Written By Umesh_K
Last Modified 25-01-2023

Hydrolysis of Salts and pH of Their Solutions: Meaning, Equations

When the salt of a weak acid, \({\rm{HA}}\), and a strong base dissolve in water, the anion \({{\rm{A}}^ – }\) is produced. The \({{\rm{A}}^ – }\) anion reacts with water to form the unionized molecule by removing a proton \(\left( {{{\rm{H}}^ + }} \right)\) from its molecule.

Similarly, the cation \({{\rm{B}}^ + }\) is formed when the salt of a weak base, \({\rm{BOH}}\), and a strong acid dissolve in water. The cation \({{\rm{B}}^ + }\) interacts with water and, in the process, accepts \({\rm{O}}{{\rm{H}}^ – }\) ion. In this article, let’s learn everything about the Hydrolysis of Salts and the pH of their solutions in detail.

Hydrolysis of Salts and pH of Their Solutions

Hydrolysis of salt is the reaction of an anion or cation with water that results in the cleavage of the \({\rm{ O – H }}\) bond.

The phrase hydrolysis is derived from the words hydro, which means water, and lysis, which means breaking. Due to the formation of extra \({\rm{HO}}\)^– ions, the solution becomes slightly basic \(({\rm{pH}} > 7)\) in anionic hydrolysis indicated in \((1)\). The excess of \({{\rm{H}}^ + }\) ions in cationic hydrolysis indicated in \((2)\) renders the solution somewhat acidic \(({\rm{pH}} < 7)\).

Examples of Salt Hydrolysis

According to their hydrolytic behaviour, various salts can be categorized into the following types:

Salts of Weak acids and Strong bases
Salts of Weak bases and Strong acids
Salts of Weak acids and Weak bases
Salts of Strong acids and Strong bases

Salts of Weak Acids and Strong Bases

\({\rm{C}}{{\rm{H}}_3}{\rm{COONa}}\), or sodium acetate, is a salt of the weak acid \({\rm{C}}{{\rm{H}}_3}{\rm{COOH}}\) and the strong base \({\rm{NaOH}}\). In an aqueous solution, it ionizes to create the anion \({\rm{C}}{{\rm{H}}_3}{\rm{CO}}{{\rm{O}}^ – }\). It is a reasonably strong base because it is the conjugate base of a weak acid, \({\rm{C}}{{\rm{H}}_3}{\rm{COOH}}.{\rm{C}}{{\rm{H}}_3}{\rm{CO}}{{\rm{O}}^ – }\) takes the \({{\rm{H}}^ + }\) ion from the water and undergoes hydrolysis.

\({\rm{C}}{{\rm{H}}_3}{\rm{CO}}{{\rm{O}}^ – } + {{\rm{H}}_2}{\rm{O}} \to {\rm{C}}{{\rm{H}}_3}{\rm{COOH}} + {\rm{O}}{{\rm{H}}^ – }\)

Due to the presence of extra \({\rm{OH}} – \) ions, the resultant solution is basic.

Salts of Weak Bases and Strong Acids

\({\rm{N}}{{\rm{H}}_4}{\rm{Cl}}\) or ammonium chloride is a salt of the weak base \({\rm{N}}{{\rm{H}}_4}{\rm{OH}}\) and the strong acid \({\rm{HCl}}\). In an aqueous solution, it ionizes to create the cation, \({\rm{NH}}_4^ + \). It is a reasonably strong acid because it is the conjugate acid of a weak base, \({\rm{N}}{{\rm{H}}_4}{\rm{OH}}.{\rm{NH}}_4^ + \) takes the \({\rm{O}}{{\rm{H}}^ – }\) ion from the water and undergoes hydrolysis.

\({\rm{NH}}_4^ + + {{\rm{H}}_2}{\rm{O}}⇌{\rm{N}}{{\rm{H}}_4}{\rm{OH}} + {{\rm{H}}^ + }\)

Due to the presence of extra \({{\rm{H}}^ + }\) ions, the resultant solution is acidic.

