This is a comprehension for questions 20-23
A sample of 20 ml 0.1 M MnO2 solution is taken. In this solution 10 ml of KI solution and 3 millimole of K2Cr2O7 is added. KI in the solution is reduced by both MnO2 & K2Cr2O7. The Mn2+ formed again is oxidised by K2Cr2O7. The MnO2 left is oxidised to form Mn+7. No KI, MnO2 and K2Cr2O7 is left
So the reactions taking place are
MnO2+KI→I2+Mn2+
K2Cr2O7+KI→I2+Cr3+
K2Cr2O7+MnO2→Mn7++Cr3+
K2Cr2O7+Mn2+→Mn7++Cr3+
The I2 formed requires 30 ml of 0.1 M sodium thiosulfate solution in another titration
The molarity of KI in the original KI sample is
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UP 0 DOWN 0 0 0
