In this article, we shall study the determination of the equivalent mass of acid, base, and salt.

Equivalent Mass of Acids:

One gram equivalent mass of an acid is that mass of it which contains one gram equivalent mass of replaceable hydrogen atoms.

Thus the equivalent mass of an acid depends on the replaceable hydrogen atoms it contains per mole. The number of replaceable hydrogen atoms present in a molecule of acid is called the basicity of the acid.

Illustration – 1:

Molecular mass of HCl = 1 + 35.5 = 36.5

Illustration – 2:

Molecular mass of H₂SO₄ = 2  + 32 + 64 = 98

Equivalent Mass of Base:

One gram equivalent mass of a base is that mass of it which contains one gram equivalent mass of the hydroxyl radical.

Thus the equivalent mass of a base depends on the number of hydroxyl radicals it contains per mole. The number of hydroxyl radical present in a molecule of a base is called the acidity of the base.

Illustration – 1:

Molecular mass of NaOH = 23 + 16 + 1 = 40

Illustration – 2:

Molecular mass of Ca(OH)₂= 40 + (16+1) x 2 = 74

Equivalent-Mass of Salts:

Equivalent mass of a simple salt is that mass of it which contains one gram equivalent of the metal or a radical

Illustration – 1:

Molecular mass of KCl = 39 + 35.5 = 74.5

In this case, KCl contains  1 gram equivalent of K and 1 gram equivalent of Cl. Hene equivalent mass of KCl is 74.5 / 1 = 74.5.

Illustration – 2:

Molecular mass of AlCl₃ = 27 + 35.5 x 3  = 133.5

In this case AlCl₂ contains  1 gram equivalent of Al and 3 gram equivalent of Cl. Hene equivalent mass of KCl is 133.5 / 3 = 44.5.

Illustration – 3:

Equivalent mass of a salt is also that mass of it, which will combine with one gram equivalent of another substance.

To find equivalent mass of Na₂CO₃

Na₂CO₃  reacts with HCl as

Na₂CO₃   + 2HCl  → 2 NaCl  + CO₂ + H₂O

Molecular mass of Na₂CO₃ = 23 x 2  + 12 x 1  + 16 x 3 = 106

one gram equivalent of Na₂CO₃ reacts with 2 gram equivalent of HCl. Hence equivalent mass of Na₂CO₃ is 106 / 2 = 53.

Equivalent Mass of Oxidising and Reducing Agents:

Note:

Metals with variable valency show variable equivalent masses depending upon their valency in the compound. For e.g. in oxides FeO, Fe₂O₃ and Fe₃O₄ the equivalent masses of Fe are 28.18.6 and 21 respectively.

Gram Equivalent:

The equivalent mass expressed in grams is called gram equivalent mass (GEM)

Milliequivalent:

A milliequivalent is one-thousandth of an equivalent mass of any substance is the equivalent mass expressed in milligrams. It is the unit which is used to express the concentration of electrolytes in tissue fluids of animals and plants.

Science > Chemistry > Concept of Atomic Mass and Equivalent Mass > Equivalent Masses of Acids, Bases, and Salts

The post Equivalent Masses of Acids, Bases, and Salts appeared first on The Fact Factor.

In the last few articles, we have studied the hydrogen displacement method, oxide formation method, reduction method, chloride formation method, and double displacement method to determine the equivalent mass of metal. In this article, we shall study use of Faraday’s laws of electrolysis to determine the equivalent mass of a substance.

The equivalent mass of a substance is the number of parts by mass of the substance which combines with or displaces or contains 1.008 parts by mass of hydrogen, 8 part by mass of oxygen, or 35.5 part by mass of chlorine. If the equivalent mass is expressed in grams then it is called gram equivalent mass (GEM).

Illustration: 

1.008 parts by weight of Hydrogen combines with 35.5 parts by weight of chlorine to give 36.5 parts by weight of HCI. Thus the equivalent mass of chlorine is 35.5.

Equivalent mass has no unit because it is a pure ratio. Some important equivalent masses are H = 1, O = 8, Cl = 35.5

Method – VII (Faraday’s First Law of Electrolysis):

Statement :

The mass of any substance deposited or liberated or dissolved at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.

Steps Involved:

A known quantity of electricity (Q = i t) is passed through electrolyte solution and mass (w) of the substance deposited or liberated during electrolysis is measured.

Using the following relation value of electrochemical equivalent (z) is calculated.

w = z i t

Then equivalent mass is calculated by the formula

Equivalent mass = 96500 x z

Numerical Problems on faraday’s First Law of Electrolysis

Example – 01:

On passing a current of 0.5 A through a solution of a salt of a metal for 32 minutes 0.3158 g of the metal was deposited. What is the equivalent mass of the metal?

