EMF (electromotive force) of a cell

The maximum potential difference that exists between two electrodes of a cell is called the electromotive force (EMF) of the cell.

In other words, electromotive force is difference in potential which causes the current to flow from an electrode of higher potential to an electrode of a lower potential.

• It is also known as cell potential
• It is measured in volts.

The emf of a cell measured under standard conditions is called standard emf. It is denoted by ‘E⁰_cell’. The standard conditions are:

1. Concentration of electrolytic solution is 1M.
2. Temperature of the system is 25⁰C.
3. Pressure of the gas is 1 atmosphere.

Q) A galvanic cell is represented as

Zn/Zn⁺⁺//Cd⁺⁺/Cd

Write the ell reaction and calculate the standard emf of the cell.

Given that E⁰_Zn⁺⁺_/Zn = -0.70V and E⁰_Cd⁺⁺_/Cd = -0.40V

Nernst equation

The Nernst equation provides the relation between the cell potential of an electrochemical cell, the standard cell potential, temperature and the equilibrium constant.

Mathematically, the Nernst equation can be expressed as:

For an oxidation half-cell reaction, when the metal electrode M gives Mⁿ⁺ ion,

The Nernst equation takes the form:

The concentration of solid metal [M] is equal to zero. Hence,

At 25⁰C:

On putting the values of R, F and T at 25⁰C, the quantity 2.303RT/F comes to be 0.0591.

Thus, at 25⁰C, the Nernst equation can be written as:

• This equation is for half cell in which oxidation occurs.
• In case of reduction half reaction, the sign of E is reversed.
• If the Nernst equation is applied to the complete cell as a whole, then it will be in the form:

Q) Calculate the emf of the cell.

Zn/Zn⁺⁺(0.001M)//Ag⁺(0.1M)/Ag

The standard electrode potential E0 of Ag/Ag⁺ is 0.80V and Zn/Zn⁺⁺ is -0.76V.

Ans….

Nernst equation applications

Nernst equation can be used to calculate the following:

• Single electrode potential (oxidation or reduction potential) at any conditions.
• Standard electrode potential.
• Comparing the relative ability as an oxidizing or reducing agent.
• Emf of an electrochemical cell
• Unknown ionic concentrations.
• pH of a solution also can be measured using Nernst equation.

References

• Atkins, Peter, Paula, de Julio, Atkin’s Physical Chemistry, Seventh Edition, Oxford University Press, (Printed in India, 2002).
• Gurtu, J.N., Snehi, H., Advanced Physical Chemistry, Seventh Edition, Pragati Prakashan India, 2000.
• Madan, R.L., tuli, G.D., Physical Chemistry, S. Chand and company, New Delhi, 2012.

The post EMF of a cell and Nernst equation: appeared first on Online Chemistry notes.

• Ionic strength of a solution can be defined as the total concentration of ions present in the solution.
• In another word, ionic strength measures the concentration of ionic atmosphere in a solution.
• It is dimensionless quantity, it has no unit.
• It is denoted by ‘µ’.

Mathematically,

Q) Calculate the ionic strength of 0.1M solution of aluminium sulphate.

Ans.

Q) Calculate the ionic strength of 0.1M KCl and 0.1M CaCl₂ solution.

Activity and Activity coefficient

In electrolytic solution, the experimentally determined value of concentration of ions is less than the actual concentration. The effective concentration of ions or electrolyte in a solution is called as activity. It is denoted by a symbol ‘a’.

Mathematically, activity ‘a’ is taken as the product of actual concentration in molarity or molality and activity coefficient ‘f’.

i.e. a = Cf ——– (i)

Where,

C = Concentration in molarity or molality

f = activity coefficient.

• For very dilute solution, activity coefficient is nearly equal to one. So, the activity becomes equal to the actual concentration.

i.e. a = c

• For concentrated solution, activity coefficient is less than one.

Rearranging equation (i),

F = a/C

Thus, activity coefficient is defined as the ratio of the activity to the actual concentration.

The activity ‘a’ of the electrolyte is taken as the product of activities of cation and anion.

i.e. a = a₊ a_–

Where,

a₊ = activity of cation

a_–= activity of anion

Similarly, activity coefficient of an electrolyle is taken as the product of activity coefficients of cation and anion.

i.e. f = f₊ f_–

where,

f₊ = activity coefficient of cation

f_– = activity coefficient of anion

The activity and activity coefficient can’t be measured experimentally but their mean value can be determined.

Debye-Huckel limiting law- Expression for the activity coefficient of electrolyte in terms of ionic strength

• Debye-Huckel limiting law relates the mean activity coefficient of an electrolyte with valency of ions and ionic strength of the solution.
• This law is applicable only for very dilute solution. So, this law is called limiting law.
• Mathematically, this law can be expressed as:

If -log is plotted against , a straight line passing through the origin having slope equal to AZ₊Z_– is obtained.

Application of Debye-Huckel limiting law:

Debye-Huckel limiting law can be used to calculate the mean activity coefficient of an electrolyte if the ionic strength or concentration of the solution and the value of ‘A’ is known.

References

• Atkins, Peter, Paula, de Julio, Atkin’s Physical Chemistry, Seventh Edition, Oxford University Press, (Printed in India, 2002).
• Gurtu, J.N., Snehi, H., Advanced Physical Chemistry, Seventh Edition, Pragati Prakashan India, 2000.
• Madan, R.L., tuli, G.D., Physical Chemistry, S. Chand and company, New Delhi, 2012.

The post Debye-Huckel limiting law, Ionic strength, Activity and Activity coefficient appeared first on Online Chemistry notes.

An acid-base indicator is a weak organic acid or base that indicates the end point of acid base titration by a visual change in colour. They are weak acid or base having different colour in basic and acidic solution.

Phenolphthalein and methyl orange are two common examples of acid base indicator.

Mechanism of acid-base indicator action – Ostwald’s theory

According to Ostwald’s theory,

1. Acid-base indicators are regarded as organic weak acid or weak base.
2. Acid-base indicators undergo partial ionization in water to furnish H⁺ ions and OH^– ions.
3. Undissociated form of indicator has different colour than the dissociated form of indicator.

The colour produced by indicator depends on the pH of the solution.

• Phenolphthalein is a weak organic acid. It is colourless at its unionized state while it has pink colour in its ionized state.

In acidic medium, equilibrium shifts to the left and solution is colourless. In alkaline medium, equilibrium shifts to the right forming pink coloured solution.

• Similarly, methyl orange is a weak organic base. Its unionized molecule is light yellow in colour but its ions are red in colour.

In alkaline medium, equilibrium shifts to the left and solution becomes light yellow. In acidic medium, equilibrium shifts to the right and solution becomes red.

Selection of indicator in acid-base titration

Indicators are the chemical species that indicates the end point by changing its own colour. Each pH indicator has its own pH range for its colour change. In order to determine the accurate end point of acid-base titration, the indicator should be selected in such a way that the pH range for the colour change of the indicator must coincide with the pH change (jump) at the end point of reaction.

Mainly two types of indicators i.e. methyl orange and phenolphthalein are used during acid-base titration. Methyl orange has pH range 3.1-4.4 and phenolphthalein has 8.2-10.

There are four types of acid base titrations.

1.Strong acid vs strong base titration:

The indicators pH range of both methyl orange (i.e. 3.1-4.4) and phenolphthalein (i.e. 8.2-10) coincide with the pH jump (i.e. 3-11) at end point. Hence, either methyl orange or phenolphthalein can be used as indicator during strong acid vs strong base titration.

2. Strong acid vs weak base titration:

The indicators pH range of only methyl orange (i.e. 3.1-4.4) coincides with the pH jump (i.e. 3-8). Hence, only methyl orange can be used as indicator during strong acid vs weak base titration.

3. Weak acid vs strong base titration:

The indicators pH range of phenolphthalein (i.e. 8.2-10) only coincides with the pH jump (i.e. 6-11). Hence, only phenolphthalein can be used as indicator during weak acid vs strong base titration.

4.Weak acid vs weak base titration:

The indicators pH range of neither methyl orange (3.1-4.4) nor phenolphthalein (i.e. 8.2-10) coincides with the pH jump ( 6-8). Hence, weak acid vs weak base cannot be usually titrated due to lack of suitable indicators.

Note : Some acid and base examples

Strong acid – HCl, HNO₃, H₂SO₄, etc.

Strong base – NaOH, KOH, Ca(OH)₂, etc.

Weak acid – formic acid, acetic acid, oxalic acid, etc.

Weak base – NH₄OH, Cu(OH)₂, Fe(OH)₃, etc.

References

• Negi, A.S., Anand, S.C., A Text Book of Physical Chemistry, Seventh Edition, New Age International Pvt. Ltd. Publishers, 1999.
• Atkins, Peter, Paula, de Julio, Atkin’s Physical Chemistry, Seventh Edition, Oxford University Press, (Printed in India, 2002).
• Gurtu, J.N., Snehi, H., Advanced Physical Chemistry, Seventh Edition, Pragati Prakashan India, 2000.
• Madan, R.L., tuli, G.D., Physical Chemistry, S. Chand and company, New Delhi, 2012.

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A solution which can resist the change in pH even on addition of small amount of acid or base is called buffer solution.

