Acids, Bases, and Salts: Definition, Key Features & Class 12 Notes

Acids, bases, and salts are fundamental concepts in chemistry. The definition of acids, bases and salts is as follows:

• Acids: A substance which releases hydrogen ions (H⁺) in aqueous solution and donates a proton (H⁺) to another substance or accepts a pair of electrons is called an acid. Examples: Sulfuric acid (H₂SO₄), Acetic acid (CH₃COOH), and  Hydrochloric acid (HCl).
• Bases: A substance that releases hydroxide ions (OH⁻) in aqueous solution and accepts a proton (H⁺) or donates an electron pair is called a base. Examples: Calcium hydroxide (Ca(OH)₂), Sodium hydroxide (NaOH), and Ammonia (NH₃).
• Salt: An ionic compound formed by neutralization reaction between an acid and a base. It comprises a positive ion (cation) from the base and a negative ion (anion) from the acid. Examples: Ammonium sulfate ((NH₄)₂SO₄), Sodium chloride (NaCl), Potassium nitrate (KNO₃).

Also Read:

NCERT Class12 notes	Class 12 Chemistry NCERT notes
NCERT Class 12 Maths notes	Physics Class 12 NCERT Notes

As per the NCERT Class 11 Chemistry, Arrhenius theory explains acids as a substance which increases the concentration of hydrogen ions (H⁺) in aqueous solution, such as H₂SO₄ and bases that increase hydroxide ion (OH⁻) concentration, such as NH₃.

While NCERT provides a qualitative understanding, exams like JEE Main extend this to quantitative analysis, particularly in ion concentration and pH calculations. For example, a 0.01 M solution of HCl dissociates completely to give [H⁺] = 0.01 M, so pH = –log(0.01) = 2.

Also Read: NCERT Solutions 

A typical JEE question may involve calculating the pH of a weak base, such as 0.01 M NH₃, given K_b = 1.8 × 10⁻⁵. This problem requires setting up an equilibrium expression and solving for [OH⁻], followed by calculating pOH and then pH. 

According to Brønsted-Lowry theory, acids are proton donors and bases are acceptors. This led to the concept of conjugate acid-base pairs. 
Examples: HCl + H₂O ⇌ H₃O⁺ + Cl⁻
Conjugate pairs: HCl/Cl⁻ and H₂O/H₃O⁺

In JEE Main, students have to calculate the concentration of H₃O⁺ using the Ka value of a weak acid like NH₄⁺, using the principles of chemical equilibrium.

Lewis Theory: It defines acids as electron acceptors and bases as electron donors. NCERT explains this using BF₃ (Lewis acid) and NH₃ (Lewis base).

The JEE level application might ask to classify AlCl₃ in the reaction:
AlCl₃ + Cl⁻ → AlCl₄⁻
Here, AlCl₃ acts as a Lewis acid because it accepts an electron pair from Cl⁻.

As per the NCERT definition, Neutralization reaction occurs when an acid reacts with bases from the salt and water. Example: HCl + NaOH → NaCl + H₂O.

Below are the key features of acids, bases and salts.

Features	Acids	Bases	Salts
Taste	Sour	Bitter and slippery to touch	salty
Ionization	Release H⁺ (hydrogen ions) in aqueous solution	Release OH⁻ (hydroxide ions)	-
Formation	-	-	Neutralization reaction between acids and bases
PH value	Less than 7	Greater than 7	neutral, acidic, or basic, based on the acid and base used
Reactivity	React with metal to produce hydrogen gas	React with acid to form salt water	-
Effect on Indicators	Methyl orange turns red  Blue litmus turns red	Red litmus to blue  Methyl orange turns yellow
Examples	Sulfuric acid (H₂SO₄), Acetic acid (CH₃COOH)	Sodium hydroxide (NaOH), Ammonium hydroxide (NH₄OH)

JEE Main questions test theory application, and calculations. Strategies include:

• Identify acid/base using Arrhenius, Brønsted-Lowry, or Lewis definitions.
• Calculate pH for strong acids/bases directly; use $K_a$ or $K_b$ for weak ones.
• Determine conjugate pairs in Brønsted-Lowry reactions.
• Balance neutralization reactions and calculate resulting ion concentrations.
• Practice Lewis acid-base reactions for complex molecules.

Problem 1: Calculate the pH of a 0.01 M HCl solution.

$\left[{\mathrm{H}}^{+}\right]=0.01\mathrm{M},\mathrm{}\mathrm{p}\mathrm{H}=-\mathrm{l}\mathrm{o}\mathrm{g}\left(0.01\right)=2$

Problem 2: Find the pH of $0.01\mathrm{M}{\mathrm{N}\mathrm{H}}_3\left(K_b=1.8\times {10}^{-5}\right)$ .

$\begin{array}{c}\left[{\mathrm{O}\mathrm{H}}^{-}\right]\approx \sqrt[]{K_b\cdot c}=\sqrt[]{1.8\times {10}^{-5}\times 0.01}\approx 4.24\times {10}^{-4}\mathrm{M}\\ \mathrm{p}\mathrm{O}\mathrm{H}=-\mathrm{l}\mathrm{o}\mathrm{g}\left(4.24\times {10}^{-4}\right)\approx 3.37,\mathrm{}\mathrm{p}\mathrm{H}=14-3.37=10.63\end{array}$

Problem 3: Identify the conjugate base of ${\mathrm{N}\mathrm{H}}_4{}^{+}$ in ${\mathrm{N}\mathrm{H}}_4{}^{+}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{N}\mathrm{H}}_3+$ ${\mathrm{H}}_3{\mathrm{O}}^{+}$ . Calculate $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$ for $0.1\mathrm{M}{\mathrm{N}\mathrm{H}}_4\mathrm{C}\mathrm{l}\left(K_b\right.$  for $\left.{\mathrm{N}\mathrm{H}}_3=1.8\times {10}^{-5}\right)$ . 

Conjugate base: ${\mathrm{N}\mathrm{H}}_3.K_a=\frac{K_w}{K_b}=\frac{{10}^{-14}}{1.8\times {10}^{-5}}\approx 5.56\times {10}^{-10}$ .

$\begin{array}{c}\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\approx \sqrt[]{K_a\cdot c}=\sqrt[]{5.56\times {10}^{-10}\times 0.1}\approx 7.46\times {10}^{-6}\mathrm{M}\\ \mathrm{p}\mathrm{H}=-\mathrm{l}\mathrm{o}\mathrm{g}\left(7.46\times {10}^{-6}\right)\approx 5.13\end{array}$

Problem 4: 20 mL of 0.1 M HCl reacts with 10 mL of 0.2 M NaOH. Determine the pH of the resulting solution.

Moles: $\mathrm{H}\mathrm{C}\mathrm{l}=0.1\times 0.02=0.002,\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}=0.2\times 0.01=0.002$ . Neutralization is complete, forming 0.002 mol NaCl in 30 mL.

$\left[\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l}\right]=\frac{0.002}{0.03}\approx 0.0667\mathrm{M}$

Neutral solution (since NaCl is neutral).