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Inverse Trigonometric Functions Class 12 Notes Maths Chapter 2

July 19, 2021

By going through these CBSE Class 12 Maths Notes Chapter 2 Inverse Trigonometric Functions, students can recall all the concepts quickly.

Inverse Trigonometric Functions Notes Class 12 Maths Chapter 2

Inverse Function Calculator is an online tool by Protonstalk to quickly and easily find out the inverse of any given function.

As we have learned in class XI, the domain and range of trigonometric functions are given below:
Functions Domain Range
Inverse Trigonometric Functions Class 12 Notes Maths 1
→ Inverse Function: We know that if f: X →Y such that y = f(x) is one-one and onto, then we define another function g : Y → X such that x = g(y), where x ∈ X and y ∈ Y, which is also one-one and onto.

In such a case, Domain of g = Range of f and Range of g = Domain of f.
g is called the inverse of f or g = f^-1.
∴ Inverse of g = g^-1 = (f^-1)^-1 =f.

The graph of the sin^-1 function is shown here.

Principal Value Branch of Function sin^-1:
It may be noted that for the domain [-1,1}, the range could be any one of the intervals. …..,
[\(\frac{-3 \pi}{2}\), \(\frac{-\pi}{2}\)], [\(\frac{-\pi}{2}\), \(\frac{\pi}{2}\)] or [\(\frac{\pi}{2}\), \(\frac{3\pi}{2}\)] …….
Inverse Trigonometric Functions Class 12 Notes Maths 2
Corresponding to each interval, we get a branch of the function sin^-1.
The branch with range [\(\frac{-\pi}{2}\), \(\frac{\pi}{2}\)] is called the principal value branch.
Thus, sin^-1: [-1,1] → [\(\frac{-\pi}{2}\), \(\frac{\pi}{2}\)]

Principal Value Branch of Function cos^-1:
The graph of the function cos^-1 is as shown here.
Inverse Trigonometric Functions Class 12 Notes Maths 3
The domain of the function cos^-1 is [-1,1]. Its range is one of the intervals. ………., (-π, 0), (0, π), (π, 2π), ………… It is one-one and onto with the range [-1, 1]. The branch with range (0, π) is called the principal value branch of the function cos-1. Thus, cos-1: [-1,1] → [0, π].

Principal Value Branch of Function tan^-1:
The function tan^-1 is defined whose domain is set of real numbers and range is one of the intervais. [ \(\frac{-3 \pi}{2}\), \(\frac{-\pi}{2}\)], [\(\frac{-\pi}{2}\), \(\frac{\pi}{2}\)] or [\(\frac{\pi}{2}\), \(\frac{3 \pi}{2}\)]…………

The graph of the function is as shown in the adjoining figure.
Inverse Trigonometric Functions Class 12 Notes Maths 4
The branch with range [\(\frac{-\pi}{2}\), \(\frac{\pi}{2}\)] is called the principal value branch of function tan1. Thus, tan^-1: R → [\(\frac{-\pi}{2}\), \(\frac{\pi}{2}\)].

Principal Value Branch of Function cosec^-1:
The graph of function cosec^-1 is as shown here. The function cosec^-1 defined whose domain is R — (-1, 1) and the range is any one of the intervals …….., [ \(\frac{-3 \pi}{2}\), \(\frac{-\pi}{2}\)] – {π}, [\(\frac{-\pi}{2}\), \(\frac{\pi}{2}\)] – {0}, [\(\frac{\pi}{2}\), \(\frac{3 \pi}{2}\)] – {π}……..
Inverse Trigonometric Functions Class 12 Notes Maths 5
The function corresponding to the range [\(\frac{-\pi}{2}\), \(\frac{\pi}{2}\)] – {0} is called the principal value branch of cosec^-1.
Thus, cosec^-1: R – (-1, 1) → [\(\frac{-\pi}{2}\), \(\frac{\pi}{2}\)] – {0}.

Principal Value Branch of Function sec^-1:
The graph of function sec^-1 is shown in adjoining figure. The sec^-1 is defined as a function whose domain is R – (-1, 1) and the range is one of the intervals……. [- π, 0] – {\(\frac{-\pi}{2}\)}, [0, π] – {\(\frac{\pi}{2}\)}, [π, 2π] – {\(\frac{3 \pi}{2}\)}, ……………
Inverse Trigonometric Functions Class 12 Notes Maths 6
The function corresponding to [0, π] – {\(\frac{\pi}{2}\)} is known as the range [0, π] – {\(\frac{\pi}{2}\)} is known as the principal value branch of sec-1.

Thus, sec^-1: R – (-1,1) → [0, π] – {\(\frac{\pi}{2}\)}

Principal Value Branch of Function cot^-1:
The graph of function cor’ is as shown here.
Inverse Trigonometric Functions Class 12 Notes Maths 7
The cot^-1 function is defined as a function whose domain is R and the range is any of the intervals, …………
(-π, 0), (0, π), (π, 2π), …………

The function corresponding to (0, π) is called the principal value branch of the function cot^-1.
Thus, cot^-1: R → (0, π).

The principal value branch of trigonometric inverse functions are as follows:
Inverse Trigonometric Functions Class 12 Notes Maths 8
→ Some Important Results:
1. (a) sin^-1\(\frac{1}{x}\) = cosec^-1x, x ≥ 1, x ≤ -1
(b) cos^-1 \(\frac{1}{x}\) = sec^-1x, x ≥ l, x ≤ -1
(c) tan^-1 \(\frac{1}{x}\) = cot^-1x, x > 0
(d) cosec^-1 \(\frac{1}{x}\) = sin^-1x, x ∈ [-1,1]
(e) sec^-1 \(\frac{1}{x}\) = cos^-1x, x ∈ [-1,1]
(f) cot^-1 \(\frac{1}{x}\) = tan^-1x, x > 0.