Example: In a biological system, blood is an example of buffer and its pH remains almost constant to 7.4 (i.e. 7.35-7.45). Blood contains a buffer of carbonic acid (H₂CO₃) and bicarbonate anion (HCO₃^–).

Types of buffer solution

1. Acidic buffer solution: The buffer solution prepared by mixing equimolar quantities of weak acid and its salt with strong base is known as acidic buffer solution.

The pH value of acidic buffer solution is less than 7.

Example: A solution containing equimolar quantities of acetic acid (CH₃COOH) and sodium acetate (CH₃COONa) is acidic buffer.

2. Basic buffer solution: The buffer solution prepared by mixing equimolar quantities of weak base and its salt with strong acid is known as basic buffer solution.

The pH value of acidic buffer solution is more than 7.

Example: A solution containing equimolar quantities of ammonium hydroxide (NH₄OH) and ammonium chloride (NH₄Cl) is basic buffer.

Mechanism of buffer action

Mechanism of buffer solution can be explained by considering acidic buffer solution as:

• Suppose, an acidic buffer prepared by adding equimolar quantities of CH₃COOH and CH₃COONa, where CH₃COOH is weakly ionized and CH₃COONa is strongly ionized.

• If we add few drops of acid (HCl) to the buffer solution then it provides H⁺ ions to the buffer. These H⁺ ions would combine with CH₃COO^– ions present in the buffer solution to form weakly ionized acetic acid as,

• Similarly, if we add small amount of NaOH to the buffer solution then it provides OH^– ions to the buffer. These OH^– ions would combine with H+ ions present in the buffer solution to form undissociated H₂O molecules.

• Thus, addition of H⁺ ions is neutralized by CH₃COO^– ions and OH^– is neutralized by H⁺ ions. Therefore, pH of the solution remains unchanged.

Buffer capacity and Buffer range

Buffer capacity : Buffer capacity can be defined as the ability of a solution to resist rapid changes in pH.

In other words, buffer capacity is defined as the number of moles of acid or base added per litre of the buffer required to cause a unit change in pH.

Buffer range: A buffer solution can be designed for any pH range, but a given buffer solution will work effectively only over a particular range.

The pH range over which a buffer solution is effective is termed as buffer range. For acidic buffer solution approximate pH range is pH=pK_a±1 and pOH=pK_b±1 for basic buffer solution.

Applications of buffer solution

Buffer solutions are important in various fields of science and industry due to their ability to maintain a stable pH level. Some of their key applications are:

1. Laboratory work: Buffers are essential in chemical and biological laboratories to maintain constant pH during experiments and reactions. This helps to ensure the accuracy and reproducibility of results.
2. Analytical chemistry: Instruments like ph meter depend on buffer solutions to calibrate and accurately measure the pH of other solutions.
3. Medicinal and clinical diagnostics: Buffers are used in various medical diagnoses, including blood tests.
4. Pharmaceuticals: Buffers are used in the formulation of drugs to ensure the stability and effectiveness of drugs, especially in intravenous medicines.
5. Water treatment: Buffer solutions are sometimes used to adjust and maintain the pH of water in swimming pools, aquariums and industrial water treatment processes.

Handerson-Hasselbalch equation – Determination of pH of buffer

Handerson-Hasselbalch equation is a formula used to calculate the pH of a buffer solution. It is expressed as:

Derivation of Handerson-Hasselbalch equation :

Preparation of buffer solution

Handerson-Hasselbalch equation can be used for the calculation while preparing buffer solution. Example:

Q) The dissociation constant of acetic acid is 1.75×10^-5. How will you prepare a buffer solution of 4.4 by mixing sodium acetate and acetic acid.

Ans..

References

• Atkins, Peter, Paula, de Julio, Atkin’s Physical Chemistry, Seventh Edition, Oxford University Press, (Printed in India, 2002).
• Gurtu, J.N., Snehi, H., Advanced Physical Chemistry, Seventh Edition, Pragati Prakashan India, 2000.

The post Buffer solution: Types,Mechanism,Uses and Preparation appeared first on Online Chemistry notes.

Redox reaction can be balanced by either of the following two methods:

• Oxidation number method
• Ion-electron method

Oxidation number method to balance redox reaction

The oxidation number of an oxidizing agent decreases and that of the reducing agent increases in a redox process. This change in oxidation number should be balanced to balance a redox reaction by the oxidation number method.

Balancing of redox reaction by oxidation number method involves the following steps:

Step I : Identifying the oxidizing and reducing agents.

Oxidizing and reducing agents are identified by observing the change in oxidation number of main elements present in oxidizing and reducing agents on reactant and product sides. Eg.

Here, Zn is oxidized and HNO₃ is reduced to Zn(NO₃)₂ and N₂O respectively.

So, Zn is reducing agent and HNO₃ is oxidizing agent.

Step II : Calculating the change in oxidation number.

The total change in oxidation number in oxidation and also in reduction is calculated after equalizing the number of atoms undergoing O.N. change on both sides. Eg.

Oxidation: total change in O.N. = 2-0 = 2

Reduction: Total change in O.N. = 1-5 = 4×2 =8

Step III : Balancing of oxidizing and reducing agents.

Number of atoms undergoing oxidation number change are balanced by criss-cross multiplication with change in oxidation number on reactant as well as product side. Eg.

Step IV : Balancing of other elements.

All the other atoms or radicals are balanced by hit and trial method.

• Balancing the equation with respect to all other atoms except hydrogen and oxygen.

• Finally, balancing hydrogen and oxygen.

It is balanced chemical equation.

• Note:

• For reactions taking place in acidic solutions, add H⁺ ions to the side deficient in hydrogen atoms.

• For reactions taking place in basic solutions, add H₂O molecules to the side deficient in hydrogen atoms and simultaneously add equal number to OH^– ions on the other side of the equation.

More examples:

Example – 1

Step I : Identifying the oxidizing and reducing agents.

Step II : Calculating the change in oxidation number.

Step III : Balancing of oxidizing and reducing agents.

Step IV : Balancing of other elements.

• Balancing the equation with respect to all other atoms except hydrogen and oxygen.

• Finally, balancing hydrogen and oxygen.

It is balanced chemical equation.

Example – 2

Step I : Identifying the oxidizing and reducing agents.

Step II : Calculating the change in oxidation number.

Step III : Balancing of oxidizing and reducing agents.

Step IV : Balancing of other elements.

It is balanced chemical equation.

Example-3 :

Step I : Identifying the oxidizing and reducing agents.

Here, Cl₂ undergoes self oxidation and reduction (i.e. disproportionation reaction). So, Cl₂ is written twice to calculate the change in oxidation number.

Step II : Calculating the change in oxidation number.

Step III : Balancing of oxidizing and reducing agents.

Step IV : Balancing of other elements.

It is balanced chemical equation.

Example-4 :

Step I : Identifying the oxidizing and reducing agents.

Step II : Calculating the change in oxidation number.

Here, more than two elements have changed in the oxidation number.

Change in O.N. of As = +2 X 2 = +4

Change in O.N. of S = +2 X 3 = +6

Change in O.N. of N = – 3

Thus, total increase in O.N. = 4+6=10

Total decrease in O.N. = 3

Step III : Balancing of oxidizing and reducing agents.

Step IV : Balancing of other elements.

It is balanced chemical equation.

Ion-Electron method [ Half reaction method] to balance reaction

This method is based on the fact that number of electrons lost during oxidation half reaction of redox reaction is equal to number of electrons gained during reduction half reaction.

Following steps are involved in balancing redox rection by ion-electron method:

• Write the given equation and identify the oxidized and reduced species by analyzing change in oxidation number.
• Write the oxidation and reduction half reactions (ionic if required)
• Balance two half reactions separately.

1. Balance main element (i.e. atom whose O.N. is changed) by hit and trial.
2. Balance oxygen by adding H₂O.
3. Balance hydrogen by adding H⁺.
4. Balance charge by adding electrons.
5. In basic medium, OH^– ions are added on either side to cancel out H⁺ ions.

• Add two half reactions by making electrons gained equal to electrons lost in two half reactions.
• Finally, convert the balanced ionic equation to balanced molecular equation (if required).

Some Examples:

Example-1 :

Counting of oxidation number:

Identification and balancing of two half reactions:

Oxidation half reaction:

Balancing charge by adding electrons:

This is balanced oxidation half reaction.

Reduction half reaction:

Balancing main element by hit and trial:

Balancing oxygen by adding H₂O:

Balancing hydrogen by adding H⁺:

Balancing charge by adding electrons:

This is balanced reduction half reaction.

Adding two half reactions:

Balanced two half reactions are added after multiplying oxidation half by 4 to cancel electrons.

This is balanced ionic equation. To convert it into molecular form, 8 NO₃^– ions are added on both side of equation.

This is balanced molecular equation.

Example-2:

Counting of oxidation number:

Identification and balancing of two half reactions:

Oxidation half reaction:

Balancing charge by adding electrons:

This is balanced oxidation half reaction.

Reduction half reaction:

Balancing main element by hit and trial:

Balancing oxygen by adding H₂O:

Balancing hydrogen by adding H⁺:

Balancing charge by adding electrons:

This is balanced reduction half reaction.

