Ionization of Acids and Bases: The substances are classified usually on the basis of how they behave in their aqueous solutions. They either dissociate completely into the solution as their individual ions, or partially dissociate, or they do not dissociate at all. Ionization is a process when a neutral molecule, such as \({\rm{HCl,}}\) which is a polar, covalent compound, dissolves and splits into two ions in their aqueous solution. The difference between dissociation and ionization is that any ionic compound dissociates in an aqueous solution, and is termed as ‘dissociation’. Ionization, on the other hand, is splitting of any neutral molecule like \({\rm{HCl}}\) too. So, when such ionization happens, the law of equilibrium is applied to it to find the ionization constant.

An example is as shown below:

For the ionization of a weak electrolyte, \({\rm{AB:}}\)
\({\text{AB}}\left( {\text{s}} \right) + {\text{aq}} \rightleftharpoons {{\text{A}}^ + }\left( {{\text{aq}}} \right) + {{\text{B}}^ – }\left( {{\text{aq}}} \right)\)

The ionic equilibrium for the reaction is an equilibrium established between the ions and the undissociated electrolyte. Hence, when the law of equilibrium is applied to the above reaction:

\(\frac{{\left[ {{{\text{A}}^ + }} \right]\left[ {{{\text{B}}^ – }} \right]}}{{\left[ {{\text{AB}}} \right]}} = {{\text{K}}_{\text{i}}}\)
The \({{\text{K}}_{\text{i}}}\) is called the ionization constant and can be applied to the ionization equilibrium of all weak electrolytes. In case of strong electrolytes, the splitting of ions is complete.

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Ionization of Acids and Bases: Overview

The Arrhenius concept of acids and bases can be applied for the ionization of acids and bases since it takes into account their ionizations in the aqueous medium.

Strong acids like \({\text{HCl,}}\) sulphuric acid \(\left( {{{\text{H}}_2}{\text{S}}{{\text{O}}_4}} \right),\) perchloric acid \(\left( {{\text{HCl}}{{\text{O}}_4}} \right),\) nitric acid \(\left( {{\text{HN}}{{\text{O}}_3}} \right),\) Hydroiodic acid \(\left( {{\text{HI}}} \right),\) etc., are strong acids because they completely dissociate into their ions in the aqueous medium. Due to this fact, they act as ‘proton donors.’ Also, the strong bases such as \({\text{NaOH}},{\text{LiOH}},\) etc., dissociate completely to give hydroxyl ions in the aqueous medium. So, as Arrhenius concept suggests, these are acids and bases which give hydronium ions and hydroxyl ions in their aqueous solution.

The dissociation can be a method of identifying the strength or on the other hand, the Bronsted-Lowry concept can be used to ascertain a strong acid (a proton donor) or a strong base (proton acceptor).

The dissociation equilibrium for acid-base dissociation of weak acid \({\text{HA}}\) is given by:
\(\mathop {{\text{HA}}}\limits_{{\text{Acid}}} \left( {{\text{aq}}} \right) + \mathop {{{\text{H}}_2}{\text{O}}\left( {\text{l}} \right)}\limits_{{\text{Base}}} \rightleftharpoons \mathop {{{\text{H}}_3}{{\text{O}}^ + }}\limits_{{\text{Conjugate}}\,{\text{Acid}}} \left( {{\text{aq}}} \right) + \mathop {{{\text{A}}^ – }\left( {{\text{aq}}} \right)}\limits_{{\text{Conjugate}}\,{\text{base}}} \)

The acid-base equilibrium is dynamic, wherein there is a transfer of protons on both sides, as seen in the above equation. The questions like which direction the equation will go with the passage of time or how it is decided can be worked out by comparing the strengths of the two acids in the dissociation equilibrium.

