InterviewSolution
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InterviewSolution
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`H_(2) SO_(2)` solution `(20 mL)` reacts quantitatively with a solution of `KMnO_(4) (20 mL)` acidified is just dilute `H_(2) SO_(4)`. The same volume of the `KMnO_(4)` solution is just decolourised by `10 mL` of `MnSO_(4)` in neutral medium. simulataneously forming a dark brown precipitate of hydrated `MnO_(2)`. The brown precipitate is dissolved in `10 mL` of `0.2 M` sodium oxalate under boiling condition in the presence of dilute `H_(2) SO_(4)`. Write the balanced equations involved in the reactions and calculate the molarity of `H_(2) O_(2)` solution. |
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Answer» Take the balanced equations as follows: `2MnO_(4)^(ɵ) + 6 H^(o+) + 5 H_(2) O_(2) rarr 2 Mn^(2+) + 8 H_(2) O + 5 O_(2)` `2 MnO_(4)^(ɵ) + 2 Mn^(2+) rarr 5 MnO_(2)` `MnO_(2) + C_(2) O_(4)^(2-) + 4 H^(o+) rarr Mn^(2+) + 2 H_(2) O + 2CO_(2)` To find molarity of `H_(2) O_(2)` Given: `20 mL H_(2) O_(2) -= 20 mL KMnO_(4)//H^(o+)` `20 mL KMnO_(4) -= mL MnSO_(4) rarr MnO_(2)` `MnO_(2) -= 10 mL (0.2 M) Na_(2) C_(2) O_(4)` From stoichiometry of above reactions: 1 mmol of `Na_(2) C_(2) O_(4) -= 1 mmol MnO_(2)` `implies 0.2 xx 10 mmol of Na_(2) C_(2) O_(4) -= 2 mmol of MnO_(2)` Also, 5 mmol of `MnO_(2) -= 3 mmol of Mn^(2+)` `implies 2 mmol of MnO_(2) -= (3)/(5) xx 2` `-= (6)/(5) mmol of Mn^(2+) (MnSO_(4))` Also, 3 mmol of `MnSO_(4) -= 2 mmol KMnO_(4)` `implies (6)/(5) mmol of MnSO_(4) -= (2)/(3) xx (6)/(5) = mmol KMnO_(4)` Now same amount of `KMnO_(4)` reacts completely with `H_(2) O_(2)`. `2 mmol of H_(2) SO_(4) -= 5 mmol of H_(2) O_(2)` `implies 2 mmol of KMnO_(4) -= (5)/(2) xx (4)/(5) = 2 mmol of H_(2) O_(2)` Now mmol of `H_(2) O_(2) = MV_(mL)`. `implies 2 = M xx 20 implies M = 0.1` |
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