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