Salt hydrolysis is when a salt reacts with water to release the acid and base. In Hydrolysis of Salt, the solution is neutral when the acid and the base are either strong or weak. Salts dissociate completely to give ions in aqueous solutions. Ions present in aqueous solutions have a marked influence on the nature of solutions. Ionic equilibria play an essential role in discussing such influences. Two such aspects are discussed in the present section. 

The salt ions that react with water may produce either weak acid or weak base or weak acid or weak base. As a result, the \({\rm{pH}}\) of solutions may become less or more than \(7\). This article will learn about salt hydrolysis, hydrolysis constant and determine if the salt is basic, acidic, or neutral.

What is Hydrolysis of Salts?

The hydrolysis of salts are explained below:

Salt Hydrolysis

Salt hydrolysis may be defined as the reaction of cation or anion or both of a salt with water to produce a basic or an acidic solution or both.

The process of salt hydrolysis is the reverse process of neutralization.

\({\rm{Salt }} + {\rm{ Water }} \to {\rm{ Acid }} + {\rm{ Base}}\)

\({\rm{(Or) BA}} + {{\rm{H}}_2}{\rm{O}} \to {\rm{HA}} + {\rm{BOH}}\)

All salts are strong electrolytes and thus ionize completely in the aqueous solution. If the acid \({\rm{(HA)}}\) produced is strong and the base \({\rm{(BOH)}}\) produced is weak, we can write the above equation as

\({{\rm{B}}^{\rm{ + }}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{O}} \to \mathop {{\rm{BOH}}{\mkern 1mu} }\limits_{{\rm{weak}}{\mkern 1mu} {\rm{base}}} {\rm{ + }}{{\rm{H}}^{\rm{ + }}}\)

Thus, in this case the cation reacts with water to give an acidic solution. This is called cationic hydrolysis.

Again, if the acid produced is weak and the base produced is strong, we can write

\({{\rm{A}}^{\rm{ – }}} + {{\rm{H}}_2}{\rm{O}} \to \mathop {{\rm{HA}}}\limits_{{\rm{weak}}\,{\rm{acid}}} {\mkern 1mu}  + {\rm{O}}{{\rm{H}}^ – }\)

Hence, the anion reacts with water to give the basic solution. This is called anionic hydrolysis.

Types of Salts Based on Their Hydrolysis

Salts are classified into four types based on their hydrolytic behaviour. They are

1. Salts of strong acid and strong base
2. Salts of a weak acid and strong base 
3. Salts of strong acid and weak base 
4. Salts of a weak acid and weak base

Salts of Strong Acid and Strong Base

These salts do not hydrolyse. In aqueous solutions, the cations (Example \({\rm{N}}{{\rm{a}}^ + },{{\rm{K}}^ + },{\rm{C}}{{\rm{a}}^{2 + }},{\rm{B}}{{\rm{a}}^{2 + }}\),etc.) of strong bases and anions (Example: \({\rm{C}}{{\rm{l}}^ – },{\rm{B}}{{\rm{r}}^ – },{\rm{NO}}_3^ – \), etc.) of strong acids get easily hydrated but do not hydrolyse and therefore the solutions of salts formed from strong acids and bases are neutral.

Examples: \({\rm{NaCl}},{\rm{NaN}}{{\rm{O}}_3},{\rm{KCl}},{\rm{KN}}{{\rm{O}}_3},\) etc.

Let us consider the aqua solution of \({\rm{NaCl}}\).

\({\rm{NaCl}}\) dissociates in water to give \({\rm{N}}{{\rm{a}}^{\rm{ + }}}\) and \({\rm{C}}{{\rm{l}}^{\rm{ – }}}\) ions.

If these ions react with water, the products would be \({\rm{NaOH}}\) and \({\rm{HCl}}\), which are strong electrolytes and undergo complete ionisation. Then \({{\rm{H}}^{\rm{ + }}}\) and  \({\rm{O}}{{\rm{H}}^{\rm{ – }}}\) ions recombine to form an undissociated water molecule. So, it is neutral. 

