20 public questions tagged with this topic.
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
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
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
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
Total volume = 3V, [HX] = (0.1V/3V) = 0.0333 M , [X-] = (0.05 × 2V/3V) = 0.0333 M , pH = 4 + log (0.0333/0.0333) = 4 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
[NH₄+] = 0.01 M , pKb = 4.74 , pOH = 4.74 + log (0.01/0.05) = 4.74 - 0.7 = 4.04 , pH = 14 - 4.04 = 9.96 .
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
Total volume = 4V, HZ initial = 0.06 × 4V = 0.08 M × 3V , NaOH = 0.02V , [Z-] = (0.02V/4V) = 0.005 M . pH = 4.5 + log (0.005/0.06) = 4.5 - 1.08 = 3.42 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
Total volume = 3V, [X-] = (0.04V/3V) = 0.0133 M , pKa = 4.5 , pH = 4.5 + log (0.0133/0.06) = 4.5 - 0.655 = 4.85 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant
After mixing: [HX] = (0.2/2) = 0.1 M , [X-] = (0.1/2) = 0.05 M . pH = pKa + log ([X-]/[HX]) = 5 + log (0.05/0.1) = 5 - 0.301 = 4.7 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant
[Y-] = 0.1 M , pKa = 4 , pH = pKa + log ([Y-]/[HY]) = 4 + log (0.1/0.2) = 4 + log 0.5 = 4 - 0.301 = 3.7 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Acid-Base Theories - Arrhenius Bronsted-Lowry Lewis and Salts Hydrolysis
[NH₄+] = 0.05 M , pKb = 4.74 , pOH = 4.74 + log (0.05/0.05) = 4.74 , pH = 14 - 4.74 = 9.26 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Acid-Base Theories - Arrhenius Bronsted-Lowry Lewis and Salts Hydrolysis
