## September 2019

### Ionization energy trend in the periodic table

The study of ionization energy is an important article in inorganic chemistry for school and college-level courses.

The electrons are raised to higher energy levels by absorption of energy from external sources. If energy supplied to electrons sufficient, electrons go completely out of the influence of the nucleus of an atom.

The amount of energy required to remove the most loosely bound electron or the outermost electron from an isolated gaseous atom of an element in its lowest energy state or ground state to produce a cation is known ionization energy.

M (g) + Ionization energy → M⁺ (g) + e

The process of ionization is an endothermic process since energy is supplied during ionization. Ionization energy generally represented I or IE and measured in electron volt or kilocalories per gram atom.

#### What is electron volt simple definition?

One electron volt is the energy consumption by an electron falling through a potential difference of one volt. Electron volt simply represented eV.

∴ 1 eV = charge of an electron × 1 volt
= (1.6 × 10⁻¹⁹ coulomb) × (1 volt)
= 1.6 × 10⁻¹⁹ Joule

1 eV = 1.6 × 10⁻¹² erg

#### Removal of an electron from the hydrogen atom

The energy required for removing an electron from energy levels of the hydrogen atom called the ionization energy. Simply the energy corresponding to the transition from n = ∞ to n = 1 gives the ionization energy of the hydrogen atom.

The ionization energy of the hydrogen atom
= (2Ï€²me⁴/h²)[(1/n₁²) - (1/n₂²)]
where n₁ = 1 and n₂ = ∞.

∴ EH = 2.179 × 10⁻¹¹ erg
= 2.179 × 10⁻¹⁸ Joule
= (2.179 × 10⁻¹⁸)/(1.6 × 10⁻¹⁹) eV.

∴ EH = 13.6 eV.

#### Second and third ionization energy

1. The electrons are removed in stages one by one from an atom. The amount of energy required to remove the first electron from a gaseous atom called its first ionization energy.
2. M (g) + IE₁ → M⁺ (g) + e
3. The energy required to remove the second electron from a cation called second Ionization energy.
M⁺ (g) + IE₂ → M⁺² (g) + e
4. Similarly, we have third, fourth ionization.
M⁺² (g) + IE₃ → M⁺³ (g) + e
M⁺ (g) + IE₄ → M⁺⁴ (g) + e
Question
How to calculate the second ionization energy of helium atom if ionization energy of hydrogen is13.6 eV?

The ground state electronic configuration of helium 1S². The second ionization potential means the removal of the second electron from the 1S orbital against the nuclear charge of +2.

∴ IEHe = (2Ï€²mZ²e⁴/h²)[(1/n₁²) - (1/n₂²)]
= Z² × IEH.

∴ Second Ionization energy of helium = 2² × 13.6
= 54.4 eV.

### First ionization energy trend in the periodic table

 Ionization energy and the periodic table
The trend of ionization energy in periodic table influenced by the following factors
• Change on the nucleus or atomic number.
• Completely-filled and half-filled orbitals.
• Shielding effect of the inner electrons.
• Overall charge on the ionizing species.

#### How are the atomic radius and ionization energy related?

Greater the atomic radius or the distance of an electron from the positive charge nucleus of an element, the weaker will be the attraction. Hence the energy required to remove the electron lower.

An atom raised to an excited state by promoting one electron to a higher energy level than the excited electron was more easily detached because of the distance between the electron and nucleus increases.

The Atomic radius decreases from left to right along a period of the periodic table because of the increasing charge on the nucleus of an atom. Thus when we move left to right along with period normally ionization energy increases.

When we moving from top to bottom in a group the ionization energy of the elements decreases with the increasing size of the atom.

#### Atomic number and ionization energy relationship

With the increasing atomic number change on the nucleus increases and more difficult to remove an electron from an atom. Hence grater would be the value of ionization.

Normally the value of ionization increases in moving from left to right in a period since with the increasing atomic number the change on the nucleus also increases.

The increase in the magnitude of ionization due to the increase in the electrostatic attraction between the outermost electrons and the nucleus of an atom. Thus it becomes more difficult to remove an electron.

#### Half filled and completely filled orbitals

According to Hund's rule, an atom having half-filled or completely filled orbital comparatively more stable and hence more energy consumption to remove an electron from such atom.

The ionization of such an atom is therefore relatively difficult than expected normally from their position in the periodic table.
Few exceptions in the value of ionization energy in the periodic table can be explained on the basis of the half-filled and completely filled orbitals.

The ionization energy of group-15 elements is higher than the group-16 elements and group-2 elements are higher than the group-3 elements in the periodic table.
Boron and nitrogen in the second period and magnesium and phosphorus in the third period have a slightly higher value of ionization energy than those normally expected.

Nitrogen and phosphorus in group-15 elements with atomic number 7 and 15 have the electronic configuration

1S² 2S² 2P³
1S² 2S² 2P⁶ 3S² 3P³.

Removal of an electron from half-filled 2P and 3P suborbital of nitrogen and phosphorus required more energy.

Removal of an electron from the group-2 element of beryllium and magnesium with completely-filled S-subshell required more energy.
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#### Shielding effect of the inner electrons

The study of electrostatic attraction between the electrons and nucleus shows that an outer electron attracted by the nucleus and repelled by the electrons of the inner shell.
The combined effect of this attractive and repulsive force acting on the outer electron experiences less attraction from the nucleus. This is known as the shielding effect.
Thus a larger number of electrons in the inner shell, lesser the attractive force for holding outer electron.

The radial distribution functions of the S, P, d subshell show that for the same principal quantum number the S-subshell most shielding than the p-subshell and least shielding the d - orbital.

∴ Shielding efficiency: S〉P〉d.

As we move down a group, the number of inner-shells increases and hence the ionization tends to decreases.

Group-2 elements
Be〉Mg〉Ca〉Sr〉Ba.

#### What trend in ionization energy occurs across a period?

 Ionization energy trend
The greater the charge on the nucleus of an atom the more energy consumed for removing an electron from the atom.

More energy consumed means more energy required for removing an electron from the atom. With the increase atomic number electrostatic attraction between the outermost electrons and the nucleus of an atom increases. Thus removal of an electron from an atom more difficult.

The values of ionization energy generally increase in moving left to right in a period since the nuclear charge of an element also increases in the same direction.

Due to the presence of a completely filled and half-filled orbital of beryllium and nitrogen, the ionization energy of beryllium and nitrogen slightly higher than the neighbor element boron and oxygen.
Ionization energy trend for the second period in the periodic table

Liã„‘Bã„‘Beã„‘Cã„‘Oã„‘N〈Fã„‘Ne.

#### The overall charge of an atom

An increase in the overall charge on the ionizing species (M⁺, M⁺², M⁺³, etc) will enormously influence the ionization since electron withdrawal from a positively charged species more difficult than from a neutral atom.

