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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
Initial: PN₂ = (1 × 0.0831 × 600/2) = 24.93 bar , PH₂ = 74.79 bar . Let 2x be PNH₃ , PN₂ = 24.93 - x , PH₂ = 74.79 - 3x . Kp = ((PNH₃)²/PN₂ (PH₂)³) = ((2x)²/(24.93 - x)(74.79 - 3x)³) = 4.0 × 10⁻³ . Solving, x ≈ 0.8 , PNH₃ = 2 × 0.8 = 1.6 bar .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
Initial: [X₂] = (0.4/2) = 0.2 M . Let α be the degree of dissociation, [X₂] = 0.2(1 - α) , [X] = 0.2 × 2α = 0.4α . Kc = ([X]²/[X₂]) = ((0.4α)²/0.2(1 - α)) = (0.16α²/0.2(1 - α)) = 0.25 , 0.8α² = 0.25(1 - α) , 0.8α² + 0.25α - 0.25 = 0 , α ≈ 0.5 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
α = 0.02 , [OH-] = 0.05 × 0.02 = 1.0 × 10⁻³ , pOH = 3 , pH = 14 - 3 = 11 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
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
Kw = Ka × Kb = 1.0 × 10⁻¹⁴ , Kb = (Kw/Ka) = (1.0 × 10⁻¹⁴/6.8 × 10⁻⁴) ≈ 1.47 × 10⁻¹¹ .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
Total pressure = PA + PB + PC = 4 , PB + PC = 3 . Kp = ((PC)²/PA (PB)²) = ((PC)²/1 · (3 - PC)²) = 0.25 , PC = 0.5 (3 - PC) , PC = 1.5 - 0.5 PC , 1.5 PC = 1.5 , PC = 1 atm .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
Let [HA] = C , degree of ionization α = 0.02 , [H+] = C α = 0.02C . Ka = ([H+][A-]/[HA]) = ((0.02C)²/C(1 - 0.02)) ≈ (0.0004C²/0.98C) = 4.08 × 10⁻⁴C = 1.0 × 10⁻⁵ . Solving, C = (1.0 × 10⁻⁵/4.08 × 10⁻⁴) ≈ 0.0245 M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
Let 2x be PAB , PA₂ = 1 - x , PB₂ = 1 - x , total pressure = (1 - x) + (1 - x) + 2x = 2 . Kp = ((PAB)²/PA₂ PB₂) = ((2x)²/(1 - x)²) = 9 , (4x²/(1 - x)²) = 9 , (2x/1 - x) = 3 , 2x = 3 - 3x , 5x = 3 , x = 0.6 , PAB = 2 × 0.6 = 1.2 atm .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
pH = 10.8 , pOH = 14 - 10.8 = 3.2 , [OH-] = 10⁻³.² ≈ 6.31 × 10⁻⁴ . [BOH] = 0.02 - 6.31 × 10⁻⁴ ≈ 0.0194 . Kb = ([OH-]²/[BOH]) = ((6.31 × 10⁻⁴)²/0.0194) ≈ 2.05 × 10⁻⁵ , but given Kb = 4.0 × 10⁻⁴ , ionization α = ([OH-]/[BOH]initial) = (6.31 × 10⁻⁴/0.02) = 0.03155 , % = 3.155%.
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
pOH = 14 - 10.5 = 3.5 , [OH-] = 10⁻³.⁵ ≈ 3.16 × 10⁻⁴ , α = ([OH-]/[BOH]) = (3.16 × 10⁻⁴/0.1) = 3.16 × 10⁻³ = 0.00316 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
Let PA₂ = PB₂ = x , PAB = 4 - 2x , total pressure = 4 - 2x + 2x = 4 . Kp = (PA₂ PB₂/(PAB)²) = (x²/(4 - 2x)²) = 0.25 , (x/4 - 2x) = 0.5 , x = 2 - x , 2x = 2 , x = 1 atm .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
