CBSE Class 11 Chemistry Notes For Degree Of Ionisation

Degree Of Ionisation And Ionisation Constant Of Weak Electrolyte

Weak electrolytes partially dissociate into ions in solutions and there always exists a dynamic equilibrium involving the dissociated ions and the undissociated molecules.

This state of equilibrium is called an ionic equilibrium. The equilibrium constant associated with an ionic equilibrium is known as the ionization or dissociation constant of the weak electrolyte.

Degree Of Ionisation

During the ionization of a solution of a weak electrolyte, the fraction of its total number of molecules that get dissociated at equilibrium is called the degree of ionization or dissociation of the electrolyte.

Degree of ionization of an electrolyte (α)

Number of dissociated molecules of the electrolyte at equilibrium/ Total number of molecules of the electrolyte.

Suppose, a fixed volume of solution contains 0.5 mol of a dissolved electrolyte. If 0.2 mol of this electrolyte gets dissociated at equilibrium, then the degree of dissociation (a) O₂ of the electrolyte

⇒  \(\frac{0.2}{0.5}=0.4\), i.e., 40 % of the electrolyte exists in an ionized state in the solution.

As strong electrolytes dissociate completely in solutions, their degree of dissociation (a)= 1 but, the degree of dissociation of weak electrolytes is always less than 1.

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Ionization or dissociation constant of weak electrolytes:

Let us consider the following equilibrium which is established by a weak AB because of its partial ionization in water

⇒ \(\mathrm{AB}(a q) \rightleftharpoons \mathrm{A}^{+}(a q)+\mathrm{B}^{-}(a q)\)

Applying the law of mass action to the equilibrium we have the equilibrium constant, K,

⇒ K= \(\frac{\left[\mathrm{A}^{+}\right]\left[\mathrm{B}^{-}\right]}{[\mathrm{AB}]}\)

Where [A⁺, [B^–], and [AB] is the molar concentrations (mol L^-1) of A⁺, B^–, and AB, respectively at equilibrium.

K represents the ionization or dissociation constant of the weak electrolyte AB.

The value of equilibrium constant (K) changes well with a variation of temperature,  at a constant uimporniure, It has a fixed value.

Oslwald’s dilution law

Let the initial concentration (before dissociation) of an aqueous solution of a weak electrolyte AB c mol^-1,

The following equilibrium Is established due to the partial dissociation of ΔH in an aqueous solution:

AB(aq) ⇌  A⁺+ (aq)+ B^–(aq)

Suppose, at equilibrium, the degree of ionization or dissociation of ΔH is a. So, a mol of ΔH on Its ionization will result In a mol of Δ1 Ion and a mol of H- Ions. The number of moles of AB that remain unlonloniscd is (1 — α).

Hence at equilibrium, the molar concentrations of different species will be as follows:

By applying the law of mass action to this equilibrium, we get an equilibrium constant

⇒ \((K)=\frac{\left[\mathrm{A}^{+}\right] \times\left[\mathrm{B}^{-}\right]}{[\mathrm{AB}]}=\frac{\alpha c \times \alpha c}{c(1-\alpha)}=\frac{\alpha^2 c}{1-\alpha}\) ……………………..(1)

Equation (1) represents the mathematical expression of Ostwald’s dilution law. The equilibrium constant (K) is called the Ionisation constant of the weak electrolyte, AB. At ordinary concentration, the value of the degree of dissociation (a) of a weak electrolyte is generally very small. So, (1 – α)≈1

.With this approximation equation (1) can be written as,

K = \(\alpha^2 c \quad \text { or, } \quad \alpha=\sqrt{\frac{K}{\boldsymbol{c}}}\)……………………..(2)

Equation (2) is the simplified mathematical expression of Ostwald’s dilution law Conclusion,’ From equation (2), it can be concluded that—

The degree of dissociation (or) of a weak electrolyte in a solution is inversely proportional to the square root of the concentration of the solution (since at constant temperature, K has a definite value).