The first ionization of the elements varies with their positions in the periodic table. In each of the tables, the noble gas has the highest value and the alkali metals the lowest value for the ionization energy.

#### Describe how ionization energy impacts ionic bonding

The study of the ionization energy of the element in a particular group of the periodic table is essential for the properties of the elements.

Lithium, sodium, potassium, rubidium, and cesium or alkali metal in the periodic table with a low value of ionization energy, point to the high reactivity of alkali metals for the formation of the ionic bonding.

### Examples of crystalline and amorphous solids

All the branches of chemical science and all the science, technology depend on the structure of atom or molecules. Thus study crystalline solids is an important article for school or college level students who want to study this article in different books or websites.

Solids are characterized by their definite shape and also their considerable mechanical strength and rigidity. The rigidity due to the absence of translatory motion of the structural units (atoms, ions, etc) of the solids. Here we study crystalline solids and amorphous solids.

These properties are due to the existence of very strong forces of attraction amongst the molecules or ions. It is because of these strong forces that the structural units (atoms, ions, etc) of the solids do not possess any translatory motion but can only have the vibrational motion about their mean position.

Liquids can be obtained by heating up to or beyond their melting points. In solids, molecules do not possess any translatory energy but posses only vibrational energy. The forces of attraction amongst them are very strong.

The effect of heating is to impart sufficient energy to molecules so that they can overcome these strong forces of attraction. Thus solids are less compressible than liquids and denser than the liquid.
Solids are generally classified into two broad categories: crystalline and amorphous substances.

#### What is the definition of crystalline solid?

The definition of crystalline solids is the solid which posses a definite structure, sharp melting point, and the constituents may be atoms, ions, molecules have order arrangement of the constituents extends in long-range order called crystalline solids.

Sodium chloride, potassium chloride, sugar, and ice, quartz are examples of crystalline solids possess a sharp melting point.

The pattern of such crystal having observed in some small crystal region to predict accurately the position of the particle in any region under observation.

#### Properties of crystalline solids

In the crystalline, the constituents may be atoms, ions, molecules.
1. Crystalline solids are a sharp melting point, flat faces and sharp edges which is a well-developed form, are usually arranged symmetrically.
2. Definite and the ordered arrangement of the constituents extends over a large distance in the crystal and called the long-range order.
3. Crystalline solids those belonging to the cubic class are enantiotropic in nature. The magnitude of the enantiotropic property depends on the direction along which is measured.

#### Why amorphous solids called supercooled liquid?

The solids which do not possess a definite structure, sharp melting point, and the constituents may be atoms ions, molecules do not have order arrangement of the constituents extends over a short-range the solids called amorphous solids.

Amorphous solids such as glass, pitch, rubber, plastics possessing many characteristics of crystalline such as definite shape rigidity and hardness, do not have this ordered arrangement and melt gradually over a range of temperatures. For this reason, they are not considered as solids but rather highly supercooled liquids.

#### Difference between crystalline and amorphous solids

1. Crystalline solids possess definite structure and sharp melting point but amorphous solids that do not possess a definite structure and sharp melting point.
2. Crystalline solids constituents(atoms, molecules) have order arrangement of the constituents extends over a long-range in solids but amorphous solids the constituents may be atoms, molecules do not have order arrangement.
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#### Classify of crystalline on the basis of forces

Study the basics of the nature of force operating between constituent particles or atoms, ions, molecules of matter, crystalline solids are classified into four categories.
 Types of crystalline solids

#### Molecular crystals types

Forces that hold the constituents of molecular crystals are of Van der Waals types. These are weaker forces because of which molecular crystals are soft and possess low melting points.

Carbon dioxide, carbon tetrachloride, argon, and most of the organic compounds are examples of these types of crystals.
This class further classified into three category
1. Non-polar binding crystals
The constituent particles of these types of crystalline solids are non-directional. Hydrogen, helium atom or non-polar hydrogen, oxygen, chlorine, carbon dioxide, methane molecules are examples of these types of crystal. The force operating between constituent particles atoms or molecules is a weak London force of attraction.
2. Polar binding crystals
The polarity of bond shows in these types of crystalline solids. Sulfur dioxide and ammonia are examples where force operating between constituent particles is the dipole-dipole attraction force.
3. Hydrogen-bonded crystals
The constituent molecule of these types of crystalline solids is polar molecule and these molecules are bounded each other by hydrogen bonding. An example of this type of crystal ice.

#### Ionic crystal structure

The forces involved here are of electrostatic forces of attraction. These are stronger than the non-directional type. Therefore ionic crystals strong and likely to be brittle.

They have little electricity with high melting and boiling point and can not be bent. The melting point of the ionic crystal increases with the decreasing size of the constituent particles.

In ionic crystals, some of the atoms may be held together by covalent bonds to form ions having a definite position and orientation in the crystal lattice. Calcium carbonate is an example of these types of crystalline solids.

#### Covalent crystal or covalent bonding crystal

The forces involved here are chemical nature or covalent bonds extended in three dimensions. They are strong and consequently, the crystals are strong and hard with high melting points. Diamond, graphite, silicon are examples of these types of crystalline solid.

#### Metallic crystalline solids

Electrons are held loosely in these types of crystals. Therefore they are good conductors of electricity. Metallic crystalline solids can be bent and are also strong.

Since the forces have non-directional characteristics the arrangement ao atoms frequently correspond to the closet packing of the sphere.

### Crystalline allotropes of carbon

Carbon has several crystalline isotropic forms only two of them are common diamond and graphite. There are four other rare and poorly understood allotropes, Î²-graphite, Lonsdaleite or hexagonal diamond, Chaoite (very rare mineral) and carbon VI.
The last two forms appear to contain -C≡C-C≡C- and are closer to the diamond in their properties.

#### Structure of graphite crystal

The various amorphous forms of carbon like carbon black, soot, etc. are all microcrystalline forms of graphite.
Graphite consists of a layer structure in each layer the C-atoms are arranged in hexagonal planner arrangement with SP² hybridized with three sigma bonds to three neighbors and one Ï€-bonds to one neighbor.
The resonance between structures having an alternative mode of Ï€ bonding makes all C-C bonds equal, 114.5 pm equal, consistent with a bond order of 1.33.
The Ï€ electrons are responsible for the electrical conductivity of graphite. Successive layers of Carbon-atoms are held by weak van der Waals forces at the separation of 335pm and can easily slide over one another.

#### Structure of diamond crystal

In diamond, each SP³ hybridized carbon is tetrahedrally surrounded by four other carbon atoms with C-C bond distance 154 pm. These tetrahedral belong to the cubic unit cell.
Natural diamond commonly contains traces of nitrogen or sometimes very rarely through traces of al in blue diamonds.