Adding two half reactions:

Balanced two half reactions are added after multiplying oxidation half by 6 to cancel electrons.

This is balanced ionic equation. To convert it into molecular form, suitable number of K⁺ and SO₄^2- have to be added on both side of equation.

This is balanced molecular equation.

Example -3:

Counting of oxidation number:

Identification and balancing of two half reactions:

Oxidation half reaction:

Balancing main element by hit and trial:

Balancing charge by adding electrons:

This is balanced oxidation half reaction.

Reduction half reaction:

Balancing oxygen by adding H₂O:

Balancing hydrogen by adding H⁺:

Balancing charge by adding electrons:

This is balanced reduction half reaction.

Adding two half reactions:

This is balanced equation.

Example-4: [ Basic medium ]

Counting of oxidation number: Here, chlorine is oxidized as well as reduced.

Identification and balancing of two half reactions:

Oxidation half reaction:

Balancing oxygen by adding H₂O:

Balancing hydrogen by adding H⁺.

The reaction is carried out in basic medium, so, all H⁺ ions must be neutralized by adding same number of OH^– ions on both sides.

Balancing charge by adding electrons:

This is balanced oxidation half reaction.

Reduction half reaction:

Balancing oxygen by adding H₂O:

Balancing hydrogen by adding H⁺:

The reaction is carried out in basic medium, so, all H⁺ ions must be neutralized by adding same number of OH^– ions on both sides.

Balancing charge by adding electrons:

This is balanced reduction half reaction.

Adding two half reactions:

This is balanced ionic equation. In order to convert it into molecular form, 8K⁺ ions are added on both sides of equation.

This is balanced molecular equation.

Example-5:

Counting of oxidation number: Here, chlorine is oxidized as well as reduced.

Identification and balancing of two half reactions:

Oxidation half reaction:

Balancing main element by hit and trial:

Balancing oxygen by adding H₂O:

Balancing hydrogen by adding H⁺:

The reaction is carried out in basic medium, so, all H⁺ ions must be neutralized by adding same number of OH^– ions on both sides.

Balancing charge by adding electrons:

This is balanced oxidation half reaction.

Reduction half reaction:

Balancing main element by hit and trial:

Balancing charge by adding electrons:

This is balanced reduction half reaction.

Adding two half reactions:

This is balanced ionic equation. In order to convert it into molecular form, 6 Na⁺ ions are added on both sides of equation.

This is balanced molecular equation.

Self Practice 

Balance the following equations by the oxidation number method or ion-electron method.

References

• https://chemicalnote.com/oxidation-and-reduction-oxidants-and-reductants-and-redox-reaction/
• Atkins, Peter, Paula, de Julio, Atkin’s Physical Chemistry, Seventh Edition, Oxford University Press, (Printed in India, 2002).
• Gurtu, J.N., Snehi, H., Advanced Physical Chemistry, Seventh Edition, Pragati Prakashan India, 2000.
• Sthapit, M.K., Pradhananga, R.R., Foundations of Chemistry, Vol 1 and 2, Fourth edition, Taleju Prakashan, 2005.

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Thermodynamics is the branch of physical science that deals with the relationships between heat and other forms of energy (such as mechanical, electrical or chemical energy).

In broad terms, thermodynamics deals with the transfer of energy from one place to another and from one form to another.

Spontaneous and non-spontaneous process

Spontaneous process:

A process which may take place by itself or by initiation is called a spontaneous process.

In other words, a process which can take place by itself or has a tendency to take place is called spontaneous process.

Spontaneous process is simply a process which is feasible.

Examples of spontaneous process :

• Dissolution of common salt in water.
• Flow of water down a hill.
• Flow of heat from hot body to a cold body.
• Combination of H₂ and I₂ to form HI.

• Combination of hydrogen and oxygen to form water (when initiated by passing an electric spark).
• Reaction between CH₄ and O₂ to form CO₂ and H₂O (when initiated by ignition).

Non-spontaneous process:

A process which can neither take place by itself nor by initiation is called a non-spontaneous process.

Examples of non-spontaneous process:

• Dissolution of sand in water.
• Flow of water up a hill.
• Flow of heat from low pressure to a high pressure.

Entropy

Entropy is a thermodynamic state quantity that is a measure of the randomness or disorderness of the system. It is denoted by the letter ‘S’.

Greater the disorderness in molecules, greater will be the entropy. Generally, entropy is inversely proportional to the intermolecular force of attraction. Hence,

∆S_gas > ∆S_liq > ∆S_solid

Mathematically,

Standard entropy: Entropy at standard condition i.e. 25⁰C, 1 atm pressure and 1 molar concentration is called standard entropy. It is denoted by ∆S⁰.

Types of entropy:

1. Entropy of vaporization: The entropy change when one mole of a liquid changes into vapour (gas) is called entropy of vaporization.

2. Entropy of fusion: The entropy change when one mole of a solid changes into liquid is called entropy of fusion.

3. Entropy of sublimation: The entropy change when one mole of a solid changes into vapour (gas) is called entropy of sublimation.

4. Entropy of combustion or oxidation: The entropy change when one mole of any substance is completely burnt or oxidized is called entropy of combustion or oxidation.

Limitations of first law of thermodynamics:

> This law only explain about the total heat content in the system but does not say anything about the direction of flow of heat.

> This law does not say whether the process (reaction) is spontaneous or not.

Second law of thermodynamics

Second law of thermodynamics can be defined in a number of ways as follows:

Kelvin Plank statement: It is impossible to construct an engine operating in a complete cycle which will convert all the heat energy into work i.e. no any system (or engine) will have 100% efficiency.

Clausius statement: It is impossible for a self acting machine to transfer heat from a body at lower temperature to a higher temperature without any external actions (agencies).

In terms of entropy: “In any spontaneous process, there is always an increase in entropy of the universe”.

In other words- “It is not possible to have a process in which the entropy of an isolated system is decreased”.

Entropy and spontaneity

According to second law of thermodynamics, a process will occur spontaneously if entropy of the universe increases. Therefore;

Gibb’s free energy

Amount of energy available( required) to perform a certain work is called free energy. Generally, the amount of heat energy required to perform a certain work is called Gibb’s free energy. It is denoted by letter ‘G’.

Mathematically,

G = H – TS

Where,

H= Enthalpy(heat content)

S= Entropy of the system

T = The absolute temperature.

Gibb’s- Helmholtz equation

According to definition of Gibb’s free energy,

G = H – TS —(i)

Let us consider G₁, H₁ and S₁ be the initial and G₂, H₂ and S₂ be the final free energy, enthalpy and entropy respectively. Then, for initial state:

G₁ = H₁ – TS₁ —(ii)

Similarly, for final state:

G₂ = H₂ – TS₂ —(iii)

From equation (ii) and (iii),

G₂ – G₁ = (H₂-TS₂) – (H₁-TS₁)

Or, G₂ – G₁ = H₂-TS₂ – H₁ + TS₁

Or, G₂ – G₁ = (H₂– H₁) – T(S₂ – S₁)

Or, ∆G = ∆H – T∆S —-(iv)

This equation (iv) is called Gibb’s-Helmholtz equation.

Standard free energy

Free energy change in between reactant and product at standard conditions i.e. 25⁰C, 1 atm pressure and 1M concentration is called standard free energy change. It is denoted by ∆G⁰.

Mathematically,

∆G⁰ = ∆H⁰ – T∆S⁰

For chemical reaction,

∆G⁰ = Ʃ∆G⁰_product – Ʃ∆G⁰_reactant

Free energy is also defined as thermodynamic function of a system which indicates the capacity of maximum useful work that can be done by the system.

Spontaneity (feasibility) of reaction on the basis of Enthalpy, Entropy and Gibb’s free energy

A reaction which occurs itself is called spontaneous or feasible reaction.

Numerical Problems

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The various units of energy are Joules(J), Ergs, Calories(Cal), etc.

1 J = 10⁷ ergs

1 Cal = 4.2 J

Some thermo – chemical terms

1. System : A system is the specific portion of universe in which the energy change is to be studied. Water taken in a beaker is an example of a system.

2. Surroundings : The remaining portion of universe which is not the part of system is called surrounding. System interacts with the surroundings.

3. Boundary : The part that separates system from the surrounding is called boundary. Boundary is that portion of universe through which system and surrounding can interact with each other. Boundary may be real or imaginary.

Different types of system

1. Open system : A system which can exchange matter as well as energy with the surrounding is called an open system. Eg. Coffee held in a cup.

2. Closed system : A system which can exchange energy but not matter with its surrounding is called a closed system. Eg. Coffee held in a closed metallic (steel) flask.

3. Isolated system: A system which can neither exchange energy nor any matter with the surrounding is called an isolated system.

State functions and path functions

A physical quantity is said to be state function if its value depends upon the initial and final state of the system only and does not depend upon the path by which this state has been attained . For example: temperature , pressure , volume , enthalpy, entropy, free energy, etc.

On the other hand, a physical quantity is said to be path function if its value depends upon the path by which this state has been attained . For example: work , heat, molar heat capacity, etc.

