Salt Hydrolysis: Definition, Equation, Formula, Questions and Examples

The idea of salt hydrolysis was discoverd by several scientist with time, but important contributions include Svante Arrhenius in 1887 Arrhenius made important contributions to the understanding of acids and bases, which contain the idea of hydrolysis of salt in water. He tells that salts breaks into ions in solution and that these ions can react with water. And the otherr one include Jacobus Henricus van 't Hoff* in 1887. Van 't Hoff formulate the understanding of ionic equilibrium and its contribution in hydrolysis, which connect to the colligative properties. And the last but not the least is Gilbert N. Lewis in 1920s. Lewis elborates the theories of acid-base chemistry, which helps in understanding how salts hydrolyze.The need for this discovery is to understanding salt hydrolysis is important for several reasons. And the reason includes

Chemical Reactions- It helps to tell the pH of solutions formed from salts, which is importantl in many chemical processes and formulations. Industrial Processes- It impacts various industrial applications, including pharmaceuticals, agriculture, and manufacturing, where controlling pH is essential.Environmental Science- It affects environmental chemistry, influencing soil pH and the behavior of pollutants. Biochemistry- It is important in biological systems where pH control is vital for enzyme activity and other biochemical processes.

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This Story also Contains

1. Hydrolysis Of Salts
2. Hydrolysis Of Weak Base And Strong Acid
3. Some Solved Examples
4. Summary

Hydrolysis Of Salts

Salt Hydrolysis

When a salt is added in water ions of the salt interact with water to cause acidity or basicity in aqueous solution. This ionic interaction is called salt hydrolysis. Interaction of cation is cationic hydrolysis and interaction of anion is anionic hydrolysis.

• Hydrolysis is reverse of neutralization and an endothermic process.
• If hydrolysis constant is K_h and neutralization constant is Kn. Then,
  Kn = 1/K_h
• A solution of the salt of strong acid and weak base is acidic and for it pH < 7 or [H⁺] > 10^-7. For example, FeCl₃ (weak base + strong acid): Solution is acidic and involves cationic hydrolysis.
• A solution of the salt of strong base and weak acid is basic and for it pH > 7 or [H⁺] < 10^-7. For example, KCN (strong base + weak acid): Solution is basic and involves anionic hydrolysis.
• A solution of the salt of weak acid and weak
  base, then:
  If K_a > K_b, it is acidic
  If K_a < K_b, it is basic
  If K_a = K_b, it is neutral
• CH₃COONH₄ (weak acid + weak base): Solution is neutral and involves both cationic and anionic hydrolysis.
• A solution of the salt of strong acid and strong base is neutral or pH = 7 or [H⁺] = 10^-7
• A salt of strong acid and strong base is never hydrolyzed however, ions are hydrated. For example, K₂SO₄ (however, strong base + strong acid).

Degree of Hydrolysis: It is defined as the fraction of total salt that has undergone hydrolysis on attainment of equilibrium. It is denoted by h.
Let c be the concentration of salt and h be its degree of hydrolysis.

$\mathrm{A}^{-}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{OH}^{-}+\mathrm{HA}$
c
c - ch ch ch

$\mathrm{K}_{\mathrm{h}}=\frac{\left[\mathrm{OH}^{-}\right][\mathrm{HA}]}{\left[\mathrm{A}^{-}\right]}=\frac{(\mathrm{ch})(\mathrm{ch})}{\mathrm{c}-\mathrm{ch}}=\frac{\mathrm{ch}^2}{1-\mathrm{h}}$

Hydrolysis Of Weak Acid And Strong Base

Such salts give alkaline solutions in water. Some of such salts are CH₃COONa, Na₂CO3, K₂CO₃, KCN, etc. For our discussion, we consider CH₃COONa (sodium acetate) in water. When CH₃COONa is put in water, it completely ionizes to give CH₃COO^- (acetate) ions and Na⁺ ions. Now acetate ions (CH₃COO^-) absorb some H⁺ ions from weakly dissociated H₂O molecules to form undissociated CH₃COOH. Na⁺ remains in the ionic state in water. Now for K_w (ionic product) of water to remain constant, H₂O further ionizes to produce more H⁺ and OH^- ions. H⁺ ions are taken up by CH₃COO^- ions leaving OH^- ions in excess and hence an alkaline solution is obtained.

Let BA represents such a salt. As it is put in water;
$\mathrm{BA}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{aq}) \rightleftharpoons \mathrm{BOH}(\mathrm{aq})+\mathrm{HA}(\mathrm{aq})$

BA dissociates into ions and BOH being strong base also ionises.