2. (a) sin^-1(-x) = – sin^-1x, x ∈ [-1,1]
(b) tan^-1(-x) = – tan^-1x, x ∈ R
(c) cosec^-1(-x) = – cosec^-1x, | x | >1
(d) cos^-1(-x) = π – cos^-1x, x ∈ [-1,1]
(e) sec^-1(-x) = π – sec^-1x, x ∈ R
(f) cot^-1(-x) = π – cot^-1x, x ∈ R

3. (a) sin^-1x + cos^-1x = \(\frac{x}{2}\), x ∈ [-1,1]
(b tan^-1x + cot^-1x = \(\frac{x}{2}\), x ∈ R
(c) cosec^-1x + sec^-1x = \(\frac{x}{2}\), | x | ≥ 1

4. (a) tan^-1x + tan^-1y = tan-1\(\frac{x+y}{1-x y}\), xy < 1
(b) tan^-1x – tan^-1y = tan-1\(\frac{x-y}{1+x y}\), xy > -1
(c) 2 tan^-1x = tan^-1\(\frac{2 x}{1-x^{2}}\), | x | <1
= sin^-1\(\frac{2 x}{1+x^{2}}\), | x | <0
= cos^-1\(\frac{1-x^{2}}{1+x^{2}}\) , x ≥ 0
(d) 2sin^-1x = sin^-1[2x\(\sqrt{1-x^{2}}\)]
(e) 2cos^-1x = cos^-1(2x² – 1)

1. Table

Function	Principal Value Branch
	Domain	Range
(i) y = sin^-1 x	– 1 ≤ x ≤ 1	\(-\frac{\pi}{2}\) ≤ y ≤ \(\frac{\pi}{2}\)
(ii) y = cos^-1 x	– 1 ≤ x ≤ 1	0 ≤ y ≤ π
(iii) y = tan^-1x	– ∞ < x < ∞	\(-\frac{\pi}{2}\) < y < \(\frac{\pi}{2}\)
(iv) y = cot^-1 x	– ∞ < x ≤ -1	0 < y < \(\frac{\pi}{2}\)
(v) y = sec^-1 x	1 ≤ x < ∞ -∞ < x ≤ -1	0 ≤ y < \(\frac{\pi}{2}\) \(\frac{\pi}{2}\)< y < π
(vi) y = cosec^-1 x	1 ≤ x < ∞ -∞ < x ≤ ∞	0 < y ≤ \(\frac{\pi}{2}\) \(-\frac{\pi}{2}\) ≤ y < 0

2. PROPERTIES :

(i) x = sin^-1(sin x) = cos^-1(cos x) = tan^-1 (tan x); etc.
(ii) (a) cosec^-1x = sin^-1 \(\frac { 1 }{ x }\) , x ≥ 1 or x ≤ -1
(b) sec^-1 x = cos^-1\(\frac { 1 }{ x }\) ,x ≥ 1 or x ≤ -1
(c) cot^-1x = tan^-1\(\frac { 1 }{ x }\), x > 0.

(iii) (a) sin^-1 (- x) = – sin^-1x, x ∈ [-1,11
(b) cos^-1 (-x) = π – cos^-1x, x ∈ [- 1, 1]
(c) tan^-1(- x) = – tan^-1 x, x ∈ R
(d) cot^-1(-x) = π – cot^-1 x, x ∈ R
(e) sec^-1 (-x) = π – sec^-1 x, |x| ≥ 1
(f) cosec^-1 (- x) = – cosec^-1 x, |x| ≥ 1

(iv) (a) sin^-1 x + cos^-1 x = \(\frac{\pi}{2}\), x ∈ (-1,1)
(b) tan ^-1 x + cot^-1x = \(\frac{\pi}{2}\) x ∈ R
(c) sec^-1x + cosec^-1= \(\frac{\pi}{2}\), |x| ≥ 1
(d) tan^-1 x + tan^-1 = tan^-1 \(\frac{x+y}{1-x y}\), xy > -1
(e) tan^-1x – tan^-1y = tan^-1 \(\frac{x-y}{1+x y}\), xy > -1
(f) 2tan^-1 x = sin^-1 \(\frac{2 x}{1+x^{2}}\) = cos^-1 \(\frac{2 x}{1-x^{2}}\) = tan^-1

(v) (a) sin ^-1x ± sin^-1 y = sin ^-1 \(\left(x \sqrt{1-y^{2}} \pm y \sqrt{1-x^{2}}\right)\)
(b) cos ^-1x ± cos^-1y = cos ^-1 \(\left(x y \mp \sqrt{1-x^{2}} \sqrt{1-y^{2}}\right)\)

Conductivity of Electronic Solution

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Conductivity of Electronic Solution

We have already learnt that when an electrolyte such as sodium chloride, potassium chloride etc is dissolved in a solvent like water, the electrolyte is completely dissociated to give its constituent ions (namely cations and anions). When an electric field is applied to such an electrolytic solution, the ions present in the solution carry charge from one electrode to another electrode and thereby they conduct electricity. The conductivity of the electrolytic solution is measured using a conductivity cell. (Fig 9.1)

Conductivity of Electronic Solution img 1

A conductivity cell consists of two electrodes immersed in an electrolytic solution. It obeys Ohm’s law like metallic conductor. i.e., at a constant temperature, the current flowing through the cell (I) is directly proportional to the voltage across the cell (V). i.e., I α V (or) I = \(\frac{V}{R}\)
⇒ V = IR …………….. (9.1)

Where ‘R’ is the resistance of the solution in ohm (Ω). Here the resistance is the opposition that a cell offers to the flow of electric current through it.

Resistivity (ρ)

Let us consider a conductivity cell in which the electrolytic solution is confied between the two electrodes having cross sectional area (A) and are separated by a distance ‘l’. Like the metallic conductor, the resistance of such an electrolytic solution is also directly proportional to the length (l) and inversely proportional to the cross sectional area (A).

R α \(\frac{l}{A}\)
R = ρ\(\frac{l}{A}\) (9.2)

Where ρ (rho) is called the specific resistance or resistivity, which depends on the nature of the electrolyte.