So, at a fixed temperature, the degree of dissociation of a weak electrolyte in its solution increases with the decrease in the concentration of the solution and decreases with the caraway in the concentration of the solution, It roof of weak electrolyte remains dissolved In 6, of the solution, then the molar concentration of the solution,

c=  \(\frac{1}{V}.\) Substituting

c= \(\frac{1}{V} .\) In equation (2),

we have an α = √KV Thin equation showing that when v Increases (as happens when the solution Is diluted), the degree of dissociation of the electrolyte also Increases.

Ostwald’s dilution law:

At a certain temperature, the degree of Ionisation of a weak electrolyte in a solution Is Inversely proportional to the square root of the molar concentration of the solution.

Or, At a certain temperature, the degree of ionization of a weak electrolyte In a solution is directly proportional to the square root of the volume of the solution containing mol of the electrolyte.

Limitation of Ostwald’s dilution law:

Ostwald’s dilution law applies to weak electrolytes only. As strong electrolytes ionize almost completely at all concentrations, this law does not apply to them.

Ionization or dissociation constant of a weak acid and concentration of H30+ ions in its aqueous solution

Weak acids partially dissociate into ions in aqueous solutions, and there always exists a dynamic equilibrium involving the dissociated ions and the undissociated molecules. Like any other equilibrium, such type of equilibrium also has an equilibrium constant, known as the ionization or dissociation constant of the corresponding weak acid.

The ionization constant of a weak acid is designated by the Ionisation constant of a weak monobasic acid:

Let HA be a weak monobasic acid which on partial ionization in an aqueous solution forms the following equilibrium

HA(aq) + H₂O(l) ⇌H₃O⁺(aq) + A^–(aq)

Using the law of mass action to the above equilibrium, we get equilibrium contract

⇒ \(K=\frac{\left[\mathrm{H}_3\mathrm{O}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]\left[\mathrm{H}_2 \mathrm{O}\right]}\)

where [H₃O⁺], [A^–], [HA], and [H₂O] represent the molar concentrations (mol-L^-1) of H₃O⁺, A⁺, HA, and H₂O, respectively, at equilibrium in solution.

In the solution, the concentration of H₂O is much higher than that of HA and its concentration does not change significantly due to partial ionization of HA.

The concentration of H₂O, therefore, remains constant.

So, \(K \times\left[\mathrm{H}_2 \mathrm{O}\right]=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\)

Since [H₂O] = constant, the K ×  (H₂O] constant is known as the ionization or dissociation constant of the weak acid and Is designated by ‘Ka’.

∴ \(K_a=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+} \| \mathrm{A}^{-}\right]}{[\mathrm{II} \mathrm{A}]}\)

Significance of ionization constant of a weak acid:

Like any other equilibrium constant, the value of the ionization constant (Ka) of a weak acid is constant at a fixed temperature.

The higher the tendency of a weak acid HA to donate proton in water, the higher the concentration of H₃O⁺ and A^– ions at equilibrium in the solution.

Hence from equation (1), it can be said that the stronger the acid, the larger the value of its ionization constant.

If the solutions of two monobasic acids have the same molar concentration, then the solution containing the acid with a larger value of Ka will have a higher concentration of H30+ ions than the other solution.

Ionization constants (Ka) of some weak monobasic acids at 25°C [in water]

The concentration of H₃O⁺ ions in an aqueous solution of a weak monobasic acid:

Let a weak acid HA on its partial ionization water form the following equilibrium:

HA(aq) + H₂O(l)⇌ H₃O⁺(aq) + A^–(aq)

If the initial concentration of HA in its aqueous solution is C mol.L^-1 and the degree of ionization of HA

At equilibrium is then the concentrations of different species at equilibrium will be as follows:

(During ionization of a weak acid, the concentration of H20 remains unchanged.) Therefore, the Ionisation constant of the weak add HA,

⇒ \(K_a=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \times\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}=\frac{c \alpha \times c \alpha}{c(1-\alpha)}=\frac{\alpha^2 c}{1-\alpha}\)

At ordinary concentration, the degree of dissociation (a) of a weak acid is negligible. So, (1 – α)≈1