Salts of Weak Acids and Weak Bases

The salt of weak acid \({\rm{C}}{{\rm{H}}_3}{\rm{COOH}}\) and weak base \({\rm{N}}{{\rm{H}}_4}{\rm{OH}}\) is ammonium acetate, \({\rm{C}}{{\rm{H}}_3}{\rm{COON}}{{\rm{H}}_4}\). It forms the anion \({\rm{C}}{{\rm{H}}_3}{\rm{CO}}{{\rm{O}}^ – }\) and the cation \({\rm{N}}{{\rm{H}}_4}^ + \) in an aqueous solution. Because both the acid and the base are weak, the conjugate base \(\left( {{\rm{C}}{{\rm{H}}_3}{\rm{CO}}{{\rm{O}}^ – }} \right)\) and conjugate acid \(\left( {{\rm{NH}}_4^ + } \right)\) are relatively strong. They take in \({{\rm{H}}^ + }\) and \({\rm{OH}} – \) ions from water, respectively, and go through hydrolysis.

\(\mathop {{\rm{C}}{{\rm{H}}_3}{\rm{CO}}{{\rm{O}}^ – }}\limits_{{\rm{conjugate}}\,{\rm{base}}} + {{\rm{H}}_2}{\rm{O}}⇌{\rm{C}}{{\rm{H}}_3}{\rm{COOH}} + {\rm{O}}{{\rm{H}}^ – }\)

\(\mathop {{\rm{NH}}_4^ + }\limits_{{\rm{conjugate}}\,{\rm{acid}}} + {{\rm{H}}_2}{\rm{O}}⇌{\rm{N}}{{\rm{H}}_4}{\rm{OH}} + {{\rm{H}}^ + }\)

The overall hydrolysis reaction may be represented as

\({\rm{C}}{{\rm{H}}_3}{\rm{CO}}{{\rm{O}}^ – } + {\rm{NH}}_4^ + + {{\rm{H}}_2}{\rm{O}}⇌{\rm{C}}{{\rm{H}}_3}{\rm{COOH}} + {\rm{N}}{{\rm{H}}_4}{\rm{OH}}\)

As previously stated, the pH of the resultant solution is determined by the relative amount of anionic and cationic hydrolysis. \(\left[ {{\rm{O}}{{\rm{H}}^ – }} \right] = \left[ {{{\rm{H}}^ + }} \right]\), and the solution is neutral if both ions react to the same extent. The solution becomes slightly acidic when the cation reacts to a greater extent. The solution will be basic if the anion is a little more reactive. As a result, a \({\rm{C}}{{\rm{H}}_3}{\rm{COON}}{{\rm{H}}_4}\) solution is neutral, an \({\rm{N}}{{\rm{H}}_4}{\rm{CN}}\) solution is slightly basic, and an \({\rm{N}}{{\rm{H}}_4}\;{\rm{F}}\) solution is slightly acidic.

Salts of Strong Acids and Bases

\({\rm{KCl}}\) or potassium chloride is the salt of strong acid \({\rm{HCl}}\) and strong base \({\rm{KOH}}\). In this case, neither the cation nor the anion undergoes hydrolysis. Therefore, the resulting solution is neutral.

Quantitative Aspect of Hydrolysis

The reaction of hydrolysis is reversible. The Hydrolysis constant or Hydrolytic constant is the equilibrium constant obtained by applying the Law of Mass action to a hydrolysis (or hydrolytic) reaction. The hydrolysis constant is denoted by the letter \({{\rm{K}}_{\rm{h}}}\). We will now move on to discussing the mathematics of salt hydrolysis.