In the above equation, consider the two acids \({\text{HA}}\) and \({{\text{H}}_3}{{\text{O}}^ + }.\) Among these two, to consider one as a stronger acid than the other, it should exceed the tendency to donate proton than the other. The equilibrium will then shift in the direction of the weaker acid. So, for instance, here, if \({\text{HA}}\) is the stronger acid than hydronium \(\left( {{{\text{H}}_3}{{\text{O}}^ + }} \right)\) ion, then \({\text{HA}}\) will donate proton to hydronium ion and not the other way round. The equilibrium in this case will move towards the formation of weak acid and weak base since the strong acid can donate proton to the strong base.
So, when strong acid dissociates in water, the resulting base that is formed will be a weaker one, making it such that the stronger acids will have very weak conjugate bases. For example, the stronger acids, \({\text{HCl}},{\text{HI}},{\text{HCl}}{{\text{O}}_4},\) etc., will have weak conjugate bases such as \({\text{C}}{{\text{l}}^ – },{\text{I}},{\text{ClO}}_4^ – ,\) etc.
On the other hand, a very strong base will have a weak conjugate acid. Also, since the weak acid, \({\text{HA}}\) will only be partially dissociated in aqueous medium, and hence, the solution largely contains undissociated \({\text{HA}}\) molecules.
Weak acids, such as hydrofluoric acid \(\left( {{\text{HF}}} \right),\) nitrous acid \(\left( {{\text{HN}}{{\text{O}}_2}} \right)\) and acetic acid \(\left( {{\text{C}}{{\text{H}}_3}{\text{COOH}}} \right)\) have very strong conjugate bases, such as the \({\text{NH}}_2^ – ,{{\text{O}}^{2 – }}\) and \({{\text{H}}^ – }.\) These are very good proton acceptors, making them much stronger bases in a reaction than water.

Ionization Constant of Water and Ionic Product

Water, as we are aware, can act as both an acid and a base, depending upon the reactants in a reaction. For example, in the presence of \({\text{HA,}}\) it accepts a proton, while in the presence of a base, \({{\text{B}}^ – },\) it donates a proton to act as an acid.

Hence, in pure water, while one water molecule acts as an acid by donating proton, other acts as a base by accepting the proton donated. Thus, there is an equilibrium that exist, such as the one below:

\(\mathop {{{\text{H}}_2}{\text{O}}\left( {\text{l}} \right)}\limits_{{\text{Acid}}} + \mathop {{{\text{H}}_2}{\text{O}}\left( {\text{l}} \right)}\limits_{{\text{Base}}} \rightleftharpoons \mathop {{{\text{H}}_3}{{\text{O}}^ + }\left( {{\text{aq}}} \right)}\limits_{{\text{Conjugate}}\,{\text{Acid}}} + \mathop {{\text{O}}{{\text{H}}^ – }\left( {{\text{aq}}} \right)}\limits_{{\text{Conjugate}}\,{\text{base}}} \)

The dissociation constant for the reaction is represented by:

\(\frac{{\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right]\left[ {{\text{O}}{{\text{H}}^ – }} \right]}}{{\left[ {{{\text{H}}_2}{\text{O}}} \right]}} = {\text{K}}\)

Since the water is a pure liquid, the concentration of water is omitted in the dissociation constant, thereby making the ‘ionic product of water’ as:

\({{\text{K}}_{\text{w}}} = \left[ {{{\text{H}}^ + }} \right]\left[ {{\text{O}}{{\text{H}}^ – }} \right]\)

The ionic product of water is experimentally calculated, and the value was found to be:

\({{\text{K}}_{\text{w}}} = 1 \times {10^{ – 14}}{{\text{M}}^2}\)

Ionization Constants of Weak Acids

Weak acids do not ionize completely in the aqueous solution. Hence, the equilibrium can be expressed as:

\({\text{HX}}\left( {{\text{aq}}} \right) + {{\text{H}}_2}{\text{O}}\left( {\text{l}} \right) \rightleftharpoons {{\text{H}}_3}{{\text{O}}^ + }\left( {{\text{aq}}} \right) + {{\text{X}}^ – }\left( {{\text{aq}}} \right)\)

The concentration – initial and change can be expressed as:

