JEE Main & Advanced Chemistry Analytical Chemistry Volumetric Analysis

Volumetric Analysis

Category : JEE Main & Advanced

Volumetric analysis is a quantitative analysis. It involves the measurement of the volume of a known solution required to bring about the completion of the reaction with a measured volume of the unknown solution.

Titration : The process of addition of the known solution from the burette to the measured volume of solution of the substance to be  estimated until the reaction between the two is just complete, is termed as titration. Thus, a titration involves two solutions;

(i) Unknown solution : The solution consisting the substance to be estimated is termed unknown solution. The substance is termed titrate.

(ii) Standard solution : The solution in which an accurately known amount of the reagent (titrant) has been dissolved in a known volume of the solution is termed standard solution. There are two types of reagents (titrants) :

(a) Primary standards : These can be accurately weighed and their solutions are not to be standardised before use. Oxalic acid $({{H}_{2}}{{C}_{2}}{{O}_{4}}.2{{H}_{2}}O)$, potassium dichromate $({{K}_{2}}C{{r}_{2}}{{O}_{7}})$, silver nitrate $(AgN{{O}_{3}})$, copper sulphate $(CuS{{O}_{4}}.5{{H}_{2}}O)$, ferrous ammonium sulphate $[FeS{{O}_{4}}{{(N{{H}_{4}})}_{2}}S{{O}_{4}}.6{{H}_{2}}O]$, sodium thiosulphate $(N{{a}_{2}}{{S}_{2}}{{O}_{3}}.5{{H}_{2}}O)$, etc., are the examples of primary standards.

(b) Secondary standards : The solutions of these reagents are to be standardised before use as these cannot be weighed accurately. The examples are sodium hydroxide $(NaOH)$, potassium hydroxide $(KOH)$, hydrochloric acid $(HCl)$, sulphuric acid $({{H}_{2}}S{{O}_{4}})$, potassium permanganate $(KMn{{O}_{4}})$, iodine, etc.

Law of equivalence : It is applied in all volumetric estimations. According to it, the chemical substances react in the ratio of their chemical equivalent masses.

$\frac{\text{Mass of substance }A}{\text{Mass of substance }B}=\frac{\text{Chemical equivalent mass of }A}{\text{Chemical equivalent mass of }B}$

or $\frac{\text{Mass of substance }A}{\text{Chemical equivalent mass of }A}$$=\frac{\text{Mass of substance }B}{\text{Chemical equivalent mass of }B}$

or gram equivalent of $A=$gram equivalent of $B$

or milli-gram equivalent of  $A=$ milli-gram equivalent of $B$

The point at which the amounts of the two reactants are just equivalent is known as equivalence point or end point. An auxiliary substance which helps in the usual detection of the completion of the titration or equivalence point or end point is termed as indicator, i.e., substances which undergo some easily detectable changes at the equivalence point are used as indicators.

Methods of expressing concentrations of solutions

The concentration of a solution can be expressed in various ways.

(1) Percent by mass

(2) Molarity

(3) molality

(4) Mole fraction

(5) Normality

Types of titrations : Titrations can be classified as :

(1) Acid base titrations or acidimetry and alkalimetry

(2) Oxidation reduction titrations or redox titrations

(3) Precipitation titrations

(4) Complexometric titrations.

(1) Acid-base titrations : When the strength of an acid is determined with the help of a standard solution of base, it is known as acidimetry. Similarly, when the strength of a base (alkali) is determined with the help of a standard solution of an acid, it is known as alkalimetry.  Both these titrations involve neutralisation of an acid with an alkali. In these titrations ${{H}^{+}}$ ions of the acid combine with $O{{H}^{-}}$ ions of the alkali to form unionised molecules of water.