Salt of a Weak acid and Strong base Hydrolysis Constant

The general hydrolysis reaction of salt of a weak acid (HA) and strong base can be written as

\({{\rm{A}}^ – } + {{\rm{H}}_2}{\rm{O}}⇌{\rm{HA}} + {\rm{O}}{{\rm{H}}^ – }\)

The expression for hydrolysis constant becomes

\({\rm{Kh = }}\frac{{[{\rm{HA}}]\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]}}{{\left[ {{{\rm{A}}^{\rm{ – }}}} \right]\left[ {{{\rm{H}}_2}{\rm{O}}} \right]}}\)

The concentration of water is assumed to be constant. So, we have

\({{\rm{K}}_{\rm{h}}}{\rm{ = }}\frac{{{\rm{[HA]}}\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]}}{{\left[ {{{\rm{A}}^{\rm{ – }}}} \right]}}\,\,\,\,\,……..{\rm{(1)}}\)

Ionic product of water, \({{\rm{K}}_{\rm{w}}}\), is expressed as

\({{\rm{K}}_{\rm{w}}}{\rm{ = }}\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]\,\,\,\,\,\,\,…….(2)\)

The dissociation of a weak acid is expressed as

\({\rm{HA}}⇌{{\rm{H}}^ + } + {{\rm{A}}^ – }\)

The acid dissociation constant, \({{\rm{K}}_{\rm{a}}}\), is expressed as

\({{\rm{K}}_{\rm{a}}}{\rm{ = }}\frac{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\left[ {{{\rm{A}}^{\rm{ – }}}} \right]}}{{{\rm{[HA]}}}}\,\,\,\,\,…..(3)\)

Dividing \((2)\) by \((3)\)

\(\frac{{{{\rm{K}}{\rm{w}}}}}{{{{\rm{K}}{\rm{a}}}}}{\rm{ = }}\frac{{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]{\rm{[HA]}}}}{{\left[ {{{\rm{A}}^{\rm{ – }}}} \right]}}{\rm{ = }}{{\rm{K}}_{\rm{h}}}\)

\(\frac{{{{\rm{K}}{\rm{w}}}}}{{{{\rm{K}}{\rm{a}}}}}{\rm{ = }}{{\rm{K}}_{\rm{h}}}\)

It is clear that the hydrolysis constant of the salt is inversely proportional to the dissociation constant of the weak acid. Therefore weaker the acid, the greater is the hydrolysis constant of the salt.

Relation Between Hydrolysis constant and degree of hydrolysis

When equilibrium is reached, the degree of hydrolysis is the fraction of the salt that has been hydrolyzed. It is commonly denoted by the symbol \(\alpha \). Let’s say we have one mole of salt dissolved in V litres of the solution to begin with. The equilibrium concentrations are as follows:

\({{\rm{A}}^ – } + {{\rm{H}}_2}{\rm{O}}⇌{\rm{HA}} + {\rm{O}}{{\rm{H}}^ – }\)

\({\rm{Equilibrium \, concentrations}}\,\,\,\,\frac{{{\rm{1 – \alpha }}}}{{\rm{V}}}\quad \frac{{\rm{\alpha }}}{{\rm{V}}}\quad \frac{{\rm{\alpha }}}{{\rm{V}}}\)

We can write,

\({{\rm{K}}_{\rm{h}}}{\rm{ = }}\frac{{{\rm{[HA]}}\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]}}{{\left[ {{{\rm{A}}^{\rm{ – }}}} \right]}}{\rm{ = }}\frac{{{\rm{\alpha /V \times \alpha /V}}}}{{{\rm{1 – \alpha /V}}}}{\rm{ = }}\frac{{{{\rm{\alpha }}^{\rm{2}}}}}{{{\rm{(1 – \alpha )V}}}}\)

If \(\alpha \) is too small, we can write

\({{\rm{K}}_{\rm{h}}}{\rm{ = }}\frac{{{{\rm{\alpha }}^{\rm{2}}}}}{{\rm{V}}}\)

\({{\rm{\alpha }}^2} = {{\rm{K}}_{\rm{h}}}{\rm{V}} = \frac{{{{\rm{K}}_{\rm{w}}}}}{{{{\rm{K}}_{\rm{\alpha }}}}}{\rm{V}}\)

\(\alpha {\rm{ = }}\sqrt {\frac{{{{\rm{K}}_{\rm{w}}}{\rm{V}}}}{{{{\rm{K}}_{\rm{a}}}}}} \)

