In neutral or faintly alkaline medium, MnO_4^-oxidizes I^-to iodate. What is the volume (in L ) of 0.02M KMnO_4 required to completely convert 1L of 0.5 MKI solution to iodate in neutral or faintly a…

Question

In neutral or faintly alkaline medium, MnO_4^-oxidizes I^-to iodate. What is the volume (in L ) of 0.02M KMnO_4 required to completely convert 1L of 0.5 MKI solution to iodate in neutral or faintly alkaline medium?

Accepted Answer

Correct answer: B. (b) 2 KMnO_4+H_2O+ KI 2 MnO_2+2 KOH+ KIO_3 ⇒1 mole ofI- requires 2 moles of MnO_4^- . Now , number of moles of 1L 0.5M=0.5 mol . ⇒ Number of moles of MnO_4^- required =2 0.5=1.0 mol . ⇒ Volume = nM= 1.00.02=50L

$In neutral or faintly alkaline medium, MnO_{4}^{-} oxidizes I^{-} to iodate. What is the volume (in L) of 0.02 M KMnO_{4} required to completely convert 1 L of 0.5 MKI solution to iodate in neutral or faintly alkaline medium?$

Solution

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