$\underset{\text{Acid}}{\mathop{HA}}\,+\underset{\text{Alkali}}{\mathop{BOH}}\,\xrightarrow{{}}\underset{\text{Salt}}{\mathop{BA}}\,+\underset{\text{Water}}{\mathop{{{H}_{2}}O}}\,$

or ${{H}^{+}}+{{A}^{-}}+{{B}^{+}}+O{{H}^{-}}\xrightarrow{{}}{{B}^{+}}+{{A}^{-}}+{{H}_{2}}O$

or ${{H}^{+}}+O{{H}^{-}}\xrightarrow{{}}{{H}_{2}}O$

The end point in these titrations is determined by the use of organic dyes which are either weak acids or weak bases. These change their colours within a limited range of hydrogen ion concentrations, i.e., $pH$ of the solution. Phenolphthalein is a suitable indicator in the titrations of strong alkalies (free from carbonate) against strong acids or weak acids. Methyl orange is used as an indicator in the titrations of strong acids against strong and weak alkalies. As no indicator gives correct results in the titrations of weak acids against weak bases, such titrations are performed by some other methods (physical methods).

(2) Oxidation reduction titrations : The titrations based on oxidation-reduction reactions are called redox titrations. The chemical reactions proceed with transfer of electrons (simultaneous loss or gain of electrons) among the reacting ions in aqueous solutions. Sometimes these titrations are named after the reagent used, as:

(i) Permanganate titrations : These are titrations in which potassium permanganate is used as an oxidising agent in acidic medium. The medium is maintained by the use of dilute sulphuric acid. Potassium permanganate acts as a self-indicator. The potential equation, when potassium permanganate acts as an oxidising agent, is :

$2KMn{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}{{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+3{{H}_{2}}O+5[O]$

or $MnO_{4}^{-}+8{{H}^{+}}+5e\xrightarrow{{}}M{{n}^{2+}}+4{{H}_{2}}O$

Before the end point, the solution remains colourless (when $KMn{{O}_{4}}$ solution is taken in burette) but after the equivalence point only one extra drop of $KMn{{O}_{4}}$ solution imparts pink colour, i.e., appearance of pink colour indicates the end point. Potassium permanganate is used for the estimation of ferrous salts, oxalic acid, oxalates, hydrogen peroxide, etc. The solution of potassium permanganate is always first standardised before its use.

(ii) Dichromate titrations : These are titrations in which, potassium dichromate is used as an oxidising agent in acidic medium. The medium is maintained acidic by the use of dilute sulphuric acid. The potential equation is

${{K}_{2}}C{{r}_{2}}{{O}_{7}}+4{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}{{K}_{2}}S{{O}_{4}}+C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+4{{H}_{2}}O+3[O]$

or $C{{r}_{2}}O_{7}^{2-}+14{{H}^{+}}+6{{e}^{-}}\xrightarrow{{}}2C{{r}^{3+}}+7{{H}_{2}}O$

The solution of potassium dichromate can be directly used for titrations. It is mainly used for the estimation of ferrous salts and iodides. In the titration of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ versus ferrous salt either an external indicator (potassium ferricyanide) or an internal indicator (diphenyl amine) can be used.

(iii) Iodimetric and iodometric titrations : The reduction of free iodine to iodide ions and oxidation of iodide ions to free iodine occurs in these titrations.

${{I}_{2}}+2{{e}^{-}}\xrightarrow{{}}2{{I}^{-}}$                (reduction)

$2{{I}^{-}}\xrightarrow{{}}{{I}_{2}}+2{{e}^{-}}$                (oxidation)

These are divided into two types :

(a) Iodimetric titrations : These are the titrations in which free iodine is used. As it is difficult to prepare the solution of iodine (volatile and less soluble in water), it is dissolved in potassium iodide solution.

$KI+{{I}_{2}}\xrightarrow{{}}\underset{\text{Potassium tri-iodide}}{\mathop{K{{I}_{3}}}}\,$

This solution is first standardised before use. With the standard solution of ${{I}_{2}}$. Substances such as sulphite, thiosulphate, arsenite, etc., are estimated.

(b) Iodometric titrations : In iodometric titrations, an oxidising agent is allowed to react in neutral medium or in acidic medium, with excess of potassium iodide to liberate free iodine.

$KI+$oxidising agent $\xrightarrow{{}}{{I}_{2}}$

Free iodine is titrated against a standard reducing agent usually with sodium thiosulphate. Halogens, oxyhalogens, dichromates, cupric ion, peroxides, etc., can be estimated by this method.

