Question

For the reaction:
\[ Cl_2 + KOH \rightarrow KCl + KClO + H_2O \] 1 mole of Cl$_2$ is passed into 2 litre, 2 M KOH solution. Determine the molarity of Cl$^-$, ClO$^-$ and OH$^-$ respectively.

• A. 1 M, 0.5 M, 0.5 M
• B. 0.5 M, 0.5 M, 1 M
• C. 1 M, 1 M, 0.5 M
• D. 0.5 M, 1 M, 0.5 M

Hint:

Always check whether the reaction occurs in hot or cold alkali — products change accordingly.

Answer 1

The Correct Option is B

Step 1: Write the balanced chemical equation

Cl₂ + 2KOH → KCl + KClO + H₂O

From the equation:
1 mole of Cl₂ produces 1 mole of Cl⁻ and 1 mole of ClO⁻.

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Step 2: Calculate initial moles of KOH

Volume of KOH solution = 2 L
Molarity of KOH = 2 M

Initial moles of KOH = 2 × 2 = 4 moles

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Step 3: Use stoichiometry of the reaction

According to the reaction:

2 moles of KOH react with 1 mole of Cl₂

Given 1 mole of Cl₂ is passed,

KOH consumed = 2 moles

Remaining KOH = 4 − 2 = 2 moles

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Step 4: Calculate molarity of remaining OH⁻

Remaining OH⁻ ions = 2 moles

Molarity of OH⁻ = 2 / 2 = 1 M

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Step 5: Calculate molarity of Cl⁻ and ClO⁻

Moles of Cl⁻ formed = 1 mole
Moles of ClO⁻ formed = 1 mole

Total volume of solution = 2 L

Molarity of Cl⁻ = 1 / 2 = 0.5 M

Molarity of ClO⁻ = 1 / 2 = 0.5 M

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Final Answer:

Molarity of Cl⁻ = 0.5 M
Molarity of ClO⁻ = 0.5 M
Molarity of OH⁻ = 1 M