⇒ \(\text { Hence, } K_a=\alpha^2 c \text { or, } \alpha=\sqrt{\frac{K_a}{c}}\)

Therefore, if the concentration (c) of a weak monobasic acid and its ionization constant (Ka) are known, then the degree of ionization (a) of the acid can be calculated by using the equation (1)

Concentration of H₃O⁺ ion [H₃O⁺] = αc…………………………(2)

⇒ \(\text { or, }\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\alpha c=\sqrt{\frac{K_a}{c}} \times c=\sqrt{c \times K_a}\)

In an aqueous solution of a weak acid, if the concentration of the acid and its degree of ionization (or) are known, then the concentration of H₃O+ ions in the solution can be calculated by using equation (2)

Alternatively, if the concentration (c) of the acid and its ionization constant (Ka) is known, then the concentration of H₃O⁺ ions can be calculated by using the equation (3).

Relative strengths of two weak monobasic acids:

Let us consider the aqueous solutions of two weak acids HA and HA-, each with the same molar concentration of cool-L^–

At a given temperature, if the ionization constants of HA and HA’ are K_a and K’_a respectively and a and a are their degrees of ionization in their respective solutions, then

⇒ \(\alpha=\sqrt{\frac{K_a}{c}} \quad \text { and } \quad \alpha^{\prime}=\sqrt{\frac{K_a^{\prime}}{c}}\)

∴ \(\frac{\alpha}{\alpha^{\prime}}=\sqrt{\frac{K_a}{K_a^{\prime}}}\)

Therefore, at a particular temperature, if the molar concentrations of the solutions of two weak monobasic acids are the same, then the acid having a larger ionization constant will have a higher degree of ionization than the other.

The degree of dissociation (a) of a weak acid in its solution of a given concentration is a measure of its strength (or proton donating tendency). The higher the degree of dissociation of an acid in its solution, the stronger the acid.

It means that the strength of an acid in its solution is proportional to its degree of dissociation in the solution.

Hence, at a particular temperature, for a definite molar concentration

⇒ \(\frac{\text { Strength of acid } \mathrm{HA}}{\text { Strength of acid } \mathrm{HA}^{\prime}}=\sqrt{\frac{K_a}{K_a^{\prime}}}\)

Ionization constant of weak polybasic acids:

An acid with more than one replaceable H-atom is called a polybasic acid,

Example:

H₂CO₃, H₃PO₄, and H₂S. H₃SO₄ and H₂S are dibasic acids as they have two replaceable H-atoms, whereas H₃PO₄ is a tribasic acid as it has three replaceable hydrogen atoms.

These acids dissociate in a series of steps, each of which attains an equilibrium and has its characteristic equilibrium constant.

Example: Ionisation of H₃PO₄ in water:

In water, H₃PO₄ ionizes in the following three steps as it contains three replaceable hydrogen atoms:

1. H₃PO₄(aq) + H₂O ⇌ H₃O⁺(aq) + H₂PO₄^–  (aq)

Ionization constant,

Ka₁ = \(\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \times\left[\mathrm{H}_2 \mathrm{PO}_4^{-}\right]}{\left[\mathrm{H}_3 \mathrm{PO}_4\right]}\)

2. H₂PO₄^–(aq) + H₂O ⇌ H₃O⁺(aq) + HPO₄^2- (aq)

Ionization constant

Ka₂ = \(\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \times\left[\mathrm{HPO}_4^{2-}\right]}{\left[\mathrm{H}_2 \mathrm{PO}_4^{-}\right]}\)

H₂PO₄^2-  + H₂O ⇌ H₃O⁺(aq) +PO₄^3- (aq)

Ionization constant,

Ka₃= \(\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \times\left[\mathrm{PO}_4^{3-}\right]}{\left[\mathrm{HPO}_4^{2-}\right]}\)

Overall ionization constant, K_a = Ka₁ × Ka₂ × Ka₃

At constant temperature, Ka₁ > Ka₂ > Ka₃ for a weak tribasic acid. Due to the electrostatic force of attraction, a singly charged anion (for example- HS^– or H₂PO^–4) has less tendency to lose a proton than a neutral molecule (for example H₂S ).

