0.4 moles of Cl ₂( g ) is passed into 200 mL of a cold NaOH solution of unknown initial molarity. After the disproportionation reaction is complete, it is found that the final concentration of unreacted OH ⁻ ions is equal to the final concentration of Cl ⁻ ions formed. Assuming the volume of the solution remains constant, the initial molarity of the NaOH solution is

Step-by-step solution

In cold and dilute alkali, chlorine undergoes disproportionation as follows: Cl ₂ + 2 OH ⁻ Cl ⁻ + ClO ⁻ + H ₂ O Let the initial moles of NaOH be n . Moles of Cl ₂ reacted = 0.4 mol From the stoichiometry, 0.4 moles of Cl ₂ consume 0.8 moles of OH ⁻ and produce 0.4 moles of Cl ⁻ . Moles of unreacted OH ⁻ = n - 0.8 Given that the final concentration of OH ⁻ equals the final concentration of Cl ⁻ , and the volume is constant, their moles must be equal: n - 0.8 = 0.4 n = 1.2 mol The initial volume of the solution is 20