#weak acid

Variation of Conductivity with Concentration and Measurement

A weak acid has Lambdam° = 350 S cm² mol⁻¹ and Lambdam = 17.5 S cm² mol⁻¹ at 0.05 M. What is the dissociation constant K

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

View Details

Corrosion and Applications of Electrochemistry

A weak acid has Lambdam° = 380 S cm² mol⁻¹ and Lambdam = 19 S cm² mol⁻¹ at 0.02 M. What is the dissociation constant Ka

α = (Lambdam/Lambdam°) = (19/380) = 0.05 . Ka = (α² c/1 - α) = (0.05² × 0.02/1 - 0.05) = (0.0025 × 0.02/0.95) = 5.263 × 10⁻⁵ .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry

View Details

Corrosion and Applications of Electrochemistry

A weak acid has Lambdam° = 380 S cm² mol⁻¹ and Lambdam = 19 S cm² mol⁻¹ at 0.02 M. What is the dissociation constant Ka

α = (Lambdam/Lambdam°) = (19/380) = 0.05 . Ka = (α² c/1 - α) = (0.05² × 0.02/1 - 0.05) = (0.0025 × 0.02/0.95) = 5.263 × 10⁻⁵ .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry

View Details

Conductance - Conductivity Molar Conductivity and Kohlrausch Law

The degree of dissociation of a weak acid is 0.02, and its molar conductivity at infinite dilution is 400 S cm² mol⁻¹. W

α = (Lambdam/Lambdam°) , 0.02 = (Lambdam/400) , Lambdam = 8 S cm² mol⁻¹ .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law

View Details

Conductance - Conductivity Molar Conductivity and Kohlrausch Law

The degree of dissociation of a weak acid is 0.02, and its molar conductivity at infinite dilution is 400 S cm² mol⁻¹. W

α = (Lambdam/Lambdam°) , 0.02 = (Lambdam/400) , Lambdam = 8 S cm² mol⁻¹ .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law

View Details

Conductance - Conductivity Molar Conductivity and Kohlrausch Law

The degree of dissociation of a weak acid is 0.1, and its molar conductivity at infinite dilution is 380 S cm² mol⁻¹. Wh

α = (Lambdam/Lambdam°) , 0.1 = (Lambdam/380) , Lambdam = 38 S cm² mol⁻¹ .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law

View Details

Conductance - Conductivity Molar Conductivity and Kohlrausch Law

The degree of dissociation of a weak acid is 0.1, and its molar conductivity at infinite dilution is 380 S cm² mol⁻¹. Wh

α = (Lambdam/Lambdam°) , 0.1 = (Lambdam/380) , Lambdam = 38 S cm² mol⁻¹ .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law

View Details

Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier

A weak acid HX ( Ka = 2.0 × 10⁻⁵ ) is mixed with 0.06 M NaX in a 1:2 volume ratio (acid:salt). If [HX] = 0.04 M after mi

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

View Details

Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier

A weak acid HY ( Ka = 1.0 × 10⁻⁵ ) has a 0.2 M solution mixed with 0.1 M NaY in a 1:1 volume ratio. What is the pH?

After mixing: [HY] = 0.1 M , [Y-] = 0.05 M , pKa = 5 , pH = 5 + log (0.05/0.1) = 5 - 0.301 = 4.7 .

Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier

View Details

Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law

A weak acid HZ has Ka = 2.0 × 10⁻⁴ . What is the pH of a 0.1 M solution of HZ ?

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

View Details

Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law

A weak acid HX ( Ka = 4.0 × 10⁻⁶ ) is mixed with 0.03 M NaX in a 2:1 volume ratio (acid:salt). If [HX] = 0.08 M after mi

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

View Details

Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law

The Ka of a weak acid HX is 2.5 × 10⁻⁵ . What is its pKa ?

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

View Details