Question:
0.2M, 500ml MnO\(_4^-\) solution is treated with 1.5M, 500ml KI solution in basic medium. The liberated \(I_2\) required ‘X’M, 300ml hypo solution. The value of ‘X’ is -
0.2M, 500ml MnO\(_4^-\) solution is treated with 1.5M, 500ml KI solution in basic medium. The liberated \(I_2\) required ‘X’M, 300ml hypo solution. The value of ‘X’ is -
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For titrations, always balance the reaction and use stoichiometry to calculate the number of moles involved. The relationship between reactants and products is key.
Updated On: Apr 7, 2026
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Correct Answer: 1.67
Solution and Explanation
Step 1: Write the balanced reaction.
The reaction between potassium permanganate (MnO\(_4^-\)) and potassium iodide (KI) in a basic medium can be written as: \[ 2MnO_4^- + 16OH^- + 10I^- \rightarrow 2Mn^{2+} + 5I_2 + 8H_2O \]
Step 2: Calculate moles of MnO\(_4^-\).
The molarity of MnO\(_4^-\) is given as 0.2M, and the volume is 500ml or 0.5L. The moles of MnO\(_4^-\) are calculated as: \[ \text{moles of MnO}_4^- = 0.2 \, \text{mol/L} \times 0.5 \, \text{L} = 0.1 \, \text{mol} \]
Step 3: Determine the moles of I\(^-\) reacted.
From the balanced equation, 2 moles of MnO\(_4^-\) react with 10 moles of I\(^-\). Therefore, the moles of I\(^-\) that reacted are: \[ \text{moles of I}^- = \frac{10}{2} \times 0.1 = 0.5 \, \text{mol} \]
Step 4: Calculate the molarity of I\(_2\).
The moles of I\(^-\) correspond to the moles of \(I_2\) formed, and the volume of \(I_2\) formed is 300ml or 0.3L. The molarity of \(I_2\) is: \[ \text{Molarity of I}_2 = \frac{\text{moles of I}_2}{\text{volume}} = \frac{0.5}{0.3} = 1.67 \, \text{M} \]
The reaction between potassium permanganate (MnO\(_4^-\)) and potassium iodide (KI) in a basic medium can be written as: \[ 2MnO_4^- + 16OH^- + 10I^- \rightarrow 2Mn^{2+} + 5I_2 + 8H_2O \]
Step 2: Calculate moles of MnO\(_4^-\).
The molarity of MnO\(_4^-\) is given as 0.2M, and the volume is 500ml or 0.5L. The moles of MnO\(_4^-\) are calculated as: \[ \text{moles of MnO}_4^- = 0.2 \, \text{mol/L} \times 0.5 \, \text{L} = 0.1 \, \text{mol} \]
Step 3: Determine the moles of I\(^-\) reacted.
From the balanced equation, 2 moles of MnO\(_4^-\) react with 10 moles of I\(^-\). Therefore, the moles of I\(^-\) that reacted are: \[ \text{moles of I}^- = \frac{10}{2} \times 0.1 = 0.5 \, \text{mol} \]
Step 4: Calculate the molarity of I\(_2\).
The moles of I\(^-\) correspond to the moles of \(I_2\) formed, and the volume of \(I_2\) formed is 300ml or 0.3L. The molarity of \(I_2\) is: \[ \text{Molarity of I}_2 = \frac{\text{moles of I}_2}{\text{volume}} = \frac{0.5}{0.3} = 1.67 \, \text{M} \]
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