### How much Van't Hoff equation - effect on temperature?

Van't Hoff equation proposed equilibrium of a chemical reaction is constant at a given temperature. The equilibrium constant, Kp values can be changed with the change of temperature.

This will be evident from the study of Kp values at different temperatures of a chemical reaction.
N₂ + O₂ ⇆ 2NO

 Temperature Kp × 10⁴ 2000° K 4.08 2200° K 11.00 2400° K 25.10 2600° K 50.30

The quantitative relation, known Van't Hoff equation connecting chemical equilibrium and temperature can be derived thermodynamically starting from Gibbs - Helmholtz equation.
The Gibbs - Helmholtz equation
Î”G⁰ = Î”H⁰ + T[d(Î”G⁰)/dT]p
Zero superscripts are indicating stranded values.

or,- (Î”H⁰/T² )= -(Î”G⁰/T² )+(1/T)[d(Î”G⁰)/dT]p
or, - (Î”H⁰/T² ) = [d/dT(Î”G⁰/T)]p.

Van't Hoff isotherm

- RT lnKp = Î”G⁰

or, - R lnKp = Î”G⁰/T.

Differentiating with respect to temperature at constant pressure
- R [dlnKp/dT]p = [d/dT(Î”G⁰/T)]p.

Comparing the above two-equation

dlnKp/dT = Î”H⁰/T²

This is the differential form of Van't Hoff equation.

#### The integrated form of van't Hoff equation

The greater the value of standard enthalpy of a reaction, the faster the chemical reaction reaching an equilibrium point.

dlnKp/dT = Î”H⁰/T²

Separating the variables and integrating
∫ dlnKp = (Î”H⁰/R)∫ (dT/T²)

Î”H⁰ independent of temperature.
or, lnKp = - (Î”H⁰/R)(1/T) + C
where C = integrating constant.

The integration constant can be evaluated and identified the value of Î”S⁰/R, using the relation
Î”G⁰ = Î”H⁰ - TÎ”S⁰.

∴ Van't hoff equation

lnKp = - (Î”H⁰/RT) + (Î”S⁰/R).

Problem
The standard free energy of chemical reaction 2A + B ⇆ 2C at 500 K = 2 KJ mol⁻¹. Calculate Kp at 500 K for the chemical reaction A + ½B ⇆ C.

Solution
Standard free energy at 500 K for the chemical reaction A + ½B ⇆ C
= (2 kJ mol⁻¹)/2
= 1 kJ mol⁻¹.

Î”G⁰ = - RT lnKp.
∴ 1 = - 8.31 × 10⁻³ × 500 × lnKp
or, lnKp = 1/(8.31 × 0.5)
= 0.2406.

∴ Kp = 1.27.

#### Heat absorption and emission in a chemical reaction

Heat absorption and emission in a chemical reaction can be studied from the following analysis.

Reactants → Products

We may study the following possibilities
• Heat absorption, Î”H⁰ = positive.
Low enthalpy side → High enthalpy side.

If heat absorbed, the forward direction of the chemical reaction is the endothermic direction and the reverse one is the exothermic direction. The dissociation of hydrogen iodide into hydrogen and Iodine is an example of this type of chemical reaction.

 HI ⇆ H₂ + I₂ Î”H⁰ = (+) ve
• Heat emission, Î”H⁰ = negative.
High enthalpy side → Low enthalpy side.

If heat emission, the forward direction of the chemical reaction is the exothermic direction and the reverse one is the endothermic direction.
The formation of ammonia from hydrogen and nitrogen is an example of this type of chemical reaction.

 N₂ + H₂ ⇆ 2NH₃ Î”H⁰ = (-) ve
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#### Endothermic and exothermic chemical reaction

Van't Hoff equation drowns the relation between equilibrium constant and endothermic and exothermic chemical reactions.
1. For an endothermic reaction, Î”H⁰ 〉 0 and the right-hand side of the equation positive. This leads to the fact that lnKp increases with increasing temperature.
2. For an exothermic reaction, Î”H⁰ 〈 0 and the right-hand side of the equation negative. This leads to the fact that lnKp decreases with increasing temperature.
 Van't Hoff equation

#### Heat change of a chemical reaction in two temperature

For the ideal system, the heat change or enthalpy change is not a function of pressure.

∴ Standard enthalpy change = enthalpy change
or, Î”H⁰ = Î”H

dlnKp/dt = Î”H/RT²

lnKp = (Î”H/RT) + Î”S/R

However, the entropy of an ideal gas depends strongly on pressure and Î”S and Î”G per mole of reaction in the mixture differ quite substantially from Î”S⁰ and Î”G⁰.

The integrated form of the Van't Hoff equation at two temperature

ln(Kp₂/Kp₁) = (Î”H/R){(T₂ - T₁)/T₁T₂)}

where Kp₁ and Kp₂ are the equilibrium constants of the reaction at two different temperatures T₁ and T₂ respectively.

Determination of Kp₁ and Kp₂ at two temperatures helps to calculate the value of change of enthalpy of the chemical reaction.

The above relation called Van't Hoff reaction isobar since pressure remains constant during the change of temperature.

Problem
Show that the equilibrium point for any chemical reaction given by Î”G = 0.

Solution
Van't Hoff reaction isotherm
Î”G = - RT lnKa + RT lnQa.
When the reaction attains equilibrium point,
Qa = Ka.

∴ Î”G = 0

#### Assumptions from Van't Hoff equation

1. The reacting system of the chemical reaction behaves ideally.
2. Î”H has taken independent of temperature for a small range of temperature change.
Due to the assumption involved Î”H and Î”U do not produce the precise value of these reactions. As Î”G independent of pressure for ideal system Î”H and Î”U also independent of pressure.

Î”G⁰ = - RT lnKp
or, [dlnKp/dT]T = 0.

Problem
Use Gibbs - Helmholtz equation to derive the Van't Hoff reaction isochore. What condition do you expect a linear relationship between log k and 1/T?

Solution
Again Kp = Kc (RT)Î”Æ”
or, lnKp = lnKc + Î”Æ” lnR + Î”Æ” lnT
Differentiating with respect to T,
dlnKp/dT = dlnKc/dT + Î”Æ”/T.

dlnKp/dT = Î”H⁰/RT²
or, dlnKc/dT = Î”H⁰/RT² - Î”Æ”/T
or, dlnKc/dT = (Î”H⁰ - Î”Æ”RT)/RT²

∴ dlnKc/dT = Î”U⁰/RT²

Î”U⁰ = standers heat of reaction at constant volume. This is the differential form of Van't Hoff reaction isochore.

The integrated form of the equation at two temperature
ln(Kc₂/Kc₁) = (Î”U⁰/R){(T₂ - T₁)/T₁T₂)}.

For ideal system Î”U⁰ = Î”U. Since the reaction occurs at constant volume and the equation called Van't Hoff reaction isochore.

