Question

A solution is prepared by reacting \(500\) mL of \(0.2\) M \(\mathrm{KMnO_4}\) with \(500\) mL of \(0.2\) M \(\mathrm{KBr}\) solution in basic medium. What are the concentrations of \(\mathrm{KBr}\) and \(\mathrm{KBrO_3}\) respectively in the resultant solution?

• A. \(0.025\) M, \(0.025\) M
• B. \(0.05\) M, \(0.025\) M
• C. \(0.05\) M, \(0.05\) M
• D. \(0.025\) M, \(0.05\) M

Hint:

Always balance the redox reaction first, calculate the limiting reagent, and then divide the remaining moles by the final solution volume to obtain the concentration.

Answer 1

The Correct Option is C

Step 1: Write the balanced reaction in basic medium. The balanced reaction is \[ 2\mathrm{MnO_4^-} + \mathrm{Br^-} + \mathrm{H_2O} \rightarrow 2\mathrm{MnO_2} + \mathrm{BrO_3^-} + 2\mathrm{OH^-}. \] Thus, \[ 2\text{ mol of }\mathrm{MnO_4^-} \] react with \[ 1\text{ mol of }\mathrm{Br^-}. \]

Step 2: Calculate the initial moles. Moles of \[ \mathrm{KMnO_4} = 0.5\times0.2 = 0.1. \] Moles of \[ \mathrm{KBr} = 0.5\times0.2 = 0.1. \] According to the stoichiometry, \[ 0.1\text{ mol of }\mathrm{MnO_4^-} \] consume \[ 0.05\text{ mol of }\mathrm{Br^-}. \] Therefore, \[ \text{Unreacted KBr} = 0.1-0.05 = 0.05\text{ mol}. \] Also, \[ \text{KBrO}_3\text{ formed} = 0.05\text{ mol}. \]

Step 3: Find the concentrations. The total volume after mixing is \[ 500+500 = 1000\text{ mL} = 1\text{ L}. \] Hence, \[ [\mathrm{KBr}] = \frac{0.05}{1} = 0.05\text{ M}, \] and \[ [\mathrm{KBrO_3}] = \frac{0.05}{1} = 0.05\text{ M}. \] Therefore, \[ \boxed{0.05\text{ M},\ 0.05\text{ M}}. \] Thus, \[ \boxed{(C)} \] is the correct answer.