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

Step 1: Writing the reaction.
Cold and dilute KOH reacts with chlorine as:
\[ \text{Cl}_2 + 2\text{OH}^- \rightarrow \text{Cl}^- + \text{ClO}^- + \text{H}_2\text{O} \]
Step 2: Calculating moles.
Initial moles of KOH = $2 \times 2 = 4$ moles
1 mole of Cl$_2$ consumes 2 moles of OH$^-$, leaving 2 moles OH$^-$.
Step 3: Final concentrations.
Moles of Cl$^-$ = 1, ClO$^-$ = 1, OH$^-$ = 2
Volume = 2 L
\[ [\text{Cl}^-] = 0.5\,\text{M},\; [\text{ClO}^-] = 0.5\,\text{M},\; [\text{OH}^-] = 1\,\text{M} \]