### The heat of a reaction and Le-Chatelier principal

Van't Hoff equations give a quantitative expression of the Le-Chatelier principle.
lnKp = - (Î”H/R)(1/T) + C.
1. Endothermic reaction, Î”H〉0, an increase in temperature increases the value of lnKp of the chemical reactions.
2. An exothermic reaction, Î”Hã„‘ 0, with rising in temperature, lnKp decreased.
Change of Kp provides the calculation of the quantitative change of equilibrium point yield of products.

According to the Le - Chatelier Principle, whenever stress placed on any system in a state of the equilibrium point, the system always reacts in a direction to reduce the applied stress.
1. The temperature increased in a chemical reaction, the system at equilibrium point will try to move in a direction in which heat absorbed, which is endothermic reaction favors.
2. When the temperature decreased in a chemical reaction, the system at equilibrium point will try to move in a direction in which heat emitted, which is exothermic reaction favors.

### What is the zero-order kinetics reactions?

Chemical kinetics is the branch of physical chemistry that deals with a study of the speed of chemical reaction. Such studies also enable us to understand the mechanism by which the reaction occurs.
Study of chemical equilibrium only initial and final states was considered, the energy relation between reactants and product is governed by thermodynamics where the time or the intermediate states were of no concern.
The velocity of a reaction is the amount of chemical change occurring per unit time. The rate is generally expressed as the decrease in the concentration of the reactant or increase in concentration per unit time.
Some surface reactions the rate has been found to be independent of concentration. These are zero-order kinetic reactions.

#### The concentration of zero-order reactions

Let us take a reaction represented as
A → Product
Let the initial concentration of the reactant a and product is zero. After the time interval t, the concentration of the reactant is (a-x) and the concentration of the product is x. Thus x is decreased concentration in zero-order reaction.
Question
If the rate of the reaction is equal to the rate constant. What is the order of the reaction?
Zero-order reaction.

#### The zero-order reaction in terms of product

The mathematical equation of zero-order kinetics in terms of product,
dx/dt = k₀
Where k₀ is the rate constant of the zero-order reaction.
or, dx = k₀dt

Integrating the above reaction,
∫dx = k₀ ∫dt
or, x = k₀t + c
where c is the integration constant of the reaction.

When t = o, x is also zero thus, C = o Thus the above equation is,
 x = k₀ t
This is the relationship between decreases of concentration of the reactant(x) within time(t).

#### Zero-order kinetics reactions in terms of reactant

Rate equation in terms of reactant,
-d[A]/dt = k₀ [A]⁰ = k₀
Where [A] is the concentration of the reactant at the time t.
or, - d[A] = k₀dt

Integrating the above equation,
We have - ∫d[A] = k₀ ∫ dt
or, - [A] = k₀t + c
where c is the integration constant of the reaction.

If initial at the time t = 0 concentration of the reactant [A]₀ Then from the above equation,
- [A]₀ = 0 + c
or, c = -[A]₀
Putting the value on the above equation,
 - [A] = kt - [A]₀
This is another form of the rate equation in zero-order kinetics.

### Half-life and zero-order kinetics reactions

The time required for half of the reaction to be completed is known as the half-life of the zero-order reaction. It means 50% of reactants disappear in that time interval.
If in a chemical reaction initial concentration is [A]₀ and after t time interval the concentration of the reactant is [A].
Then, [A]₀ - [A] = kt
Thus when t = t½, that is the half-life of the reaction, the concentration of the reactant [A] = [A]₀/2. Putting the value on the above equation,
We have [A]₀ - [A]₀/2 = k t½
or, k t½ = [A]₀/2
 t½ = [A]₀/2k
Thus for the zero-order kinetics the half-life of the reaction proportional to its initial concentration.

### Examples of zero-order of reactions

The only heterogeneous catalyzed reactions may have zero-order kinetics.
 Zero-order kinetics reaction
Question
The half-life of a zero-order reaction is x and the reaction is completed on t₁ time. What is the relation between x and t₁?
2x = t₁
Question
For the reaction H₂ + Cl₂ → 2HCl on sunlight and taking place on the water. What is the order of the reaction?
This is a zero-order reaction.
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### Unit of the rate constant in a chemical reaction

The rate equation in terms of product for the nth-order reaction is,
d[A]/dt = k [A]n
or, k = (d[A]/dt) × (1/[A]n)
Thus the unit of rate constant(k) = (unit of concentration)/{unit of time × (unit of concentration)n}
= (unit of concentration)1-n/unit of time
Zero-order kinetics reaction the concentration is expressed in lit mole⁻¹ and time in sec
Then the rate constant = (lit mol⁻¹)/sec
= mol lit⁻¹sec⁻¹
Question
For the reaction N₂O₅ → 2NO₂ + ½ O₂, the rate of disappearance of N₂O₅ is 6.25 × 10⁻³ mol lit⁻¹sec⁻¹, what is the rate of formation of NO₂ and O₂ respectively?
1.25 × 10⁻² and 3.125 × 10⁻³ mol lit⁻¹sec⁻¹
Question
The rate constant of a chemical reaction is 5 × 10⁻⁸ mol lit⁻¹sec⁻¹. What is the order of this reaction? How long does it take to change concentration from 4 × 10⁻⁴ moles lit⁻¹ to 2 × 10⁻² moles lit⁻¹?
The reaction is a zero-order reaction and 3.92 × 10⁵ sec takes to change concentration from 4 × 10⁻⁴ moles lit⁻¹ to 2 × 10⁻² moles lit⁻¹.
Question
For a reaction, N₂ + 3 H₂ → 2NH₃, if d[NH₃]/dt = 2 × 10⁻⁴mol lit⁻¹sec⁻¹, what is the order and the value of - d[H₂]/dt of this reaction?
Zero-order reaction and the value of d[H₂]/dt = 3 × 10⁻⁴ mol lit⁻¹sec⁻¹

### Characteristics of zero-order kinetics

1. The rate of the reaction is independent of concentration.
2. Half-life is proportional to the initial concentration of the reactant.
3. The rate of the reaction is always equal to the rate constant of the reaction at all concentrations.

### Online college courses for soft and hard acids bases

Soft and hard acids and bases (SHAB) principles are very helpful for study the stability of the complex between acid and bases and it is very useful for study online college courses. This principle was proposed in 1963 by Ralph G. Pearson.

A + :B → A : B

According to the SHAB principle, the complex most stable when participating acid and base are either both soft or both hard and least stable when one of the reactants (namely acids and base) is very hard and the other one is very soft.

To study the donner properties of different bases and preferences of a particular base to bind acid. Hydrogen ion and methyl mercury (II) ion is used for this comparison study.

Proton and methyl mercury cation have only one coordination position to accommodate only one coordinate bond but the two cations vary widely in their preferences to bases.