Salt of Strong Acid and Weak Base 

The aqueous solutions of this type’s salts are acidic because the cation of the salt is reactive, and this type of salt undergoes cationic hydrolysis from a weak base and \({{\rm{H}}^{\rm{ + }}}\) ions.

Example: \({\rm{N}}{{\rm{H}}_4}{\rm{Cl}},{\rm{C uSO}}{{\rm{ }}_4},{\rm{Ca}}{({\rm{OH}})_2},\) etc.

As an illustration, the hydrolysis of \({\rm{N}}{{\rm{H}}_{\rm{4}}}{\rm{Cl}}\) may be represented as follows:

\({\rm{N}}{{\rm{H}}_4}{\rm{Cl}} + {{\rm{H}}_2}{\rm{O}} \leftrightarrow {\rm{N}}{{\rm{H}}_4}{\rm{OH}} + {\rm{HCl}}\)

Or \({\rm{NH}}_4^ + + {\rm{C}}{{\rm{l}}^ – } + {{\rm{H}}_2}{\rm{O}} \leftrightarrow {\rm{N}}{{\rm{H}}_4}{\rm{OH}} + {{\rm{H}}^ + } + {\rm{C}}{{\rm{l}}^ – }\)

As it produces \({{\rm{H}}^{\rm{ + }}}\) ions, the solution of such a salt is acidic in character.

Salt of a Weak Acid and Strong Base

The aqueous solutions of these types of salts are basic because the anion of the salt is reacting, and the salt undergoes anionic hydrolysis to form weak acid and \({\rm{O}}{{\rm{H}}^ – }\) ions.

Example: \({\rm{C}}{{\rm{H}}_3}{\rm{COONa}},{\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}\), etc.

Let us consider the aqueous solution of \({\rm{C}}{{\rm{H}}_3}{\rm{COONa}}\)

\({\rm{C}}{{\rm{H}}_3}{\rm{COONa}} \to {\rm{C}}{{\rm{H}}_3}{\rm{CO}}{{\rm{O}}^ – } + {\rm{N}}{{\rm{a}}^ + }\)

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

\(\therefore \left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right] > \left[ {{{\rm{H}}^{\rm{ + }}}} \right]\), hence the solution is basic in nature

Salts of a Weak Acid and Weak Base

Maximum hydrolysis occurs in this type of salt because both the cation and anion are reactive and react with water to produce \({{\rm{H}}^{\rm{ + }}}\) and \({\rm{O}}{{\rm{H}}^ – }\) ions.

Then the solution may be neutral, or acidic or basic.

The solution is neutral if the rate of cationic hydrolysis \(=\) the rate of anionic hydrolysis.
The solution is acidic if the rate of cationic hydrolysis \(>\) the rate of anionic hydrolysis.

The solution is basic if the rate of cationic hydrolysis \(<\) anionic hydrolysis.

Example: \({\rm{C}}{{\rm{H}}_3}{\rm{COON}}{{\rm{H}}_4},{\rm{N}}{{\rm{H}}_4}\;{\rm{F}},{\rm{AlP}}{{\rm{O}}_4},\) etc.

Let us consider the aqueous solutions of \({\rm{C}}{{\rm{H}}_3}{\rm{COON}}{{\rm{H}}_4}.\)

It is neutral in nature because

\({\rm{C}}{{\rm{H}}_3}{\rm{COON}}{{\rm{H}}_4} \to {\rm{C}}{{\rm{H}}_3}{\rm{CO}}{{\rm{O}}^ – } + {\rm{NH}}_4^ + \)

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

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

Both the reactions take place at the same speed so, the solution is neutral in nature.

Hydrolysis Constant \(\left( {{{\rm{K}}_{\rm{h}}}} \right)\)

The equilibrium constant derived by applying the law of mass action to a hydrolysis reaction is called hydrolysis constant (or) hydrolytic constant. It is denoted by \({{\rm{K}}_{\rm{h}}}\). 

\({\rm{BA}} + {{\rm{H}}_2}{\rm{O}} \leftrightarrow {\rm{HA}} + {\rm{BOH}}\)

Using the law of mass action, the equilibrium constant \({\rm{K = }}\frac{{{\rm{[HA][BOH]}}}}{{{\rm{[BA]}}\left[ {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right]}} \Rightarrow {\rm{K}}\left[ {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right]{\rm{ = }}\frac{{{\rm{[HA][BOH]}}}}{{{\rm{[BA]}}}}\)

Since water is present in excess, then its concentration is taken as constant.