Given: i = 0.5 A, t = 32 min = 32 x 60 s , W =  0.3158

Solution:

By Faraday’s first law of electrolysis,

W = z i t

z = W/ (i t) = 0.3158/(0.5 x 32 x 60)

z = 3.29 x 10^-4 g/C

Now, Equivqlent mass = 96500 x z

Equivalent mass = 96500 x 3.29 x 10^-4 = 31.74

Ans: The equivalent mass of metal is 31.74

Example – 02:

On passing a current of 0.6 A through a solution of a salt of copper for 20 minutes 0.24 g of the copper was deposited. What is the equivalent mass of copper?

Given: i = 0.6 A, t = 20 min = 20 x 60s, W =  0.24 g

Solution:

By Faraday’s first law of electrolysis,

W = z i t

z = W/ (i t) = 0.24/(0.6 x 20 x 60)

z = 3.33 x 10^-4 g/C

Now, Equivalent mass = 96500 x z

Equivalent mass = 96500 x 3.33 x 10^-4 = 31.84

Ans: The equivalent mass of copper is 31.84

Method – VIII (Faraday’s Second Law of Electrolysis):

Statement:

When the same quantity of electricity is passed through different electrolytes (generally connected in series), the masses of different substances deposited or liberated or dissolved at the respective electrodes are directly proportional to their chemical equivalents (equivalent masses).

Steps Involved:

The same quantity of electricity is passed through the solution of different electrolytes, the masses of different substances liberated or evolved as a result of electrolysis are noted.

Then equivalent mass is calculated by the formula.

Where W₁ = mass of the first substance deposited

W₂ = mass of the second substance deposited

E₁ = Equivalent mass of the first substance

E₂ = Equivalent mass of the second substance

Numerical Problems on faraday’s Second Law of Electrolysis

Example – 01:

An electric current is passed through two cells containing CuSO₄ and AgNO₃ solutions respectively connected in series. The masses of copper and silver deposited are 0.424 g and 1.44 g respectively. Find the equivalent mass of copper if that of silver is 108.

Given: W_Cu = 0.424 g,  W_Ag = 1.44 g, E_Ag = 108

To Find: E_Cu =?

Solution:

By Faraday’s second law of electrolysis,

Ans: The equivalent mass of copper is 31.8

Example – 02:

The same quantity of electricity that liberated 2.158 g of silver was passed through a solution of a gold salt and 1.314 g of gold was deposited. The equivalent mass of silver is 107.9. Calculate the equivalent mass of gold. also find oxidation state and valency of gold.

Given: W_Ag = 2.158 g, W_Au = 1.314 g, E_Ag = 107.9

To Find: E_Au =?

Solution:

By Faraday’s second law of electrolysis,

Ans: The equivalent mass of gold is 65.7 and its oxidation state is+3. Valency is 3.

Science > Chemistry > Concept of Atomic Mass and Equivalent Mass > Use of Laws of Electrolysis

The post Equivalent Mass Using Faraday’s Laws of Electrolysis appeared first on The Fact Factor.

In the last two articles, we have studied the hydrogen displacement method, oxide formation method, reduction method, and chloride formation method to determine the equivalent mass of metal. In this article, we shall study double displacement method and metal displacement method.

The equivalent mass of a substance is the number of parts by mass of the substance which combines with or displaces or contains 1.008 parts by mass of hydrogen, 8 part by mass of oxygen, or 35.5 part by mass of chlorine. If the equivalent mass is expressed in grams then it is called gram equivalent mass (GEM).

Illustration: 

1.008 parts by weight of Hydrogen combines with 35.5 parts by weight of chlorine to give 36.5 parts by weight of HCI. Thus the equivalent mass of chlorine is 35.5.

Equivalent mass has no unit because it is a pure ratio. Some important equivalent masses are H = 1, O = 8, Cl = 35.5

Method – V (Double Displacement Method):

Procedure:

In this method, a known mass of a compound (say AB) is treated with a known mass of another compound say (CD). By the exchange of radicals, the new compound is formed. The mass of the new compound formed is found.

AB    +   CD  →  AD   +   CB

Then equivalent mass is calculated using following formula.

Numerical Problems on Double Displacement Method

Example – 01:

0.106 g of sodium carbonate was treated with an excess of calcium carbonate and the mass of calcium carbonate was found to be 0.1 g. Find the equivalent mass of calcium carbonate if that of sodium carbonate is 53.