This preference was estimated from the experimental determination of equilibrium constants for the exchange reactions.

BH⁺ + [CH₃Hg(H₂O)]⁺ ⇄ [CH₃HgB]⁺ + H₃O⁺

Experimental results indicate that bases in which nitrogen be a donor atom, oxygen or fluorine prefer to coordinate with the hydrogen ion. Bases which has donor atom phosphorus, sulfur, iodine, bromine, chlorine or carbon prefers to coordinate with mercury.

#### The neutralization reaction of acids and bases

Lewis defined, an acid-base neutralization reaction involves an interaction of a vacant orbital of an acid and a filled or unshared orbital of a base.

 A + : B ⇄ A: B Lewis acid Lewis base Adduct

The species A is called Lewis acid or a generalized acid and B is called Lewis base or a generalized base. Strong acid and a strong base will form a stable complex.

#### What are soft and hard bases?

The donor atoms of the second category are of low electronegativity, high polarizability, and are easy to oxidize. Such donors have been called ‘soft bases' by Pearson since they are holding on to their valence electrons rather loosely.

The donor atoms in the first group have high electronegativity, low polarisability and hard to oxidize. Such donors have been named ‘hard bases' by Pearson since they hold on to their electrons strongly.

Question
Classify the following as soft and hard acids and bases.

1. Hydride ion.
2. Nickel (IV) ion.
3. Iodine (+1) ion.
4. Hydride ion.

1. The hydride ion has a negative charge and too large in size compared to the hydrogen atom. Electronegativity of hydride ion quite low. Due to high polarizability and large size, the valence electron is loosely bound in hydride ion and it is a soft base.
2. The nickel (IV) has quite a high positive charge and small size compared to the nickel (II). Electronegativity of nickel (IV) will be very high and polarisability will low. Hence it is hard acid.
3. Mono-positive iodine has a low positive charge and has a large size. It has a low electronegativity and high polarizability.
4. Hydrogen ion has the smallest size with a high positive charge density and absence of unshared pair of electrons in its valence shell. All these will give a high electronegativity and very low polarizability. Hence hydrogen ion is a hard acid.

#### Study the properties of soft and hard bases

In simple terms, hardness associated with a tightly held electron shell with little tendency to polarize. On the other hand, softness associated with a loosely bound polarizable electron shell.

It will be seen that within a group of the periodic table softness of the Lewis bases increases with the increase in the size of the donor atoms. Among the halide ions, softness increases in the order.
F⁻ã„‘Cl⁻ã„‘Br⁻ã„‘I⁻
 Soft and hard acids and bases
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#### What are soft and hard acids?

After having gone through a classification of bases, a classification of Lewis acids is necessary. The preferences of a given Lewis acid towards ligands of different donor atoms are usually determined from the stability constant values of the respective complexes or from some other useful equilibrium constant measurements.

When this is done, metal complexes with different donor atoms can be classified into two sets based on the sequences of their stability.

Hard acids have small acceptor atoms, are of high positive charge and do not contain unshared pair of electrons in their valence shell, although all these properties may not appear in one and the same acid.
These properties lead to high electronegativity and low polarizability. In keeping with the naming of the bases, such acids are termed as 'hard acids'.

N≫P; O≫S; F〉Cl〉Br〉I

Soft acids have large acceptor atoms, are of low positive charge and contain ushered pairs of electrons in their valence shell. These properties lead to high polarizability and low electronegative. Again in keeping with the naming of the bases, such acids are termed 'Soft acids'.

N≫P; O≫S; F〉Cl〉Br〉I
 Hard Acids Borderline acids Soft Acids H⁺, Li⁺, Na⁺ K⁺, Be⁺², Mg⁺², Ca⁺², Sr⁺², Mn⁺², Al⁺³, Ga⁺³, In⁺³, La⁺³, Lu⁺³, Cr⁺³, Co⁺³ Fe⁺³, As⁺³, Si⁺⁴, Ti⁺⁴, U⁺⁴, Ce⁺³, Sn⁺⁴, VO⁺², UO₂⁺², MoO⁺³ Fe⁺, Co⁺², Ni⁺², Cu⁺², Zn⁺², Pb⁺², Sn⁺², Sb⁺², Bi⁺², Rh⁺², B(CH₃)₃, SO₂, NO⁺, GaH₃ Cu⁺, Ag⁺, Au⁺, Tl⁺, Hg⁺, Pd⁺, Cd⁺, Pt⁺, CH₃Hg⁺, Tl⁺³, BH₃, GaCl₃, InCl₃, I⁺, Br⁺, I₂, Br₂, and zerovalent metal atom

#### SHAB principle for soft and hard acids and bases

This principle also means that if there is a choice of the chemical reaction between an acid and two bases and two acids and a base, A hard acid will prefer to combine with a hard base and a soft acid will prefer to combine with soft base and thus a more stable product will be obtained.

The hard acid - hard base may interact with strong ionic forces. Hard acids have small acceptor atoms and positive charge while the hard bases have small-donor atoms but often with a negative charge. Hence a strong ionic interaction will lead to the hard acid-base combination.

On the other hand, a soft acid - soft base combination mainly a covalent interaction. Soft acids have large acceptor atoms, are of low positive charge and contain ushered pairs of electrons in their valence shell.

#### Application of SHAB principle in acid chemistry

The SHAB concept is extremely useful in elucidating many properties of chemical elements in acid chemistry and will be often referred to at appropriate places.
1. Boron trifluoride and boron trihydride the behavior of boron is different in these two compounds.
The presence of hard fluoride ions in boron trifluoride easy to add hard bases.
The presence of soft hydride ions in boron trihydride easy to add soft bases.
2. [CoF₆]⁻³ more stable than [CoI₆]⁻³. It will be seen that cobalt (III) ion is a hard acid, fluoride ion is a hard base and iodide ion is a soft base.
In [CoF₆]⁻³, hard acid, and hard base form a stable complex than the [CoI₆]⁻³ form by a hard acid and soft base.
3. The existence of certain ores can also be rationalized by applying the SHAB principle. Thus hard acids such as magnesium, calcium, and aluminum occur in nature as magnesium carbonate, calcium carbonate, and aluminum oxide but not as magnesium calcium and aluminum sulfides
Since the anion CO₃⁻² and O⁻² are hard bases and S⁻² is a soft base. Soft acids such as Cu⁺, Ag⁺ and Hg⁺², on the other hand, occur in nature as sulfides.
4. The borderline acids such as Ni⁺², Cu⁺², and Pb⁺² occur in nature both as carbonates and sulfides. The combination of hard acids and hard bases occurs mainly through ionic bonding as in Mg(OH)₂ and that of soft acids and soft bases occurs mainly by covalent bonding as in HgI₂.
Question
Why AgI2- stable, but AgF2- does not exist

We know that mono positive silver ion is a soft acid, fluoride ion is hard to base and iodide ion is the soft base.
Hence AgI₂⁻ (soft acid + soft base) is a stable complex and AgF₂⁻ (soft acid + hard base) does not exist.