\({\rm{K}}\left[ {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right]{\rm{ =  constant  = }}{{\rm{K}}_{\rm{h}}}\)

\({{\rm{K}}_{\rm{h}}}{\rm{ = }}\frac{{{\rm{[HA][BOH]}}}}{{{\rm{[BA]}}}}\)

Degree of Hydrolysis

The degree of hydrolysis of salt is defined as the fraction (or percentage) of the total salt which is hydrolysed, i.e.,

\({\rm{h}} = \frac{{{\rm{ number \, of \, moles \, of \, the \, salt \, hydrolysed }}}}{{{\rm{ Total \, number \, of \, moles \, of \, salt \, taken }}}}\)

Degree of Hydrolysis and Ph of Salt Solutions

As already explained, salts of strong acids and strong bases do not undergo hydrolysis; therefore, to talk of \({{\rm{K}}_{\rm{h}}}\) or \({\rm{h}}\) of such salts is meaningless. The cases of the other three types of salts are discussed below:

Salts of a Weak Acid and Strong Base

1. Hydrolysis constant: Representing the salt by BA, as usual, the hydrolysis may be represented as follows:

i.e. it is a case of anion hydrolysis.

The hydrolysis constant \({{\rm{K}}_{\rm{h}}}\) for the above reaction will be given by \({{\rm{K}}_{\rm{h}}}{\rm{ = }}\frac{{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]{\rm{[HA]}}}}{{\left[ {{{\rm{A}}^{\rm{ – }}}} \right]}}\quad {\rm{ \ldots \ldots (1)}}\)

For the weak acid HA, the dissociation equilibrium is: \({\rm{HA}} \leftrightarrow {{\rm{H}}^{\rm{ + }}}{\rm{ + }}{{\rm{A}}^{\rm{ – }}}\)

\(\therefore\) The dissociation constant \({{\rm{K}}_{\rm{a}}}\) of the acid HA will be given by \({{\rm{K}}_{\rm{a}}}{\rm{ = }}\frac{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\left[ {{{\rm{A}}^{\rm{ – }}}} \right]}}{{[{\rm{HA}}]}}\,\,….(2)\)

Further, the ionic product of water, \({{\rm{K}}_{\rm{W}}}\), is given by \({{\rm{K}}_{\rm{W}}}{\rm{ = }}\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]……{\rm{(3)}}\)

Multiplying equation \((1)\) with \((2)\) and dividing by \((3)\), we get

\(\frac{{{{\rm{K}}{\rm{h}}} \times {{\rm{K}}{\rm{a}}}}}{{{{\rm{K}}_{\rm{W}}}}}{\rm{ = }}\frac{{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right][{\rm{HA}}]}}{{\left[ {{{\rm{A}}^{\rm{ – }}}} \right]}} \times \frac{{\left[ {{{\rm{H}}^ + }} \right]\left[ {{{\rm{A}}^{\rm{ – }}}} \right]}}{{[{\rm{HA}}]}} \times \frac{{\rm{1}}}{{\left[ {{{\rm{H}}^ + }} \right]\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]}}{\rm{ = 1}}\) or \({{\rm{K}}{\rm{h}}}{\rm{ = }}\frac{{{{\rm{K}}{\rm{W}}}}}{{{{\rm{K}}_{\rm{a}}}}}……(4)\)

b. Degree of hydrolysis: Assume the salt’s original concentration in the solution is \({\rm{C}}\) moles/litre, and \({\rm{h}}\) is its degree of hydrolysis at this concentration. Then we have

\({{\rm{A}}^{\rm{ – }}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{O}} \to {\rm{O}}{{\rm{H}}^{\rm{ – }}}{\rm{ + HA}}\)

