One mole of Cl₂ (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)

One mole of Cl2 (g) was passed into 2 L of cold 2 M KOH solution. After the reaction, the concentrations of Cl−, ClO− and OH− are respectively

Q. One mole of Cl₂ (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)

(A) 0.5 M, 0.5 M, 0.5 M

(B) 1 M, 1 M, 1 M

(D) 0.75 M, 0.75 M, 1 M

Correct Answer: 0.5 M, 0.5 M, 1 M

Explanation

Chlorine reacts with cold and dilute KOH to give chloride and hypochlorite ions.

Cl₂ + 2KOH → KCl + KClO + H₂O

From the balanced equation:

1 mole Cl₂ consumes 2 moles KOH 1 mole Cl₂ produces 1 mole Cl⁻ and 1 mole ClO⁻

Initial moles of KOH:

Moles = M × V = 2 × 2 = 4 moles

Moles of KOH consumed = 2 moles

Remaining moles of KOH:

4 − 2 = 2 moles

Now calculate concentrations (volume = 2 L):

[Cl⁻] = 1 / 2 = 0.5 M [ClO⁻] = 1 / 2 = 0.5 M [OH⁻] = 2 / 2 = 1 M

Hence, the concentrations are 0.5 M, 0.5 M and 1 M.

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