If \(\frac{l}{A}\) = 1 m^-1, then, ρ = R. Hence the resistivity is defined as the resistance of an electrolyte confined between two electrodes having unit cross sectional area and are separated by a unit distance. The ratio (\(\frac{l}{A}\)) is called the cell constant, Unit of resitivity is ohm metre (Ωm).

Conductivity

It is more convenient to use conductance rather than resistance. The reciprocal of the resistance (\(\frac{1}{R}\)) gives the conductance of an electrolytic solution. The SI unit of conductance is Siemen (S).

C = \(\frac{1}{R}\) ………….. (9.3)

Substitute (R) from (9.2) in (9.3)

Conductivity of Electronic Solution img 2

⇒ i.e., C = \(\frac{1}{ρ}\).\(\frac{A}{l}\) ………….. (9.4)

The reciprocal of the specific resistance (\(\frac{1}{ρ}\)) is called the specific conductance (or) conductivity. It is represented by the symbol κappa(κ).

Substitute \(\frac{1}{ρ}\) = κ in equation (9.4) and rearranging

⇒ κ = C.(\(\frac{l}{A}\)) ……………. (9.5)

Conductivity of Electronic Solution img 3

If A = 1m² and l = 1m; then κ = C.

The specific conductance is defined as the conductance of a cube of an electrolytic solution of unit dimensions (Fig 9.2). The SI unit of specific conductance is Sm^-1.

Example

A conductivity cell has two platinum electrodes separated by a distance 1.5 cm and the cross sectional area of each electrode is 4.5 sq cm. Using this cell, the resistance of 0.5 N electrolytic solution was measured as 15 Ω. Find the specific conductance of the solution.

Solution

Conductivity of Electronic Solution img 4

Molar Conductivity (Λ_b)

Solutions of different concentrations have different number of electrolytic ions in a given volume of solution and hence they have different specific conductance. Therefore a new quantity called molar conductance (Λ_m) was introduced.

Let us imagine a conductivity cell in which the electrodes are separated by 1m and having V m³ of electrolytic solution which contains 1 mole of electrolyte. The conductance of such a system is called the molar conductance (Λ_m)

We have just learnt that the conductance of 1 m³ electrolytic solution is called the specific conductance (κ). Therefore, the conductance of the above mentioned V m³ solution (Λ_m) is given by the following expression.

(Λ_m) = κ × V ……….. (9.6)

Therefore, Volume of the solution containing one mole of solute = \(\frac{1}{M}\)(mol^-1L)

Conductivity of Electronic Solution img 6

The above relation defines the molar conductance in terms of the specific conductance and the concentration of the electrolyte.

Example

Calculate the molar conductance of 0.025 M aqueous solution of calcium chloride at 25°C. The specific conductance of calcium chloride is 12.04 × 10^-2Sm^-1.

Conductivity of Electronic Solution img 7

Equivalent Conductance (Λ)

Equivalent conductance is defined as the conductance of ‘V’ m³ of electrolytic solution containing one gram equivalent of electrolyte in a conductivity cell in which the electrodes are one metre apart. The relation between the equivalent conductance and the specific conductance is given below.

………….. (9.9)

Where κ the specific conductance and N is the concentration of the electrolytic solution expressed in normality.

Factors Affecting Electrolytic Conductance

If the interionic attraction between the oppositely charged ions of solutes increases, the conductance will decrease.

Solvent of higher dielectric constant show high conductance in solution. Conductance is inversely proportional to the Viscosity of the medium. i.e., conductivity increases with the decrease in viscosity.

If the temperature of the electrolytic solution increases, conductance also increases. Increase in temperature increases the kinetic energy of the ions and decreases the attractive force between the oppositely charged ions and hence conductivity increases.

Molar conductance of a solution increases with increase in dilution. This is because, for a strong electrolyte, interionic forces of attraction decrease with dilution. For a weak electrolyte, degree of dissociation increases with dilution.

Measurement of Conductivity of Ionic Solutions

We have already learnt to measure the specific resistance of a metallic wire using a metre bridge in your physics practical experiment. We know that it works on the principle of wheatstone bridge. Similarly, the conductivity of an electrolytic solution is determined by using a wheatstone bridge arrangement in which one resistance is replaced by a conductivity cell filled with the electrolytic solution of unknown conductivity.

Conductivity of Electronic Solution img 9

In the measurement of specific resistance of a metallic wire, a DC power supply is used. Here, if we apply DC current through the conductivity cell, it will lead to the electrolysis of the solution taken in the cell. So, AC current is used for this measurement to prevent electrolysis.

A wheatstone bridge is constituted using known resistances P, Q, a variable resistance S and conductivity cell (Let the resistance of the electrolytic solution taken in it be R) as shown in the figure 9.3. An AC source (550 Hz to 5 KHz) is connected between the junctions A and C.

Connect a suitable detector E (Such as the telephone ear piece detector) between the junctions ‘B’ and ‘D’. The variable resistance ‘S’ is adjusted until the bridge is balanced and in this conditions there is no current flow through the detector.

Under Balanced Condition,

\(\frac{P}{Q}\) = \(\frac{R}{S}\)
∴R = \(\frac{P}{Q}\) × S ………….. (9.10)

The resistance of the electrolytic solution (R) is calculated from the known resistance values P, Q and the measured ‘S’ value under balanced condition using the above expression (9.10).

Conductivity Calculation

Specific conductance (or) conductivity of an electrolyte can be calculated from the resistance value using the following expression.

κ = \(\frac{1}{R}\)(\(\frac{l}{A}\)) [∵equation 9.5]

The value of the cell constant \(\frac{l}{A}\) is usually provided by the cell manufacturer. Alternatively the cell constant may be determined using KCl solution whose concentration and specific conductance are known.

Example

The resistance of a conductivity cell is measured as 190 Ω using 0.1M KCl solution (specific conductance of 0.1M KCl is 1.3 Sm^-1). When the same cell is filled with 0.003M sodium chloride solution, the measured resistance is 6.3KΩ. Both these measurements are made at a particular temperature. Calculate the specific and molar conductance of NaCl solution.

