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# Calculations based on Principle of Volumetric Analysis Volumetric Analysis : An important method for determining the amount of a particular substance is based on measuring the volume of reactant solution. Suppose substance A in solution reacts with substance B. If you know the volume and concentration of a solution of B that just reacts with substance A in a sample, you can determine the amount of A.

Calculations based on Principle of Volumetric Analysis

Volumetric Analysis

An important method for determining the amount of a particular substance is based on measuring the volume of reactant solution. Suppose substance A in solution reacts with substance B. If you know the volume and concentration of a solution of B that just reacts with substance A in a sample, you can determine the amount of A.

Titration is a procedure for determining the amount of substance A by adding a carefully measured volume of a solution of A with known concentration of B untill the reaction of A and B is just completed. Volumetric analysis is a method of analysis based on titrations.

Law

"Equal volume of equinormal solutions exactly neutralise the other solution having same concentration and volume".

V1 N1  =  V2 N2

V1, V2  -  Volume of solutions.

N1, N2  -  Strength of solutions.

Solved problem

Calculating the volume of reactant solution needed

1. What volume of 6M HCl and 2M HCl should be mixed to get one litre of 3M HCl?

Solution

Suppose the volume of 6M HCl required to obtain 1L of 3M HCl = XL

Volume of 2M HCl required  =  (1-x)L

Applying the molarity equation

M1V1  +   +   M2V2  =   =   M3V3

6MHCl+   2MHCl=   3MHCl

6x+2(1-x)      =  3x1

6x+2-2x  =  3

4x  =  1

x  =  0.25L

hence, volume of 6M HCl required  =  0.25L

Volume of 2M HCl required      =  0.75L

2. How much volume of 10M HCl should be diluted with water to prepare 2.00L of 5M HCl.

Solution

N1V1  =   N2V2

10N HCl  =  5N HCl

10xV1  =  5 x 2.00

V1  =  (5 x 2.00 )/ 10

=  1.00L

Problems for Practice

1. NiSO4 reacts with Na3PO4 to give a yellow green precipitate of Ni3(PO4)2 and a solution of Na2SO4.

3NiSO4(aq) + 2Na3PO4(aq) Ni3(PO4)2 (s) + 3Na2SO4(aq)

How many mL of 0.375M NiSO4 will react with 45.7 mL of 0.265M Na3PO4?

2. What volume of 0.250M HNO3 reacts with 42.4 mL of 0.150M Na2CO3 in the following reaction?

2HNO3(aq)  +  Na2CO3(aq) 2NaNO3(aq)  +  H2O(aq)+CO2(g)

3. A flask contains 53.1 mL of 0.150M Ca(OH)2 solution. How many mL of 0.350M Na2CO3 are required to react completely with Ca(OH)2 in the following reaction.

Na2CO3(aq) +  Ca(OH)2(aq) CaCO3(aq)  + 2NaOH(aq)

Determination of equivalent masses of elements

Equivalent masses can be determined by the following methods:

1.     Hydrogen displacement method

2.     Oxide method

3.     Chloride method

4.     Metal displacement method

1.Hydrogen displacement method

This method is used to determine the equivalent mass of those metals such as magnesium, zinc and aluminium which react with dilute acids and readily displace hydrogen.

Mg + 2HCl -> MgCl2 + H2

Zn + H2SO4 - > ZnSO4 + H2

2Al + 6HCl - > 2AlCl3 + 3H2

From the mass of the metal and the volume of hydrogen displaced, the equivalent mass of the metal can be calculated.

2.Oxide Method

This method is employed to determine the equivalent mass of those elements which readily form their oxides eg. magnesium, copper etc. Oxide of an element can be formed by direct method or by indirect method.

Magnesium forms its oxide directly on heating

2Mg + O2  2 MgO

In the case of copper, its oxide is obtained in an indirect manner i.e. copper is dissolved in concentrated nitric acid and the copper(II) nitrate formed after evaporation is strongly heated to give copper (II) oxide.