${{I}_{2}}+N{{a}_{2}}{{S}_{2}}{{O}_{3}}\xrightarrow{{}}2NaI+N{{a}_{2}}{{S}_{4}}{{O}_{6}}$

$2CuS{{O}_{4}}+4KI\xrightarrow{{}}C{{u}_{2}}{{I}_{2}}+2{{K}_{2}}S{{O}_{4}}+{{I}_{2}}$

${{K}_{2}}C{{r}_{2}}{{O}_{7}}+6KI+7{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}$$C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+4{{K}_{2}}S{{O}_{4}}+7{{H}_{2}}O+3{{I}_{2}}$

In iodimetric and iodometric titrations, starch solution is used as an indicator. Starch solution gives blue or violet colour with free iodine. At the end point the blue or violet colour disappears when iodine is completely changed to iodide.

(3) Precipitation titrations : The titrations which are based on the formation of insoluble precipitates, when the solutions of two reacting substances are brought in contact with each other, are called precipitation titrations. For example, when a solution of silver nitrate is added to a solution of sodium chloride or a solution of ammonium thiocyanate, a white precipitate of silver chloride or silver thiocyanate is formed.

$AgN{{O}_{3}}+NaCl\xrightarrow{{}}AgCl+NaN{{O}_{3}}$

$AgN{{O}_{3}}+N{{H}_{4}}CNS\xrightarrow{{}}AgCNS+N{{H}_{4}}N{{O}_{3}}$

Such titrations involving silver nitrate are called argentometric titrations.

(4) Complexometric titrations : A titration, in which an undissociated complex is formed at the equivalence point, is called complexometric titration. These titrations are superior to precipitation titrations as there is no error due to co-precipitation.

$H{{g}^{2+}}+2SC{{N}^{-}}\xrightarrow{{}}Hg{{(SCN)}_{2}}$

$A{{g}^{+}}+2C{{N}^{-}}\xrightarrow{{}}{{[Ag{{(CN)}_{2}}]}^{-}}$

EDTA (ethylenediamine tetra-acetic acid) is a useful reagent which forms complexes with metals. In the form of disodium salt, it is used to estimate $C{{a}^{2+}}$ and $M{{g}^{2+}}$ ions in presence of eriochrome black-$T$ as an indicator.

Equivalent masses of acids and bases : Equivalent masses of some acids and bases are given in the following table

 Acid Basicity Mol. Mass Eq. Mass $HCl$ 1 36.5 $\frac{36.5}{1}=36.5$ $HN{{O}_{3}}$ 1 63 $\frac{63}{1}=63.0$ ${{H}_{2}}S{{O}_{4}}$ 2 98 $\frac{98}{2}=49.0$ $C{{H}_{3}}COOH$ 1 60 $\frac{60}{1}=60.0$ ${{H}_{2}}{{C}_{2}}{{O}_{4}}.2{{H}_{2}}O$ 2 126 $\frac{126}{2}=63.0$ ${{H}_{3}}P{{O}_{4}}$ 3 98 $\frac{98}{3}=32.7$ ${{H}_{3}}P{{O}_{3}}$ 2 82 $\frac{82}{2}=41.0$ ${{H}_{3}}P{{O}_{2}}$ 1 66 $\frac{66}{1}=66.0$

 Alkali Acidity Mol. Mass Eq. Mass $NaOH$ 1 40 $\frac{40}{1}=40$ $KOH$ 1 56 $\frac{56}{1}=56$ $Ca{{(OH)}_{2}}$ 2 74 $\frac{74}{2}=37$ $N{{H}_{4}}OH$ 1 35 $\frac{35}{1}=35$

Calculations of Volumetric analysis

The following points should be kept in mind while making calculations of volumetric exercises.

(i) $1g$ equivalent mass of a substance reacts completely with $1g$ equivalent mass of any other substance. $1g$ equivalent mass of a substance means equivalent mass of the substance in grams. For example,

$1g$ equivalent mass of $NaOH=40g$ of $NaOH$

$1g$ equivalent mass of ${{H}_{2}}S{{O}_{4}}=49g$ of ${{H}_{2}}S{{O}_{4}}$

$1g$ equivalent mass of $KMn{{O}_{4}}$ in acidic medium

$=31.6\,g$ of $KMn{{O}_{4}}$

$1g$ equivalent mass of hydrated oxalic acid

$=63g$ of hydrated oxalic acid

Note : Equivalent mass is a variable quantity and depends on the reaction in which the substance takes part. The nature of the reaction should be known before writing the gram equivalent mass of the substance. For example in the reactions.