The hydrolysis constant \({{\rm{K}}_{\rm{h}}}\) will, therefore, be given by if \({{\rm{K}}_{\rm{h}}}{\rm{ = }}\frac{{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right][{\rm{HA}}]}}{{\left[ {{{\rm{A}}^{\rm{ – }}}} \right]}}{\rm{ = }}\frac{{{\rm{ch}} \cdot {\rm{ch}}}}{{{\rm{c}}\left( {{\rm{1 – h}}} \right)}}{\rm{ = }}\frac{{{\rm{c}}{{\rm{h}}^{\rm{2}}}}}{{{\rm{1 – h}}}}\)

If \({\rm{h}}\) is very small as compared to \(1\), we can take \({\rm{1 – h}} \approx {\rm{1}}\) so that the above expression becomes

\({\rm{c}}{{\rm{h}}^{\rm{2}}}{\rm{ = }}{{\rm{K}}_{\rm{h}}}\) or \({{\rm{h}}^{\rm{2}}}{\rm{ = }}\frac{{{{\rm{K}}_{\rm{h}}}}}{{\rm{c}}}\) or \({\rm{h = }}\sqrt {\frac{{{{\rm{K}}_{\rm{h}}}}}{{\rm{c}}}} \)

Substituting the value of \({{\rm{K}}_{\rm{h}}}\) from equation \((4)\), we get \({\rm{h = }}\sqrt {\frac{{{{\rm{K}}_{\rm{w}}}}}{{{{\rm{K}}_{\rm{a}}}{\rm{.c}}}}} ……{\rm{(5)}}\)

c. \({\rm{pH}}\) In the present case, we have

i.e. \(\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right] = {\mathop{\rm ch}\nolimits} \therefore \left[ {{{\rm{H}}^{\rm{ + }}}} \right]{\rm{ = }}\frac{{{{\rm{K}}_{\rm{w}}}}}{{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]}}{\rm{ = }}\frac{{{{\rm{K}}_{\rm{w}}}}}{{{\rm{ch}}}}\)

substituting the value of \({\rm{h}}\) from equation \(5\), we get 

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

\(\therefore {\rm{pH}} =  – \log {{\rm{H}}^ + } =  – {\rm{log}}\sqrt {\frac{{{{\rm{K}}_{\rm{w}}}.{{\rm{K}}_{\rm{a}}}}}{{\rm{c}}}}  = {\rm{\;}} – {\rm{\;}}\frac{1}{2}{\rm{log}}\left( {\frac{{{{\rm{K}}_{\rm{w}}}.{{\rm{K}}_{\rm{a}}}}}{{\rm{c}}}} \right)\)

This equation may be written as

\({\rm{pH}} = \frac{{\rm{1}}}{{\rm{2}}}\left[ {{\rm{ – log}}{{\rm{K}}_{\rm{w}}}{\rm{ – log}}{{\rm{K}}_{\rm{a}}}{\rm{ + logc}}} \right]\)

At \(298\;{\rm{K}},{\rm{p}}{{\rm{K}}_{\rm{w}}}{\rm{ = 14}}\) Hence at \(298\;{\rm{K}}\) we have

\({\rm{pH = 7 + }}\frac{{\rm{1}}}{{\rm{2}}}\left[ {{\rm{p}}{{\rm{K}}_{\rm{a}}}{\rm{ + logc}}} \right]……(6)\)

The \({\rm{pH}}\) of the solution can thus be calculated by using molar concentration ‘c’ of the solution and the dissociation constant \({{\rm{K}}_{\rm{a}}}\) of the weak acid involved.

Salts of Strong Acid and Weak Base 

a. Hydrolysis constant: For the salt BA, the hydrolysis may be represented as:

i.e. it is a case of cationic hydrolysis.

