The Aluminium Hydroxide Formula is written as Al(OH)3, representing a compound where one aluminium ion bonds with three hydroxide groups. This formula is a key topic in NCERT Class 10 and Class 12 Chemistry, appearing in chapters on metals, non-metals, and p-block elements. It is equally relevant for NEET and JEE Main aspirants who must understand amphoteric oxides and hydroxides. This article covers the formula expression, molar mass, physical and chemical properties, a complete formula sheet, solved examples, CBSE exam tips, and common mistakes to avoid.
Key Aluminium Hydroxide Formulas at a Glance
Quick reference for the most important formulas related to Aluminium Hydroxide.
- Molecular formula: \( \text{Al(OH)}_3 \)
- Molar mass: \( M = 78.00 \text{ g/mol} \)
- Reaction with acid (acidic behaviour): \( \text{Al(OH)}_3 + 3\text{HCl} \rightarrow \text{AlCl}_3 + 3\text{H}_2\text{O} \)
- Reaction with base (basic behaviour): \( \text{Al(OH)}_3 + \text{NaOH} \rightarrow \text{NaAlO}_2 + 2\text{H}_2\text{O} \)
- Thermal decomposition: \( 2\text{Al(OH)}_3 \xrightarrow{\Delta} \text{Al}_2\text{O}_3 + 3\text{H}_2\text{O} \)
- Formation from salt solution: \( \text{AlCl}_3 + 3\text{NaOH} \rightarrow \text{Al(OH)}_3 + 3\text{NaCl} \)
What is Aluminium Hydroxide Formula?
The Aluminium Hydroxide Formula, Al(OH)3, represents the chemical compound formed by the combination of aluminium (Al) and three hydroxide (OH−) ions. It is also called aluminium(III) hydroxide or hydrated alumina. In nature, it occurs as the mineral gibbsite (also known as hydrargillite). Other naturally occurring polymorphs include bayerite, doyleite, and nordstrandite.
Aluminium Hydroxide is an amphoteric compound. This means it can react with both acids and bases. This dual nature makes it a very important compound in NCERT Class 10 Chemistry (Chapter 3 — Metals and Non-metals) and Class 12 Chemistry (Chapter 6 — General Principles and Processes of Isolation of Elements, and Chapter 8 — The p-Block Elements).
It appears as an amorphous white powder or gelatinous solid. Its molar mass is approximately 78.00 g/mol. It is practically insoluble in water but dissolves readily in both strong acids and strong alkalis. This property is central to many CBSE board exam questions and competitive exam problems.
Aluminium Hydroxide Formula — Expression and Variables
\[ \text{Al(OH)}_3 \]
The formula tells us that one aluminium atom is bonded to three hydroxide groups. The oxidation state of aluminium in this compound is +3, and each hydroxide ion carries a charge of −1, making the compound electrically neutral overall.
| Symbol / Group | Quantity / Ion | Charge / Unit |
|---|---|---|
| Al | Aluminium atom | +3 (oxidation state) |
| OH | Hydroxide ion | −1 each |
| Al(OH)3 | Aluminium Hydroxide molecule | Neutral (overall) |
| M | Molar mass | 78.00 g/mol |
| Density | Physical property | 2.42 g/cm³ |
Key Physical Properties
| Property | Value |
|---|---|
| Molecular Formula | Al(OH)3 |
| Molar Mass | 78.00 g/mol |
| Appearance | White amorphous powder or gelatinous solid |
| Density | 2.42 g/cm³ |
| Melting Point | 300 °C (decomposes) |
| Solubility in Water | Practically insoluble (0.0001 g/100 mL) |
| Nature | Amphoteric |
Derivation of Molar Mass
We can calculate the molar mass of Al(OH)3 step by step using atomic masses from the periodic table.
Step 1: Atomic mass of Al = 27 g/mol
Step 2: Atomic mass of O = 16 g/mol; atomic mass of H = 1 g/mol. So each OH group = 16 + 1 = 17 g/mol.