Similarly, it is more difficult for a doubly charged anion (For example:  HPO₄^2- ) to lose a proton than a singly charged anion.

Ionization or dissociation constant of a weak base and concentration of OH^– ions in its aqueous solution

When a weak base is dissolved in water, it reacts with water to form its conjugate acid and OH- ions. Eventually, an equilibrium involving the conjugate acid and unreacted base is established. Such an equilibrium has its characteristic equilibrium constant known as the ionization constant of the weak base. The ionization constant of weak basis is denoted by.

Ionisation Constant of a weak monoacid lc base:

Let us consider an aqueous solution of a weak monoacidic base B. Since B is a weak base, a small nili fiber of molecules reacts with an equal number of H₂O molecules to form the conjugate acid, BH⁺, and OH^– ions.

A dynamic equilibrium is thus established between BH⁺, OH^– and unionized molecules as follows:

⇒ \(\mathrm{B}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{BH}^{+}(a q)+\mathrm{OH}^{-}(a q)\)

Applying the law of mass action to the above equilibrium, we have, an equilibrium constant

⇒  \(K=\frac{\left[\mathrm{BH}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\mathrm{B}]\left[\mathrm{H}_2 \mathrm{O}\right]}\)

Where [BH⁺], [OH^–], [B], and [H₂O] represent the molar concentrations of BH⁺, OH⁺, B, and H₂O respectively, at equilibrium. In an aqueous solution, the concentration of H₂O is much higher than that of B. Thus, any change in concentration that occurs because of the reaction of H₂O with B can be neglected. So, [H₂O] remains essentially constant.

So, \(K \times\left[\mathrm{H}_2 \mathrm{O}\right]=\frac{\left[\mathrm{BH}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\mathrm{B}]}\)

Since [H₂O] = constant, K × [H₂O] = constant This constant is known as the ionization constant of the weak base B and is denotedby’Kb ‘

Therefore \(K_b=\frac{\left[\mathrm{BH}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\mathrm{B}]}\)

Equation (1) expresses the ionization constant for the weak monoacidic base, B.

Significance of ionization constant of a weak base:

As in the case of other equilibrium constants, the ionization constant of a weak base (Kb) has a definite value at a particular temperature.

If the weak base, B, has a high tendency to accept protons in water, it reacts with water to a greater extent. This results in high concentrations of BH⁺ and OH^– in the solution and gives rise to a large value of Kb for the base. Therefore, the larger the value of Kb for a weak base, the stronger the base.

At a certain temperature, if the aqueous solution of two weak bases has some molar concentration, then the solution containing the base with a larger value of Kb will have a higher concentration of OH” ions at equilibrium in the solution.

Ionization constants (kb) of some weak mono-acidic bases at 25 °C :

The concentration of OH^– in an aqueous solution of a weak monoacidic base: Let a weak base, B on partial ionization in an aqueous solution form the following equilibrium:

B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH^–(aq)

Let the initial concentration of B in the aqueous solution be c mol.L^-1. According to the above equation, at equilibrium, if the concentration of OH^– is x mol.L^-1, then the concentrations of BH⁺ & B will be × mol.L^-1and (c-x) mol.L^-1 respectively because according to the given equation, 1 molecule of B and 1 molecule of H20 react to form one BH⁺ and one OH^– ion.

Therefore, ionization constant ofthe weak base, B is:

⇒ \(K_b=\frac{\left[\mathrm{BH}^{+}\right] \times\left[\mathrm{OH}^{-}\right]}{[\mathrm{B}]}=\frac{x \times x}{\mathrm{c}-x}=\frac{x^2}{c-x}\)

Since the base is weak, a very small amount of it reacts with water. So, x is negligible in comparison to c, and hence c-xx.