The hydrolysis constant \({{\rm{K}}_{\rm{h}}}\) will be given by \({{\rm{K}}_{\rm{h}}}{\rm{ = }}\frac{{{\rm{[BOH]}}\left[ {{{\rm{H}}^{\rm{ + }}}} \right]}}{{\left[ {{{\rm{B}}^{\rm{ + }}}} \right]}}{\rm{ \ldots \ldots }}{\rm{.(7)}}\)

For the weak base \({\rm{BOH}}\), the dissociation equilibrium is: \({\rm{BOH}} \leftrightarrow {{\rm{B}}^{\rm{ + }}}{\rm{ + O}}{{\rm{H}}^{\rm{ – }}}\)

∴ The dissociation constant \({{\rm{K}}_{\rm{b}}}\) of the weak base \({\rm{BOH}}\) will be given by \({{\rm{K}}_{\rm{b}}}{\rm{ = }}\frac{{\left[ {{{\rm{B}}^ + }} \right]\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]}}{{[{\rm{BOH}}]}}\,\,\,….\left( 8 \right)\)

The ionic product of water \({{\rm{K}}_{\rm{W}}}\) is given by

\({{\rm{K}}_{\rm{W}}}{\rm{ = }}\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\left[ {\begin{array}{*{20}{l}}
{{\rm{O}}{{\rm{H}}^{\rm{ – }}}}\end{array}} \right]……(9)\)

Multiplying equation \(7\) with \(8\) and dividing by \(9\) we get 

\(\frac{{{{\rm{K}}_{\rm{h}}}{{\rm{K}}_{\rm{b}}}}}{{{{\rm{K}}_{\rm{W}}}}}{\rm{ = 1}}\)

 \({{\rm{K}}_{\rm{h}}}{\rm{ = }}\frac{{{{\rm{K}}_{\rm{W}}}}}{{{{\rm{K}}_{\rm{b}}}}}\quad  \ldots  \ldots  \ldots  \ldots  \ldots ..(10)\)

b. Degree of hydrolysis: Proceeding as in the earlier case, we get the same relationship, viz,

\({\rm{h = }}\sqrt {\frac{{{{\rm{K}}_{\rm{h}}}}}{{\rm{c}}}} \)

Substituting the value of \({{\rm{K}}_{\rm{h}}}\) from equation \(10\) we get  \({\rm{h = }}\sqrt {\frac{{{{\rm{K}}_{\rm{W}}}}}{{{{\rm{K}}_{\rm{b}}}{\rm{c}}}}} …….(11)\)

c. \({\rm{pH:}}\) In this case, we have 

substituting the value of \({\rm{h}}\) from equation \(11\), we get

\(\left[ {{{\rm{H}}^{\rm{ + }}}} \right]{\rm{ = ch = c}} \cdot \sqrt {\frac{{{{\rm{K}}_{\rm{W}}}}}{{{{\rm{K}}_{\rm{b}}}{\rm{c}}}}} {\rm{ = }}\sqrt {\frac{{{{\rm{K}}_{{\rm{W}} \cdot }}{\rm{c}}}}{{{{\rm{K}}_{\rm{b}}}}}} {\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left( {\frac{{{{\rm{K}}_{\rm{W}}} \cdot {\rm{c}}}}{{{{\rm{K}}_{\rm{b}}}}}} \right)\)

 \(\therefore {\rm{pH =  – log}}\left[ {{{\rm{H}}^{\rm{ + }}}} \right]{\rm{ =  – }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{log}}\left( {\frac{{{{\rm{K}}_{\rm{w}}}{\rm{c}}}}{{{{\rm{K}}_{\rm{b}}}}}} \right)\) or  \({\rm{pH =  – }}\frac{{\rm{1}}}{{\rm{2}}}\left[ {{\rm{log}}{{\rm{K}}_{\rm{w}}}{\rm{ – log}}{{\rm{K}}_{\rm{b}}}{\rm{ + logc}}} \right]{\rm{ \ldots (12)}}\)

As before, this equation can be rewritten as  \({\rm{pH = }}\frac{{\rm{1}}}{{\rm{2}}}\left[ {{\rm{p}}{{\rm{K}}_{\rm{w}}}{\rm{ – p}}{{\rm{K}}_{\rm{b}}}{\rm{ – logc}}} \right]\)

At \(298\;{\rm{K}},{\rm{p}}{{\rm{K}}_{\rm{w}}}{\rm{ = 14}}\) Hence, at \(298\;{\rm{K}}\) we have \({\rm{pH = 7 – }}\frac{{\rm{1}}}{{\rm{2}}}\left[ {{\rm{p}}{{\rm{K}}_{\rm{b}}}{\rm{ + logc}}} \right]\)

