Skip to content

#pH calculation

53 public questions tagged with this topic.

The Ksp of Mg(OH)2 is 1.8 × 10⁻¹¹ . What is the pH at which [Mg²⁺] = 1.0 × 10⁻⁴ M in a saturated solution?

For Mg(OH)2 Mg²⁺ + 2OH- , Ksp = [Mg²⁺][OH-]² = 1.8 × 10⁻¹¹ , (1.0 × 10⁻⁴)[OH-]² = 1.8 × 10⁻¹¹ , [OH-]² = 1.8 × 10⁻⁷ , [OH-] = 1.34 × 10⁻⁴ , pOH = 3.87 , pH = 14 - 3.87 = 10.13 .

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

The Ksp of CuS is 6.3 × 10⁻³⁶ . What is the pH at which [Cu²⁺] = 1.0 × 10⁻¹² M in a saturated solution, given Ka of H₂S

For CuS Cu²⁺ + S²⁻ , Ksp = [Cu²⁺][S²⁻] = 6.3 × 10⁻³⁶ , [S²⁻] = 6.3 × 10⁻²⁴ . For H₂S 2H+ + S²⁻ , K = Ka₁ × Ka₂ = 9.5 × 10⁻²⁷ , [S²⁻] = (K [H₂S]/[H+]²) , assume [H₂S] = 0.1 M , 6.3 × 10⁻²⁴ = (9.5 × 10⁻²⁷ × 0.1/[H+]²) , [H+]² = 1.51 × 10⁻⁴ , [H+] = 1.23 × 10⁻² , pH ≈ 1.91 .

Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law

A weak acid HY ( Ka = 2.0 × 10⁻⁵ ) is mixed with 0.01 M NaOH in a 2:1 volume ratio (acid:base). If the final [HY] = 0.04

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

The ionization constant of a weak acid HA is 1.0 × 10⁻⁵ . What is the pH of a 0.1 M solution of this acid?

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