Question
Mercury hydroxide dissolved readily in acidic solution but mercury sulfide does not?

In the case of mercury hydroxide and mercury sulfide, mercury is a soft acid and hydroxide and sulfide are a hard base and soft base respectively.

Mercury sulfide formed by soft acid and soft base will be more stable than mercury hydroxide formed by soft acid and hard base.

More stability of mercury sulfide than that of mercury hydroxide explains why mercury hydroxide dissolved readily in acidic solution but mercury sulfide does not dissolve readily in acidic solution.

### f block elements names and symbols

The f-block in the periodic table appears in two series characterized by the filling of 4f and 5f sublevel in the respective third inner principal quantum number from outermost.

The 4f block contains fourteen elements cerium to lutetium with the atomic number from 58 to 71 and is called Lanthanum's as they appear after lanthanum.

The 5f block contains fourteen elements thorium to lawrencium with the atomic number from 90 to 103 and is called actinides as they appear after actinium.

#### General electronic configuration of 4f block elements

The 4f block in the periodic table has been variously called rare earth, lanthanum’s, and lanthanum. The lanthanum atoms and their trivalent ions have the following general electronic configuration.

Lanthanum atoms
[Pd] 4fn 5S² 5P⁶ 5d¹ 6S²
where n has values 1 to 14.

Lanthanum (M⁺³) ions
[Pd] 4fn 5S² 5P⁶
where n has values 1 to 14.
 4f-block elements in the periodic table
4f-block or inner transition elements with increasing atomic number, electrons are added to the deep-seated 4f-orbital. The outer electronic configuration of 4f-elements is 6S² and inner orbitals contain f -electrons.
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#### Cerium, Gadolinium, and Lutetium electronic configuration

Only Cerium, Gadolinium, and Lutetium contain one electron in 5d orbital with atomic number 58, 64 and 87. The electronic configuration of these elements is outside of the general electronic configuration.

Electrons of similar spin developed an exchange interaction which leads to the stabilization of the system. For the electrons of similar spin, repulsion is less by an amount called the exchange energy of an electron.

The greater the number of electrons with parallel spins the greater is exchanged interaction and the greater stability. The basis of Hound's rules of maximum spin municipality.

Cerium
[Pd] 4f¹ 5S² 5P⁶ 5d¹ 6S²

Gadolinium atom contains one electron in a 5d orbital. 4f and 5d are very close in terms of the energy of 4f and 5d orbital. In such a case, the half-filled orbital is slightly more stable than orbital with one additional electron by increasing exchange energy.

Maximum stability of f-shell when there are seven electrons with parallel spins in the seven f-orbital. This half-filled energy level stabilizes by exchange energy.

[Pd] 4f⁷ 5S² 5P⁶ 5d¹ 6S²

Lutetium also has the f¹⁴d¹ configuration where the last electrons have added the capacity of the f-shell.

Lutetium
[Pd] 4f¹⁴ 5S² 5P⁶ 5d¹ 6S².

#### Electronic configuration of praseodymium

Praseodymium possesses electronic configuration 4f³ 6s² instead of the expected one 4f² 5d¹ 6s². This can be explained by (n + l) rules, the orbital which has a higher value of (n + l) is the higher energy orbitals.

n + l = 4+3 = 7 for 4f-orbital.
n + l = 5+7 =7 for 5d-orbital
.

4f and 5d-shell sum of principal and azimuthal quantum number the same. In this case, the highest number of principal quantum numbers is the higher energy quantum level of an atom and 5d-orbital is the higher energy quantum shell.

Again electrons are fed into orbitals in order of increasing energy until all the electrons have been accommodated.

Electron filling process for praseodymium f-electron filling first and possess electronic configuration 4f³ 6s².

Question
Why +3 oxidation number is so common and stable in lanthanum?

The nature of lanthanum's elements is such that three electrons are removed comparatively easy to give the normal trivalent state.

The ground state electronic configuration of the natural lanthanum's atoms and trivalent ions

Lanthanum's [Pd] 4fn 5S² 5P⁶ 5d¹ 6S²

Lanthanum ions [Pd] 4fn 5S² 5P⁶
where n is 0 to 14 from Lanthanum to Lutetium.

As a consequence, the f-electrons can not participate in the chemical reactions thus +3 oxidation numbers common and stable in lanthanums.

### General electronic configuration of 5f block elements

 5f-block elements in the periodic table
The second series f block elements result from the filling of 5f-orbital and consist of elements thorium to lawrencium with atomic number 90 to 103.

All of them are radioactive but most abundant isotopes of thorium and uranium have very long half-lives.
General electronic configuration of these elements and ion are

Actinides atoms
[Rn] 4fn 5d1-2 6S²
where n has values 1 to 14

Actinides (M⁺³) ions
[Rn] 4fn
where n has values 1 to 14.

### Learn quantum physics online

Spectral fine-structure lines in quantum physics suggested that the energy levels of an electron are more complex than a consideration of the electrostatic interactions from the Bohr hydrogen energy level and the Sommerfeld model. In this article, we can study four quantum numbers for study online college courses.

Magnetic interactions comparatively small in magnitude are responsible for the fine structure defects of an atom.

The closer examination of the atomic spectra demanded further refinement of the atomic orbitals. It was found that under high resolution the spectral lines of alkali metals had a fine structure of an atom.

Study the fine structure of an atom four quantum number needed to explain the various spectra.
These quantum numbers are the identification numbers for an individual electron in an atom to describe the position and energy level of an atom.

In order to study the size, shape, orientation of energy levels or orbital four quantum numbers are necessary. These quantum numbers are
• Principal quantum number
• Azimuthal quantum number
• Magnetic quantum number
• Spin quantum number.

#### How to find the principal quantum number?

The principal quantum number describes the energy level or the principal shell to which an electron can stay. The principal quantum number denoted by n.

The primary importance of the principal quantum number for determining the size of an atom and energy of an electron. From Bohr's model, the energy value of the hydrogen energy level is fixed by the fixed value of n.

Another multielectron atom energy value of each electron depends mostly on the principal quantum number of an atom.

As the value of the principal quantum number increases the atomic radius or nucleus electron separation increases and the energy also raised.

The principal quantum number is always an integer and can assume the value, 1, 2, 3, 4.... but not zero.

n = 1, 2, 3, 4, 5, ............... ∞.

#### Azimuthal quantum number formula

This quantum number was introduced by Sommerfeld in his atomic model and gives the angular momentum of an electron in its elliptical movement around the nucleus of an atom and fine structure of the hydrogen spectrum.