$2NaCl+2{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}2NaHS{{O}_{4}}+2HCl$         …..(i)

$2NaCl+{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}N{{a}_{2}}S{{O}_{4}}+2HCl$                …..(ii)

The value of $g$ equivalent mass of ${{H}_{2}}S{{O}_{4}}$ in reaction (i) is $98g$ and in reaction (ii) $49g$.

(ii) Number of $g$ equivalents $=\frac{\text{Mass of the substance in }g}{\text{Equivalent mass of the substance}}$

Number of $g$ moles $=\frac{\text{Mass of the substance in }g}{\text{Molecular mass of the substance}}$

$=\frac{\text{Volume in litres of the substance at N}\text{.T}\text{.P}\text{.}}{22.4}$(only for gases)

Number of milli-equivalent $=\frac{\text{Mass in }g\times 1000}{\text{Equivalent mass}}$

Number of milli-moles $=\frac{\text{Mass in }g\times 1000}{\text{Molecular mass}}$

(iii) Molarity $=\frac{\text{No}\text{. of moles of the solute}}{\text{No}\text{. of litres of the solution}}$ $=\frac{w}{m\times V}$

Molarity $\times$ molecular mass = strength of the solution $(g/L)$ No. of moles of the solute = Molarity $\times$ No. of litres of solution Mass of the solute in $g(w)=$ molarity $\times$ No. of litres of solution $\times$ mol. mass of solute

Normality $=\frac{\text{No}\text{. of }g\text{ equivalent of the solute}}{\text{No}\text{. of litres of the solution}}$ $=\frac{w}{E\times V}$

Normality $\times$ equivalent mass = strength of the solution (g/L)

No. of equivalents of the solute = Normality $\times$ No. of litres of solution

Mass of the solute in $g(w)=$ Normality $\times$ No. of litres of solution $\times$ Eq. mass of the solute

$\frac{\text{Molecular mass}}{\text{Equivalent mass}}=n=\frac{\text{Normality}}{\text{Molarity}}$

Normality $=n\times$ Molarity

(iv) Normality equation : When solutions $A$ and $B$ react completely.

${{N}_{A}}{{V}_{A}}={{N}_{B}}{{V}_{B}}$

Normality of $A\times$ volume of $A=$Normality of $B\times$ volume of $B$

or $\frac{\text{Strength }A}{\text{Eq}\text{. mass }A}\times {{V}_{A}}=\frac{\text{Strength }B}{\text{Eq}\text{. mass }B}\times {{V}_{B}}$

$\frac{\text{Wt}\text{. of metal hydroxide}}{\text{wt, of metal oxide}}=\frac{\text{Eq}\text{. wt}\text{. of metal hydroxide}}{\text{Eq}\text{. wt}\text{. of metal oxide}}$

$=\frac{\text{Eq}\text{. wt of metal }+\text{ Eq}\text{. wt of }OH}{\text{Eq}\text{. wt}\text{. of metal }+\text{ Eq}\text{. wt of }{{O}^{2-}}}$

(v) When the solution is diluted, the following formulae can be applied :

${{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}$ or ${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$ or ${{S}_{1}}{{V}_{1}}={{S}_{2}}{{V}_{2}}$

Before dilution = After dilution

(vi) If a number of acids are mixed, the combined normality of the mixture, ${{N}_{x}}$, is given

${{N}_{x}}{{V}_{x}}={{N}_{1}}{{V}_{1}}+{{N}_{2}}{{V}_{2}}+{{N}_{3}}{{V}_{3}}+......$

Where ${{V}_{x}}$ is the total volume of the mixture, ${{N}_{1}}$ and ${{V}_{1}}$ are the normality and volume respectively of one acid, ${{N}_{2}}$ and ${{V}_{2}}$ of the second acid and so on.

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