Given: Mass of Na₂CO₃ =  0.106 g. Mass of CaCO₃ = 0.1 g, Eq. Mass of  Na₂CO₃ = 53,

To Find: Eq. Mass of  CaCO₃ = ?

Solution:

Ans: The equivalent mass of CaCO3 is 50.

Example – 02:

0.194 g of chloride of a certain metal, when dissolved in water and treated with an excess of silver nitrate yield 0.50 g of silver chloride. Calculate the equivalent mass of the metal. (Ag = 108, Cl = 35.5)

Given: Mass of metal chloride =  0.194 g, Mass of silver chloride = 0.50 g,

To Find: Eq. Mass of  Metal = ?

Solution:

Let the equivalent mass of metal be E.

∴  0.194 × (108 + 35.5) = 0.50 ( E + 35.5)

∴  0.194 × 143.5 =0.50 E + 17.75

∴  27.839 – 17.75 = 0.50 E

∴  10.089 = 0.50 E

∴  E = 10.089 / 0.50 = 20.17

Ans: The equivalent mass of metal is 20.17.

Example – 03:

1.520 g of the hydroxide of metal gave 0.995 g of its oxide. Calculate the equivalent mass of the metal.

Given: Mass of hydroxide =  1.520 g, Mass of oxide = 0.995 g

Solution:

Let the equivalent mass of metal be E.

∴  1.520 × (E + 8) = 0.995 ( E + 17)

∴  1.520 E +  12.16  = 0.995 E + 16.915

∴  1.520 E – 0.995 E   =  16.915 – 12.16

∴  0.525 E = 4.755

∴  E = 4.755 / 0.525  = 9.06

Ans: The equivalent mass of metal is 9.06.

Example – 04:

1.0 g of an acid when completely acted upon by magnesium gave 1.301 g of anhydrous magnesium salt. Find the equivalent mass of the acid. Mg = 24, H = 1.

Solution:

Atomic mass of Mg = 24

Equivalent mass of Mg = Atomic mass /valency =  24 / 2 = 12

Mass of acid =  1.0 g

Mass of magnesium salt = 1.301 g

Let equivalent mass of acid be E.

∴  1 × (E + 12) = 1.301 ( E + 1)

∴  E + 12  = 1.301 E + 1.301

∴  12 – 1.301  = 1.301 E – E

∴  0.301 E = 10.699

∴  E = 10.699 / 0.301  = 35.54

Equivalent mass of acid = 1 + 35.54 = 36.54

Ans: Hence equivalent mass of acid is 36.54.

Example – 05:

Chloride of a metal ‘M’ contains 47.23% of the metal. 1.00 g of this metal displaced from a compound 0.88 g of another metal N. Find equivalent masses of M and N respectively.

Solution:

In chloride % of metal = 47.23, hence % of chlorine = 100 – 47.23 = 52.77

Let us consider 100 g of chloride

Mass of metal = 47.23 g, Mass of chlorine = 52.77 g

Hence equivalent mass of the metal M is 31.77.

Given that, 1.00 g of metal M displaced from a compound 0.88 g of another metal N.

Hence equivalent mass of N = 31.77 x 0.88 = 27.96

Ans: The equivalent mass of M is 31.77 and that of N is 27.96..

Example – 06:

4.215 g of metallic carbonate was heated in a hard glass tube and CO2 evolved was found to measure 1336 ml at 27 °C and 700 mm of pressure. What is the equivalent mass of the metal? (IITJEE 1976).

Given: V = 1336 ml, P = 700 mm of Hg, T = 27 °C = 27 + 273 = 300 K, P_O = 760 mm of Hg, T_O = 273 K

Solution:

Molecular mass of CO₂ = 12 + 16 x 2 = 44 g

1 mole of CO₂ at STP occupies 22.4 dm³ by volume.

44 g of CO₂ at STP occupies 22.4 dm³ by volume.

Mass of metal carbonate =  4.215 g

Mass of CO₂ evolved = 2.2 g

Mass of metal oxide  = 4.215 – 2.2 = 2.015 g

Let the equivalent mass of metal be E.

∴  4.215 × (E + 8) = 2.015 ( E + 30)

∴ 4.215 E + 33.72  = 2.015 E + 60.45

∴ 4.215 E –  2.015 E =  60.45 – 33.72

∴  2.2 E = 26.73

∴  E = 26.73 / 2.2  = 12.15

Ans: The equivalent mass of metal is 12.15.

Example – 07:

For dissolution of 1.08 g of metal 0.49 g of sulphuric acid was required. If the specific heat of metal is 0.06. Find the atomic mass of the metal.

Solution:

Mass of metal =  1.08 g

Mass of sulphuric acid = 0.49 g

Let the equivalent mass of metal be E.