Step 3: Three OH groups = 3 × 17 = 51 g/mol.
Step 4: Total molar mass = 27 + 51 = 78 g/mol.
\[ M_{\text{Al(OH)}_3} = 27 + 3 \times (16 + 1) = 27 + 51 = 78 \text{ g/mol} \]
Complete Chemistry Formula Sheet — Aluminium Compounds & Related Reactions
| Formula / Reaction Name | Expression | Variables | SI Units / Notes | NCERT Chapter |
|---|---|---|---|---|
| Aluminium Hydroxide Formula | \( \text{Al(OH)}_3 \) | Al = aluminium, OH = hydroxide | Molar mass = 78 g/mol | Class 10, Ch 3; Class 12, Ch 8 |
| Reaction with HCl (acidic) | \( \text{Al(OH)}_3 + 3\text{HCl} \rightarrow \text{AlCl}_3 + 3\text{H}_2\text{O} \) | HCl = hydrochloric acid, AlCl3 = aluminium chloride | Neutralisation reaction | Class 10, Ch 3 |
| Reaction with NaOH (basic) | \( \text{Al(OH)}_3 + \text{NaOH} \rightarrow \text{NaAlO}_2 + 2\text{H}_2\text{O} \) | NaOH = sodium hydroxide, NaAlO2 = sodium aluminate | Amphoteric behaviour | Class 10, Ch 3; Class 12, Ch 8 |
| Thermal Decomposition | \( 2\text{Al(OH)}_3 \xrightarrow{\Delta} \text{Al}_2\text{O}_3 + 3\text{H}_2\text{O} \) | Al2O3 = aluminium oxide (alumina) | Occurs above 300 °C | Class 12, Ch 6 |
| Preparation from AlCl3 | \( \text{AlCl}_3 + 3\text{NaOH} \rightarrow \text{Al(OH)}_3 + 3\text{NaCl} \) | AlCl3 = aluminium chloride, NaCl = sodium chloride | Double displacement | Class 10, Ch 3 |
| Preparation from Al2(SO4)3 | \( \text{Al}_2(\text{SO}_4)_3 + 6\text{NaOH} \rightarrow 2\text{Al(OH)}_3 + 3\text{Na}_2\text{SO}_4 \) | Al2(SO4)3 = aluminium sulphate | Double displacement | Class 10, Ch 3 |
| Aluminium Oxide Formula | \( \text{Al}_2\text{O}_3 \) | Al = aluminium, O = oxygen | Molar mass = 102 g/mol | Class 12, Ch 8 |
| Aluminium Chloride Formula | \( \text{AlCl}_3 \) | Al = aluminium, Cl = chlorine | Molar mass = 133.5 g/mol | Class 12, Ch 8 |
| Sodium Aluminate Formula | \( \text{NaAlO}_2 \) | Na = sodium, Al = aluminium, O = oxygen | Molar mass = 81.97 g/mol | Class 12, Ch 8 |
| Alum Formula | \( \text{KAl(SO}_4)_2 \cdot 12\text{H}_2\text{O} \) | K = potassium, Al = aluminium, SO4 = sulphate | Molar mass = 474.39 g/mol | Class 12, Ch 8 |
Aluminium Hydroxide Formula — Solved Examples
Example 1 (Class 9-10 Level) — Calculating Molar Mass
Problem: Calculate the molar mass of Aluminium Hydroxide, Al(OH)3, using standard atomic masses.
Given: Atomic mass of Al = 27 u, O = 16 u, H = 1 u
Step 1: Write the formula: \( \text{Al(OH)}_3 \). Count atoms: 1 Al, 3 O, 3 H.
Step 2: Mass contribution of Al = 1 × 27 = 27 g/mol.
Step 3: Mass contribution of O = 3 × 16 = 48 g/mol.