For example, a person standing on the fifth floor of a building has a fixed value of potential energy, irrespective of the fact whether he reached there by stairs or by a lift. Thus, the potential energy of the person is a state function. On the other hand, the work done by the legs of the person to reach the same height is not same in the two cases. Hence, work done is a path function.

Extensive and Intensive properties

Extensive properties: The property whose value depends upon the amount of the substance present in the system is called extensive property. For example: mass, volume, energy, etc.

Intensive properties: The property whose value does not depend upon the amount of the substance present in the system, but depends upon the nature of the substance is called intensive property. For example: temperature, pressure, density, concentration, surface tension, viscosity, etc.

Different types of thermodynamic process

i) Isothermal process: A thermodynamic process in which the temperature of the system remains constant is called an isothermal process. For an isothermal process, ∆T=0.

ii) Adiabatic process: A thermodynamic process in which heat of the system remains constant is called an adiabatic process. For an adiabatic process, ∆q=0.

iii) Isobaric process : A thermodynamic process in which pressure of the system remains constant is called an isobaric process. For an isobaric process, ∆p=0.

iv) Isochoric process : A thermodynamic process in which volume of the system remains constant is called an isochoric process. For an isobaric process, ∆v=0.

v) Cyclic process : When a system in a given state goes through a number of different processes and finally returns to its initial state, then the overall process is called cyclic process. For a cyclic process, ∆E=0 and ∆H=0.

Internal Energy (E)

The total amount of energy stored in a system (substance) under a given set of conditions is called internal energy. The internal energy is the sum of the different types of energies associated with atoms or molecules, such as electronic energy (E_e), nuclear energy (E_n), chemical bond energy (E_c), potential energy (E_p), and kinetic energy (E_k).

E = E_e + E_n + E_c + E_p + E_k

If the internal energy of a system in the initial state is E₁ and in the final state is E₂, then the change in internal energy (∆E) is :

∆E = E₂ – E₁

Enthalpy (H)

Enthalpy is the total heat content of a system. It is equivalent to the sum of the internal energy and the product of the pressure and volume of the system.

H = E + PV

Where,

H = Enthalpy, E = Internal Energy, P = Pressure and V = Volume of the system.

Enthalpy change(∆H) :

∆H = ∆E + P∆V

Different types of Heat of reaction or Enthalpy of reaction

1. Heat of combustion: The heat of combustion of a substance is defined as the heat change (usually heat evolved) when one mole of a substance is completely burnt or oxidized in oxygen. Eg.

Completely oxidized means:

For example, carbon may be oxidized to CO and CO₂ . Completely oxidation means oxidation to CO₂ and not to CO.

2. Heat of formation : The heat of formation of a substance is defined as the heat change i.e. heat evolved or absorbed when one mole of substance is formed from its constituent elements under a given conditions of temperature and pressure. It is usually represented by ∆H_f.

Standard heat of formation: The heat change i.e. heat evolved or absorbed when one mole of substance is formed from its constituent elements in standard states of temperature and pressure ( i.e. 298K and 1 atm) is called standard heat of formation. It is usually represented by ∆H_f⁰. Eg.

When one mole of CO₂ is formed from its elements – C and O₂ in standard state, 393.5KJ heat is produced. Hence, the standard heat of formation of CO₂ is 393.5KJ.

3. Heat (enthalpy) of neutralization: The heat change i.e enthalpy change when one gram equivalent of an acid is neutralized by base in dilute aqueous solution is called heat of neutralization.

For example: when one gram equivalent of HCl is neutralized by one gram equivalent of NaOH in dilute aqueous solution, 57.1KJ of heat is produced. Thus, heat of neutralization of HCl with NaOH is 57.1KJ.

• Heat of vaporization– Enthalpy change when one mole of a substance changes from liquid state into vapour state.
• Heat of solution– Enthalpy change when one mole of a solute is dissolved in an excess of solvent at a given temperature so that further dilution involves no heat change.
• Heat of sublimation– Enthalpy change when one mole of solid converts into vapour at a given temperature and pressure.
• Heat of fusion– Enthalpy change when one mole of solid is converted into liquid state at its melting point at one atmospheric pressure.

Exothermic and Endothermic reaction and their energy profile diagram

Exothermic reaction:

The reaction in which heat is liberated /released from the system is called exothermic reaction.

For example:- Combustion of methane in oxygen is an exothermic process.

Endothermic reaction :

The reaction in which heat is absorbed from the surrounding is called endothermic reaction. For example:- Combination of H₂ and I₂ to give HI is an endothermic process.

First law of thermodynamics (Law of Conservation of Energy)

This law state that, “the total energy of the universe i.e. system and surrounding always remains constant, however it may changes from one form to another.” i.e. Energy can neither be created nor destroyed.

Let us consider a system having initial internal energy (E₁). If heat (q) is supplied to the system and work (w) is done on the system then the internal energy in the final stage i.e. E₂ is given as,

E₂ = E₁ + q + w

E₂-E₁ = q + w

Therefore, ∆E = q + w ………..(i)

This equation (i) is the mathematical statement of first law of thermodynamics which shows the relationship among internal energy, work and heat.

In thermodynamics, w=P∆V, then,

∆E = q + P∆V

Where, ∆V is the change in volume and P is the external pressure.

Hess’s law (of constant heat summation)

“The total amount of heat evolved or absorbed in a reaction (i.e. enthalpy change) in a reaction is always same whether the reaction takes place in one step or in a number of steps. In other words, the total amount of heat change in a reaction depends only upon the nature of the initial reactants and the final products and is independent of the path or the manner by which this change is brought about.”

Consider a general reaction,

Suppose the heat evolved in this reaction is Q₁ Joules.

Now suppose the same reaction takes place in three steps as follows:

Suppose the heat evolved in these three steps are q₁, q₂ and q₃ Joules respectively.

Thus the total heat evolved (suppose Q₂) = q₁ + q₂ + q₃ Joules.

Then, according to Hess’s law, we must have Q₁ = Q₂

To illustrate the Hess’s law of constant heat summation, let us take the example in which carbon is burnt to CO₂.

It is also possible to carry out this reaction in two steps as,

From this example, it is clear that :

Numerical problems

The post Energetics of Chemical Reactions appeared first on Online Chemistry notes.

Rate of reaction

Unit of rate of reaction:

Average rate and instantaneous rate of a reaction

Average rate of reaction: Rate of reaction measured over an interval of time is called average rate of reaction.

i.e. the average rate of reaction is the change in concentration of a reactant or product in a given interval of time.

Instantaneous rate of a reaction:

The rate of reaction at any particular moment of time during the course of chemical reaction is called the instantaneous rate of reaction.

i.e. the instantaneous rate of reaction is the change in concentration of a reactant or product at particular instant of time.

Note:

If we want instantaneous rate at any particular time, then ∆t should be infinitesimally small tending to zero.

Rate law equation, Rate constant and it’s units

Rate of reaction depends upon the concentration of reactants.

Let us consider the general reaction,

The rate of reaction may not depend upon all the ‘a’ concentration terms of A and all the ‘b’ concentration terms of B. suppose by experiment, it is found that the rate of reaction depends upon ‘x’ concentration terms of A and ‘y’ concentration terms of B. Then,

If [A] = [B] = 1. The, Rate = k

So, rate constant (k) can be defined as the rate of reaction when the initial concentration of reactant is unit.

Units of Rate constant:

Order and Molecularity of a Reaction

Order of reaction:

The order of reaction is defined as the sum of powers of all concentration terms in the rate law equation.

Order of reaction may also be defined as “the total number of concentration variables which affect the rate of reaction.

1. Zero order reaction:

The reaction whose rate does not depend upon the concentration of the any reactant is known as zero order reaction.

2.First order reaction :

The reaction whose rate depends upon the concentration of only one reactant variable is known as first order reaction.

3. Second order reaction:

The reaction whose rate depends upon two concentration variables is known as second order reaction.

4. Third order reaction:

The reaction whose rate depends upon three concentration variables is known as third order reaction.

Molecularity of Reaction :

The total number of chemical species involved in the rate determining step of the reaction is called molecularity of the reaction. For example,

If the molecularity of reaction is 1, the reaction is unimolecular. If the molecularity is 2, the reaction is bimolecular and so on.

Differences between molecularity and order of reaction:

Rate expression for first order reaction

Half life period (t_1/2)

Time required to decompose exactly half of the initial concentration of reactant is known as half life period. It is denoted by t_1/2.

Half life period of first order reaction is:

Pseudo first order reaction (Pseudo unimolecular reaction)

The reaction of higher order which follows the kinetics of first order under special conditions is called pseudo first order reaction or pseudo unimolecular reaction.

In other words, the reaction in which molecularity is more than one and is found to be first order is known as pseudo unimolecular reaction.

For example, acidic hydrolysis of ester is a pseudo first order reaction.

Activation Energy (E_a) and Activated complex

The minimum amount of energy required by the reactant to change into product is called activation energy.

OR, The minimum amount of energy which must be supplied to the reactants to enable them to cross over the energy barrier is called activation energy.

If the activation energy is higher in a reaction then slower will be the rate of reaction.

The highest energy state in a chemical reaction in which old bonds are broken and new bonds are formed is called transition state.

The chemical complex formed at highest energy state (i.e. transition state) in a chemical reaction and is highly unstable in nature is called activated complex.

