Q72
jee_main_2026_24_january_morning
Precipitation and pH Dependence
Consider two Group IV metal ions X²⁺ and Y²⁺. A solution containing 0.01 M X²⁺ and 0.01 M Y²⁺ is saturated with H₂S. The pH at which the metal sulphide YS will form as a precipitate is ____ (Nearest integer)
(Given : Kₛₚ(XS) = 1 × 10⁻²² at 25°C, Kₛₚ(YS) = 4 × 10⁻¹⁶ at 25°C, [H₂S] = 0.1 M in solution, Kₐ₁ × Kₐ₂(H₂S) = 1.0 × 10⁻²¹, 2 = 0.30, 3 = 0.48, 5 = 0.70)
Numerical Answer. Answer: 4 to 4
Solution
Related Formula
For precipitation: [M²⁺][S²⁻] ≥ Kₛₚ [S²⁻] = Kₐ₁ · Kₐ₂ · [H₂S][H^+]²Core Logic
For precipitation of YS(s): [Y²⁺][S²⁻] ≥ Kₛₚ(YS) [S²⁻] ≥ 4 × 10⁻¹⁶0.01 = 4 × 10⁻¹⁴ M
Since H₂S dissociates in water: H₂S leftharpoons 2H^+ + S²⁻ [S²⁻][H^+]²[H₂S] = Kₐ₁ × Kₐ₂ = 1.0 × 10⁻²¹
Substitute the [S²⁻] threshold: [S²⁻] = 1.0 × 10⁻²¹ × [H₂S][H^+]² ≥ 4 × 10⁻¹⁴
Step 1: Calculate pH
Using [H₂S] = 0.1 M: 10⁻²¹ × 0.1[H^+]² = 4 × 10⁻¹⁴ 10⁻²²[H^+]² = 4 × 10⁻¹⁴ [H^+]² = 10⁻²²4 × 10⁻¹⁴ = (1)/(4) × 10⁻⁸ = 0.25 × 10⁻⁸ = 25 × 10⁻¹⁰ [H^+] = 5 × 10⁻⁵ M
pH = - (5 × 10⁻⁵) = 5 - 5 = 5 - 0.70 = 4.3 Rounding to the nearest integer, pH ≈ 4.
Pattern Recognition
Equating the S²⁻ concentration required for Kₛₚ with the S²⁻ provided by the H₂S equilibrium directly links solubility product to pH.
Chapter Mix
Class 11 Chemistry: Equilibrium Class 12 Chemistry: Qualitative Analysis