Given that

Conductivity of Electronic Solution img 10

Solubility Product

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Solubility Product

We have come across many precipitation reactions in inorganic qualitative analysis. For example, dil HCl is used to precipitate Pb²⁺ions as PbCl₂ which is sparingly soluble in water.

Kidney stones are developed over a period of time due to the precipitation of Ca²⁺(as calcium oxalate etc.). To understand the precipitation, let us consider the solubility equilibria that exist between the undissociated sparingly soluble salt and its constituent ions in solution.

For a general salt X_mY_m

The equilibrium constant for the above is

Solubility Product img 2

In solubility equilibria, the equilibrium constant is referred as solubility product constant (or) Solubility product. In such heterogeneous equilibria, the concentration of the solid is a constant and is omitted in the above expression

K_{sp = [Xⁿ⁺]^m[Y^m]ⁿ}

The solubility product of a compound is defined as the product of the molar concentration of the constituent ions, each raised to the power of its stoichiometric co – efficient in a balanced equilibrium equation.

Solubility product finds useful to decide whether an ionic compound gets precipitated when solution that contains the constituent ions are mixed.

When the product of molar concentration of the constituent ions i.e., ionic product, exceeds the solubility product then the compound gets precipitated.

The expression for the solubility product and the ionic product appears to be the same but in the solubility product expression, the molar concentration represents the equilibrium concentration and in ionic product, the initial concentration (or) concentration at a given time ‘t’ is used.

In general we can summarise as,

Ionic product > K_sp, precipitation will occur and the solution is super saturated. Ionic product < K_sp, no precipitation and the solution is unsaturated.

Ionic product = K_sp, equilibrium exist and the solution is saturated.

Determination of Solubility Product from Molar Solubility

Solubility product can be calculated from the molar solubility i.e., the maximum number of moles of solute that can be dissolved in one litre of the solution. For a solute X_mY_n,

X_mY_n(s) ⇄ mXⁿ⁺(aq) + nY^m-(aq)

From the above stoichiometrically balanced equation we have come to know that 1 mole of X_mY_n(s) dissociated to furnish ‘m’ moles of Xⁿ⁺ and ‘n’ moles of Y^m- is ‘s’ is molar solubility of X_mY_n, then

[Xⁿ⁺]=ms and [Y^m+]=ns
∴ K_sp = [Xⁿ⁺]^m[Y^m+]ⁿ

K_sp=(ms)^m(ns)ⁿ
K_sp=(m)^m(n)ⁿ(s)^m+n

Salt Hydrolysis

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Salt Hydrolysis

When an acid reacts with a base, a salt and water are formed and the reaction is called neutralization. Salts completely dissociate in aqueous solutions to give their constituent ions. The ions so produced are hydrated in water. In certain cases, the cation, anion or both react with water and the reaction is called salt hydrolysis. Hence, salt hydrolysis is the reverse of neutralization reaction.

Salts of Strong Acid and a Strong Base

Let us consider the reaction between NaOH and nitric acid to give sodium nitrate and water.
NaOH(aq) + HNO₃(aq) → NaNO₃(aq) + H₂O(l)

The salt NaNO₃ completely dissociates in water to produce Na⁺ and NO₃^– ions.
NaNO₃(aq) → Na⁺(aq) + NO₃^–(aq)

Water dissociates to a small extent as
H₂O(l) ⇄ H⁺(aq) + OH^–(aq)

Since [H⁺] = [OH^–], water is neutral
NO₃^– ion is the conjugate base of the strong acid HNO₃ and hence it has no tendency to react

with H⁺. Similarly, Na⁺ is the conjugate acid of the strong base NaOH and it has no tendency to react with OH^–.

It means that there is no hydrolysis. In such cases [H⁺] = [OH^–] pH is maintained and, therefore, the solution is neutral.

Hydrolysis of Salt of Strong Base and Weak Acid (Anionic Hydrolysis)

Let us consider the reactions between sodium hydroxide and acetic acid to give sodium acetate and water.
NaOH (aq) + CH₃COOH(aq) ⇄ CH₃COONa(aq) + H₂O(l)

In aqueous solution, CH₃COONa is completely dissociated as below

CH₃COONa (aq) → CH₃COO^–(aq) + Na⁺(aq)

CH₃COO^– is a conjugate base of the weak acid CH₃COOH and it has a tendency to react with H⁺
from water to produce unionised acid.

There is no such tendency for Na⁺ to react with OH^–.

CH₃COO^–(aq) + H₂O(l) ⇄ CH₃COOH(aq) + OH^–(aq) and therefore [OH^–]>[H⁺], in such cases, the solution is basic due to hydrolysis and the pH is greater than 7. Let us find a relation between the equilibrium constant for the hydrolysis reaction (hydrolysis constant) and the dissociation constant of the acid.

Salt Hydrolysis img 1

K_h value in terms of degree of hydrolysis (h) and the concentration of salt (C) for the equilibrium can be obtained as in the case of ostwald’s dilution law. K_h = h²C and i.e [OH^–] = \(\sqrt{\mathrm{K}_{\mathrm{h}} \cdot \mathrm{C}}\) and i.e [OH^–] = \(\sqrt{\mathrm{K}_{\mathrm{h}} \cdot \mathrm{C}}\).

pH of salt solution in terms of K_a and the concentration of the electrolyte

pH + pOH = 14
pH = 14 – p OH = 14 – {- log [OH^–]}
= 14 + log[OH^–]

Salt Hydrolysis img 2

Hydrolysis of Salt of Strong Acid and Weak Base (Cationic Hydrolysis)

Let us consider the reactions between a strong acid, HCl, and a weak base, NH₄OH, to produce a salt, NH₄Cl, and water.

HCl (aq) + NH₄OH(aq) ⇄ NH₄Cl(aq) + H₂O(l)
NH₄Cl(aq) → NH₄⁺ + Cl^–(aq)

NH₄⁺ is a strong conjugate acid of the weak base NH₄OH and it has a tendency to react with with OH^– from water to produce unionised NH₄OH shown below.