⇒ \(K_b=\frac{x^2}{c} \quad \text { or, } x=\sqrt{K_b \times c}\)

Therefore, in an aqueous solution of the weak base, B

⇒ \(\left[\mathrm{OH}^{-}\right]=\sqrt{K_b \times c}\)

If the ionization constant (Kb) of a monoacidic weak base and the concentration (c) of its aqueous solution are known, then the concentration of OH⁺ ions in the solution can be determined with the help of equation (1)

Determination of [H₃O⁺] in a solution of strong acid and [OH-] in a solution of a strong base

Strong acids (For example HCl, HBr, HClO₄, HNOs, H₂SO₂) and strong bases [e.g., NaOH, KOH, Ca(OH)₂] completely ionize in their aqueous solutions. Hence, the concentration of H₃O⁺ ions (or, OH^– ions) in the aqueous solution of a strong acid (or a strong base) can be calculated from the initial concentration of the acid (or base) in the solution.

When the concentration of the solution is in ‘molar’ unit:

Suppose, the molarity of an aqueous solution of a strong acid (or base) is M. If each acid (or base) molecule in the solution produces × H₃O⁺ (or OH^–) ions, then in the case of solution of an acid, [H₃O⁺] = x M and in the case of solution of a base [OH^–] = x x M.

Examples:

1. 1 molecule of strong monobasic acid on its ionization (HCl, HBr, HClO₄, HNO₃, etc.) gives an H₃O⁺ ion. Hence, in such a solution, the molar concentration of H₃O⁺ ions = the molar concentration of the acid solution.

For example, the molar concentration of H₃O⁺ ions in 0.1(M) HCl or HNO₃ solution = 0.1(M).

2. Each molecule of a strong dibasic acid (for example H₂SO₄) on its ionization produces two H₃O⁺ ions.

Therefore, in such a solution, [HsO⁺] =2x molar concentration of the solution. For example, in 0.1(M) H₂SO₄ solution, [H₃O⁺] =2 × 0.1

= 0.2(M).

3.Similarly, in 0.1(M) NaOH solution, [OH^–] =0.1(M) and in O.l(M) Ca(OH)₂ solution, [OH^–] =2 × 0.1

= 0.2(M).

When the concentration of the solution is in ‘normal unit:

If the concentration of a solution of strong acid (or base) is given in the ‘normal’ unit, then the concentration of H₃O⁺ (or OH^– ) ion in that solution will be equal to the normal concentration of the solution.

Example:

1. If The concentration of H₃O⁺ ionizing.l(N) H₂SO₄ solution= 0.1(N). Since the H₃O+ ion is monovalent, the molar concentration of the H₂O+ ion in 0.1(N) H₂SO₄ solution is 0.1(M).

The concentration of OH^– ion in 0.1(N) Ca(OH)₂ solution= 0.1(N). Since OH^– is monovalent, the molar concentration of OH^– ion in 0.1(N) Ca(OH)₂ solution= 0.1(M) Normality of a solution =nx Molarity; where n= basicity in case of an acid; acidity in case of a base; total valency location (or anion) per formula unit in case of a salt.

Examples:

1. 0.1  (M) HCl = 0.1(N) HCl solution.

∵ Basicity of HCl =1

2. 0.1(M) H₂SO₄ = 2 × 0.1 = 0.2 (N)H₂SO₄ solution.

∵  Basicity Of H₂SO₄ =2

3. 0.1(M) H₂SO₄ = 2 × 0.1 = 0.2 (N)H₂SO₄ solution.

∵  Acidity Of CO(OH₂)=2

4. 0.1(M) Ca₂+ = 2 × 0.1= 0.2 (N) Ca₂⁺

∵ Valency Of Ca²⁺ in its salt =2

5. 0.1(M)Al₂(SO₄)₃ solution = 6 × 0.1 = 0.6(N) Al₂(SO₄)₃ solution

∵ In each molecule of Al₂(SO₄)₃,

Total valency of cation (Al³⁺) or anion (SO₄^2-) = Number of ions of Al³⁺or SO₄^2- x Valency of Al³⁺ or SO₄^2- =6]