\({\rm{ = }}\sqrt {\frac{{{{\rm{K}}_{\rm{w}}}}}{{{{\rm{K}}_{\rm{a}}}{\rm{C}}}}} \)

Where \({\rm{C = }}\) Initial concentration of the salt

Derivation of pH

We know that,

\(\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]{\rm{ = }}\frac{{\rm{\alpha }}}{{\rm{V}}}{\rm{ = \alpha C}}\)

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

\(\left[ {{{\rm{H}}^{\rm{ + }}}} \right]{\rm{ = }}\frac{{{{\rm{K}}_{\rm{w}}}}}{{{\rm{\alpha C}}}}\)

\({\rm{\alpha }} = \sqrt {\frac{{{{\rm{K}}_{\rm{w}}}}}{{{{\rm{K}}_{\rm{a}}}{\rm{C}}}}} \)

\(\left[ {{{\rm{H}}^ + }} \right] = \frac{{{{\rm{K}}_{\rm{w}}}}}{{\rm{C}}}\sqrt {\frac{{{{\rm{K}}_{\rm{a}}}{\rm{C}}}}{{{{\rm{K}}_{\rm{w}}}}}} = \sqrt {\frac{{{{\rm{K}}_{\rm{w}}}{{\rm{K}}_{\rm{a}}}}}{{\rm{c}}}} \)

After taking logarithms, the above equation can be written as

\( – \log \left[ {{{\rm{H}}^ + }} \right] = \, – \frac{1}{2}\log \,{{\rm{K}}_{\rm{w}}} – \log {{\rm{K}}_{\rm{a}}} + \frac{1}{2}\log {\rm{C}}\)

\({\rm{pH}} = \frac{1}{2}{\rm{p}}{{\rm{K}}_{\rm{w}}} + \frac{1}{2}{\rm{p}}{{\rm{K}}_{\rm{a}}} + \frac{1}{2}\log {\rm{C}}\)

\( = 7 + \frac{1}{2}{\rm{p}}{{\rm{K}}_{\rm{a}}} + \frac{1}{2}\log {\rm{C}}\)

It is evident that the pH of the solution will always be greater than \(7\). Thus, an aqueous solution of the salt of a weak acid and strong base will always be alkaline.

Salts of Weak Bases and Strong Acids
Hydrolysis Constant

The general hydrolysis reaction of salt of a weak base \({\rm{(BOH)}}\) and strong acid can be written as

\({{\rm{B}}^ + } + {{\rm{H}}_2}{\rm{O}}⇌{\rm{BOH}} + {{\rm{H}}^ + }\)

The expression for hydrolysis constant becomes

\({\rm{Kh = }}\frac{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\left[ {{\rm{BOH}}} \right]}}{{\left[ {{{\rm{B}}^{\rm{ + }}}} \right]\left[ {{{\rm{H}}_2}{\rm{O}}} \right]}}\)

The conc of water is assumed to be constant. So, we have

\({{\rm{K}}_{\rm{h}}}{\rm{ = }}\frac{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]{\rm{[BOH]}}}}{{\left[ {{{\rm{B}}^{\rm{ + }}}} \right]}}\,\,\,\,\,\,…..(1)\)

Ionic product of water, \({{\rm{K}}_{\rm{W}}}\), is expressed as

\({{\rm{K}}_{\rm{w}}}{\rm{ = }}\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]\,\,\,\,\,\,….(2)\)

The dissociation of a weak base is expressed as

\({\rm{BOH}}⇌{{\rm{B}}^ + } + {\rm{O}}{{\rm{H}}^ – }\)

The base dissociation constant, \({{\rm{K}}_{\rm{b}}}\), is expressed as

\({{\rm{K}}_{\rm{b}}}{\rm{ = }}\frac{{\left[ {{{\rm{B}}^{\rm{ + }}}} \right]\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]}}{{{\rm{[BOH]}}}}\,\,\,\,…..(3)\)

Dividing \((2)\) by \((3)\)

\(\frac{{{{\rm{K}}_{\rm{w}}}}}{{{{\rm{K}}_{\rm{b}}}}} = \frac{{\left[ {{{\rm{H}}^ + }} \right]\left[ {{\rm{BOH}}} \right]}}{{\left[ {{{\rm{B}}^ + }} \right]}}{{\rm{K}}_{\rm{h}}}\)

\(\frac{{{{\rm{K}}_{\rm{w}}}}}{{{{\rm{K}}_{\rm{b}}}}} = {{\rm{K}}_{\rm{h}}}\)

It is clear that the hydrolysis constant of the salt is inversely proportional to the dissociation constant of the weak base. Therefore weaker the base, the greater is the hydrolysis constant of the salt.