NH₄⁺(aq) + H₂O(l) ⇄ NH₄OH(aq) + H⁺(aq)

There is no such tendency shown by Cl^– and therefore [H⁺]>[OH^–]; the solution is acidic and the pH
is less than 7.

As discussed in the salt hydrolysis of strong base and weak acid. In this case also, we can establish a relationship between the K_h and k_b as K_h.K_b = K_w

Let us calculate the K_h value in terms of degree of hydrolysis (h) and the concentration of salt

Salt Hydrolysis img 3

Hydrolysis of Salt of Weak Acid and Weak Base (Anionic & Cationic Hydrolysis)

Let us consider the hydrolysis of ammonium acetate.
CH₃COONH₄(aq) → CH₃COO^–(aq) + NH₄⁺(aq)

In this case, both the cation (NH₄⁺) and anion (CH₃COO^–) have the tendency to
react with water

CH₃COO^– + H₂O ⇄ CH₃COOH + OH^–
NH₄⁺ + H₂O ⇄ NH₄OH + H⁺

The nature of the solution depends on the strength of acid (or) base i.e, if K_a> K_b; then the solution is acidic and pH < 7, if K_a< K_b; then the solution is acidic and pH < 7, if K_a < K_b; then the solution is basic and pH > 7, if K_a = K_b; then the solution is basic and pH > 7, if K_a = K_b; then the solution is neutral.

The relation between the dissociation constant (K_a, K_b) and the hydrolysis constant is given by the following
expression.

K_a.K_b.K_h = K_w

pH of the Solution

pH of the solution can be calculated using the following expression,
pH = 7 + 1/2 pK_a – 1/2 pK_b.

Buffer Solution

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Buffer Solution

Do you know that our blood maintains a constant pH, irrespective of a number of cellular acid – base reactions. Is it possible to maintain a constant hydronium ion concentration in such reactions? Yes, it is possible due to buffer action.

Buffer is a solution which consists of a mixture of a weak acid and its conjugate base (or) a weak base and its conjugate acid. This buffer solution resists drastic changes in its pH upon addition of a small quantities of acids (or) bases, and this ability is called buffer action. The buffer containing carbonic acid (H₂CO₃) and its conjugate base HCO^–₃ is present in our blood. There are two types of buffer solutions.

1. Acidic Buffer Solution:

A solution containing a weak acid and its salt.
Example: Solution containing acetic acid and sodium acetate

2. Basic Buffer Solution:

A solution containing a weak base and its salt.
Example: Solution containing NH₄OH and NH₄Cl

Buffer Action

To resist changes in its pH on the addition of an acid (or) a base, the buffer solution should contain both acidic as well as basic components so as to neutralize the effect of added acid (or) base and at the same time, these components should not consume each other. Let us explain the buffer action in a solution containing CH₃COOH and CH₃COONa. The dissociation of the buffer components occurs as below.

Buffer Solution img 1

If an acid is added to this mixture, it will be consumed by the conjugate base CH₃COO^– to form the undissociated weak acid i.e, the increase in the concentration of H⁺ does not reduce the pH significantly.

CH₃COO^–(aq) + H⁺(aq) → CH₃COOH(aq)

If a base is added, it will be neutralized by H₃O⁺, and the acetic acid is dissociated to maintain the equlibrium. Hence the pH is not significantly altered.

Buffer Solution img 2

These neutralization reactions are identical to those reactions that we have already discussed in common ion effect.

Les us analyse the effect of the addition of 0.01 mol of solid sodium hydroxide to one litre of a buffer solution containing 0.8 M CH₃COOH and 0.8 M CH₃COONa. Assume that the volume change due to the addition of NaOH is negligible. (Given: K₄ for CH₃COOH is 1.8 × 10^-5).

Buffer Solution img 3

The dissociation constant for CH₃COOH is given by

Buffer Solution img 4

The above expression shows that the concentration of H⁺ is directly proportional to

Buffer Solution img 5

Let the degree of dissociation of CH₃COOH be α then,
[CH₃OOH] = 0.8 – α and [CH₃COO^–]
= α + 0.8

Buffer Solution img 6

Given that

K_afor CH₃COOH is 1.8 × 10^-5
∴ [H⁺] = 1.8 × 10^-5; pH = – log (1.8 × 10^-5)
= 5 – log 1.8
= 5 – 0.26
pH = 4.74

Calculation of pH after adding 0.01 mol NaOH to 1 litre of buffer.

Given that the volume change due to the addition of NaOH is negligible
∴[OH^–] = 1.8 × 10^-5; pH = – log (1.8 × 10^-5)
= 5 – log 1.8
= 5 – 0.26
pH = 4.74

Caluculation of pH after adding 0.01 mol NaOH to 1 litre of buffer.

Given that the volume change due to the addition of NaOH is negligible
∴ [OH^–] = 0.01 M.

The consumption of OH^– are expressed by the following equations.

Buffer Solution img 7

The addition of a strong base (0.01 M NaOH) increased the pH only slightly ie., from 4.74 to 4.75 . So, the buffer action is verified.

Buffer Capacity and Buffer Index

The buffering ability of a solution can be measured in terms of buffer capacity. Vanslyke introduced a quantity called buffer index, β , as a quantitative measure of the buffer capacity. It is defined as the number of gram equivalents of acid or base added to 1 litre of the buffer solution to change its pH by unity.

Here, β = \(\frac{dB}{d(pH)}\) …………. (8.19)
dB = number of gram equivalents of acid / base added to one litre of buffer solution.
d(pH) = The change in the pH after the addition of acid / base.

Henderson – Hassel Balch Equation

We have already learnt that the concentration of hydronium ion in an acidic buffer solution depends on the ratio of the concentration of the weak acid to the concentration of its conjugate base present in the solution i.e.,

Buffer Solution img 8 ………….. (8.20)

The weak acid is dissociated only to a small extent. Moreover, due to common ion effect, the dissociation is further suppressed and hence the equilibrium concentration of the acid is nearly equal to the initial concentration of the unionised acid. Similarly, the concentration of the conjugate base is nearly equal to the initial concentration of the added salt.