Salts of a Weak Acid and Weak Base

Hydrolysis constant: Representing the salt by \({\rm{BA}}\) as before, we have

i.e. it involves both cation hydrolysis as well as anion hydrolysis. The equation for the hydrolysis constant will be;

\({{\rm{K}}_{\rm{h}}}{\rm{ = }}\frac{{{\rm{[BOH][HA]}}}}{{\left[ {{{\rm{B}}^{\rm{ + }}}} \right]\left[ {{{\rm{A}}^{\rm{ – }}}} \right]}}……..(13)\)

For the weak acid, \({\rm{HA}} \leftrightarrow {{\rm{H}}^{\rm{ + }}}{\rm{ + }}{{\rm{A}}^{\rm{ – }}}\) So, that \({{\rm{K}}_{\rm{a}}}{\rm{ = }}\frac{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\left[ {{{\rm{A}}^{\rm{ – }}}} \right]}}{{{\rm{[HA]}}}}…….(14)\)

For the weak base \({\rm{BOH}} \leftrightarrow {{\rm{B}}^{\rm{ + }}}{\rm{ + O}}{{\rm{H}}^{\rm{ – }}}\) so that \({{\rm{K}}_{\rm{b}}}{\rm{ = }}\frac{{\left[ {{{\rm{B}}^{\rm{ + }}}} \right]\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]}}{{{\rm{[BOH]}}}}……..(15)\)

Also, we know that \({{\rm{K}}_{\rm{W}}}{\rm{ = }}\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]………(16)\)

Multiplying equations \(13, 14, 15\) and dividing by equation \(16\) we get

\(\frac{{{{\rm{K}}_{\rm{h}}}{{\rm{K}}_{\rm{a}}}{{\rm{K}}_{\rm{b}}}}}{{{{\rm{K}}_{\rm{w}}}}}{\rm{ = 1}}\) or  \({{\rm{K}}_{\rm{h}}}{\rm{ = }}\frac{{{{\rm{K}}_{\rm{K}}}}}{{{{\rm{K}}_{\rm{a}}}{{\rm{K}}_{\rm{b}}}}}……..(17)\)

Degree of hydrolysis: In this case, we have

 \(\therefore {{\rm{K}}_{\rm{h}}}{\rm{ = }}\frac{{[{\rm{BOH}}][{\rm{HA}}]}}{{\left[ {{{\rm{B}}^ + }} \right]\left[ {{{\rm{A}}^{\rm{ – }}}} \right]}}{\rm{ = }}\frac{{{\rm{chch}}}}{{{\rm{c}}\left( {{\rm{1 – h}}} \right){\rm{c}}\left( {{\rm{1 – h}}} \right)}}{\rm{ = }}\frac{{{{\rm{h}}^{\rm{2}}}}}{{{{\left( {{\rm{1 – h}}} \right)}^{\rm{2}}}}}\,\,….\left( {18} \right)\)

Here, the relationship between \({{\rm{K}}_{\rm{h}}}\) and \({\rm{h}}\) does not involve c. Thus, the degree of hydrolysis of such salt is independent of the concentration of the solution.

If \({\rm{h}}\) is very small in comparison to \(1\), we can take \(1 – {\rm{h}} \approx 1\) so that equation \(18\) becomes

\({{\rm{K}}_{\rm{h}}}{\rm{ = }}{{\rm{h}}^{\rm{2}}}\) or  \({\rm{h}} = \sqrt {{{\rm{K}}_{\rm{h}}}} {\rm{ = }}\sqrt {\frac{{{{\rm{K}}_{\rm{W}}}}}{{{{\rm{K}}_{\rm{a}}}{{\rm{K}}_{\rm{b}}}}}} …….(19)\)

c. \({\rm{pH}}\) For the weak acid \({\rm{HA}}\),

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

\(\therefore {{\rm{K}}_{\rm{a}}}{\rm{ = }}\frac{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\left[ {{{\rm{A}}^{\rm{ – }}}} \right]}}{{{\rm{[HA]}}}}\) or  \(\left[ {{{\rm{H}}^{\rm{ + }}}} \right]{\rm{ = }}{{\rm{K}}_{\rm{a}}}\frac{{{\rm{[HA]}}}}{{\left[ {{{\rm{A}}^{\rm{ – }}}} \right]}}{\rm{ = }}{{\rm{K}}_{\rm{a}}}\frac{{{\rm{ch}}}}{{{\rm{c(1 – h)}}}}{\rm{ = }}{{\rm{K}}_{\rm{a}}}\frac{{\rm{h}}}{{{\rm{1 – h}}}}\)

