acid × volume of acid soln in litre = Normality of base × volume of basic solution in litre

Na × Va = Nb × Vb

Normality of mixture

a) For similarly soln of mixture

NmixVmix = N1V1 + N2V2 + …
Vmix = V1 + V2 + …

b) For different soln of mixture

NmixVmix = NaVa − NbVb
Vmix = Va + Vb

If Nmix +ve then the resulting soln is acidic.

If Nmix −ve then the resulting soln is basic.

If Nmix = 0 then the resulting soln is neutral.

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