Question

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)

• A. 1 M, 1 M, 1 M
• B. 0.5 M, 0.5 M, 0.5 M
• C. 0.5 M, 0.5 M, 1 M
• D. 0.75 M, 0.75 M, 1 M

Hint:

Cold dilute alkali gives hypochlorite, while hot concentrated alkali gives chlorate.

Answer 1

The Correct Option is C

Let's solve this problem step-by-step. The reaction given time involves chlorine gas reacting with a cold potassium hydroxide solution.

When Cl₂(g) is passed into cold and dilute KOH solution, it leads to a disproportionation reaction. This reaction can be represented as follows:

\(Cl_2(g) + 2KOH(aq) \rightarrow KCl(aq) + KClO(aq) + H_2O(l)\)

In cold and dilute conditions, Cl₂ is converted to Cl^- and ClO^- as the main products.

Initial conditions:

• The solution is 2 M, and the volume of KOH solution is 2 L.
• Using the molarity and volume, the initial moles of KOH are: \(2 \text{ M} \times 2 \text{ L} = 4 \text{ moles}\).
• 1 mole of Cl₂ is used.

Since 1 mole of Cl₂ reacts with 2 moles of KOH, the reaction will consume 2 of the 4 moles of KOH available, leaving 2 moles unreacted.

From the reaction, for every 1 mole of Cl₂ used, 1 mole each of KCl and KClO is produced:

• Moles of Cl^- formed = 1 mole
• Moles of ClO^- formed = 1 mole

Since the volume of the solution is 2 L, we calculate the concentrations as follows:

• Concentration of Cl^- = \(\frac{1 \text{ mole}}{2 \text{ L}} = 0.5 \text{ M}\)
• Concentration of ClO^- = \(\frac{1 \text{ mole}}{2 \text{ L}} = 0.5 \text{ M}\)
• Concentration of remaining OH^- ions = \(\frac{2 \text{ moles}}{2 \text{ L}} = 1 \text{ M}\)

Therefore, the concentrations of Cl^-, ClO^-, and OH^- are 0.5 M, 0.5 M, and 1 M, respectively.

The correct answer is: 0.5 M, 0.5 M, 1 M