Step 4: Mass contribution of H = 3 × 1 = 3 g/mol.
Step 5: Total molar mass = 27 + 48 + 3 = 78 g/mol.
\[ M = 27 + 48 + 3 = 78 \text{ g/mol} \]
Answer
The molar mass of Aluminium Hydroxide [Al(OH)3] = 78 g/mol.
Example 2 (Class 11-12 Level) — Amphoteric Reaction with Excess NaOH
Problem: 7.8 g of Aluminium Hydroxide is dissolved in excess NaOH solution. Write the balanced equation and calculate the moles of sodium aluminate (NaAlO2) produced.
Given: Mass of Al(OH)3 = 7.8 g; Molar mass of Al(OH)3 = 78 g/mol.
Step 1: Write the balanced equation for the reaction of Al(OH)3 with NaOH:
\[ \text{Al(OH)}_3 + \text{NaOH} \rightarrow \text{NaAlO}_2 + 2\text{H}_2\text{O} \]
Step 2: Calculate moles of Al(OH)3:
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{7.8}{78} = 0.1 \text{ mol} \]
Step 3: From the balanced equation, 1 mol Al(OH)3 produces 1 mol NaAlO2. Therefore, 0.1 mol Al(OH)3 produces 0.1 mol NaAlO2.
Step 4: Molar mass of NaAlO2 = 23 + 27 + 32 = 82 g/mol. Mass of NaAlO2 = 0.1 × 82 = 8.2 g.
Answer
Moles of NaAlO2 produced = 0.1 mol; Mass of NaAlO2 = 8.2 g.
Example 3 (JEE/NEET Level) — Percentage Composition and Amphoteric Behaviour
Problem: (a) Calculate the percentage by mass of aluminium in Al(OH)3. (b) A student adds excess dilute H2SO4 to 15.6 g of Al(OH)3. Calculate the mass of Al2(SO4)3 formed. [Molar mass of Al2(SO4)3 = 342 g/mol]
Given: Mass of Al(OH)3 = 15.6 g; M[Al(OH)3] = 78 g/mol; M[Al2(SO4)3] = 342 g/mol.
Part (a) — Percentage of Al:
Step 1: Mass of Al in one mole of Al(OH)3 = 27 g/mol.
\[ \% \text{Al} = \frac{27}{78} \times 100 = 34.62\% \]
Part (b) — Mass of Al2(SO4)3:
Step 2: Write the balanced equation:
\[ 2\text{Al(OH)}_3 + 3\text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 6\text{H}_2\text{O} \]
Step 3: Moles of Al(OH)3 = 15.6 / 78 = 0.2 mol.
Step 4: From the equation, 2 mol Al(OH)3 gives 1 mol Al2(SO4)3. So 0.2 mol Al(OH)3 gives 0.2/2 = 0.1 mol Al2(SO4)3.
Step 5: Mass = 0.1 × 342 = 34.2 g.
Answer
(a) Percentage of Al in Al(OH)3 = 34.62%. (b) Mass of Al2(SO4)3 formed = 34.2 g.
CBSE Exam Tips 2025-26
- Memorise the formula correctly: Always write Al(OH)3 with the hydroxide group in brackets. Writing AlOH3 is a common error that costs marks in CBSE board exams 2025-26.
- Understand amphoteric nature: The CBSE board frequently asks students to write both the acidic and basic reactions of Al(OH)3. Practise both equations until they are automatic.
- Molar mass calculation: In stoichiometry questions, always verify your molar mass. The value 78 g/mol is standard. Double-check by counting 1 Al + 3 O + 3 H = 27 + 48 + 3 = 78.
- Thermal decomposition: Remember that Al(OH)3 decomposes on heating to give Al2O3 and water. This reaction is asked in the context of the Bayer process in Class 12.
- Applications in MCQs: We recommend noting that Al(OH)3 is used as an antacid, a fire retardant, and in water purification. These application-based questions appear in CBSE 2025-26 one-mark and two-mark sections.