Species	\({\text{HX}}\)	\({{\text{H}}_3}{{\text{O}}^ + }\)	\({{\text{X}}^ – }\)
Initial Concentration	\({\text{c}}\)	\({\text{0}}\)	\(0\)
lonization extent \(\left( \alpha \right)\)	\({\text{ – c}}\alpha \)	\( + {\text{c}}\alpha \)	\( + {\text{c}}\alpha \)
Equilibrium Concentration (M)	\({\text{c – c}}\alpha \)	\({\text{c}}\alpha \)	\({\text{c}}\alpha \)

The equilibrium constant for the acid dissociation can be derived from the above data as:

\({{\text{K}}_{\text{a}}} = \left[ {{{\text{H}}^ + }} \right]\left[ {{{\text{X}}^ – }} \right]/\left[ {{\text{HX}}} \right]\)
The \({{\text{K}}_{\text{a}}}\) is the measure of strength of an acid. The larger the \({{\text{K}}_{\text{a}}}\) value, the stronger the acid. The standard state concentration of all species is \({\text{1M,}}\) and \({{\text{K}}_{\text{a}}}\) is a dimensionless quantity.

The ionization constants of some of the weak acids are as given below:

Acid	Ionization Constant, \({{\text{K}}_{\text{a}}}\)
Hydrocyanic acid, \({\text{HCN}}\)	\(4.9 \times {10^{ – 10}}\)
Hydrofluoric acid, \({\text{HF}}\)	\(3.5 \times {10^{ – 4}}\)
Nitrous acid, \({\text{HN}}{{\text{O}}_2}\)	\(4.5 \times {10^{ – 4}}\)
Acetic Acid, \({\text{C}}{{\text{H}}_3}{\text{COOH}}\)	\(1.74 \times {10^{ – 5}}\)
Phenol, \({{\text{C}}_6}{{\text{H}}_5}{\text{OH}}\)	\(1.3 \times {10^{ – 10}}\)

The \({\text{pH}}\) scale can be used for calculating the \({{\text{K}}_{\text{a}}}\) value or the \({\text{p}}{{\text{K}}_{\text{a}}}\) value, the equilibrium concentration of any species in the reaction, and the degree of ionization of the acid as well as the \({\text{pH}}\) of the solution.

The equation is given by:

\({\text{p}}{{\text{K}}_{\text{a}}} = – \log \left( {{{\text{K}}_{\text{a}}}} \right)\)
Where, \({{\text{K}}_{\text{a}}} = \) ionization constant of the acid

Ionization Constants of Weak Bases

A weak base, \({\text{MOH,}}\) partially ionizes into its ions. The equation can be represented as:
\({\text{MOH}}\left( {{\text{aq}}} \right) \rightleftarrows {{\text{M}}^ + }\left( {{\text{aq}}} \right) + {\text{O}}{{\text{H}}^ – }\left( {{\text{aq}}} \right)\)
The equilibrium constant for base ionization is called as the ‘base ionization constant’ and is represented by \({{\text{K}}_{\text{b}}}.\) This can be calculated using the concentration in molarity of the species present, as follows:
\ \({{\rm{K}}_{\rm{b}}}{\rm{ = }}\frac{{\left[ {{{\rm{M}}^{\rm{ + }}}} \right]\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]}}{{\left[ {{\rm{MOH}}} \right]}}\)
Here, if \({\text{C}}\) is the initial concentration, and \(\alpha = \) degree of ionization of the base MOH, then, the equilibrium constant can be written as:
\({{\text{K}}_{\text{b}}} = \frac{{{\text{C}}{\alpha ^2}}}{{\left( {1 – \alpha } \right)}}\)

The values of some of the weak bases are as follows:

Weak Base	Equilibrium Constant, \({{\text{K}}_{\text{b}}}\)
Ammonia, \({\text{N}}{{\text{H}}_3}\)	\(1.77 \times {10^{ – 5}}\)
Urea, \({\text{N}}{{\text{H}}_2}{\text{CON}}{{\text{H}}_2}\)	\(1.3 \times {10^{ – 14}}\)
Aniline, \({{\text{C}}_6}{{\text{H}}_5}{\text{N}}{{\text{H}}_2}\)	\(4.27 \times {10^{ – 10}}\)
Pyridine, \({{\text{C}}_5}{{\text{H}}_5}\;{\text{N}}\)	\(1.77 \times {10^{ – 9}}\)
Dimethlyamine, \({\left( {{\text{C}}{{\text{H}}_3}} \right)_2}{\text{NH}}\)	\(5.4 \times {10^{ – 4}}\)