…………… (8.21)

Here [acid] and [salt] represent the initial concentration of the acid and salt, respectively used to prepare the buffer solution. Taking logarithm on both sides of the equation

……………. (8.22)

Reverse the sign on both sides

…………….. (8.23)

We know that

pH = – log [H₃O⁺] and pK_a = – logK_a

Buffer Solution img 11

Common ion Effect

July 3, 2021

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Common ion Effect

When a salt of a weak acid is added to the acid itself, the dissociation of the weak acid is suppressed further. For example, the addition of sodium acetate to acetic acid solution leads to the suppression in the dissociation of acetic acid which is already weakly dissociated. In this case, CH₃COOH and CH₃COONa have the common ion, CH₃COO^–.

Let us analyse why this happens. Acetic acid is a weak acid. It is not completely dissociated in aqueous solution and hence the following equilibrium exists. CH₃COOH(aq) ⇄ H⁺(aq) + CH₃COO^–(aq).

However, the added salt, sodium acetate, completely dissociates to produce Na⁺ and CH₃COO^– ion.

CH₃COONa(aq) → Na⁺(aq) + CH₃COO^–(aq)

Hence, the overall concentration of CH₃COO^– is increased, and the acid dissociation equilibrium is disturbed. We know from Le chatelier’s principle that when a stress is applied to a system at equilibrium, the system adjusts itself to nullify the effect produced by that stress.

So, in order to maintain the equilibrium, the excess CH₃COO^– ions combines with H⁺ ions to produce much more unionized CH₃COOH i.e, the equilibrium will shift towards the left. In other words, the dissociation of a weak acid (CH₃COOH) is suppressed. Thus, the dissociation of a weak acid (CH₃COOH) is suppressed in the presence of a salt (CH₃COONa) containing an ion common to the weak electrolyte. It is called the common ion effect.

The common-ion effect refers to the decrease in solubility of an ionic precipitate by the addition to the solution of a soluble compound with an ion in common with the precipitate. This behaviour is a consequence of Le Chatelier’s principle for the equilibrium reaction of the ionic association/dissociation.

The common ion effect is the phenomenon in which the addition of an ion common to two solutes causes precipitation or reduces ionization. An example of the common ion effect is when sodium chloride (NaCl) is added to a solution of HCl and water.

The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium. The common ion effect suppresses the ionization of a weak base by adding more of an ion that is a product of this equilibrium. The reaction is put out of balance, or equilibrium.

Aquifers which contain chalk and limestone, a common ion effect is used in these to obtain drinking water. Calcium carbonate is sparingly soluble in water, and it can be precipitated out by adding sodium chloride in the solution. In this way, the common ion effect is used in treatment of water.

According to Le Chatelier’s principle, addition of more ions alters the equilibrium and shifts the reaction to favor the solid or deionized form. In the case of an an acidic buffer, the hydrogen ion concentration decreases, and the resulting solution is less acidic than a solution containing the pure weak acid.

Le Chatelier’s principle is an observation about chemical equilibria of reactions. It states that changes in the temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system in order to achieve a new equilibrium state.

At a given temperature, the product of water concentrations and ions is known as the ionic product of water. With the rise in temperature, the value increases i.e. the concentration of H⁺ and OH^– ions increases with temperature increases.

If you have a solution and solute in equilibrium, adding a common ion (an ion that is common with the dissolving solid) decreases the solubility of the solute. This is because Le Chatelier’s principle states the reaction will shift toward the left (toward the reactants) to relieve the stress of the excess product.

pH gives us the measure of acid/base strength of any solution. pH scale ranges from 0-14, 7 is neutral, below 7 it represents acidic nature i.e. the compound is acidic and above 7 it represents basic nature i.e. the compound is basic i.e. pH is the negative logarithm of hydrogen ion.

The solubility of slightly soluble substances can be decreased by the presence of a common ion. This effect is known as common ion effect. Addition of a common ion to a slightly soluble salt solution will add up to the concentration of the common ion.

Le Chatelier’s principle (also known as “Chatelier’s principle” or “The Equilibrium Law”) states that when a system experiences a disturbance (such as concentration, temperature, or pressure changes), it will respond to restore a new equilibrium state.

Explanation:

This phenomenon is called the common-ion effect. When a compound dissolves in water it dissociates into ions. Increasing the concentration of one of these ions will shift the equilibrium towards the compound, thereby making it hard for the compound to dissolve in water (decreases solubility of compound).

When there is an increase in pressure, the equilibrium will shift towards the side of the reaction with fewer moles of gas. When there is a decrease in pressure, the equilibrium will shift towards the side of the reaction with more moles of gas.

The classic example of the practical use of the Le Chatelier principle is the Haber-Bosch process for the synthesis of ammonia, in which a balance between low temperature and high pressure must be found.

Le Chatelier’s Principle helps to predict what effect a change in temperature, concentration or pressure will have on the position of the equilibrium in a chemical reaction. This is very important, particularly in industrial applications, where yields must be accurately predicted and maximised.

Ionisation of Weak Acids

July 2, 2021

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Ionisation of Weak Acids

We have already learnt that weak acids are partially dissociated in water and there is an equilibrium between the undissociated acid and its dissociated ions. Consider the ionisation of a weak monobasic acid HA in water.

Applying law of chemical equilibrium, the equilibrium constant K_c is given by the expression

Ionisation of Weak Acids img 1 …………. (8.9)

The square brackets, as usual, represent the concentrations of the respective species in moles per litre. In dilute solutions, water is present in large excess and hence, its concentration may be taken as constant say K. Further H₃O⁺ indicates that hydrogen ion is hydrated, for simplicity it may be replaced by H⁺. The above equation may then be written as,

………….. (8.10)

The product of the two constants K_C and K gives another constant. Let it be K_a

…………. (8.11)

The constant K_a is called dissociation constant of the acid. Like other equilibrium constants, K_a also varies only with temperature. Similarly, for a weak base, the dissociation constant can be written as below.