Factors affecting the rate of reaction

The following factors influence the rate of a reaction:

1. Concentration of reactants: The rate of reaction increases with the increase in concentration of reactant except zero order reactions. The chance of collision between reactant molecules to give products increases with the increase in number of reactant particles.

2. Temperature : The increase in temperature generally increase the rate of reaction because increasing temperature increases kinetic energy of reactant molecule and more fraction of molecules become able to cross activation energy barrier to give product.

Experimentally, it has been found that the rate of reaction increases by 2 or 3 times with every rise in temperature by 10⁰C.

3.Surface area of reactant: The rate of reaction increases with increase in total surface area since the reactive site increases. For example, reaction between a marble piece (CaCO₃) and dil. HCl is very slow but CaCO₃ powder with dil. HCl is very fast.

4. Catalyst : The rate of a reaction can be increased by adding catalyst. A catalyst is a substance which provides a new path for the reaction with lower activation energy and speeds up the reaction.

Collision theory of reaction rate

To occur a reaction there must be collision between the reacting molecules. But all the molecules which collide do not necessary to give products.

Only a certain fraction of total number collisions are capable to give products, such collisions are called effective collisions. Conditions for effective collisions are:

i. The reacting species should have sufficient energy to break the chemical bonds in the reacting molecules.

The minimum amount of energy which the colliding molecules must possess is known as threshold energy. This means only those collisions will give products which possesses energy greater than threshold energy.

ii. The colliding species should have proper orientation so that old bonds existing between reacting species may break and new bonds are formed.

For example: During the reaction between CO and NO₂ the products are formed only only when the colliding molecules have proper orientation at the time of collision.

Thus, the collision in which the colliding molecules do not posses the minimum energy for effective collisions (threshold energy) or proper orientation do not form products.

Numerical Problems

Use rate law equation

1. In a reaction H₂ (g) + I₂ (g) → 2HI (g), the rate of disappearance of I₂ is found to be 1×10^-6 molL^-1sec^-1. What will be the corresponding rate of appearance of HI?

2. The reaction X+Y → Product is a second order reaction. Write three different rate law expressions which may be true to the above reaction.

3. For a given reaction, A + B → Product, the rate law is found to be, Rate = K[A]¹ [B]². What happens to the rate of the reaction when

a. Concentration of both A and B are doubled.

b. Concentration of A is doubled and that of B remains constant.

c. Concentration of A is halved and that of B is doubled.

4. The following experimental data are obtained in the college laboratory for the reaction, 2A + B₂ → 2AB

a. What is the order for A and B₂ and overall order?

b. Calculate the rate constant.

c. Find the rate law.

d. Calculate the rate of formation of AB if the concentration of A and B₂ are 2.0 and 4.0 mol L^-1 respectively.

e. Why are chemists interested in obtaining an order of reaction and rate equation?

5. The experimental data for the reaction 2A + B → C is

Determine:-

a. Overall order of reaction.

b. Rate law equation

c. Calculate the rate of formation of C when concentration of [A] and [B] are 0.6 mol L^-1 and 0.3 mol L^-1 respectively.

6. The reaction between A and B is first order with respect to A and zero order with respect to B. Complete the following table.

Numerical using first order rate expression

7. A first order reaction has a rate constant of 1.15×10^-3 S^-1. Calculate the time required for 5 gm of this reactant to reduce to 3 gm.

8. The half life period of first order reaction is 3 hours. Find the time required to complete 87.5% of the reaction.

9. When 50% of reactant in the first order reaction disappears in 20 minutes, find the time taken only when 12.5% of the reactant will have remained.

10. Show that the time required for a first order reaction for 99.9% completion is almost 10 times than required for 50% completion.

11. For a first order reaction, the rate constant is 2.2×10^-5 sec^-1. Calculate the fraction of the reactant consumed in 1 hour and 30 minutes.

Numerical using Arrhenius Equation

12. Calculate the energy of activation of a reaction if its rate constant doubles when the temperature is raised from 17⁰C to 27⁰C.

13. The rate constant of a first order reaction becomes 5 times when the temperature is raised from 350 K to 400 K. Calculate the activation energy for the reaction.

The post Chemical Kinetics appeared first on Online Chemistry notes.

Non-electrolytes

The substances which do not conduct electricity in the molten state or in aqueous solution are called non-electrolytes. Non-polar organic compounds are non-electrolytes. For example: sugar, urea, etc.

Electrolytes

The substances which conduct electricity in the molten state or in aqueous solution are called electrolytes. Ionic compounds and highly polar organic ( covalent) compounds are electrolytes. For example: HCl, NaCl, NaOH, H₂SO₄, CH₃COOH, etc.

Electrolytes are further classified as:

Strong electrolytes : The electrolytes which are almost completely ionized in aqueous solution are called strong electrolytes. For example: HCl, H₂SO₄, HNO₃, NaOH, KOH, NaCl, Na₂SO₄, etc.

Weak electrolytes : The electrolytes which ionize to small extent in aqueous solution are called weak electrolytes. For example: CH₃COOH, H₂CO₃, NH₄OH, H₃PO₄, HCN, etc.

Ionic equilibrium

The weak electrolytes are ionized only partially (feebly). The ions produced as a result of ionization of weak electrolytes are present in dynamic equilibrium with the unionized molecules. Such equilibrium is called ionic equilibrium.

Thus, the ionic equilibrium is the equilibrium which is established between the unionized molecules and the ions in solution of weak electrolytes.

Degree of ionization (dissociation)

The fraction of the total number of molecules of an electrolyte which ionizes into ions is called degree of ionization or degree of dissociation. It is denoted by α .

Factors affecting degree of ionization :

• Concentration : Degree of ionization is inversely proportional to the concentration of solution.
• Nature of electrolyte: strong electrolytes are totally ionized in their solution, hence degree of ionization is 1. But weak electrolytes are partially ionized and degree of ionization is always very less than 1.
• Dilution or volume : Degree of ionization is directly proportional to the dilution (volume) of solution.
• Temperature : Degree of ionization of weak electrolytes increases with increase in temperature.

Ostwald’s dilution law

Statement : At constant temperature, degree of ionization of a weak electrolyte is directly proportional to the square root of dilution(volume) and inversely proportional to the square root of concentration.

Mathematically ;

Derivation : Let us consider the ionization of weak electrolyte AB in water. Suppose C mole-1 be the concentration of the electrolyte and α be the degree of ionization. Then,

From Ostwald’s dilution law,

Limitation of Ostwald’s dilution law : This law is applicable only to weak electrolytes and it fails in case of strong electrolyte.

Different Concepts of Acids and Bases

1. Arrhenius Concept of Acids and Bases :

An acid is a compound which gives H⁺ ions in water. Eg.

A base is a compound which gives OH^– ions in water. Eg.

Limitations of Arrhenius concept :

• Arrhenius concept is limited to aqueous medium only. It fails to explain the behavior of acids and bases in non-aqueous medium.
• It fails to explain the acidic nature of CO_2, SO₂, NO₂ etc. which do not contain any hydrogen.
• It fails to explain the basic nature of CaO, NH₃, Na₂CO₃, etc. which do not contain hydroxyl group.
• CH₄ contains hydrogen but is not an acid. Similarly, CH₃OH have OH group but is not a base.

2. Bronsted – Lowry Concept of acids and bases :

An acid is a substance that can donates a proton.

A base is a substance that accepts a proton. Eg.

Here HCl donates proton and H₂O accepts proton. Hence HCl is an acid and H₂O is base.

Limitations of Bronsted – Lowry Concept :

• Substances like BF₃, AlCl₃, etc. do not contain any hydrogen and hence cannot give a proton but behave as acids.
• It can not explain the reaction between acidic oxide like CO₂, SO₂, SO₃, etc and the basic oxides like CaO, MgO, etc. where is no proton transfer.

3. Lewis concept of acids and bases:

• An acid is a substance which can accept a pair of electrons.
• A base is a substance which can donate a pair of electrons.

Limitations of Lewis concept:

• It does not explain the acidic behavior of well known protic acids such as HCl, H₂SO₄, etc. which do not form coordinate bond with bases.
• It does not explain the relative strength of acids and bases , as it is not based on ionization.

Conjugate Acid- Base pair

A pair of acid and base that differs from each other by a single proton(H⁺) is known as conjugate acid – base pair.

In this reaction Hydrochloric acid and chloride ion are conjugate acid- base pair. Similarly, hydronium ion (H₃O⁺) and water are another conjugate acid – base pair.

Note:

Some examples of Acids and Bases :

Strong acids : HI > H₂SO₄ > HCl > HNO₃ , etc.

Strong bases : NaOH > KOH > Mg(OH)₂ > Ca(OH)₂ , etc.

Weak acids : HCOOH, CH₃COOH, HF, H₂S, H₂CO₃, etc.

Weak bases : NH₃, NH₄OH, etc.

Salt and it’s types

A salt is an ionic compound that results from the neutralization reaction of an acid and a base.

Eg. NaCl, KCl, Na₂CO₃, CuSO₄ etc.

Arrhenius concept: Those substances which gives cation except H⁺ and anion except OH^– when dissolves in water is Arrhenius salt. Eg. NaCl.

