JEE Main202624 January 2026Evening ShiftChemistryRedox ReactionsActual
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)
Options
- A0.5 M , 0.5 M , 1 M
- B0.75 M , 0.75 M , 1 M
- C0.5 M , 0.5 M , 0.5 M
- D1 M , 1 M , 1 M
Correct answer
A. 0.5 M , 0.5 M , 1 M
Step-by-step solution
Cl₂ undergoes disproportionation in cold KOH: Cl₂ + 2OH^- Cl^- + ClO^- + H₂O 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