14 public questions tagged with this topic.
Phenol is more acidic due to resonance stabilization of the phenoxide ion, allowing it to form a soluble salt with NaOH, unlike ethanol, which is a weaker acid.
Ref: NCERT Class 12 Chemistry > Chapter 7: Alcohols, Phenols and Ethers > Topic: Physical Properties and Acidity of Alcohols and Phenols
In Kolbe’s reaction, phenol reacts with CO₂ and NaOH under heat and pressure, followed by acidification, to form salicylic acid.
Ref: NCERT Class 12 Chemistry > Chapter 7: Alcohols, Phenols and Ethers > Topic: Chemical Reactions of Phenols and Ethers and Uses
In the Reimer-Tiemann reaction, phenol reacts with CHCl₃ and NaOH to form salicylaldehyde (2-hydroxybenzaldehyde) after acidification.
Ref: NCERT Class 12 Chemistry > Chapter 7: Alcohols, Phenols and Ethers > Topic: Chemical Reactions of Phenols and Ethers and Uses
This is the Reimer-Tiemann reaction, where phenol reacts with CHCl₃ and NaOH to form salicylaldehyde (2-hydroxybenzaldehyde) via ortho substitution.
Ref: NCERT Class 12 Chemistry > Chapter 7: Alcohols, Phenols and Ethers > Topic: Chemical Reactions of Phenols and Ethers and Uses
SN₂ substitution with NaOH favors primary halides. CH₃CH₂I (iodide) reacts fastest due to the best leaving group.
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Nature of C-X Bond and Methods of Preparation
Let volumes be 2V and V, total volume = 3V. Moles: HY = 0.04 × 3V , initial [HY] = 0.06 M , moles NaOH = 0.01V , [Y-] = (0.01V/3V) = 0.00333 M , remaining [HY] = 0.04 . pH = 4.7 + log (0.00333/0.04) = 4.7 - 1.08 = 3.62 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law
HB + OH- → B- + H₂O , moles: HB = 0.05 , OH- = 0.025 , after reaction: [HB] = 0.025 M , [B-] = 0.025 M . Buffer: pH = pKa + log ([B-]/[HB]) = -log(4.0 × 10⁻⁶) + log (0.025/0.025) = 5.4 + 0 = 5.4 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
Total volume = 3V, initial [HA] = 0.09 M , [A-] = (0.04V/3V) = 0.0133 M , pKa = 5.3 , pH = 5.3 + log (0.0133/0.06) = 5.3 - 0.655 = 4.65 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
For NaOH , [OH-] = 0.001 M = 10⁻³ M , so pOH = -log(10⁻³) = 3 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Acid-Base Theories - Arrhenius Bronsted-Lowry Lewis and Salts Hydrolysis
Reaction: NaOH + HCl → NaCl + H₂O. Moles of NaCl = 11.7/58.5 = 0.2 mol. 1 mol NaCl from 1 mol NaOH; mass = 0.2 × 40 = 8 g.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Reactions in Solutions and Mass Percentage and Numericals
Reaction: 2 NaOH + H₂SO₄ → Na₂SO₄ + 2 H₂O. Moles of NaOH = 14.2 / 40 = 0.355 mol. 2 mol NaOH produce 1 mol Na₂SO₄; 0.355 mol produce 0.1775 mol. Mass = 0.1775 × 142 ≈ 25.2 g.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Mole Concept and Molar Masses and Percentage Composition
M₁V₁ = M₂V₂. 0.5 × 200 = M₂ × 1000. M₂ = (0.5 × 200)/1000 = 0.1 M.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Importance and Nature of Chemistry and Properties of Matter