Relation Between Hydrolysis Constant and Degree of Hydrolysis

When equilibrium is reached, the degree of hydrolysis is the fraction of the salt that has been hydrolyzed. It is commonly denoted by the symbol \({\rm{\alpha }}\). Let’s say we have one mole of salt dissolved in V litres of the solution to begin with. The equilibrium concentrations are as follows:

\({{\rm{B}}^ + } + {{\rm{H}}_2}{\rm{O}}⇌{\rm{BOH}} + {{\rm{H}}^ + }\)

\({\rm{Equilibrium\, concentrations}}\,\,\,\,\,\frac{{{\rm{1 – \alpha }}}}{{\rm{V}}}\quad \frac{{\rm{\alpha }}}{{\rm{V}}}\quad \frac{{\rm{\alpha }}}{{\rm{V}}}\)

We can write,

\({{\rm{K}}_{\rm{h}}}{\rm{ = }}\frac{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]{\rm{[BOH]}}}}{{\left[ {{{\rm{B}}^{\rm{ + }}}} \right]}}{\rm{ = }}\frac{{{\rm{\alpha /V \times \alpha /V}}}}{{{\rm{(1 – \alpha )/V}}}}{\rm{ = }}\frac{{{{\rm{\alpha }}^{\rm{2}}}}}{{{\rm{(1 – \alpha )V}}}}\)

If \({\rm{\alpha }}\) is too small, we can write,

\({{\rm{K}}_{\rm{h}}}{\rm{ = }}\frac{{{{\rm{\alpha }}^{\rm{2}}}}}{{\rm{V}}}\)

\({{\rm{K}}_{\rm{h}}}{\rm{ \times V = }}{{\rm{\alpha }}^{\rm{2}}}\)

\({\rm{\alpha = }}\sqrt {{{\rm{K}}_{\rm{h}}}{\rm{ \times V}}} \)

\({{\rm{K}}_{\rm{h}}} = \frac{{{{\rm{K}}_{\rm{w}}}}}{{{{\rm{K}}_{\rm{b}}}}}\)

\({\rm{\alpha }} = \sqrt {\frac{{{{\rm{K}}_{\rm{w}}}}}{{{{\rm{K}}_{\rm{b}}}}} \times {\rm{V}}} \)

\( = \sqrt {\frac{{{{\rm{K}}_{\rm{w}}}}}{{{{\rm{K}}_{\rm{b}}} \times {\rm{C}}}}} \)

Where \({\rm{C = }}\) Initial concentration of the salt

Derivation of pH

We know that,

\(\left[ {{{\rm{H}}^{\rm{ + }}}} \right]{\rm{ = }}\frac{{\rm{\alpha }}}{{\rm{V}}}{\rm{ = \alpha \times C}}\)

Or we can write

\(\left[ {{{\rm{H}}^ + }} \right] = \frac{1}{{\rm{V}}}\sqrt {\frac{{{{\rm{K}}_{\rm{w}}} \times {\rm{V}}}}{{{{\rm{K}}_{\rm{b}}}}}} = \sqrt {\frac{{{{\rm{K}}_{\rm{w}}}}}{{{{\rm{K}}_{\rm{b}}}{\rm{V}}}}} = \sqrt {\frac{{{{\rm{K}}_{\rm{w}}} \times {\rm{C}}}}{{{{\rm{K}}_{\rm{b}}}}}} \)

After taking logarithms, the above equation can be written as

\({\rm{ – log}}\left[ {{{\rm{H}}^{\rm{ + }}}} \right]{\rm{ =\, – }}\frac{{\rm{1}}}{{\rm{2}}}{{{\rm{K}}_{\rm{w}}}}{\rm{ – }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{logC + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{p}}{{{\rm{K}}_{\rm{b}}}}\)

\({\rm{pH = 7 – }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{p}}{{{\rm{K}}_{\rm{b}}}}{\rm{ – }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{logC}}\)

In this case, it is evident that pH will always be less than \(7\). Thus, the solution of a salt of weak base and strong acid will always be acidic.