1. Acidic salt : Salt formed by the reaction between strong acid and weak base is called acidic salt.

2. Basic salt : Salt formed by the reaction between strong base and weak acid is called basic salt.

3. Neutral salt : Salt formed by the reaction between strong acid and strong base is called neutral salt.

Ionic product of water, P^H and P^OH

Pure water is a weak electrolyte and ionizes slightly as :

Now, at equilibrium applying law of mass action,

Since, degree of ionization of weak electrolyte is very small, the concentration of unionized water can be taken as constant. Thus we can write,

Where, K_w is a constant and is known as ionic product of water. Its value is constant at particular temperature and varies with the change in temperature.

The value of ionic product of water (K_w) at 25⁰C (i.e.298K) is found to be 10^-14.

Therefore, for pure water, [H⁺] = [OH^–] = 10^-7 molL^-1 (M)

Note:

P^H and P^OH

pH of a solution is defined as the negative logarithm of hydrogen ion concentration.

i.e. pH = -log[H⁺]

pOH of a solution is defined as the negative logarithm of hydroxide ion concentration.

i.e. pOH = -log[OH^–]

Relation between P^H and P^OH

We know that,

pH = -log[H⁺] and

pOH = -log[OH^–]

At 25⁰C, ionic product of water(K_w) = [H⁺][OH^–] = 10^-14

Or, [H⁺][OH^–] = 10^-14

Taking log on both sides, we get,

log[H⁺] + log[OH^–] = -14log10

or, -log[H⁺] + {-log[OH^–]} = 14

Hence, pH + pOH = 14

Hydrolysis of salt

Acid reacts with base to give salt and water. Some salts when dissolved in water interact with water and break down into parent acids and bases. such reaction is known as salt hydrolysis. For example:

When ammonium chloride is dissolved in water gives hydrochloric acid (i.e strong acid) and ammonium hydroxide (i.e weak base).

Strong acid ionizes completely in the solution where as weak base ionizes partially. Hence, the resulting solution is acidic as it contains excess of hydrogen ions.

Thus, the aqueous solution of ammonium chloride is acidic in nature.

Q) Explain why the aqueous solution of MgSO₄ is acidic in nature?

Q) Washing soda solution is alkaline. Why?

Note :

The acidic salt (salt of a strong acid with a weak base) gives acidic solution when dissolved in water.

The basic salt (salt of a strong base with a weak acid) gives basic solution when dissolved in water.

The neutral salt (salt of a strong acid with a strong base) gives neutral solution when dissolved in water.

Solubility product

“Solubility product is defined as the product of ionic concentration in saturated solution at constant temperature.”

If a pinch of sparingly soluble salts like AgCl, BaSO₄, PbCl₂, etc. is dissolved in water, a saturated solution will be formed in which there exists an equilibrium between solid salt and its ions.

For example,

Relation between solubility and solubility product:

Let, solubility of sparingly (weakly) soluble salt is S mole L^-1 , then,

1. For AB type salts :

1. For AB₂ type salt :

2. For AB₃ type salt :

3. For A₂B₃ type salt:

Difference between solubility product and Ionic product :

Both ionic product and solubility product represent the product of the concentrations of the ions in the solution. The term ionic product has a broad meaning since, it is applicable to all types of solutions, either unsaturated or saturated or supersaturated.

On the other hand, the term solubility product is applied only to a saturated solution in which there exists a dynamic equilibrium between the undissolved salt and the ions present in solution. Thus, the solubility product is in fact the ionic product for a saturated solution at a constant temperature.

Relation between ionic product and solubility product and precipitation:

A sparingly soluble salt precipitates from a solution only when its ionic product exceeds solubility product value. Thus when

Ionic product > K_sp, solution is supersaturated and hence precipitation takes place.

Ionic product < K_sp, solution is unsaturated and hence precipitation does not take place.

Ionic product = K_sp, solution is saturated.

Common ion effect

When a strong electrolyte having one ion common to weak electrolyte is added to the solution of weak electrolyte then the ionization of weak electrolyte is decreased. This effect is called common ion effect.

For example:

a. Ionization of ammonium hydroxide is decreased by the addition of ammonium chloride.

b. Ionization of acetic acid is decreased by the addition of sodium acetate.

Application of common ion effect and solubility product principle in precipitation reaction:

Precipitation of metal ions in qualitative analysis (salt analysis):

In qualitative analysis, metal ions are precipitated as their insoluble salts.

a. Precipitation of sulphides of group-II metal ions:

In group II of qualitative analysis, when H₂S gas is passed into the original solution containing HCl then metal ions such as Pb⁺⁺, Cu⁺⁺, Hg⁺⁺, Cd⁺⁺,etc. are precipitated in the form of their sulphides. The ionization of H₂S is suppressed due to the presence of common hydrogen ion(H⁺) produced by HCl.

Here, the concentration of S^{– –} ions is decreased due to common ion effect. The solubility product of sulphides of group II metal ions is relatively low. Therefore, the ionic product of sulphide ions and metal ions exceeds the solubility product and hence the the metal ions of group II are precipitated as sulphide (PbS, CuS, HgS, CdS, etc.)

But, the reduced concentration of S^{– –} ions is not sufficient to precipitate metal ions of other groups (III, IV and V) because their solubility products are comparatively high.

b. Precipitation of hydroxides of group-IIIA metal ions:

In group IIIA of qualitative analysis, when NH₄OH is passed into the original solution containing NH₄Cl then metal ions such as Fe⁺⁺⁺, Al⁺⁺⁺, Cr⁺⁺⁺,etc. are precipitated in the form of their hydroxides. The ionization of NH₄OH is suppressed due to the presence of common NH₄⁺ ion produced by NH₄Cl.

Here, the ionic product of hydroxide ions and metal ions exceeds the solubility product of metal hydroxides of group IIIA only and hence  the metal ions of group IIIA are precipitated as their hydroxides {Al(OH)₃, Fe(OH)₃, Cr(OH)₃, etc.}

But, the reduced concentration of OH ^– ions is not sufficient to precipitate metal ions of other groups ( IV and V) because their solubility products are comparatively high.

Note:

Buffer solution

A solution which can resist change in pH even on addition of small amount of acid or base is called buffer solution.

Example: In a biological system, blood is an example of buffer and its pH remains almost constant to 7.4.

Buffer solutions are of two types:

i. Acidic buffer: The buffer prepared by mixing equimolar quantities of weak acid and its salt with strong base is known as acidic buffer. The pH value of acidic buffer is less than 7.

Example: A solution containing equimolar quantities of acetic acid (CH₃COOH) and sodium acetate (CH₃COONa) is acidic buffer.

ii. Basic buffer: The buffer prepared by mixing equimolar quantities of weak base and its salt with strong acid is known as basic buffer. The pH value of basic buffer is more than 7.

Example: A solution containing equimolar quantities of ammonium hydroxide (NH₄OH) and ammonium chloride (NH₄Cl) is acidic buffer.

Numerical Problems

Calculation of pH and pOH:

Strong acid and strong base:

1. Calculate H⁺, OH^– ion concentration and pH at 25⁰C.
2. 0.001M HCl solution.
3. 0.05M H₂SO₄ solution.
4. 0.04M NaOH solution.
5. 0.020M Ba(OH)₂.
6. 0.05N H₂SO₄ solution.
7. Calculate the [H⁺], [OH^–] and pH in a solution in which 2g NaOH is dissolved in a 2 litre solution.
8. The pH of a solution is 6. Its hydrogen ion concentration is increased to 1000 times. What is the pH of the resulting solution?
9. What is the new pH if 100ml of a solution of HCl with pH=0.52 is diluted by the addition of 200ml of pure water?
10. The pH of sulphuric acid solution is 4. What is the molarity of the solution?
11. The pH of sulphuric acid solution is 3. What is the normality of the solution?

Very dilute (aq.) solution:

1. Calculate pH of 1.0×10^-8M HCl solution.
2. Calculate pH of an aqueous solution containing 10^-7 moles NaOH per litre.
3. Sodium hydroxide having pH=8 is diluted 1000 times. Calculate the pH of the diluted base.

pH of Weak acid and base:

1. Calculate pH of 0.2M solution of hydrocyanic acid (HCN). Given, ionization constant of HCN=7.2×10^-10
2. If the pH of 0.1M solution of a weak acid is 3. Calculate the value of K_a for the acid.

pH of mixture:

1. 100 ml of 0.1N HCl is mixed with 300ml of 0.05N HNO₃. Calculate the pH of the resulting mixture.
2. 10cc of N/2 hydrochloric acid, 30cc of N/10 nitric acid and 60cc of N/5 sulphuric acid are mixed together. Calculate pH of the mixture.
3. 1ml of 0.1N HCl is added to 999ml of NaCl. Calculate the pH of the resulting solution.
4. Calculate pH of a mixture obtained by mixing 75ml of 0.2M HCl and 25ml of 0.2M NaOH solution.
5. Equal volume of two solutions having pH=5 and pH=10 are mixed. Find the pH of resulting mixture.
6. Calculate pH of the solution obtained by mixing equal volume of solutions with pH=4 and pH=5.
7. Calculate pH of solution obtained by mixing 50cc solution of pH=1 and 100cc solution of pH=2.
8. Equal volume of two solutions having pH=4 and pH=9 are mixed. Find the pH of the resulting mixture.
9. The pH of the resulting solution obtained by mixing 100ml of xM NaOH and 100ml of 0.1M H₂SO₄ is 1.699. Find the value of x.
10. 85 ml of 0.1 M H₂SO₄ solution is titrated with 0.1 N KOH solution. Find the pH of the solution when the acid is one-fourth neutralized.