The general geometric shapes of an electron cloud or orbital are described by the azimuthal quantum number or angular momentum quantum number. Permitted values of l for a given value of n has 0 to (n-1).
∴ l = 0, 1, 2, 3.....(n-1)
The total number of different values of l equal to n.

n = 1, l = 0 (1S-subshell)
n = 2, l = 0, 1 (2S and 2P-subshell)
n = 3, l = 0, 1, 2 (3S, 3P, and 3d-subshell).

#### What does the magnetic quantum number determine?

Bohr's model could not explain the splitting of a single spectral line into a number of closely spaced lines in presence of magnetic field or presence of the electric field.

The presence of more lines in the spectrum of the magnetic field or electric field indicates the energy levels are further subdivided by the additional quantum number called magnetic quantum number.

The magnetic quantum number associated with the orientation of the electron cloud with respect to a given direction, usually that of a strong magnetic field. This quantum number denoted by ml.

A given value of the azimuthal quantum number, the magnetic quantum number can have any integral value between +1 to -1.
∴ ml = + l, (l - 1), ..... 0 ..... - 2, - l

#### Spin quantum number in chemistry

When spectral lines of hydrogen, lithium, sodium, and potassium observed by the instrument of high resolving power, each of the lines of the spectral series was found to consist of a pair of lines known double line structure.

To describe these double lines of the fine structure another fourth quantum number necessary and it is known as a spin quantum number.

The electron itself regarded as a small magnet. A beam of a hydrogen atom can be split into two beams by a strong magnetic field. This indicates that there are two kinds of spin that can be differentiated on the basis of their behavior in a magnetic field.

The electron can either spin clockwise or counterclockwise. The two directions of spin represent as(↑↓).

The spin quantum number independent of the other three quantum numbers. Two directions of spin are represented as (↑↓) can have two possible spin quantum number values (+ ½) and (- ½) depending on the direction of rotation of the electron about its axis.

### How to find quantum numbers of an element?

 Quantum numbers of an atom
Question
What are the four quantum numbers of the 19th electron of chromium?

The atomic number of chromium 24 and the electronic configuration of chromium
1S² 2S² 2P⁶ 3S² 3P⁶ 4S¹ 3d⁵.

19th electron means 4S¹ electron.
n = 4, l = 0, m = 0, and s = +½ for 4S electron.

∴ Quantum numbers set for 19th electron of chromium
4, 0, 0, +½.

Question
What is the correct set of four quantum numbers for the valence electron of rubidium?

The correct set of four quantum numbers for the valence electron of the rubidium atom

5, 0, 0, +½

Question
How many electrons in an atom can have the following quantum numbers n = 4 and l = 1?

6 electrons in an atom can have the following quantum numbers n = 4 and l = 1.

Question
How many possible numbers of orbitals of an atom when n = 4?

Number of possible orbitals when n = 4
[1(4S) + 3(4P) + 5(4d) + 7(4f)]
= 16.

Question
How many possible orbitals are there when n = 3, l = 1, and ml = 0 ?

The number of possible orbitals = 1.

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### How are the shapes of atomic orbitals determined?

Atomic orbitals are the basic building blocks of the atomic orbital diagram or alternatively known as the electron cloud or wave mechanics model.

According to the electron cloud model, an orbital is a region in space where the probability of finding an electron maximum.

The probability of finding the electron cloud in 1S orbital of the hydrogen atom at certain positions near the nucleus of the hydrogen atom and electron density is maximum in the region just surrounded the nucleus of an atom.

According to the electron wave model, the wave function of the electron in an atom is called orbital. The wave function is plotted against distance and space in three dimensional marked by a curve will give the orbital diagram of an atom.

The probability of finding an electron in space around the nucleus involves two aspects, radial probability, and angular probability.

It is not possible to represent completely in one diagram on paper the directional properties of electron orbital. An angular probability distribution must be combined mentally to have an overall shape of the orbital.

#### How many orbitals are in the s sublevel?

The angular probability distribution is greater interact and importance for S-subshell. The S-electron has no angular dependence because the relevant wave function is independent of angles Î¸ and Î¦.
 S-subshell of an atom
With the nucleus at the origin of the cartesian axes, the sphere of the radius represents the probability of finding the electron cloud in S-subshell.

The electron cloud distribution in S-subshell has a spherically symmetrical probability distribution.

#### Orbital diagram of p-subshell of an atom

 P-subshell of an atom
P-subshell has magnetic quantum number is 1, 0, -1 and P-subshell are three orientations in space. These orientations represented as Px, Py, and Pz.

P-subshell designated as Px, Py and Pz are mutually perpendicular and they are concentrated along the respective coordinate axis X, Y and Z.
Unlike the S-subshell, the angular part of the P wave is dependent on Î¸ and Î¦ and P-subshell is shielding by S-subshell.

#### The shape of d-subshell of an atom

 d-subshell of an atom
When n =3, the orbitals start with the 3rd main energy level of an atom.
l = 2(d - orbital),
ml = -2, -1, 0, 2, 1.

d - subshell has five orientations in space.

Absence of magnetic field all these five d-subshell are equivalent in energy level and this d-subshell said to be five-fold degenerate energy level of an atom.

### Definition of pH and pOH in chemistry

pH and pOH concept is an important article for all school and college courses and also largely used in analytical chemistry.
Hydrogen ion plays a key role in the acid-base functions. Hydrogen ion in acid chemistry represented as H₃O⁺ ion.

H⁺ + H₂O → H₃O⁺

The ionic radius of the naked hydrogen ion is ~10⁻¹³ cm = 10⁻¹⁵ Ã…, which is very small. Therefore has a very high charge/radius ratio ~10⁵. According to Fajan's rule, a hydrogen ion is most effective in polarizing other ions or molecules.

In H₃O⁺ there are assumed to coordinate bonds from water oxygen to the proton, thus giving the proton a helium electronic configuration.

#### How does dilution affect the pH of acidic solutions?

Perchloric acid reacts readily with the water solution to gives a series of hydrates.
HClO₄ (112°C)

HClO₄, H₂O (+ 50°C)

HClO₄, 2H₂O (- 17.8°C)

HClO₄, 3H₂O (- 37°C )

HClO₄, 3.5H₂O (- 41.4°C)

Most remarkable hydrates of perchloric acid are monohydrates, melting at a much higher temperature than the covalent anhydrous acid. Monohydrates of perchloric acid very stable and can be heated to around 100⁰C without decomposition.

The monohydrate about ten times vicious then anhydrous acid. It has the same crystal lattice as the ionic ammonium perchlorate and it shows an ionic compound.
[H₃O⁺][ClO₄]

#### Water act as an acid and a base - ionic product of water

Water molecule ionizing weakly to form hydrogen ion and hydroxyl ions. There will always be an equilibrium between hydrogen and hydroxyl ions in the water molecule.