Salts of Weak Acids and Weak Bases
Hydrolysis Constant

Both the weak acid anion \(\left( {{{\rm{X}}^ – }} \right)\) and the weak base cation \(\left( {{{\rm{B}}^{\rm{ + }}}} \right)\) are hydrolyzed simultaneously in this type of salt.

\({{\rm{B}}^ + } + {{\rm{X}}^ – } + {{\rm{H}}_2}{\rm{O}}⇌{\rm{BOH}} + {\rm{HX}}\)

The expression for hydrolysis constant becomes

\({{\rm{K}}_{\rm{h}}}{\rm{ = }}\frac{{\left[ {{\rm{BOH}}} \right]\left[ {{\rm{HX}}} \right]}}{{\left[ {{{\rm{B}}^{\rm{ + }}}} \right]\left[ {{{\rm{X}}^{\rm{ – }}}} \right]\left[ {{{\rm{H}}_2}{\rm{O}}} \right]}}\)

The conc of water is assumed to be constant. So, we have

\({{\rm{K}}_{\rm{h}}}{\rm{ = }}\frac{{{\rm{[BOH][HX]}}}}{{\left[ {{{\rm{B}}^{\rm{ + }}}} \right]\left[ {{{\rm{X}}^{\rm{ – }}}} \right]}}\,\,\,\,……(1)\)

We can write for ionization of weak acid, weak base, and water,

\({\rm{HX}}⇌{{\rm{H}}^{\rm{ + }}}{\rm{ + }}{{\rm{X}}^{\rm{ – }}}\quad {{\rm{K}}_{\rm{a}}}{\rm{ = }}\frac{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\left[ {{{\rm{X}}^{\rm{ – }}}} \right]}}{{{\rm{[HX]}}}}\,\,\,\,\,….(2)\)

\({\rm{BOH}}⇌{{\rm{B}}^{\rm{ + }}}{\rm{ + O}}{{\rm{H}}^{\rm{ – }}}\quad {{\rm{K}}_{\rm{b}}}{\rm{ = }}\frac{{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]\left[ {{{\rm{B}}^{\rm{ + }}}} \right]}}{{{\rm{[BOH]}}}}\,\,\,\,\,\,…..(3)\)

\({{\rm{K}}_{\rm{w}}}{\rm{ = }}\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]\,\,\,\,\,\,\,…..(4)\)

Dividing \((4)\) by \((3)\) and \((2)\), we have

\(\frac{{{{\rm{K}}_{\rm{w}}}}}{{{{\rm{K}}_{\rm{a}}} \times {{\rm{K}}_{\rm{b}}}}}{\rm{ = }}\frac{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]{\rm{[HX][BOH]}}}}{{\left[ {{{\rm{X}}^{\rm{ – }}}} \right]\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\left[ {{{\rm{B}}^{\rm{ + }}}} \right]\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]}}\)

\( = \frac{{[{\rm{HX}}][{\rm{BOH}}]}}{{\left[ {{{\rm{X}}^ – }} \right]\left[ {{{\rm{B}}^ + }} \right]}}\)

\(\frac{{{{\rm{K}}_{\rm{w}}}}}{{{{\rm{K}}_{\rm{a}}} \times {{\rm{K}}_{\rm{b}}}}} = {{\rm{K}}_{\rm{h}}}\)

Relation between Hydrolysis Constant and Degree of Hydrolysis

Let’s say we have one mole of salt dissolved in V litres of a solution to begin with. The equilibrium concentrations are as follows:

\({{\rm{B}}^ + } + {{\rm{X}}^ – } + {{\rm{H}}_2}{\rm{O}}⇌{\rm{BOH}} + {\rm{HX}}\)

\(\frac{{{\rm{1 – \alpha }}}}{{\rm{V}}}\frac{{{\rm{1 – \alpha }}}}{{\rm{V}}}\quad \frac{{\rm{\alpha }}}{{\rm{V}}}\quad \frac{{\rm{\alpha }}}{{\rm{V}}}\)

We can write,

\({{\rm{K}}_{\rm{h}}}{\rm{ = }}\frac{{{\rm{\alpha /V \times \alpha /V}}}}{{{\rm{1 – \alpha /V \times 1 – \alpha /V}}}}\)

\({{\rm{K}}_{\rm{h}}}{\rm{ = }}\frac{{{{\rm{\alpha }}^{\rm{2}}}}}{{{{{\rm{(1 – \alpha )}}}^{\rm{2}}}}}\)

If \({\rm{\alpha }}\) is too small, we can write,

\({{\rm{K}}_{\rm{h}}}{\rm{ = }}{{\rm{\alpha }}^{\rm{2}}}\)

\({\rm{\alpha = }}\sqrt {{{\rm{K}}_{\rm{h}}}} \)

\({{\rm{K}}_{\rm{h}}} = {{\rm{K}}_{\rm{w}}}/{{\rm{K}}_{\rm{a}}} \times {{\rm{K}}_{\rm{b}}}\)

\({\rm{\alpha }} = \sqrt {\frac{{{{\rm{K}}_{\rm{w}}}}}{{{{\rm{K}}_{\rm{a}}} \times {{\rm{K}}_{\rm{b}}}}}} \)

Derivation of pH

The dissociation equilibrium of the weak acid, \({\rm{HX}}\), can be used to calculate the hydrogen ion concentration in a solution of a salt of a weak acid and weak base.

\({\rm{HX}}⇌{{\rm{H}}^ + } + {{\rm{X}}^ – }\)

\({{\rm{K}}_{\rm{a}}}{\rm{ = }}\frac{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\left[ {{{\rm{X}}^{\rm{ – }}}} \right]}}{{{\rm{[HX]}}}}\)

\(\left[ {{{\rm{H}}^{\rm{ + }}}} \right]{\rm{ = }}\frac{{{{\rm{K}}_{\rm{a}}}{\rm{[HX]}}}}{{\left[ {{{\rm{X}}^{\rm{ – }}}} \right]}}\)

Also, we know that

\({\rm{[HX] = }}\frac{{\rm{\alpha }}}{{\rm{V}}}\) and \(\left[ {{{\rm{X}}^{\rm{ – }}}} \right]{\rm{ = }}\frac{{{\rm{1 – \alpha }}}}{{\rm{V}}}\)

Substituting these values, we have

\(\left[ {{{\rm{H}}^{\rm{ + }}}} \right]{\rm{ = }}\frac{{{{\rm{K}}_{\rm{a}}} \times \frac{\alpha }{{\rm{V}}}}}{{\frac{{{\rm{1 – }}\alpha }}{{\rm{V}}}}}{\rm{ = }}{{\rm{K}}_{\rm{a}}}\left( {\frac{\alpha }{{{\rm{1 – }}\alpha }}} \right)\)

If \({\rm{\alpha }}\) is too small, we can write,

\(\left[ {{{\rm{H}}^{\rm{ + }}}} \right]{\rm{ = }}{{\rm{K}}_{\rm{a}}}{\rm{ \times \alpha }}\)

\(\left[ {{{\rm{H}}^ + }} \right] = {{\rm{K}}_{\rm{a}}}\sqrt {\frac{{{{\rm{K}}_{\rm{w}}}}}{{{{\rm{K}}_{\rm{a}}}\,{{\rm{K}}_{\rm{b}}}}}} = \sqrt {\frac{{{{\rm{K}}_{\rm{w}}}{{\rm{K}}_{\rm{a}}}}}{{{{\rm{K}}_{\rm{b}}}}}} \)