Hence equivalent mass of metal is 108.

Specific heat = 0.06

Actual atomic mass = Equivalent mass x valency = 108 x 1 = 108

Ans: The atomic mass of the metal is 108.

Method – VI (Metal Displacement Method):

Procedure:

In this method known mass of metal is added to the solution of a salt of the other (Placed lower in electrochemical series).

Metal A + Salt of metal B → Salt of metal A + Metal B

The precipitate formed is washed dried and carefully weighed.

Equivalent mass is calculated using following formula.

Numerical Problems on Metal Displacement Method

Example – 01:

1.296 g of silver metal was displaced when 0.382 g of copper was added to the solution of silver sulphate. If the equivalent mass of silver metal is 108. Find that of copper.

Given: Mass of silver = 1.296 g, Mass of copper = 0.382 g, Eq. mass of silver = 108

To Find: Eq. mass of copper =?

Solutions:

Ans: The equivalent mass of copper is 31.83

Example – 02:

1.8 g of iron displaces 2.04 g copper from copper sulphate solution. If copper has an equivalent mass of 31.7. Find that of iron.

Given: Mass of iron = 1.8 g, Mass of copper = 2.04 g. Eq. mass of copper = 31.7

To Find: Eq. mass of iron =?

Solution: 

Ans: The equivalent mass of iron is 27.97.

Example – 03:

2.47 g of CuO obtained by oxidising 1.986 g of copper by nitric acid. 0.335 g of copper was precipitated by 0.346 g of zinc from CuSO4. Find the equivalent mass of copper and zinc.

Given: Mass of CuO = 2.47 g. Mass of copper = 1.986 g.

To Find: the equivalent mass of copper and zinc =?

Solution: 

Mass of oxygen = 2.47 – 1.986 = 0.484 g

Eq. mass of copper = 32.83

Ans: The equivalent mass of copper is 32.83 and that of zinc zinc is 33.91

Previous Topic: Equivalent Mass by Oxide Formation, Chloride Formation, and Reduction Method

Next Topic: Equivalent Mass Using Faraday’s Laws of Electrolysis

Science > Chemistry > Concept of Atomic Mass and Equivalent Mass > Double Displacement Method

The post Equivalent Mass by Double Displacement Method appeared first on The Fact Factor.

In the last article, we have studied the hydrogen displacement method to find the equivalent mass of an element. In this article, we shall study three more methods oxide formation method, reduction method, and chloride formation method.

The equivalent mass of a substance is the number of parts by mass of the substance which combines with or displaces or contains 1.008 parts by mass of hydrogen, 8 part by mass of oxygen, or 35.5 part by mass of chlorine. If the equivalent mass is expressed in grams then it is called gram equivalent mass (GEM).

Illustration: 

1.008 parts by weight of Hydrogen combines with 35.5 parts by weight of chlorine to give 36.5 parts by weight of HCI. Thus the equivalent mass of chlorine is 35.5.

Method – II (Oxide Formation Method):

Procedure:

A known mass of an element is reacted with oxygen. Mass of the oxide formed is measured. The mass of oxygen in the oxide is calculated using formula

Mass of oxygen = Mass of the oxide – Mass of the element

Numerical Problems on Oxide Formation Method:

Example – 01:

0.4 g of metal, when heated in air, gave 0.72 g of the metal oxide. Find the equivalent mass of the metal.

Given: Mass of metal = 0.4 g, Mass of oxide = 0.72 g

Solution:

Mass of oxygen = 0.72  – 0.4  = 0.32 g

Ans: The equivalent mass of the metal is 10.

Example – 02:

If the mass of the copper taken is 0.324 g and mass of the product on heating its nitrate is 0.406 g. calculate the chemical equivalent of copper.

Given: Mass of metal = 0.324 g, Mass of oxide = 0.406 g,

Solution:

Mass of oxygen = 0.406 – 0.324 = 0.082 g

Ans: Hence chemical equivalent of the metal is 31.61.

Example – 03:

1.08 g of metal oxide on heating decomposes to give pure metal and 56.0 ml of oxygen at NTP. What is the chemical equivalent of metal?

Given: Mass of metal oxide = 1.08 g, Volume of oxygen at NTP = 56.0 ml = 0.056 dm³.

Solution:

One mole of any gas occupies 22.4 dm³ by volume at NTP.

Mass of metal = Mass of oxide – Mass of oxygen = 1.08 – 0.08 = 1 g

Ans: The chemical equivalent of the metal is 100.