Amines, as derivative of ammonia, are weaker bases, wherein one or more hydrogen atoms are replaced with some other group. Examples include quinine, codeine, methylamine, etc. They all behave as weak bases because of their very small \({{\text{K}}_{\text{b}}}.\)
The \({\text{pH}}\) scale can also be extended here, to obtain the equation:
\({\text{p}}{{\text{K}}_{\text{b}}} = – \log \left( {{{\text{K}}_{\text{b}}}} \right)\)
where,\({{\text{K}}_{\text{b}}} = \) ionization constant of the base
The \({\text{p}}{{\text{K}}_{\text{a}}}\) and \({\text{p}}{{\text{K}}_{\text{b}}}\) can be calculated using the \({\text{p}}{{\text{K}}_{\text{w}}}\) value, as follows:
\({\text{p}}{{\text{K}}_{\text{a}}} + {\text{p}}{{\text{K}}_{\text{b}}} = {\text{p}}{{\text{K}}_{\text{w}}} = 14\,\left( {{\text{at}}\,298\;{\text{K}}} \right)\)
Also, \({\text{p}}{{\text{K}}_{\text{W}}} = {\text{pH}} + {\text{pOH}} = 14\)

Summary

Ionization is a process wherein any substance, including a neutral molecule such as \({\text{HCl}}\) splits into their individual ions in an aqueous solution. While strong acids and strong bases splits completely into their ions or dissociates, the weak acids and weak bases only partially ionize into their individual ions. The acid and base equilibrium in an acid-base reaction is dynamic, and the reaction can proceed both in forward or backward directions, depending upon the strength of the acid. The equilibrium will shift towards the weaker acid. Also, a weaker acid will have a stronger conjugate base on the other side of the reaction, and vice versa. This can help in determining the equilibrium constant for such reactions by taking into account fact that it is equal to the product of concentration of individual ions in the solution. For weaker acid and weaker base, however, the equilibrium constant also includes the undissociated acid and base in the solution since they only partially dissociate into their ions. The ionic product of water and its \({\text{pH}}\) scale can be used to determine the \({\text{pH}}\) of a solution of weak base or weak acid.

FAQs

Q1. What are the weak acids and bases?
Ans. The acids (\({\text{HN}}{{\text{O}}_2},{\text{HF}},\) etc.) and bases (Ammonia, methylamine, etc.) do not dissociate completely into their ions in their aqueous solution. They are called weak acids and weak bases. It is also confirmed using the value of their equilibrium constant.

Q2. What are examples of weak bases?
Ans. Examples of weak bases include Ammonia, Urea, Aniline, pyridine, and all ammonia derivatives such as dimethylamine, quinine, codeine, methylamine, etc.

Q3. Is \({\text{NaOH}}\) a weak base?
Ans. The strength of a base is determined by its equilibrium constant value and the way it dissociates into its ions in its aqueous solution. \({\text{NaOH}}\) is a strong base because it dissociates completely into its ions, \({\text{N}}{{\text{a}}^ + }\) and \({\text{O}}{{\text{H}}^ – }.\)

Q4. What is acid ionization?
Ans. Acid ionization is the dissociation of an acid into its ions in its aqueous solution. For example:
\({\text{HX}}\left( {{\text{aq}}} \right) + {{\text{H}}_2}{\text{O}}\left( {\text{l}} \right) \rightleftarrows {{\text{H}}_3}{{\text{O}}^ + }\left( {{\text{aq}}} \right) + {{\text{X}}^ – }\left( {{\text{aq}}} \right)\)

Q5. What is base ionization?
Ans. Base ionization is the dissociation of a base into its ions in its aqueous solution. For example:
\({\text{MOH}}\left( {{\text{aq}}} \right) \rightleftarrows {{\text{M}}^ + }\left( {{\text{aq}}} \right) + {\text{O}}{{\text{H}}^ – }\left( {{\text{aq}}} \right)\)

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