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Equilibrium appeared 36 times across 3 years — 4.2% of Chemistry. This question is from Ionic Equilibrium and Polyprotic Acids.

Year 2026 2025 2024 Total
Questions 10 18 8 36

The first and second ionization constants of H₂X are 2.5 × 10⁻⁸ and 1.0 × 10⁻¹³ respectively. The concentration of X²⁻ in 0.1 M H₂X solution is _ × 10⁻¹⁵ M. (Nearest Integer)

Numerical Answer Type:
Enter a numerical value Answer: 100 to 100 +4 marks

Solution & Explanation

Core Logic

For a weak diprotic acid H₂X, the concentration of X²⁻ in a solution of concentration C is approximately equal to the second dissociation constant Ka₂ under standard weak acid approximation conditions ([X²⁻] ≈ Ka₂ = 1.0 × 10⁻¹³ M).

Converting to the required format: 1.0 × 10⁻¹³ = 100 × 10⁻¹⁵ M.

Step 1: Final Calculation

Thus, the integer value is 100.

Pattern Recognition

Sees: polyprotic acid dissociation equilibrium calculation for secondary anion concentration. Trap: Assuming complete dissociation for the second step.

Chapter Mix

Class 11 Chemistry: Equilibrium

Reference Study Guides

More Equilibrium Previous-Year Questions — Page 8

Q62 jee_main_2024_31_jan_morning Equilibrium Constant
For the given reaction, choose the correct expression of KC from the following :- Fe(aq)³⁺ + SCN(aq)⁻ leftharpoons (FeSCN)(aq)²⁺
  • A. KC = [FeSCN²⁺][Fe³⁺][SCN⁻]
  • B. KC = [Fe³⁺][SCN⁻][FeSCN²⁺]
  • C. KC = [FeSCN²⁺][Fe³⁺]²[SCN⁻]²
  • D. KC = [FeSCN²⁺]²[Fe³⁺][SCN⁻]

Solution

Related Formula
KC = [Products][Reactants]
Core Logic
KC = Products ion conc.Reactants ion conc. KC = [FeSCN²⁺][Fe³⁺][SCN⁻]
Chapter Mix

Class 11 Chemistry: Equilibrium

More Equilibrium Questions — jee_main_2026_21_jan_evening

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