H₂O ⇄ H⁺ + OH⁻

Hydrogen ion solvated in water to form hydronium ion or simply hydrogen ion. Equilibrium for dissociation of water will have its own equilibrium constant value.

K = ([H⁺] × [OH⁻])/[H₂O]
or, K × [H₂O] = [H⁺] × [OH⁻]
where [H₂O] = concentrations of water solution.

In any dilute aqueous solution, the concentration of water molecules = 55.5 moles/liter greatly exceeds that of any other ion, the concentration of water can be taken as a constant.

∴ K × [H₂O] = Kw = [H⁺] × [OH⁻]

where Kw = dissociation constant of water or ionic product of water. The concentration of hydrogen ion and hydroxyl ion in pure water has been determined by 10⁻⁷ M each.

∴ Kw = [H⁺] × [OH⁻]
= 10⁻⁷ × 10⁻⁷
= 1.0 × 10⁻¹⁴ M
The above relation tells us that in aqueous solution the concentration of hydrogen ion and hydroxyl ion are inversely proportional to each other.

If hydrogen ion concentration increases 100 fold, that of the hydroxyl ion concentration decrease 100 fold to maintain Kw constant.

#### Ionization of water endothermic or exothermic

Ionization of water into hydrogen and hydroxyl ion is an endothermic chemical reaction. Chemical reactions in which heat absorbed by the system from the surroundings are known as endothermic reactions.
H₂O + 13.7 kcal → H⁺ + OH⁻

#### Le Chatelier principle acid dissociation

If a system in equilibrium, a change in any factors that determine the condition of equilibrium will cause the equilibrium to shift in such a way as to minimize the effect of this change.

According to Le-Chatelier’s principle, increasing temperature will facilitate dissociation and giving higher values of Kw.
 Temperature Kw Values 20⁰C 0.68 × 10⁻¹⁴ 25⁰C 1.00 × 10⁻¹⁴ 60⁰C 9.55 × 10⁻¹⁴

#### What is the pH of pure water?

The dissociation constant of pure water is a very low value and the expression of the concentrations of hydrogen ion and hydroxyl ion of a solution in terms of such low figures not much convenient and meaningful.

Sorensen proposed the use of a term known as pH to measure the concentration of hydrogen ion in acid and alkali solution.

pH = - log[H⁺]

In acid chemistry, this relation provides the relation between the concentration of hydrogen ion in the solution and pH of this solution.

The concentration of hydrogen ion = 10⁻¹ M has a pH = 1 and for a solution having hydrogen ion concentration = 10⁻¹⁴ M has a pH = 14.

For such solutions having hydrogen ion concentration in the range of 10⁻¹ M to 10⁻¹⁴ M is more convenient and meaningful to express the acidity in terms of pH rather than hydrogen ion concentrations.

#### Calculate the pH of 0.01 m sulfuric acid

Sulfuric acid is a dibasic acid. Molarity of sulfuric acid = 2 × molarity of the sulfuric acid solution. Thus 0.1 M and 0.2 N sulfuric acids are the same and the pH of the solution

pH = - log [H⁺]
= - log (0.2)
= 0.699

The study of pH identified the acidic and basic solution. With the increasing concentration of hydrogen ion pH values decreases. The more acidic the solution in which pH is lower. If the acidity of a solution goes down 100 fold, the pH goes up by the two units.

#### How to calculate pH of 0.002 M hydrochloric acid solution?

Hydrochloric acid is a strong acid or electrolyte and completely dissociated in the solution.
Concentration of H⁺ = [HCl]
= 0.002 = 2×10⁻³ M.

∴ pH = - log [H⁺]
= - log (2×10⁻³)
= (3 - log2)
= 2.7.

#### How to calculate pH of 0.002 M acetic acid solution?

The concentration of hydrogen ion in acetic acid
= √(Ka× [CH₃COOH]
= √(2 × 10-5× 2 × 10⁻³)
= 2 × 10⁻⁴.

∴ The pH of 0.002 acetic acid solution
= - log [H⁺]
= -log (2 × 10⁻⁴)
= (4 - log2)
= 3.7.

### Hydroxide ion concentration from pOH

The concentration of hydroxide ion also expressed as the concentration of hydrogen ion in the solution and it is expressed as the pOH of the alkaline solution.

pOH = - log[OH-]

If the acidity of a solution goes down 100 fold its pH goes up by two units. A solution of pH 1 has hydrogen ion concentration 100 times greater than that of pH 3. Taking the case of OH⁻ ions, the pOH will go down by two units then pOH goes from 13 to 11.

#### Show that pH + pOH = pkw = 14

The study previously the product of hydrogen ion and hydroxyl ion has constant.
∴ [H⁺] × [OH⁻] = 10⁻¹⁴
or, log[H⁺] × [OH⁻] = log10⁻¹⁴
∴ log[H⁺] + log[OH⁻] = -14
or, -log[H⁺] - log[OH⁻] = 14.

From the definition
- log [H⁺] = pH and -log[OH⁻] = pOH

∴ pH + pOH = 14

### Describe pH scale acids and bases

We can now proceed to differentiate between neutral, acidic or basic on the basis of relative concentrations of hydrogen and hydroxyl ions and on the basis of pH of the solution to constructed the pH scale of the solution.
 pH scale of acids bases
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#### What ph indicates a neutral solution?

A neutral solution is one in which the concentrations of hydrogen and hydroxyl ions are equal.
∴ [H⁺] = [OH⁻] = 10⁻⁷ M.

From the definition of pH
- log[H⁺] = -log (10⁻⁷) = 7.

The pH and pOH of the neutral solution = 7.

#### pH value of an acidic solution

An acidic solution has the concentration of hydrogen ion greater then hydroxyl ion.

∴ [H⁺] 〉[OH⁻]
or, [H⁺] 〉 10⁻⁷ M
and [OH⁻] 〈 10⁻⁷ M

From the definition of pH and pOH
- log [H⁺] 〉- log (10⁻⁷)
∴ pH 〈 7.

- log [OH⁻] 〈 - log (10⁻⁷)
∴ pOH 〉 7.

#### Basic solution properties

[H⁺]〈 [OH⁻]
or, [OH⁻] 〉10⁻⁷M
and [H⁺]〈 10⁻⁷M
In terms of pH, we have the following relations,
[H⁺] 〈 10⁻⁷ M
or, pH 〉 7

A mathematical definition of pH provides a negative value when hydrogen ion exceeds 1 M. However pH measurements of such concentrated solutions are avoided as these solutions are not likely to be dissociated fully.

From the above study of pH and pOH play the key role in acid chemistry also analytical chemistry. This study recognizes the experimental facts of chemical substances or solutions as acidic or basic and properties of acids and bases.

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