Example – 04:

0.139 g of metal, when dissolved in dilute hydrochloric acid, evolved 29.5 ml of hydrogen when collected over water at 13 °C and 741 mm pressure. What would be the mass of the oxygen present in 100 g of the oxide of the metal if the aqueous tension at 13 °C is 11.2 mm?

Given: W = 0.139 g, V = 29.5 ml, P = 741 mm of Hg, f = 11.2 mm of Hg, T = 13 °C = 13 + 273 = 286 K, P_O = 760 mm of Hg, T_O = 273 K.

Solution:

Let ‘x’ g be the mass of oxygen in the oxide

Mass of oxide = 100 g, Mass of oxygen = x g

∴ Mass of metal = (100 – x) g

∴  57.6 x = 800 – 8x

∴  65.6 x = 800

∴  x = 800/ 65.6 = 12.20 g

Ans: 100 g of metal contains 12.20 g of oxygen

Method – III (Reduction Method):

Procedure:

A known mass of a metal oxide is reduced to metal. Mass of the metal obtained is measured. The mass of oxygen in the oxide is calculated using formula

Mass of oxygen = Mass of oxide – Mass of element

The equivalent mass is calculated using the formula

Numerical Problems on Reduction Method:

Example – 01:

1.44 g of the metal oxide on reduction gave 0.8 g of metal. Find the equivalent mass of the metal.

Given: Mass of metal = 0.8 g, Mass of oxide = 1.44 g, Mass of oxygen = 1.44  – 0..8 = 0.64 g

Solution:

Ans: The equivalent mass of the metal is 10.

Example – 02:

On heating 0.8567 g of copper oxide in a current of hydrogen, the resulted in the formation of 0.6842 g of copper. Find the atomic mass of copper.

Given: Mass of copper = 0.6842 g, Mass of oxide = 0.8567 g

Solution:

Mass of oxygen = 0.8567 – 0.6842  = 0.1725 g

Ans: The equivalent mass of the metal is 31.63.

Method – IV (Equivalent Mass by Chloride Formation Method):

Procedure:

A known mass of a metal is reacted with chlorine. Mass of the chloride obtained is measured. The mass of chlorine in the chloride is calculated using formula

Mass of chlorine = Mass of chloride – Mass of the element

The equivalent mass is calculated using the formula

Example – 01:

2.00 g of metal yielded 2.656 g of its chloride. Find the equivalent mass of the metal.

Given: Mass of metal = 2.00 g, Mass of chloride = 2.656 g

Solution:

Mass of chlorine = 2.656 – 2.00 = 0.656 g

Ans: The equivalent mass of the metal is 108.2.

Example – 02:

The chloride of a metal contained 52.85 % of metal. What is the equivalent mass of the metal?

Given: % of metal = 52.85,

Solution:

% of Chlorine = 100 – 52.85 = 47.15, Consider 100 g of chloride

Mass of metal = 52.85 g, Mass of chloride = 47.15 g

Ans: The equivalent mass of the metal is 39.79.

In the next article, we shall study double displacement method and metal displacement method to determine equivalent mass of a substance.

Previous Topic: Equivalent Mass by Hydrogen Displacement Method

Next Topic: Equivalent Mass by Double Displacement Method

Science > Chemistry > Concept of Atomic Mass and Equivalent Mass > Oxide Formation, Reduction, and Chloride Formation Method

The post Equivalent Mass By Oxide Formation, Reduction, and Chloride Formation Method appeared first on The Fact Factor.

In the last few articles, we have studied the concept of atomic mass and methods to determine it. In this article, we shall study the hydrogen displacement method to determine the equivalent mass of a metal.

The equivalent mass of a substance is the number of parts by mass of the substance which combines with or displaces or contains 1.008 parts by mass of hydrogen, 8 part by mass of oxygen, or 35.5 part by mass of chlorine. If the equivalent mass is expressed in grams then it is called gram equivalent mass (GEM).

Illustration: 

1.008 parts by weight of Hydrogen combines with 35.5 parts by weight of chlorine to give 36.5 parts by weight of HCI. Thus the equivalent mass of chlorine is 35.5.

Equivalent mass has no unit because it is a pure ratio. Some important equivalent masses are H = 1, O = 8, Cl = 35.5

Hydrogen Displacement Method:

Principle:

The known mass of a metal to react with dilute acids and volume of hydrogen produced in the reaction is measured and the equivalent mass of an element is calculated using formula

This method is useful for the metals which react, or dissolves in mineral acids and liberates hydrogen gas. e.g. Mg, Zn, Al, Ca, Zn, Sn etc.

Procedure:

Clean and weigh accurately a piece of metal (Mg / Zn / Al) whose equivalent mass is to be found and place it in a conical flask containing distilled water.

