In neutral or faintly alkaline medium, MnO ₄⁻ oxidizes I ⁻ to iodate. What is the volume (in L ) of 0.02 M KMnO ₄ required to completely convert 1 ~L of 0.5 M KI solution to iodate in neutral or faintly alkaline medium?

Step-by-step solution

2 KMnO ₄+ H ₂ O + KI 2 MnO ₂+2 KOH + KIO ₃ 1 mole of I ⁻ requires 2 moles of MnO ₄⁻ . Now, number of moles of 1 ~L 0.5 M =0.5 ~mol . Number of moles of MnO ₄⁻ required =2 0.5=1.0 ~mol . Volume = n M = 1.0 0.02 =50 ~L