$\mathrm{B}^{+}+\mathrm{A}^{-}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{B}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})+\mathrm{HA}(\mathrm{aq})$

Thus, the net reaction is:$\mathrm{A}^{-}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{OH}^{-}(\mathrm{aq})+\mathrm{HA}(\mathrm{aq})$

Thus, the hydrolysis constant(K_h) is given as:$\mathrm{K}_{\mathrm{h}}=\frac{\left[\mathrm{OH}^{-}\right][\mathrm{HA}]}{\left[\mathrm{A}^{-}\right]}$

pH of Solution

pH of a basic solution is given as:

$\mathrm{pH}=14+\log \left[\mathrm{OH}^{-}\right]$and $\left[\mathrm{OH}^{-}\right]=\mathrm{ch}=\sqrt{\mathrm{K}_{\mathrm{h}} \mathrm{c}}$

Thus, substituting for K_h, we get:

$\begin{aligned} & {\left[\mathrm{OH}^{-}\right]=\sqrt{\frac{\mathrm{K}_{\mathrm{w}} \mathrm{c}}{\mathrm{K}_{\mathrm{a}}}}} \\ & \mathrm{pH}=14+\log _{10} \sqrt{\frac{\mathrm{K}_{\mathrm{w}} \mathrm{c}}{\mathrm{K}_{\mathrm{a}}}} \\ & \text { Thus, } \mathrm{pH}=\frac{1}{2}\left(\mathrm{pK}_{\mathrm{w}}+\mathrm{pK}_{\mathrm{a}}+\log _{10} \mathrm{c}\right) \\ & \text { Hence, } \mathrm{pH}=7+\frac{1}{2}\left(\mathrm{pK}_{\mathrm{a}}+\log _{10} \mathrm{c}\right)\end{aligned}$

Hydrolysis Of Weak Base And Strong Acid

Such salts give acidic solutions in water. Some of such salts are NH₄Cl, ZnCl₂, FeCl₃, etc. For the purpose of discussion, we will consider the hydrolysis of NH₄Cl. When NH₄Cl is put in water, it completely ionises in water to give NH₄⁺ and Cl^- ions. NH₄⁺ ions combine with OH^- ions furnished by weakly dissociated water to form NH₄OH (a weak base). Now for keeping K_w constant, water further ionises to give H⁺ and OH^- ions, where OH^- ions are consumed by NH₄⁺ ions leaving behind H⁺ ions in solution to give an acidic solution.
Let BA be one of such salts. When it is put into water, the reaction is as follows.$\mathrm{B}^{+}+\mathrm{A}^{-}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{BOH}+\mathrm{H}^{+}(\mathrm{aq})+\mathrm{A}^{-}(\mathrm{aq})$

Thus the net reaction of hydrolysis is as follows:

$\mathrm{B}^{+}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{BOH}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})$
c - ch ch ch

$\begin{aligned} & \mathrm{K}_{\mathrm{h}}=\frac{[\mathrm{BOH}]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{B}^{+}\right]}=\frac{(\mathrm{ch})(\mathrm{ch})}{\mathrm{c}-\mathrm{ch}}=\frac{\mathrm{ch}^2}{1-\mathrm{h}^2}=\mathrm{ch}^2 \quad(\text { assuming } \mathrm{h}<<1) \\ & \text { Thus, } \mathrm{h}=\sqrt{\frac{\mathrm{K}_{\mathrm{h}}}{\mathrm{c}}}\end{aligned}$

Considering ionisation of weak base BOH and H₂O.

$\begin{aligned} & \mathrm{BOH} \rightleftharpoons \mathrm{B}^{+}+\mathrm{OH}^{-} \quad \Rightarrow \quad \mathrm{K}_{\mathrm{b}}=\frac{\left[\mathrm{B}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\mathrm{BOH}]} \\ & \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-} \quad \Rightarrow \quad \mathrm{K}_{\mathrm{w}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]\end{aligned}$

From expressions for K_h, K_b and K_w, we have :

$\mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{b}}} \Rightarrow \mathrm{h}=\sqrt{\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{b}} \mathrm{c}}}$

pH of Solution
Now, pH = - log [H⁺]and $\left[\mathrm{H}^{+}\right]=\mathrm{ch}=\mathrm{c} \sqrt{\frac{\mathrm{K}_{\mathrm{h}}}{\mathrm{c}}}=\sqrt{\mathrm{K}_{\mathrm{h}} \mathrm{c}} \Rightarrow\left[\mathrm{H}^{+}\right]=\sqrt{\frac{\mathrm{K}_{\mathrm{W}} \mathrm{c}}{\mathrm{K}_{\mathrm{b}}}}$

$\Rightarrow \quad \mathrm{pH}=-\log _{10} \sqrt{\frac{\mathrm{K}_{\mathrm{w}} \mathrm{c}}{\mathrm{K}_{\mathrm{b}}}}$

$\Rightarrow \quad \mathrm{pH}=\frac{1}{2}\left(\mathrm{pK}_{\mathrm{w}}-\mathrm{pK}_{\mathrm{b}}-\log _{10} c\right)$

Hence, $\mathrm{pH}\left(\right.$ at $\left.25^{\circ} \mathrm{C}\right)=7-\frac{1}{2}\left(\mathrm{pK}_{\mathrm{b}}+\log _{10} c\right)$

Recommended topic video on (Salt Hydrolysis )

Some Solved Examples

Example.1

1.The definition of acid and bases as acids are substances that dissociates in water to give $H^{+}$ ions and bases are substances that produces $O H^{-}$ ions was given by

1) (correct)Arrhenius

2)Lewis

3)Bronsted Lowry

4)None of these

Solution

The definition of acid and bases as acids are substances that dissociate in water to give $H^{+}$ ions and bases are substances that produce OH^- ions was given by Arrhenius.