But from equation \({\rm{18,}}\frac{{\rm{h}}}{{{\rm{1 – h}}}}{\rm{ = }}\sqrt {{{\rm{K}}_{\rm{h}}}} \) so that we have

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

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

or  \({\rm{pH}} =  – \frac{{\rm{1}}}{{\rm{2}}}\left[ {{\rm{log}}{{\rm{K}}_{\rm{a}}}{\rm{ + log}}{{\rm{K}}_{\rm{w}}}{\rm{ – log}}{{\rm{K}}_{\rm{b}}}} \right]\)

or this equation may be written as  \({\rm{pH}} = \frac{1}{2}\left[ {{\rm{p}}{{\rm{K}}_{\rm{w}}}{\rm{ + p}}{{\rm{K}}_{\rm{a}}}{\rm{ – p}}{{\rm{K}}_{\rm{b}}}} \right]\)

or at \(298\;{\rm{K}}\) we have \({\rm{pH}} = 7 + \frac{1}{2}\left[ {{\rm{p}}{{\rm{K}}_{\rm{a}}}{\rm{ – p}}{{\rm{K}}_{\rm{b}}}} \right]\)

An Important Observation

For salts of a weak acid and weak base, from the above expression,

1. If \({\rm{p}}{{\rm{K}}_{\rm{a}}} \, < \, {\rm{p}}{{\rm{K}}_{\rm{b}}}{\rm{,pH}}\) of the solution will be less than \(7\) and the solution will be acidic.
2. If \({\rm{p}}{{\rm{K}}_{\rm{a}}}\, > \, {\rm{p}}{{\rm{K}}_{\rm{b}}}{\rm{,pH}}\) of the solution will be greater than \(7\) and the solution will be basic.
3. If \({\rm{p}}{{\rm{K}}_{\rm{a}}}{\rm{\, = \, p}}{{\rm{K}}_{{\rm{b,}}}}{\rm{pH}}\) of the solution will be equal to \(7\), and the solution will be neutral.

Summary

Neutralization reaction between base and acid results in the formation of salt. Hydrolysis is the reverse of neutralization. When a salt completely dissociates in water, it’s cation or anion reacts with water to produce hydroxide ions or hydronium ions that affect the \({\rm{pH}}\) of the solution. In this article, we learned about types of salts, hydrolysis constant, degree of hydrolysis, calculation of hydrolysis constant, degree of hydrolysis and \({\rm{pH}}\) of salt solutions.

FAQs

Q.1. What types of salts undergo hydrolysis?
Ans: Salts are classified into four types based on their hydrolytic behaviour. They are
1. Salts of a weak acid and strong base 
2. Salts of strong acid and weak base 
3. Salts of a weak acid and weak base

Q.2. What does hydrolysis mean?
Ans: Hydrolysis is a decomposition reaction where one of the reactants is water and is used to break chemical bonds in the other reactant. It may be considered the reverse of the condensation reaction in which two molecules combine, producing water as one of the products.

Q.3. How do you write hydrolysis equations?
Ans: If a compound is represented by the formula \({\rm{XY}}\) in which \({\rm{X}}\) and \({\rm{Y}}\) are groups or atoms and water is represented by the formula \({\rm{HOH}}\), the hydrolysis reaction may be represented by the reversible chemical equation

