19 public questions tagged with this topic.
α = (Lambdam/Lambdam°) = (17.5/350) = 0.05 . Ka = (α² c/1 - α) = (0.05² × 0.05/1 - 0.05) = (0.0025 × 0.05/0.95) = 1.316 × 10⁻⁴ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement
α = (Lambdam/Lambdam°) = (40/400) = 0.1 . Ka = (α² c/1 - α) = (0.1² × 0.1/1 - 0.1) = (0.01 × 0.1/0.9) = 1.111 × 10⁻³ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
Total volume = 3V, [X-] = (0.06 × 2V/3V) = 0.04 M , pKa = 4.7 , pH = 4.7 + log (0.04/0.04) = 4.7 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
Kw = Ka × Kb = 1.0 × 10⁻¹⁴ , Kb = (1.0 × 10⁻¹⁴/6.2 × 10⁻¹⁰) ≈ 1.61 × 10⁻⁵ .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
For HZ H+ + Z- , Ka = (x²/0.1 - x) ≈ (x²/0.1) = 2.0 × 10⁻⁴ . Solving, x² = 2.0 × 10⁻⁵ , x = 4.47 × 10⁻³ , pH = -log(4.47 × 10⁻³) ≈ 2.35 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law
Kw = Ka × Kb = 1.0 × 10⁻¹⁴ , Kb = (1.0 × 10⁻¹⁴/1.2 × 10⁻²) = 8.33 × 10⁻¹³ .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law
Total volume = 3V, [X-] = (0.03V/3V) = 0.01 M , pKa = 5.4 , pH = 5.4 + log (0.01/0.08) = 5.4 - 0.903 = 4.5 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law
pKa = -log Ka = -log(2.5 × 10⁻⁵) = 5 - log 2.5 ≈ 5 - 0.398 = 4.6 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law
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
Kw = Ka × Kb = 1.0 × 10⁻¹⁴ , so Kb = (Kw/Ka) = (1.0 × 10⁻¹⁴/4.9 × 10⁻¹⁰) ≈ 2.04 × 10⁻⁵ .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law
For HA H+ + A- , Ka = ([H+][A-]/[HA]) = (x²/0.1 - x) ≈ (x²/0.1) = 1.0 × 10⁻⁵ . Solving, x = sqrt1.0 × 10⁻⁶ = 1.0 × 10⁻³ , so pH = -log(1.0 × 10⁻³) = 3 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
Kw = Ka × Kb = 1.0 × 10⁻¹⁴ , Kb = (1.0 × 10⁻¹⁴/1.8 × 10⁻⁵) ≈ 5.56 × 10⁻¹⁰ .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