Hence, the answer is the option(1).

Example.2

2.The pH of 0.1 M solution of the following salts increases in the order:

1)$\mathrm{NaCl}<\mathrm{NH}_4 \mathrm{Cl}<\mathrm{NaCN}<\mathrm{HCl}$

2) (correct)$\mathrm{HCl}<\mathrm{NH}_4 \mathrm{Cl}<\mathrm{NaCl}<\mathrm{NaCN}$

3)$\mathrm{HCl}<\mathrm{NaCl}<\mathrm{NaCN}<\mathrm{NH}_4 \mathrm{Cl}$

4)$\mathrm{NaCN}<\mathrm{NH}_4 \mathrm{Cl}<\mathrm{NaCl}<\mathrm{HCl}$

Solution

Solution of HCl and $\mathrm{NH}_4 \mathrm{Cl}$ will be acidic, NaCl solution of neutral whereas solution of NaCN will be basic.

Hence, the answer is the option (2).

Example.3

3. An aqueous solution of aluminum sulphate would show:

1)Basic

2) (correct)Acidic

3)Neutral

4)Any of above

Solution

$\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3$ is a salt of $W_B / S_A \cdot\left[\mathrm{Al}(\mathrm{OH})_3+\mathrm{H}_2 \mathrm{SO}_4\right]$ it hydrolyses and gives the acidic solution.

Hence, the answer is the option (2).

Example.4

4.A certain weak acid has K_a = 1.0 x 10^-4. Calculate the equilibrium constant for its reaction with a strong base.

1)10^-10

2) (correct)10¹⁰

3)10⁴

4)10⁸

Solution

Given,

$\mathrm{HA} \rightleftharpoons \mathrm{H}^{+}+\mathrm{A}^{-}, \mathrm{K}_{\mathrm{a}}=10^{-4}$

and we know that

$\mathrm{H}^{+}+\mathrm{OH}^{-} \rightleftharpoons \mathrm{H}_2 \mathrm{O}, \mathrm{K}=\frac{1}{\mathrm{~K}_{\mathrm{w}}}=10^{14}$

Adding the above two reactions we get

$\mathrm{HA}+\mathrm{OH}^{-} \rightleftharpoons \mathrm{A}^{-}+\mathrm{H}_2 \mathrm{O}, \mathrm{K}_{\mathrm{net}}=\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{K}_{\mathrm{w}}}$

$\therefore \mathrm{K}_{\text {net }}=\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{K}_{\mathrm{w}}}=\frac{10^{-4}}{10^{-14}}=10^{10}$

Hence, the answer is the option (2).

Example.5

5.Which one of the following is not an acid salt?

1)$\mathrm{NaH}_2 \mathrm{PO}_4$

2)${ }^1 \mathrm{NaH}_2 \mathrm{PO}_3$

3)$\mathrm{Na}_2 \mathrm{SO}_3$

4) (correct)All of the above are acid salts

Solution

Salt of $S_B / W_A$ are called acid salts.
$\mathrm{NaH}_2 \mathrm{PO}_4$ ( Salt of $\left.\mathrm{NaOH}+\mathrm{H}_3 \mathrm{PO}_4\right)\left(S_B / W_A\right)$
$\mathrm{NaH}_2 \mathrm{PO}_2$ (Salt of $\left.\mathrm{NaOH}+\mathrm{H}_3 \mathrm{PO}_2\right)\left(S_B / W_A\right)$
$\mathrm{NaH}_2 \mathrm{PO}_3$ (Salt of $\left.\mathrm{NaOH}+\mathrm{H}_3 \mathrm{PO}_3\right)\left(S_B / W_A\right)$
All are acid salts.

Hence, the answer is the option (4).

Summary

The acidic or basic solution are formed by the hydrolyze of acid and bases, depending on the nature of their salt. Acidic Salts formed from strong acids and weak bases are Useful in Buffer Solutions such as Acidic salts can help create buffer solutions that resist changes in pH. pH Adjustment in which can be used to acidify solutions in various industrial processes. acid-base hydrolysis has its application in corrosion control such that the Acidic conditions can be used in controlling corrosion rates in metal processing. The basic Salts are those which are formed from weak acids and strong bases. Useful in making the Buffer Solutions such as the Basic salts also take part to buffer solutions that can stop changes in pH. pH Adjustment which can neutralize acids, which is useful in various chemical and industrial applications. The basic buffer is also used as the Cleaning Agent in which Basic salts can act as effective cleaning agents or detergents due to their ability to saponify fats and oils. Overall, the hydrolysis of these salts is beneficial in maintaining pH balance, creating buffer systems, and various industrial and cleaning applications.