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

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

Shivani PooniaUpdated on 22 Aug 2025, 11:18 AM IST

Why does a solution of sodium chloride remain neutral in water, while a solution of ammonium chloride turns acidic? What makes some salt solutions taste bitter, some sour, and others neutral? Why does the taste of some salts vary? Well, we will get these answers after reading this article on the hydrolysis of salt. Hydrolysis of salts is the process in which a salt reacts with water to produce an acidic or basic solution.

This Story also Contains

  1. Hydrolysis Of Salts
  2. Degree of Hydrolysis
  3. Hydrolysis Of Weak Acid And Strong Base
  4. Hydrolysis Of Weak Base And Strong Acid
  5. Some Solved Examples
Salt Hydrolysis: Definition, Equation, Formula, Questions and Examples
Salt Hydrolysis

The idea of salt hydrolysis was discovered by several scientists over 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 the hydrolysis of salt in water. He says that salts break into ions in solution and that these ions can react with water. Van 't Hoff formulated the understanding of Ionic Equilibrium and its contribution to hydrolysis, which connects to the colligative properties. And last but not least is Gilbert N. Lewis in the 1920s.

Hydrolysis Of Salts

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

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

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

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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 Kw (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(Kh) 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 Kh, 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 Kw 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 Kh, Kb and Kw, 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: The definition of acids and bases as acids are substances that dissociate in water to give $H^{+}$ ions, and bases are substances that produce $O H^{-}$ 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 option(1).

Example 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 option (2).

Example 3: An aqueous solution of aluminum sulfate would show:

1)Basic

2) (correct)Acidic

3)Neutral

4)Any of the 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 option (2).

Example 4: A certain weak acid has Ka = 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 option (2).

Example 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

Salts 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 option (4).

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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.

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Frequently Asked Questions (FAQs)

Q: What is the role of salt hydrolysis in the formation of stalactites and stalagmites?
A:

Salt hydrolysis plays a crucial role in the formation of stalactites and stalagmites. As water containing dissolved calcium bicarbonate (Ca(HCO3)2) drips, the bicarbonate ions undergo hydrolysis, releasing CO2 and forming carbonate ions. This process leads to the precipitation of calcium carbonate (CaCO3), forming these cave structures over time.

Q: How does salt hydrolysis relate to the concept of amphoterism?
A:

Salt hydrolysis is closely related to amphoterism, which is the ability of a substance to act as both an acid and a base. Amphoteric compounds, when dissolved in water, can form salts that

Q: Why is it important to consider salt hydrolysis in the production of paper?
A:

In paper production, understanding salt hydrolysis is crucial for controlling the pH of the pulp and paper-making process. The hydrolysis of salts used in various stages of production can affect the pH, which in turn influences factors such as fiber swelling, additive effectiveness, and the final paper quality. Proper pH control, considering hydrolysis effects, is essential for producing high-quality paper products.

Q: How does salt hydrolysis affect the taste of mineral water?
A:

Salt hydrolysis can significantly influence the taste of mineral water. Different mineral salts in water undergo varying degrees of hydrolysis, producing subtle changes in pH and releasing ions that contribute to the water's flavor profile. For example, the hydrolysis of calcium and magnesium salts can give water a slightly alkaline taste.

Q: What is the relationship between salt hydrolysis and the formation of acid rain?
A:

Salt hydrolysis contributes to the formation and effects of acid rain. When acidic pollutants like sulfur dioxide and nitrogen oxides dissolve in atmospheric water, they form salts that can undergo hydrolysis, producing strong acids. These hydrolysis reactions increase the acidity of rainwater, leading to the environmental impacts associated with acid rain.

Q: How does salt hydrolysis impact the effectiveness of fertilizers?
A:

Salt hydrolysis plays a significant role in the effectiveness of fertilizers. Many fertilizer salts undergo hydrolysis when dissolved in soil water, which can affect soil pH and nutrient availability. For example, ammonium-based fertilizers can hydrolyze to produce acidic conditions, while nitrate-based fertilizers tend to increase soil pH through hydrolysis reactions.

Q: Why is it important to consider salt hydrolysis when designing cooling systems for industrial processes?
A:

Understanding salt hydrolysis is crucial in designing cooling systems because it can affect the pH and corrosiveness of the cooling water. Hydrolysis of dissolved salts can lead to pH changes that may cause scaling or corrosion in pipes and heat exchangers. Proper treatment and monitoring of cooling water chemistry, including hydrolysis effects, is essential for system efficiency and longevity.

Q: How does salt hydrolysis contribute to the weathering of rocks and minerals?
A:

Salt hydrolysis contributes to chemical weathering of rocks and minerals. As water interacts with minerals, it can cause hydrolysis reactions that break down the mineral structure. For example, the hydrolysis of feldspar minerals in granite can lead to the formation of clay minerals and the release of ions into solution, gradually eroding the rock.

Q: What is the significance of salt hydrolysis in soil science?
A:

In soil science, salt hydrolysis is crucial for understanding soil pH and nutrient availability. Different salts in soil can hydrolyze to produce acidic or basic conditions, affecting plant growth and microbial activity. For example, the hydrolysis of aluminum salts in soil can lead to soil acidification, impacting crop productivity.

Q: How does salt hydrolysis relate to the concept of hard and soft water?
A:

Salt hydrolysis helps explain the behavior of hard and soft water. In hard water, calcium and magnesium ions can undergo hydrolysis, affecting the water's pH and its interaction with soaps and detergents. Water softeners often work by replacing these ions with sodium ions, which do not undergo significant hydrolysis.