Solubility and Solubility product constant relation:

1. The solubility of AgCl in water at 25⁰C is found to be 1.87×10^-3 gmL^-1 . Calculate the solubility product of AgCl at this temperature. ( Ag = 107.9, Cl= 35.5)
2. If the solubility of PbI₂ is 1.68×10^-3 mol per liter, considering 90% ionization find it’s solubility product.
3. The solubility of CaF₂ in water at 298 K is 1.7×10^-3 gm per 100 cm³. Calculate the solubility product of CaF₂ at given temperature.
4. The solubility product constant of BaSO₄ in water at 25⁰C is 1×10^-10 mol²L^-2. Calculate the solubility of BaSO₄ in gm per liter. ( Ba = 137 )
5. 0.00143 gm of AgCl dissolved in 1 liter of water at 25⁰C to form a saturated solution. What is the solubility product of the salt?
6. Calculate minimum volume of water required to dissolve one gram of CaSO₄ at 298 K. (K_sp= 9.1×10^-6)

Common ion Effect:

1. The Solubility product of CaCO₃ is 8.7×10^-9 at 25⁰C. Calculate its solubility in 0.1M CaCl₂.
2. The solubility product of Ferric hydroxide is 1.7×10^-18 at a given temperature. Find its solubility in 0.1 M NaOH at same temperature.
3. The solubility of AgCl in water at 298 K is 1.43×10^-3 gmL^-1. Calculate its solubility in 0.5 M KCl solution. Also, calculate the amount of AgCl precipitated.
4. Calculate solubility of AgCl(s) in (a) pure water (b) a solution of 0.1M NaCl at 25⁰C. [Ksp of AgCl = 1.0×10^-10]
5. The solubility product of Ca(OH)₂ at 298 K is 4.42×10^-5. A 500 ml of saturated solution of Ca(OH)₂ is mixed with an equal volume of 0.4 M of NaOH. What mass of Ca(OH)₂ is precipitated?

Solubility product principle:

1. Given that the solubility product of BaSO₄ is 1×10^-10. Will a precipitate form when the equal volume of 2×10^-8 M BaCl₂ solution and 2×10^-3 M Na₂SO₄ solution are mixed?

The post Ionic equilibrium appeared first on Online Chemistry notes.

Basically there are two types of chemical analysis:

• Qualitative analysis
• Quantitative analysis.

Qualitative analysis tells ‘what’ is in a sample, while quantitative analysis is used to tell ‘how much’ of a given component is in a sample.

Volumetric analysis

Volumetric analysis is one of the quantitative methods of analysis which basically involves the determination of the quantity of a substance present in a given solution by reacting a known volume of it with a solution of another substance of known concentration.

The process by which this analysis is carried out is called ‘titration’.

Equivalent weight

Equivalent weight of an element is defined as the number of parts by weight of that element which either combine or displace 1.008 parts by weight of hydrogen or 8 parts by weight of oxygen or 35.5 parts by weight of chlorine.

Mathematically :

1.Equivalent weight of an element :

2. Equivalent weight of an acid :

If the polyprotic acid gets partially neutralized, its basicity differs from the number of replaceable hydrogen atoms present in the molecule. For example,

Out of three hydrogen atoms present in H₃PO₄, two atoms are replaced and its basicity is two.

Hence, its equivalent weight (E) = mol. wt./2 = 98/2 =49

3. Equivalent weight of a base :

If a polyprotic base gets partially neutralized, its acidity also varies. For example,

Out of three hydroxyl ions present, only two are replaced and its basicity is two. Hence, its equivalent weight(E) = mol.wt./2 = 78/2 = 39

Note :

Basicity – Number of replaceable hydrogen ions present in acid.

Acidity – Number of replaceable hydroxyl ions present in base.

Basicity of CH₃COOH is 1 (not 4) i.e. no. of replaceable hydrogen only (not all hydrogen).

4. Equivalent weight of salt :

5. Equivalent weight of ions:

6. Equivalent weight of oxidizing and reducing agent :

Gram equivalent weight

Note : Equivalent weight of any substance is simply a number. It does not have any unit.

The equivalent weight expressed in gram is called gram equivalent weight.

Eg. one gram equivalent weight of NaOH is 40 gram of NaOH. Similarly, two gram equivalent weight of NaOH is 80 grams.

Different ways of expressing the concentration (strength) of solution

The amount of a solute present in the definite quantity of a solution is called concentration of solution. The concentration of a solution may be expressed in any of the following ways :

1. Gram per litre : The amount of solute in gram present in one litre of solution is known as gram per litre of solution.

2. Normality : The number of gram equivalent of solute present in a litre of solution is known as normality of solution.

Normal solution(N) : A solution containing one gram equivalent of solute in one litre of solution is known as normal solution. Example, if 53gm of Na₂CO₃ is present in one litre of solution then its strength is 1N and is said to be normal solution.

Semi-normal solution(N/2) : A solution containing half of gram equivalent of solute in one litre of solution is known as semi-normal solution. Example, if 26.5gm of Na₂CO₃ is present in one litre of solution then its strength is N/2 and is said to be semi-normal solution.

Decinormal solution(N/10) : A solution containing 1/10^th of gram equivalent of solute in one litre of solution is known as deci-normal solution. Example, if 5.3gm of Na₂CO₃ is present in one litre of solution then its strength is N/10 and is said to be decinormal solution.

Centinormal solution(N/100) : A solution containing 1/100^th of gram equivalent of solute in one litre of solution is known as centinormal solution. Example, if 0.53gm of Na₂CO₃ is present in one litre of solution then its strength is N/100 and is said to be centinormal solution.

3. Molarity : The number of moles of solute present in a litre of solution is known as molarity of solution.

Molar solution(1M) : A solution containing one mole of solute present in one litre of solution is known as molar solution. Example, if 106g of Na₂CO₃ is present in one litre of solution then its strength is 1M and is said to be molar solution.

Relation between normality and molarity

4. Molality : The number of moles of solute present in one kilogram of solvent is known as molality of solution.

Molal solution(1m) : A solution containing one mole of solute present in one kilogram of solvent is known as molal solution. Example, if 106g of Na₂CO₃ is present in one kilogram of solvent then its strength is 1m and is said to be molal solution.

5. Percentage (%) : Percentage strength means number of parts of solute in 100 parts of solvent. It is expressed in three different ways:

6. Formality(F): It is the number of gram formula weight of solute dissolved in one litre of solution.

7. Part per million (ppm): It is the weight of solute present in one million parts (10⁶) by weight of solution. It is used to express the concentration of very dilute solution.

8. Part per billion (ppb) : It is the weight of solute present in one billion parts (10⁹) by weight of solution. It is used to express the concentration of very dilute solution.

Normality equation (N₁V₁ = N₂V₂)

Calculation of the wt. of solute taken to make solution

Q) How much NaOH is required to prepare 250 ml of N/10 NaOH solution?

Amount of NaOH required (w) = NEV/100

= 0.1 X 40 X 250 / 1000 = 1 gm.

Hence, 1gm NaOH is dissolved in 250ml water to prepare N/10 solution.

Normality factor (f) : Normality factor is defined as the ratio of actual weight of substance taken to the theoretical weight of substance to be taken.

Titration

The process of determining the strength of unknown solution with the help of standard solution (i.e. the solution whose strength is known) is called titration.

Typically, the titrant (the solution of known strength) is added from a burette to a known quantity of the titrand (the unknown solution) until the reaction is completed.

Titration is also called standardization.

Titration can be classified into following types:

• Acid-base or neutralization titration
• Oxidation-reduction or redox titration
• Precipitation titration
• Complexometric titration

Acid-base or neutralization titration :

The titration which involves the acid-base reaction( i.e. neutralization reaction) is called acid-base titration.

Acidimetry: The process of determining the strength of unknown acid by titrating it with the standard solution of alkali in presence of indicator is called acidimetry.

Alkalimetry: The process of determining the strength of unknown base by titrating it with the standard solution of acid in presence of indicator is called alkalimetry.

Oxidation-Reduction or Redox titration:

The process of determining the strength of an oxidizing agent by titrating it with the standard solution of reducing agent or vice-versa is called redox titration. For example:

Titration of potassium permanganate (KMnO₄) (i.e. oxidizing agent) with standard oxalic acid (C₂H₂O₄) (i.e. reducing agent) in acidic medium is a redox titration.

Precipitation titration:

The titration in which a precipitate is formed as a result of reaction between the reacting solutions is called precipitation titration. For example:

Some terms and terminology related to volumetric analysis

Titrant (Standard solution) : The solution whose strength is known during the titration is known as titrant. The titrant is taken in a burette during titration.