Q.4. What is the hydrolysis constant of the salt?
Ans: The equilibrium constant derived by applying law of mass action to a hydrolysis reaction is called hydrolysis constant (or) hydrolytic constant. It is denoted by \({{\rm{K}}_{\rm{h}}}\). 
\({\rm{BA}} + {{\rm{H}}_2}{\rm{O}} \leftrightarrow {\rm{HA}} + {\rm{BOH}}\)
Applying law of mass action, the equilibrium constant \({\rm{K}} = \frac{{{\rm{[HA][BOH]}}}}{{{\rm{[BA]}}\left[ {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right]}} \Rightarrow {\rm{K}}\left[ {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right] = \frac{{{\rm{[HA][BOH]}}}}{{{\rm{[BA]}}}}\)
Since water is present in excess, then its concentration is taken as constant.
\({\rm{K}}\left[ {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right]{\rm{ =\, constant \, = }}{{\rm{K}}_{\rm{h}}}\)
\({{\rm{K}}_{\rm{h}}}{\rm{ = }}\frac{{{\rm{[HA][BOH]}}}}{{{\rm{[BA]}}}}\)

Q.5. Calculate the degree of hydrolysis and \({\rm{pH}}\) of \({\rm{0}}{\rm{.1MC}}{{\rm{H}}_{\rm{3}}}{\rm{COONa}}\) solution. Hydrolysis constant of \({\rm{C}}{{\rm{H}}_3}{\rm{COONa}}\) is \(5.6 \times {10^{ – 10}}\)
Ans: \({\rm{C}}{{\rm{H}}_3}{\rm{COONa}}\)  is a salt of weak acid \(\left( {{\rm{C}}{{\rm{H}}_3}{\rm{COOH}}} \right)\)  and a strong base \(({\rm{NaOH}})\)
Hence, the solutions is alkaline due to hydrolysis.
\(\mathop {{\rm{C}}{{\rm{H}}_3}{\rm{CO}}{{\rm{O}}^ – }}\limits_{0.1 \times \left( {1 – {\rm{h}}} \right)}  + {{\rm{H}}_2}{\rm{O}} \leftrightarrow {\rm{\;}}\mathop {{\rm{C}}{{\rm{H}}_3}{\rm{COOH}}}\limits_{0.1 \times {\rm{h}}}  + \mathop {{\rm{O}}{{\rm{H}}^ – }}\limits_{0.1 \times {\rm{h}}}\)
\({{\rm{K}}_{\rm{h}}}{\rm{ = }}\frac{{\left[ {{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}} \right]\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]}}{{\left[ {{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}^{\rm{ – }}}} \right]}}{\rm{ = }}\frac{{{\rm{(0}}{\rm{.1 \times h)(0}}{\rm{.1 \times h)}}}}{{{\rm{0}}{\rm{.1(1 – h)}}}}\)
\({\rm{h}}\) is small \((1 – {\rm{h}} \to 1)\)
\(5.6 \times {10^{ – 10}} = 0.1 \times {{\rm{h}}^2}\)
\({{\rm{h}}^{\rm{2}}} = \frac{{5.6 \times {{10}^{ – 10}}}}{{0.1}} = 56 \times {10^{ – 10}}\)
Degree of hydrolysis is \({\rm{h}} = 7.48 \times {10^{ – 5}}\)
\(\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right] = {\rm{Ch}} = 0.1 \times 7.48 \times {10^{ – 5}} = 7.48 \times {10^{ – 6}}{\rm{M}}\)
\(\left[ {{{\rm{H}}^{\rm{ + }}}} \right] = \frac{{{{\rm{K}}_{\rm{W}}}}}{{\left[ {{\rm{O}}{{\rm{H}}^{\rm{ – }}}} \right]}} = \frac{{{{10}^{ – 14}}}}{{7.48 \times {{10}^{ – 6}}}} = 1.34 \times {10^{ – 9}}{\rm{M}}\)
\({\rm{pH}} = – \log \left[ {{{\rm{H}}^{\rm{ + }}}} \right] = – \log \left( {1.34 \times {{10}^{ – 9}}} \right) = 8.87\)

Q.6. How do you know if salt will undergo hydrolysis?
Ans: Hydrolysis is the reverse of neutralization. When salt is added to the water, then anion, cation, or salt react with water, and if the solution becomes either basic or acidic, it is a hydrolysis process.

Examples on Effect of Hydrolysis on Solubility of Sparingly Soluble Salt

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