Degree Of Ionisation Numerical Examples

Question 1. At 25°C temperature, the molar concentrations of NH₃, NH+4 and OH- at equlibrium are 9.6 × 10^-3(M), 4.0 × 10^-4(M) and 4.0 × 10^-4(M) respectively. Determine the ionization constant of NH₃ at that temperature.
Answer:

In the aqueous solution of NH₃, the following equilibrium is established

⇒  NH₃(aq) + H₂O(l) ⇌  NH₄⁺(aq) + OH^– (aq)

∴ Ionisation Constant Of NH₃, Kb = \(=\frac{\left[\mathrm{NH}_4^{+}\right] \times\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_3\right]}\)

As [NH₃] = 9.6 × 10^-3+(m),

[NH₄⁺] = 4.0 × 10^-4(M)

[OH^–] = 4.0 × 10^-4(M)

⇒ \(K_b=\frac{\left(4 \times 10^{-4}\right) \times\left(4 \times 10^{-4}\right)}{9.6 \times 10^{-3}}=1.67 \times 10^{-5}\)

Question 2. A 0.1(M) solution of acetic acid is 1.34% ionized at 25°C Calculate the ionization constant of the acid.
Answer:

We know, the ionization constant of a weak monobasic acid example CH₃COOH is

⇒ \(K_a=\frac{\alpha^2 c}{1-\alpha}\) where- a = degree of ionization and c = initial concentration of the acid solution.

As \(\alpha=\frac{1.34}{100}=1.34 \times 10^{-2} \text { and } c=0.1(\mathrm{M})\)

⇒ \(K_a=\frac{\alpha^2 c}{1-\alpha}=\frac{\left(1.34 \times 10^{-2}\right)^2 \times 0.1}{\left(1-1.34 \times 10^{-2}\right)}=1.82 \times 10^{-5}\)

∴ At 25°C, ionisation constant of CH₃COOH = 1.82 × 10^-5

Question 3. In a 0.01(M) acetic acid solution, the degree of ionization of acetic acid is 4.2%. Determine the concentration of HgO+ ions in that solution
Answer:

Acetic acid is a weak monobasic acid. In such a solution, [H30+] = ac; where, c and a are the initial concentration of the acid and its degree of ionization respectively.

⇒ \(\text { As } \alpha=\frac{4.2}{100}=4.2 \times 10^{-2} \text { and } c=0.01(\mathrm{M}) \text {, }\) in 0.01(M) acetic acid solution.

[H₃O⁺] =ac = 4.2 × 10^-2 ×  0.01

= 4.2 × 10^-4 (M)

Question 4. The value of the ionization constant of pyridine (C₆H₆N) at 25°C is 1.6 × 10^-9. what is the concentration of OH^– ions in a 0.1(M) aqueous solution of pyridine at that temperature
Answer:

Pyridine is a monoacidic base. In aqueous solutions of such bases

⇒ \(\left[\mathrm{OH}^{-}\right]=\sqrt{c \times K_b}\)

Where c = initial concentration of the base and K_b = ionization constant of the weak base.

As c = 0.1(M) and Kb = 1.6 × 10^-9 in 0.1(M)

Aqueous pyridine solution,

⇒ \(\left[\mathrm{OH}^{-}\right]=\sqrt{0.1 \times 1.6 \times 10^{-9}}=1.26 \times 10^{-5}(\mathrm{M}) .\)

Question 5. At 25°C, the value of the ionization constant of a weak monobasic acid, HA is 1.6 × 10^-4 What is the degree of ionization of HA in its 0.1(M) aqueous solution?
Answer:

Degree of ionization of weak monobasic acid (a) \(=\sqrt{\frac{K_a}{c}}.\)

Given, c = 0.1(M) and Ka = 1.6 × 10-4

The degree of ionization of HA in its 0.1(M) aqueous solution

= \(\sqrt{\frac{K_a}{c}}=\sqrt{\frac{1.6 \times 10^{-4}}{0.1}}=0.04\)