Titrand : The solution whose strength is not known i.e. the solution whose strength is to be determined during titration is known as trand. The titrand is pipette out by the pipette and taken in the flask during titration.

Indicatorts: Indicators are those chemical substances which indicate the completion of reaction (i.e. end point) during titration by changing their own colour. Indicators are of two types:

1. Internal indicators
2. External indicators

Internal indicators: The indicator which is added to the flask is known as internal indicator. Eg. acid-base indicators, self indicators, precipitation indicators, etc.

Acid-base indicators: Indicators used in acid-base titration are called acid-base or pH indicators. Eg. methyl orange, phenolphthalein, litmus paper, methyl red, etc.

Self indicators: A self indicator is a rectant which itself indicates the end point of reaction by changing the color of reaction mixture. Eg. KMnO₄ acts as an self indicator during titration between KMnO₄ and oxalic acid because it gives pink colour in basic medium and colourless in acidic medium.

External indicators: External indicator is added in between the titration process by taking a drop of titrated mixture.

Example: potassium ferricyanide is used in titration of ferrous ammonium sulphate and K₂Cr₂O₇.

End point : The stage(point) during the titration at which indicators indicates the completion of reaction by changing its own colour is called end point.

If the titration is carried out in between acid and base then the end point is called neutral point.

Equivalence point (Theoretical end point): The stage (point) during the titration at which number of gram equivalent of titrant is equal to the number of gram equivalent of titrand is called equivalence point.

Indicator changes its colour after the equivalence point, hence equivalence point is also called theoretical end point.

Titration error : In theoretical concept, when equivalent amount of titrant is added ti the titrand the indicator should change the colour and indicate the end point. But in actual practice, the indicator changes its colour either in acidic or alkaline medium. Therefore in practice a small difference occurs between equivalence point and end point. This difference between end point and equivalence point is called titration error.

Primary and secondary standard solution/substance

Primary standard solution/substance:

The substance whose standard solution can be prepared directly by dissolving the known weight of substance in fixed volume of solution is known as primary standard substance. Examples: anhydrous sodium carbonate (Na₂CO₃), oxalic acid (C₂O₄H₂), silver nitrate (AgNO₃), etc.

The solution of primary standard substance is known as primary standard solution. Concentration of primary standard solution remains constant for certain interval.

Criteria for a substance to be primary standard:

For a substance to be a primary standard it must fulfill the following requirements:

• It must be available in pure form and should be non-toxic.
• It should be stable (i.e. composition should not be vary for sufficiently long time).
• It should not be hygroscopic or deliquescent.
• It should have high molecular or equivalent weight.
• It should be easily soluble in water and not decomposed by water.

Secondary standard solution/substance:

The substance whose solution can be standardize or strength can be determined by the help of primary standard solution is known as secondary standard substance. Examples: HCl, NaOH, FeSO₄, KMnO₄, etc.

The solution of secondary standard substance is known as secondary standard solution. Concentration of secondary standard solution changes after certain interval.

Selection of indicators during acid-base titration

Indicators are the chemical species that indicates the end point by changing its own colour. Each pH indicator has its own pH range for its colour change. In order to determine the accurate end point of acid-base titration, the indicator should be selected in such a way that the pH range for the colour change of the indicator must coincide with the pH change (jump) at the end point of reaction.

Mainly two types of indicators i.e. methyl orange and phenolphthalein are used during acid-base titration. Methyl orange has pH range 3.1-4.4 and phenolphthalein has 8.2-10.

There are four types of acid base titrations.

1.Strong acid vs strong base titration:

The indicators pH range of both methyl orange (i.e. 3.1-4.4) and phenolphthalein (i.e. 8.2-10) coincide with the pH jump (i.e. 3-11) at end point. Hence, either methyl orange or phenolphthalein can be used as indicator during strong acid vs strong base titration.

2. Strong acid vs weak base titration:

The indicators pH range of only methyl orange (i.e. 3.1-4.4) coincides with the pH jump (i.e. 3-8). Hence, only methyl orange can be used as indicator during strong acid vs weak base titration.

3. Weak acid vs strong base titration:

The indicators pH range of phenolphthalein (i.e. 8.2-10) only coincides with the pH jump (i.e. 6-11). Hence, only phenolphthalein can be used as indicator during weak acid vs strong base titration.

4.Weak acid vs weak base titration:

The indicators pH range of neither methyl orange (3.1-4.4) nor phenolphthalein (i.e. 8.2-10) coincides with the pH jump ( 6-8). Hence, weak acid vs weak base cannot be usually titrated due to lack of suitable indicators.

Note :

Strong acid – HCl, HNO₃, H₂SO₄, etc.

Strong base – NaOH, KOH, Ca(OH)₂, etc.

Weak acid – formic acid, acetic acid, oxalic acid, etc.

Weak base – NH₄OH, Cu(OH)₂, Fe(OH)₃, etc.

Numerical problems

1) Find the normality of 0.53 gm/litre Na₂CO₃ solution.

2) Calculate the normality and molarity of 5% of NaOH solution.

3) What is the normality of 20cc of 2M phosphoric acid (H₃PO₄) ?

4) Which one has higher concentration and why?

i. 80 g/lit NaOH solution and 3M NaOH solution.

ii. 5.3 g/lit Na₂CO₃ and N/10 Na₂CO₃ solution.

5) 20 cc of 2N HCl is completely neutralized by 1.5N NaOH. Calculate the volume of base.

6) X cc of 5N HCl was diluted to one litre of normal solution. Calculate the value of X.

7) What volume of water must be added to 40 ml of 0.25N acid solution in order to make it exactly decinormal solution?

8) 4 gm H₂SO₄ is mixed with 50 ml 2N NaOH solution. Predict the solution is either acidic or basic or neutral.

9) 40 ml 2N H₂SO₄ is mixed with 20 ml 2N NaOH and 20 ml 1.5N Na₂CO₃. Predict the solution is either acidic or basic or neutral.

10) 50 ml 2N H₂SO₄ is mixed with 30 ml 3N NaOH. Predict the solution is either acidic or basic or neutral. Also calculate the resulting strength of solution.

11) 25 cc of a solution of HCl reacts completely with 10 gm of CaCO₃. What is the strength of HCl in normality?

12) 100 cc of 0.1 N HCl is mixed with 50 cc 0.1 N NaOH. Calculate the concentration of resulting solution in normality, molarity and gram per litre.

13) 20 ml 2N HCl is mixed with 35ml 1N NaOH. The mixture is then diluted to 1 litre. Calculate the resulting normality of diluted solution.

14) 25 cc of an alkali solution is mixed with 8 cc of 0.75N acid solution and for complete neutralization it further requires 15 cc of 0.8N acid solution. Find the strength of the given alkali solution.

15) 20 ml of a decinormal solution of NaOH neutralizes 25 ml of solution of dibasic acid containing 6gm of acid per litre. Find the equivalent weight and molecular weight of acid.

16) 0.012 gm of a divalent metal is completely dissolved in 40 cc of N/10 NaOH for complete neutralization. Find the equivalent weight and atomic weight of metal.

17) 3 gm trivalent metal was completely dissolved with 750 ml of 1N HCl. The residual solution further required 1000 ml of N/2 NaOH solution for complete neutralization. Find the atomic mass of the metal.

18) How many moles of H₂SO₄ are required to neutralize 4 litre of 2N NaOH solution?

19) 0.715 gm of Na₂CO₃.xH₂O required 20 ml of seminormal HCl solution for complete neutralization. Find the value of x.

20) 1 gm of NaOH is added to 2 litre of H₂SO₄ solution so that the pH of resulting solution is 7. Find the molarity of acid solution.

21) 0.8 gm of a divalent metal was dissolved in 100 cc of 1.28 normal HCl and the solution was diluted to 200 cc. 50 cc of this diluted solution required 54.6 cc of 0.22 N NaOH for neutralization. Calcute the atomic mass of the metal.

22) x gm of metal having equivalent weight 12 was completely disssolved in 100 cc of seminormal HCl the volume was then made upto 500 cc. 25cc of this acid required 17.5 cc of N/10 NaOH for complete neutralization. Find the value of x.

23) What volume of 12N HCl and 3N HCl must be mixed to form one litre 6N HCl?

24) What volume of 12M NaOH and 2M NaOH should be mixed to get 2 litre of 9M NaOH solution.

25) 1.42gm of a mixture of calcium and magnesium carbonates were completely dissolved in 200 cc of N/5 HCl. The solution was made upto 250cc and 10cc of this solution required 12cc of N/30 Na₂CO₃ solution for neutralization. Calculate the percentage composition of the mixture.

26) 1 gm of an ordinary sample of limestone dissolved in 16.6cc of 0.92N HCl leaving some sandy residue. calculate the percentage of pure CaCO₃ in the sample.

Amount of salt formed:

27. Calculate the volume of 1M NaOH required to neutralize 200ml of 2N HCl. What mass of NaCl is produced from the neutralization reaction?

28. Two litre of 1M of H₂SO₄ is mixed with one litre of 1M NaOH solution. Calculate the strength of the mixture and mass of salt produced from the reaction.

29. If 2L of HCl is mixed with 1L of NaOH solution, 117g of NaCl is formed, then calculate the molarity and normality of HCl and NaOH.

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