………….. (8.12)

Ostwald’s Dilution Law

Ostwald’s dilution law relates the dissociation constant of the weak acid (K_a) with its degree of dissociation (α) and the concentration (c). Degree of dissociation (α) is the fraction of the total number of moles of a substance that dissociates at equilibrium.

We shall derive an expression for ostwald’s law by considering a weak acid, i.e. acetic acid (CH₃COOH). The dissociation of acetic acid can be represented as

CH₃COOH ⇄ H⁺ + CH₃COO^–

The dissociation constant of acetic acid is,

Ionisation of Weak Acids img 6

Substituting the equilibrium concentration in equation (8.13)

Ionisation of Weak Acids img 7 …………. (8.14)

We know that weak acid dissociates only to a very small extent. Compared to one, α is so small and hence in the denominator (1 – α) ~ 1. The above expression (8.14) now becomes,

Ionisation of Weak Acids img 8 ………….. (8.15)

Let us consider an acid with K_a value 4 × 10^-4 and calculate the degree of dissociation of that acid at two different concentration 1 × 10^-2M and 1 × 10^-4M using the above expression (8.15) For 1 × 10^-2M,

Ionisation of Weak Acids img 9
= 2 × 10^-1
= 0.2

For 1 × 10^-4M
Ionisation of Weak Acids img 10
= 2

When the dilution increases by 100 times, (Concentration decreases from 1 × 10^-2M to 1 × 10^-2M), the dissociation increases by 10 times. Thus, we can conclude that, when dilution increases, the degree of dissociation of weak electrolyte also increases. This statement is known as Ostwald’s dilution Law.

The concentration of H⁺ (H₃O⁺) can be caluculated using the K_a value as below.

[H⁺] = αC (Refer table) ………….. (8.16)

Equilibrium molar concentration of [H⁺] is equal to αC

Ionisation of Weak Acids img 11

Similarly, for a weak base

K_b = α² and α = \(\sqrt{\frac{K_{b}}{C}}\)
[OH^–] = αC
(or)
[OH^–] = \(\sqrt{\mathrm{K}_{\mathrm{b}} \mathrm{C}}\) ………….. (8.18)

The pH Scale

July 2, 2021

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The pH Scale

We usually deal with acid / base solution in the concentration range 10^-1 to 10^-2M. To express the strength of such low concentrations, Sorensen introduced a logarithmic scale known as the pH scale. The term pH is derived from the French word ‘Purissance de hydrogene’ meaning, the power of hydrogen. pH of a solution is defined as the negative logarithm of base 10 of the molar concentration of the hydronium ions present in the solution.

pH = – log₁₀[H₃O⁺] …………. (8.5)

The concentration of H₃O⁺ in a solution of known pH can be calculated using the following expression.

[H₃O⁺] = 10^-pH (or) [H₃O⁺] = antilog of (-pH) …………. (8.6)

Similarly, pOH can also be defined as follows

pOH = -log₁₀[OH^–] …………….. (8.7)

As discussed earlier, in neutral solutions, the concentration of [H₃O⁺] as well as [OH⁺] is equal to 1 × 10^-7M at 25°C . The pH of a neutral solution can be calculated by substituting this H₃O⁺concentration in the expression (8.5)

pH = – log₁₀[H₃O⁺]
= – log₁₀10^-7
= (-7)(-1)log₁₀10 = +7(1) = 7 [∵log₁₀¹⁰ = 1]

Similary, we can calculate the pOH of a neutral solution using the expression (8.7), it is also equal to 7. The negative sign in the expression (8.5) indicates that when the concentration of [H₃O⁺] increases the pH value decreases.

For example, if the [H₃O⁺] increases from to 10^-7 to 10^-5M the pH value of the solution decreases from 7 to 5. We know that in acidic solution, [H₃O⁺]>[OH^–], i.e; [H₃O⁺]>10^-7. So, we can conclude that acidic solution should have pH value less than 7 and basic solution should have pH value greater than 7.

Relation Between pH and pOH

A relation between pH and pOH can be established using their following definitions

The pH Scale img 1

The pH Scale img 2

Ionisation of Water

July 2, 2021

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Ionisation of Water

We have learnt that when an acidic or a basic substance is dissolved in water, depending upon its nature, it can either donate (or) accept a proton. In addition to that the pure water itself has a little tendency to dissociate. i.e, one water molecule donates a proton to an another water molecule. This is known as auto ionisation of water and it is represented as below.

Ionisation of Water img 1

In the above ionisation, one water molecule acts as an acid while the another water molecule acts as a base. The dissociation constant for the above ionisation is given by the following expression

…………… (8.3)

The concentration of pure liquid water is one. i.e, [H₂O]² = 1
∵ K_w = [H₃O⁺][OH^–] …………. (8.4)

Here, K_w represents the ionic product (ionic product constant) of water.

It was experimentally found that the concentration of H₃O⁺ in pure water is 1 × 10^-7 at 25°C. Since the dissociation of water produces equal number of H₃O⁺ and OH^–, the concentration of OH^–is also equal to 1 × 10^-7 at 25°C.

Therefore, the ionic product of water at 25°C is

K_W = [H₃O]⁺[OH^–] …………. (8.4)
K_W = (1 × 10^-7)(1 × 10^-7)
= 1 × 10^-14.

Like all equilibrium constants, K_w is also a constant at a particular temperature. The dissociation of water is an endothermic reaction. With the increase in temperature, the concentration of H₃O⁺ and OH^– also increases, and hence the ionic product also increases.

In neutral aqueous solution like NaCl solution, the concentration of H₃O⁺ is always equal to the concentration of OH^– whereas in case of an aqueous solution of a substance which may behave as an acid (or) a base, the concentration of H₃O⁺ will not equal to
[OH^–].

Ionisation of Water img 3

We can understand this by considering the aqueous HCl as an example. In addition to the auto ionisation of water, the following equilibrium due to the dissociation of HCl can also exist.