- Link to p-Block elements: In Class 12 Chapter 8, Al(OH)3 is discussed under Group 13 elements. Connect its amphoteric nature to the general trend of hydroxide behaviour across periods.
Common Mistakes to Avoid
- Incorrect formula writing: Many students write Al(OH)2 or AlOH3 instead of the correct Al(OH)3. Always remember aluminium has a +3 valency, so it needs three OH− groups.
- Confusing acidic and basic reactions: Students often mix up the products. When Al(OH)3 reacts with an acid, it forms a salt and water. When it reacts with a base (NaOH), it forms sodium aluminate (NaAlO2), not aluminium oxide.
- Wrong molar mass: Some students use 79 g/mol or 76 g/mol. The correct value is exactly 78 g/mol. Verify it each time using atomic masses: Al(27) + 3×O(16) + 3×H(1) = 78.
- Forgetting to balance equations: In the reaction with H2SO4, students often write 1:1 ratio. The correct balanced equation requires 2 mol Al(OH)3 for every 3 mol H2SO4.
- Ignoring the gelatinous nature: In descriptive questions, students forget to mention that freshly precipitated Al(OH)3 is gelatinous. This property is important for its use in water treatment and is asked in CBSE practical-based questions.
JEE/NEET Application of Aluminium Hydroxide Formula
In our experience, JEE aspirants frequently encounter Aluminium Hydroxide in questions on stoichiometry, amphoteric behaviour, and the chemistry of p-block elements. NEET aspirants see it in the context of antacids and biological applications. Here are the key application patterns you must master.
Application Pattern 1 — Amphoteric Nature Questions
JEE Main and NEET regularly test whether students can identify Al(OH)3 as amphoteric. A typical question asks which of the given hydroxides reacts with both HCl and NaOH. The answer is Al(OH)3. You must be able to write both reactions quickly and correctly.
\[ \text{Al(OH)}_3 + 3\text{HCl} \rightarrow \text{AlCl}_3 + 3\text{H}_2\text{O} \quad (\text{acts as base}) \]
\[ \text{Al(OH)}_3 + \text{NaOH} \rightarrow \text{NaAlO}_2 + 2\text{H}_2\text{O} \quad (\text{acts as acid}) \]
Application Pattern 2 — Stoichiometry and Mole Concept
JEE Main frequently uses Al(OH)3 in mole concept problems. Students are given a mass of Al(OH)3 and asked to find the volume of gas produced, the mass of product formed, or the number of molecules present. Always start by calculating moles using \( n = \frac{m}{M} = \frac{m}{78} \).
Application Pattern 3 — Industrial and Medical Applications (NEET)
NEET Biology and Chemistry questions sometimes reference Al(OH)3 as an antacid (used to neutralise excess stomach acid, HCl). The relevant reaction is:
\[ \text{Al(OH)}_3 + 3\text{HCl} \rightarrow \text{AlCl}_3 + 3\text{H}_2\text{O} \]
Additionally, Al(OH)3 is used as a fire retardant in polymer composites. When heated, it releases water vapour, which cools the burning material. This thermal decomposition reaction is tested in JEE Advanced problem sets on energetics and stoichiometry. Our experts suggest practising at least five numerical problems on this reaction before your exam.
FAQs on Aluminium Hydroxide Formula
Explore More Chemistry Formulas
Understanding the Aluminium Hydroxide Formula is just one step in mastering inorganic chemistry. We recommend exploring our related formula guides to build a complete picture. Visit our Complete Chemistry Formulas hub for a full list of NCERT-aligned formulas. You can also study the Silicon Dioxide Formula to understand another important oxide from the same period, or review the Bleaching Powder Formula for more practice with inorganic compound formulas. For broader stoichiometry practice, our Ammonia Formula guide provides excellent worked examples. All these topics are covered in the official NCERT syllabus, which you can verify at ncert.nic.in.