∴ Degree of ionisation of HA in its 0.1(M) aqueous solution = 0.04 x 100%= 4%

Question 6. The ionization constant of ammonia is 1.8 × 10^-5at 25°C. Calculate the degree of ionization of ammonia in its 0.1(M) aqueous solution at that temperature.
Answer:

NH₃ is a weak monoacidic base. In aqueous solutions degree of ionization (a) of such base

⇒ \(\sqrt{\frac{\kappa_b}{c}}.\)

As c = 0.1(M) and k_b = 1.8 × 10^-5 tire degree of ionisation of NH₆ in its 0.1(M) aqueous solution

⇒ \(\sqrt{\frac{1.8 \times 10^{-5}}{0.1}}=0.0134=0.0134 \times 100 \%=1.34 \%\)

Ionic Product Of Water

Pure water is a very poor conductor of electricity, indicating its very low ionization. Due to the die self-ionization of pure water, H⁺ and OH^– ions are formed and the following dynamic equilibrium involving H⁺, OH^–ions, and unionized water molecules is established:

H₂O(l) + H₂O ⇌  H₃O⁺(aq) + OH^–(aq)

Applying the mass action to this equilibrium, we get

⇒ \(K_d=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \times\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{H}_2 \mathrm{O}^2\right.}\)

where [H₃O⁺], [OH^– ], and [H₂O] are the molar concentrations of H₃O+ OH (aq) and H₂O(l) at equilibrium, respectively, and Kd is the ionization or dissociation constant of water.

As the degree of ionization of water Is very small, its equilibrium concentration is almost the same as Its concentration before Ionisation. Thus, at equilibrium, [H₂O]2 = constant.

From equation(1)  we have, kd[H₂O)² = [H₃O⁺] × [OH^–]

Since, [H₂O]² = constant, Kd × [H₂O]² = constant. This constant is called the Ionic product of water & is denoted by K_w.

Therefore K_w=[H₃O⁺] × [OH^–]

Concentrations of H₃O⁺ and OH^– in aqueous solution

Any aqueous solution, whether it is acidic or basic, always contains both H₃O⁺ and OH ions. A concentrated acid solution also contains OH^– ions although its concentration is much lower compared to H₃O⁺ ions. Likewise, a concentrated alkali solution also contains H₃O⁺ ions but with a much lower concentration than OH^– ions.

On the other hand, in a neutral solution, the concentrations of H₃O+ and OH ions are always the same. At a particular temperature, if the ionic product of water Kw, then for

⇒ \(\text { a neutral solution: }\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]=\sqrt{K_w}\)

⇒ \(\text { an acidic solution: }\left[\mathrm{H}_3 \mathrm{O}^{+}\right]>\sqrt{K_w} \text { and }\left[\mathrm{OH}^{-}\right]<\sqrt{K_w}\)

⇒ \(\text { a basic solution: }\left[\mathrm{H}_3 \mathrm{O}^{+}\right]<\sqrt{K_w} \text { and }\left[\mathrm{OH}^{-}\right]>\sqrt{K_w}\)

At 25°C, Kw = 10’14. So, at this temperature, in the case of

an acidic solution: [H₃O⁺] > 10^-7 mol.L^-1 and [OH^-1] < 10^-7 mol.L^-1

a basic solution: [OH^–] > 10^-7 mol.L^-1 and [H₃O⁺] < 10^-7 mol.L^-1

Determination of [H₂O₄] and [OH-] in aqueous solution:

At a particular temperature, if the value of and any one of [H₃O+] or [OH-] are known, then the other can be determined using either of the following equations:

⇒ \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\frac{K_w}{\left[\mathrm{OH}^{-}\right]} \text {or, }\left[\mathrm{OH}^{-}\right]=\frac{K_w}{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}\)

For example, in an aqueous solution at 25°C if [H₃O⁺] = 10^-4(M),

⇒ \(\text { then }\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{10^{-4}}=10^{-10}(\mathrm{M})\)

[since K_w(25°C)=10^-14]

PK_w = pK_w = -log10K_w At 25°C,K_w=10^-14

Therefore, at 25°C, pK_w = -log₁₀ ( 10^-14) = 14.