The mouth of the conical flask is fitted with a cork through which side tube (gas carrying tube) and a thistle tube are inserted as shown in the diagram. Using side tube the conical flask is connected to graduated (calibrated) test tube called Eudiometer. Eudiometer tube is completely filled with water and is inverted on the side tube as shown in the diagram. The bottom part of thistle tube is dipped in the water in the flask. Mineral acid like HCl is added to the flask through the thistle funnel.

The reaction between the metal and acid takes place and hydrogen gas is liberated which is collected in eudiometer by downward displacement of water.

When the evolution of gas has stopped the mouth of the eudiometer tube is closed with a thumb and carried to a tray of water where the level of water inside and outside the tube is equalised and the volume of hydrogen gas is read at atmospheric pressure.

Observations:

The weight of metal piece = W g

The volume of hydrogen collected = V dm³

Atmospheric pressure  = P mm of Hg

Aqueous tension = f mm of Hg

Room temperature = t °C

Absolute temperature = T K

Calculations:

Now the volume of hydrogen at S.T.P. from the above data using following formula is calculated.

Where Volume of hydrogen at S.T.P.  V_o dm³,

Pressure at S.T.P. P_o mm of Hg = 760 mm = 1.013 x 10⁵ N/m²

The absolute temperature at S.T.P.  T_o K = 273 K

Notes:

• Hydrogen is collected over water and it is moist. Hence, the pressure of hydrogen = (P – f) mm of Hg is used.
• This method does not give accurate results.

Numerical Problems

Example – 01:

0.205 g of a metal on treatment with a dil. acid gave 106.6 ml of hydrogen collected over water at 755 mm pressure and 17 °C. Calculate the equivalent mass of the metal if aqueous tension at 17 °C is 14.4 mm.

Given: W = 0.205 g, V = 106.6 ml, P = 755 mm of Hg, f = 14.4 mm of Hg, (P – f) = 755 – 14.4 = 740.6 mm of Hg, t = 17 °C, T = 17 + 273 = 290 K, P_o = 760 mm of Hg, T_o = 273 K.

Solution:

Ans: the equivalent mass of the metal is 23.5

Example – 02:

0.33 g of metal on dissolving in dilute sulphuric acid, liberates 113 ml of dry hydrogen at NTP. Determine the equivalent mass of the metal.

Given: W = 0.33 g, V_o = 113 ml = 0.113 dm³

Solution:

Ans: the equivalent mass of the metal is 32.71

Example – 03:

0.05 g of metal on treatment with a dilute acid gave 51 ml of hydrogen collected over water at 751 mm pressure and 27 °C. Calculate the equivalent mass of the metal.

Given: W = 0.05 g, V = 51 ml, P = 751 mm of Hg, f = 0 (not mentioned in problem),  P – f = P, t = 27 °C, T = 27 + 273 = 300 K, P_o = 760 mm of Hg, T_o = 273 K

Solution: 

Ans: the equivalent mass of the metal is 12.21

Example – 04:

0.0396 g of metal was completely dissolved in dilute hydrochloric acid and hydrogen evolved is mixed with dry oxygen. Then the mixture is sparked. 13.75 ml of dry oxygen at 27 °C and 680 mm pressure is required for complete combustion of oxygen formed. Find the equivalent mass of the metal.

Given: W = 0.0396 g, P = 680 mm of Hg, f = 0 (not mentione), t = 27O C, T = 27 + 273 = 300 K, P_o = 760 mm of Hg, T_o = 273 K

Solution: 

Combustion of hydrogen is represented as

2 H_2(g) + O_2(g)   → 2H₂O_(g))

2 vol          1 vol           2 vol

Thus  1vol of oxygen combines with 2 vol of hydrogen.

Using Gay-Lussac’s law of combining volume we can say that

13.75 ml of oxygen combines with 2 x 13.75 = 27.5 ml of hydrogen.

∴  V =  27.5 ml

Ans: the equivalent mass of the metal is 19.80

In the next article, we shall study oxide formation method, reduction method, and chloride formation method to determine equivalent mass.

Previous Topic: Atomic Mass by Dulong Petit’s Law

Next Topic: Equivalent Mass by Oxide formation, Chloride Formation, and Reduction Method

Science > Chemistry > Concept of Atomic Mass and Equivalent Mass > Hydrogen Displacement Method

The post Equivalent Mass By Hydrogen Displacement Method appeared first on The Fact Factor.

In this article, we shall study different methods of preparation of colloids.

Preparation of Lyophilic Sols:

For preparing lyophilic sol, the dispersed phase is directly added to dispersion medium in cold or by warming.

