Question
One mole of Cl _ 2 ( ~g ) was passed into 2 L of cold 2 M KOH solution. After the reaction, the concentrations of Cl ^ - , ClO ^ - and OH ^ - are respectively (assume volume remains constant)
JEE Main
Chemistry
Redox Reactions
2026
JEE Main 2026 (24 January Shift 2)
JEE Main Chemistry Question (2026) — Solution
Options
- A. 0.5 M , 0.5 M , 1 M
- B. 0.75 M , 0.75 M , 1 M
- C. 0.5 M , 0.5 M , 0.5 M
- D. 1 M , 1 M , 1 M
Answer
A. 0.5 M , 0.5 M , 1 M
Step-by-step solution
Cl₂ undergoes disproportionation in cold KOH: Cl_2 + 2OH^- Cl^- + ClO^- + H_2O Initial: 1 mole Cl₂, 2 L × 2 M KOH = 4 moles OH⁻ Molar ratio Cl₂:OH⁻ = 1:2 (stoichiometric) Products formed: 1 mole Cl⁻, 1 mole ClO⁻; Excess OH⁻ = 4 - 2 = 2 moles In 2 L solution: [Cl⁻] = 0.5 M, [ClO⁻] = 0.5 M, [OH⁻] = 1 M
Related: Chemistry — Redox Reactions · All PYQ Banks