Cu + 4HNO3 Cu(NO3)2  + 2H2O + 2 NO2

2Cu(NO3)2  2CuO + 4 NO2 + O2

Calculations

Mass of the element taken = w1 g

Mass of the oxide of the element = w2 g

Mass of oxygen = (w2-w1) g

(w2 - w1) g of oxygen has combined with w1 g of the metal.

8 g of oxygen will combine with w1 / (w2 - w1) x 8

This value represents the equivalent mass of the metal.

3.Chloride Method

The equivalent mass of those elements which readily form their chlorides can be determined by chloride method. For example, a known mass of pure silver is dissolved completely in dilute nitric acid. The resulting silver nitrate solution is treated with pure hydrochloric acid when silver chloride is precipitated. It is then filtered, dried and weighed. Thus from the masses of the metal and its chloride, the equivalent mass of the metal can be determined as follows :

4.Calculations

Mass of the metal = w1 g

Mass of the metal chloride = w2 g

Mass of chlorine = (w2 - w2) g

(w2 - w1) g of chlorine has combined with w1 of the metal

35.46 g of chlorine will combine with 35.46 x w1 / (w2 - w1) g of the metal

This value gives the equivalent mass of the metal.

Uses of volumetric laws

If the volume of the acid is represented as V1, the normality of the acid as N1, volume of base as V2 the normality of the base as N2, then from the law of volumetric analysis it follows that

V1 x N1 = V2 x N2

All volumetric estimations are based on the above relationship. If any three quantities are known, the fourth one can readily be calculated using the above expression.

Equivalent mass of acid, base, salt, oxidising agent and reducing agent

Acids contain one or more replaceble hydrogen atoms. The number of replaceble hydrogen atoms present in a molecule of the acid is referred to its basicity.

1.Equivalent mass of the acid

Equivalent mass of an acid is the number of parts by mass of the acid which contains 1.008 part by mass of replaceble hydrogen atom.

Equivalent mass of the acid  =        molar mass of an acid / No. of replaceble hydrogen atom

= molar mass of the acid / basicity of the acid

For example, the basicity of sulphuric acid is 2.

Equivalent mass of H2SO4 = Molar mass of H2SO4 / 2  = 98/2 = 49

2.Equivalent mass of the base

Equivalent mass of a base is the number of parts by mass of the base

which contains one replaceable hydroxyl ion or which completely neutralises one gram equivalent of an acid. The number of hydroxyl ions present in one mole of a base is known as the acidity of the base. Sodium hydroxide, potassium hydroxide, ammonium hydroxide are examples of monoacidic bases.

Calcium hydroxide is a diacidic base

In general,

equivalent mass of a base      = molar mass of the base / acidity of the base

Equivalent mass of KOH = 56 /1 = 56

3.Equivalent mass of a salt

Equivalent mass of a salt is a number of parts by mass of the salt that is produced by the neutralisation of one equivalent of an acid by a base.

In the case of salt like potassium chloride, the salt formed by the neutralisation of one equivalent of an acid by a base.

KOH + HCl -> KCl + H2O

Therefore, the equivalent mass of the salt is equal to its molar mass.

4.  Equivalent mass of an oxidising agent

The equivalent mass of an oxidising agent is the number of parts by mass which can furnish 8 parts by mass of oxygen for oxidation either directly or indirectly.

For example, potassium permanganate is an oxidising agent. In acid medium potassium permanganate reacts as follows

2 KMnO4 + 3 H2SO4 ->  K2SO4 + 2 MnSO4 + 3 H2O + 5 [O] 316 80

80 parts by mass of oxygen are made available from 316 parts by mass of KMnO4

8 parts by mass of oxygen will be furnished by

(316 x 8 ) / 80 = 3.16

Equivalent mass of KMnO4  = 31.6g equiv-1

5. Equivalent mass of a reducing agent

The equivalent mass of a reducing agent is the number of parts by mass of the reducing agent which is completely oxidised by 8 parts by mass of oxygen or with one equivalent of any oxidising agent.