HCl + H₂O ⇄ H₃O⁺ + Cl^–

In this case, in addition to the auto ionisation of water, HCl molecules also produces H₃O⁺ ion by donating
a proton to water and hence [H₃O⁺]>[OH^–]. It means that the aqueous HCl solution is acidic. Similarly, in basic solution such as aqueous NH₃, NaOH etc…. [OH^–]>[H₃O⁺].

Strength of Acids and Bases

July 2, 2021

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Strength of Acids and Bases

The strength of acids and bases can be determined by the concentration of H₃O⁺ (or) OH^– produced per mole of the substance dissolved in H₂O. Generally we classify the acids/bases either as strong or weak. A strong acid is the one that is almost completely dissociated in water while a weak acid is only partially dissociated in water.

Let us quantitatively define the strength of an acid (HA) by considering the following general equilibrium.

Strength of Acids and Bases img 1

The equilibrium constant for the above ionisation is given by the following expression

Strength of Acids and Bases img 2 ………… (8.1)

We can omit the concentration of H₂O in the above expression since it is present in large excess and essentially unchanged.

……………. (8.2)

Here, K_a is called the ionisation constant or dissociation constant of the acid. It measures the strength of an acid. Acids such as HCl, HNO₃ etc… are almost completely ionised and hence they have high K_a value (K_a for HCl at 25°C is 2 × 10⁶).

Acids such as formic acid (K_a = 1.8 × 10^-4 at 25°C), acetic acid (1.8 × 10^-5 at 25°C) etc.. are partially ionised in solution and in such cases, there is an equilibrium between the unionised acid molecules and their dissociated ions. Generally, acids with K_a value greater than ten are considered as strong acids and less than one considered as weak acids.

Let us consider the dissociation of HCl in aqueous solution,

As discussed earlier, due to the complete dissociation, the equilibrium lies almost 100% to the right. i.e., the Cl^– ion has only a negligible tendency to accept a proton form H₃O⁺. It means that the conjugate base of a strong acid is a weak base and vice versa. The following table illustrates the relative strength of conjugate acid – base pairs.

Strength of Acids and Bases img 5

Acids and Bases

July 2, 2021

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Acids and Bases

The term ‘acid’ is derived from the latin word ‘acidus’ meaning sour. We have already learnt in earlier classes that acid tastes sour, turns the blue litmus to red and reacts with metals such as zinc and produces hydrogen gas. Similarly base tastes bitter and turns the red litmus to blue.

These classical concepts are not adequate to explain the complete behaviour of acids and bases. So, the scientists developed the acid – base concept based on their behaviour.

Let us, learn the concept developed by scientists Arrhenius, Bronsted and Lowry and Lewis to describe the properties of acids and bases.

Arrhenius Concept

One of the earliest theories about acids and bases was proposed by swedish chemist Svante Arrhenius. According to him, an acid is a substance that dissociates to give hydrogen ions in water. For example, HCl, H₂SO₄ etc., are acids. Their dissociation in aqueous solution is expressed as

The H⁺ ion in aqueous solution is highly hydrated and usually represented as H₃O⁺, the simplest hydrate
of proton [H(H₂O)]⁺. We use both H⁺ and H₃O⁺ to mean the same.

Similarly a base is a substance that dissociates to give hydroxyl ions in water. For example, substances like NaOH, Ca(OH)₂ etc., are bases.

Limitations of Arrhenius Concept

Arrhenius theory does not explain the behaviour of acids and bases in non aqueous solvents such as acetone, Tetrahydrofuran etc.
This theory does not account for the basicity of the substances like ammonia (NH₃) which do not possess hydroxyl group.

Lowry – Bronsted Theory (Proton Theory)

In 1923, Lowry and Bronsted suggested a more general definition of acids and bases. According to their concept, an acid is defined as a substance that has a tendency to donate a proton to another substance and base is a substance that has a tendency to accept a proton from other substance. In other words, an acid is a proton donor and a base is a proton acceptor.

When hydrogen chloride is dissolved in water, it donates a proton to the later. Thus, HCl behaves as an acid and H₂O is base. The proton transfer from the acid to base can be represented as

HCl + H₂O ⇄ H₃O⁺ + Cl^–

When ammonia is dissolved in water, it accepts a proton from water. In this case, ammonia (NH₃) acts as a base and H₂O is acid. The reaction is represented as

H₂O + NH₃ ⇄ NH⁺₄ + OH^–

Let us consider the reverse reaction in the following equilibrium

H₃O⁺ donates a proton to Cl^– to form HCl i.e., the products also behave as acid and base. In general,
Lowry – Bronsted (acid – base) reaction is represented as

Acid₁ + Base₂ ⇄ Acid₂ + Base₁

The species that remains after the donation of a proton is a base (Base₁) and is called the conjugate base of the Bronsted acid (Acid₁). In other words, chemical species that differ only by a proton are called conjugate acid – base pairs.

Acids and Bases img 4

HCl and Cl^–, H₂O and H₃O⁺ are two conjugate acid – base pairs, i.e; Cl^– is the conjugate base of the acid HCl. (or) HCl is conjugate acid of Cl^–. Similarly H₃O⁺ is the conjugate acid of H₂O.

Limitations of Lowry – Bronsted Theory

(i) Substances like BF₃, AlCl₃ etc., that do not donate protons are known to behave as acids.

Lewis Concept

In 1923, Gilbert. N. Lewis proposed a more generalised concept of acids and bases. He considered the electron pair to define a species as an acid (or) a base. According to him, an acid is a species that accepts an electron pair while base is a species that donates an electron pair. We call such species as Lewis acids and bases.

A Lewis acid is a positive ion (or) an electron deficient molecule and a Lewis base is a anion (or) neutral molecule with at least one lone pair of electrons.

Les us consider the reaction between Boron tri flouride and ammonia.

Acids and Bases img 5

Here, boron has a vacant 2p orbital to accept the lone pair of electrons donated by ammonia to form a new coordinate covalent bond. We have already learnt that in coordination compounds, the Ligands act as a Lewis base and the central metal atom or ion that accepts a pair of electrons from the ligand behaves as a Lewis acid.

Acids and Bases img 6