After taking logarithms, the above equation can be written as

\({\rm{ – log}}\left[ {{{\rm{H}}^{\rm{ + }}}} \right]{\rm{ = \,- }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{log}}{{{\rm{K}}_{\rm{w}}}}{\rm{ – }}\frac{{\rm{1}}}{{\rm{2}}}{{{\rm{K}}_{\rm{a}}}}{\rm{ + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{log}}{{{\rm{K}}_{\rm{b}}}}\)

\({\rm{pH = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{p}}{{{\rm{K}}_{\rm{w}}}}{\rm{ + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{p}}{{{\rm{K}}_{\rm{a}}}}{\rm{ – }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{p}}{{{\rm{K}}_{\rm{b}}}}\)

\({\rm{pH = 7 + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{p}}{{{\rm{K}}_{\rm{a}}}}{\rm{ – }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{p}}{{{\rm{K}}_{\rm{b}}}}\)

It is clear from the above equation that the pH of the solution will depend upon the pK values of the acid and the base.

Salt of Strong Acid and Strong Base

The salt of strong acid and the strong base do not undergo hydrolysis, and the resulting solution is neutral.

Summary

Hydrolysis of salt is the reaction of an anion or cation with water that results in the cleavage of the \({\rm{O–H}}\) bond. According to their hydrolytic behaviour, various salts can be categorized into various types. In the case of salts of weak acids and strong bases, the resultant solution is basic due to the presence of extra \({\rm{O}}{{\rm{H}}^ – }\) ions. In contrast, in the case of salts of weak bases and strong acids, the resultant solution is acidic due to the presence of extra \({{\rm{H}}^ + }\) ions. In the case of salts of weak acids and weak bases, the pH of the resultant solution is determined by the relative amount of anionic and cationic hydrolysis.

In contrast, in the case of salts of strong acids and bases, neither the cation nor the anion undergoes hydrolysis. Therefore, the resulting solution is neutral. The expression for pH values for the different salt solutions can be derived.

FAQs on Hydrolysis of Salts and pH of their Solutions

Q.1. What is the hydrolysis of salt?
Ans: Hydrolysis is the reaction of an anion or cation with water that results in the cleavage of the \({\rm{O–H}}\) bond. When the salt of a weak acid, \({\rm{HA}}\), and a strong base dissolve in water, the anion \({{\rm{A}}^ – }\) is produced. The \({{\rm{A}}^ – }\) anion reacts with water to form the unionized molecule by removing a proton \(\left( {{{\rm{H}}^ + }} \right)\) from its molecule. Similarly, the cation \({{\rm{B}}^ + }\) is formed when the salt of a weak base, BOH, and a strong acid dissolve in water. The cation \({{\rm{B}}^ + }\) interacts with water and, in the process, accepts \({\rm{O}}{{\rm{H}}^ – }\) ion.

Q.2. Why the solution of strong acid and the strong base is neutral?
Ans: In the case of strong acid and strong base, neither the cation nor the anion undergoes hydrolysis. Therefore, the resulting solution is neutral.

Q.3. Which salt solution will have the lowest value of pH?
Ans: The pH of the solution of a salt of a weak base and strong acid is given as
\({\rm{pH = 7 – }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{p}}{{\rm{K}}_{\rm{b}}}{\rm{ – }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{logC}}\)
In this case, it is evident that pH will always be less than \(7\). So the solution of the salt of a weak base and strong acid will have the lowest value of pH.

Q.4. Why the solution of the salt of strong acid and strong base is neutral?
Ans: The salt of strong acid and the strong base do not undergo hydrolysis. So the resulting solution is neutral.

Q.5. What will be the pH value of the solution of the weak acid and weak base?
Ans: The pH value of the solution of the weak acid and weak base is given as
\({\rm{pH}} = 7 + \frac{1}{2}{\rm{p}}{{\rm{K}}_{\rm{a}}} – \frac{1}{2}{\rm{p}}{{\rm{K}}_{\rm{b}}}\)
It is clear from the above equation that the pH of the solution will depend upon the pK values of the acid and the base.

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