Colloidal solutions of starch, glue, gelatin, etc. in water can be prepared by this method. Solutions of colloidal electrolytes such as soaps and dyestuffs can also be prepared by this method.

Preparation of Lyophobic Sols:

For preparing lyophobic sol, the substance in bulk is broken down into particles of colloidal dimensions (Dispersion) or aggregating smaller particles into particles of colloidal dimensions (condensation). To improve the stability of sol certain substances are added to the sol, the substances added are called stabilizers.

Dispersion methods:

In the dispersion method particle of larger size are broken down to the colloidal size in the dispersion medium. Starting with the material in massive form, a colloidal solution is prepared by using suitable devices to disintegrate it into particles of colloidal size. Normally this is carried out by physical methods.

Mechanical dispersion method :

The substance which is to be dispersed is finely ground. It is then mixed with the dispersion medium, protective materials or stabilizer is also added when a coarse suspension is obtained. This suspension is then passed through a colloid mill.  A colloid mill consists of two heavy metal discs placed one above the other separated by a very small gap from each other.  They are rotated in the opposite directions at a very high speed of about 7000 r.p.m. The sol results due to the large shearing effect. Protective material used prevents particles from coagulation.

Using this method sols of indigo, sulphur, toothpaste, printer ink, paints, ointments etc. are prepared.

Electrical Dispersion or Bredig’s Arc Method:

This method used to prepared metal sols like platinum, silver, gold, copper in water. A dispersion medium (conductivity water) and a trace of sodium hydroxide (the stabilising agent) is taken in porcelain or glass (non conducting) vessel. The vessel containing dispersion medium is surrounded by a freezing mixture. Metal to be dispersed is dipped in the vessel in the form of electrodes. Electrodes are connected to the high voltage source. The ends of electrodes in the dispersion medium are very near to each other. A very high voltage is applied and then an electrical arc is struck between the tips of electrodes. This creates large heat due to which metal rods melt, evaporate and suddenly cooled due to freezing mixture gives rise to the colloidal solution of the metal.

Functions of the freezing mixtures are

• Freezing mixture helps in condensation of metal vapours forming the colloids
• It prevents vapourisation of water.
• It prevents coagulation of colloids, by keeping sol cold.

Peptization or Chemical Dispersion:

Redispersion of freshly prepared precipitate into the sol by adding an electrolyte containing common ion is called as peptization. An electrolyte used for peptization is called as the peptizing agent. Peptization is a reverse process of coagulation. The peptization action is due to the preferential adsorption of one of the ions of the electrolyte on the precipitate.

Example – 1: 

Freshly prepared Fe(OH)₃ precipitate when treated with dilute solution of FeCl₃, reddish brown ferric hydroxide sol is formed (Fe³⁺ being common ion). In this case, FeCl₃ is the peptizing agent.

Fe (OH)₃ +   FeCl₃ →  [Fe(OH)₃] Fe³⁺

Example – 2: 

Fresh Silver chloride precipitate when treated with a small amount of dilute HCl, a silver chloride sol is formed.

Example – 3:

Cadmium sulphate can be peptized with the help of hydrogen sulphate.

Ultrasonic Dispersion:

Ultrasound is a very effective processing method in the generation and application of colloidal size particles. High-intensity ultrasonic waves are used for this purpose. During sonicating of liquids the ultrasonic waves that propagate through dispersion medium result in alternating high-pressure (compression) and low-pressure (rarefaction) cycles. This mechanical stress causes Ultrasonic cavitation in liquids. It creates high-speed liquid jets of up to 1000km/h. Such jets press liquid at high pressure between the particles and separate them from each other. Smaller particles are accelerated with the liquid jets and collide at high speeds.

Various substances like oils, mercury, sulphur, sulphides and oxides of metal can be dispersed in a colloidal state by this method.

Condensation methods:

These methods involve chemical reactions. In these methods factors like temperature, pressure, concentrations.  etc. are properly maintained.  The unwanted ions present in the sol are removed by dialysis , as these ions may eventually coagulate the sol.

Chemical Methods:

Oxidation Method:

Preparation of Colloidal Sulphur: 

When H₂S in water (aqueous solution) is exposed to air, it slowly gets oxidised to sulphur. The sulphur so formed remains in water in the colloidal state and the solution so formed remains in water in the colloidal state and the solution has a slightly milkish appearance.

H₂S   +     O2  →   H₂O  +    2S (colloidal)

A sol of sulphur can also be prepared when H₂S gas is bubbled through an aqueous solution of SO₂.