(i) Ferrous sulphate reacts with an oxidising agent in acid medium according to the equation

2 FeSO4 + H2SO4 + (O) Fe2 (SO4)3 + H2O

2 x 152g  16g

16 parts by mass of oxygen oxidised 304 parts by mass of ferrous sulphate

8 parts by mass of oxygen will oxidise 304/16 x 8 parts by mass of ferrous sulphate.

The equivalent mass of crystalline ferrous sulphate, FeSO4 7H2O is 152 + 126 = 278

126 is the mass corresponding to 7 moles of water.

(ii) In acid medium, oxalic acid is oxidised according to the equation

(COOH)2 + (O) 2 CO2 + H2O

16 Parts by mass of oxygen oxidised 90 parts by mass of anhydrous oxalic acid.

8 parts by mass of oxygen will oxidise 90/16 x 8 = 45 parts by mass of anhydrous oxalic acid.

Equivalent mass of anhydrous oxalic acid = 45 g equiv-1

But equivalent mass of crystalline oxalic acid, (COOH)2. 2H2O=126/2 = 63 g equiv-1.

Determination of Molecular Mass Victor-Meyer's Method

Principle

In this method a known mass of a volatile liquid or solid is converted into its vapour by heating in a Victor-Meyer's tube. The vapour displaces its own volume of air. The volume of air displaced by the vapour is measured at the experimental temperature and pressure. The volume of the vapour at s.t.p is then calculated. From this the mass of 2.24 x 10-2m3 of the vapour at S.T.P. is calculated. This value represents the molecular mass of the substance.

method

The apparatus consists of an inner Victor-Meyer tube, the lower end of which is in the form of a bulb. The upper end of the tube has a side tube which leads to a trough of water. The Victor-Meyer tube is surrounded by an outer jacket. In the outer jacket is placed a liquid which boils at a temperature at least 30 K higher than the boiling point of the volatile substance under study. A small quantity of glass wool or asbestos fiber covers the bottom of the Victor-Meyer tube to prevent breakage when the bottle containing the substance is dropped in.

Procedure

The liquid in the outer jacket is allowed to boil and when no more air escapes from the side tube, a graduated tube filled with water is inverted over the side tube dipping in a trough full of water. A small quantity of the substance is exactly weighed in a small stoppered bottle and quickly dropped in the heated Victor-Meyer tube and corked immediately.

The bottle falls on the asbestos pad and its content suddenly changes into vapour, blow out the stopper and displace an equal volume of air which collects in the graduated tube. The volume of air in the graduated tube is measured by taking it out by closing its mouth with the thumb and dipping it in a jar full of water. When the water levels outside and inside the tube are the same, the volume of air displaced is noted. The atmospheric pressure and laboratory temperature are noted.

Calculations

Mass of the volatile substance = wg

Volume of air displaced = Volume of vapour = V1 m3

Laboratory temperature = T1 K

Let the atomospheric pressure be P

Pressure of dry vapour = Atomospheric pressure - aqueous tension at. T1 K Let the aqueous tension be p Nm-2 at that temperature.

Pressure of dry vapour = P1 = [P-p]

Standard temperature = T0 = 273 K

Standard pressure = = P0 = = 1.013 x 105 Nm-2

Let the volume of the vapour at standard temperature and pressure be V0 m3

From the gas equation, it follows

P0V0 /  T0  =  P1 V1 / T1

V0   =  (P1 V1 / T1) x (T0/ P0)

The mass of V0 m3 of vapour at s.t.p is w g.

The mass of 2.24 x 10-2 m3 of the vapour at s.t.p. is

= 2.24x10-2xW / V0

The value thus calculatd gives the molecular mass

Molecular mass  =  2 x vapour density

Vapour density = Molecular mass / 2

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