H₂S  +     SO₂ →  2 H₂O   +  3S (colloidal)

Reduction Method:

Preparation of Gold Sol: 

A number of metals like silver, gold, platinum, mercury lead can be obtained in the colloidal state by the reduction of their salt solutions (dilute) using suitable reducing agents like hydrogen sulphide, formaldehyde, stannous chloride, tannic acid etc.

Gold sol can be obtained when AuCl₃(dil) solution is treated with stannous chloride.

2 AuCl₃ +  3 SnCl₂ → 3 SnCl₄ +  2 Au (colloidal)

Similarly, silver, platinum mercury sols are prepared.

AgNO₃  +  Tannic acid  →       Ag sol

AuCl₃  +  Tannic acid  →       Au sol

Hydrolysis Method:

Preparation of Ferric Hydroxide Sol:

A colloidal solution of ferric hydroxide is obtained by boiling a dilute solution of ferric chloride.

FeCl₃ + 3H₂O →  Fe(OH)₃   + 3 HCl

Preparation of Silicic Acid Sol:

By hydrolysis of a dilute solution of sodium silicate with a hydrochloric acid, the colloidal solution of silicic acid is obtained.

Double Decomposition Method:

Preparation of Arsenious Sulphide Sol: 

Arsenious sulphide, As₂S₃ is a lyophobic colloid. It is obtained by the hydrolysis of arsenious oxide (AS₂0₃) with boiling distilled water, followed by passing H₂S gas through solution obtained. In the colloidal solution of arsenious sulphide, each particle is surrounded by HS- ions, produced by the dissociation of H₂S. This sulphide ion layer is further surrounded by the counter ion layer of H⁺ ions.

As₂O₃  + 3 H₂O  →     2As(OH)₃  (boiling)

2 As(OH)₃ + 3H₂S   →   As₂S₃  + 6H₂O

(light yellow sol)

By Exchange Solvent Method:

There are a number of substances whose colloidal solutions can be prepared by taking a solution of the substance in one solvent and pouring it into another solvent in which the substance is relatively less soluble.

Preparation of Sulphur or Phosphorous Sol:

If a solution of sulphur or phosphorus prepared in alcohol is poured into water, a colloidal solution of sulphur or phosphorus is obtained due to the low solubility of sulphur or phosphorus in water.

By change of physical state: 

A colloidal solution of certain elements such as mercury and sulphur are obtained by passing their vapours through cold water containing a stabilizer ( an ammonium salt or a citrate).

Excessive Cooling Method:

A colloidal solution of ice in an organic solvent like ether or chloroform can be prepared by freezing a solution of water in the solvent. The molecules of water which can no longer be held in solution, separately combine to form particles of colloidal size.

Purification of Colloidal Solution

Dialysis:

The process of separating the particles of colloid from those of crystalloid, by means of diffusion through a suitable membrane (animal membrane or parchment paper) is called dialysis. The apparatus used for the performing dialysis is called dialyser.

Principle:

The colloidal particles can not pass through a parchment or cellophane membrane while the ions of the electrolyte (crystalloids) can pass through it.

Process:

A bag made up of suitable semipermeable membrane containing the colloidal solution is suspended in a vessel through which fresh water is continuously flown. The molecules and ions of crystalloids diffuse through the membrane into the water and are washed away. Thus the sol in the bag is purified.

Dialysis can be used for removing HCl from the ferric hydroxide sol.

Electrodialysis:

The ordinary process of dialysis is slow. (ii) To increase the process of purification, the dialysis is carried out by applying an electric field. This process is called electrodialysis. The ions present in the colloidal solution migrate towards oppositely charged electrodes.

The important application of electrodialysis process in the artificial kidney machine used for the purification of the blood of the patients whose kidneys have failed to work.

Ultrafiltration:

The pores of ordinary filter paper are large, hence colloidal particles pass through them easily. If the pores of the ordinary filter paper are made smaller by soaking the filter paper in a solution of gelatin of colloidion (it is a mixture of 4% nitro-cellulose in alcohol and ether) and subsequently hardened by soaking in formaldehyde

The treated filter paper may retain colloidal particles and allow the true solution particles to escape. Such filter paper is known as ultrafilter and the process of separating colloids by using ultrafilters is known as ultrafiltration.

The colloidal particles left on ultrafilter paper are then washed with a fresh dispersion medium to get a pure colloidal solution.

Ultracentrifugation:

In this method, the colloidal solution is placed in a high-speed centrifugal machine. On centrifuging, the colloidal particles settle down. The impurities remain in the centrifugate and are removed. The settled colloidal particles are mixed with the dispersion medium to form the colloidal solution again.

Science > Chemistry > Colloids > Preparation of Colloids

The post Preparation of Colloids appeared first on The Fact Factor.