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0 votes
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in Chemistry by (2.8k points)

0.1L of 0.01 M KMnO4 oxidized 100 mL H2O2 in acidic medium. Volume of same KMnO4 required in alkaline medium to oxidize 0.1L of same H2O2 will be (Mn changes to Mn2+ in acidic medium and MnO2 in alkaline medium)

by (10 points)
i cant solve it and i want its solution . plz help me.
by (2.8k points)
You can know the n-factor.
 
1st case:- n-factor=5 [Mn(7) to Mn(2)]

Now, this is one of the best concepts of mole concept,
"The EQUIVALENTS(normality×volume) of all reactants and products are EQUAL."

N=M×n-factor


Thereby, (100ml)(0.05N)=(100ml)V

Do the same in the next case but n-factor in 2nd case is 3.

Solve it further by yourself, it will help you to gain confidence.

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2 Answers

0 votes
by (15 points)
ans:) d none 
btw the ans is 500 ml

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by (38.5k points)

In acidic medium

2KMn+7O+ 5H\(O^{-1}_2\) + 3H2SO4 \(\rightarrow\)5\(O^0_2\) + 2 Mn+2SO4 + K2SO4 + 8H2O

We have given,

Molarity of KMnO4 = 0.01M.

For reaction in acidic medium the normality of 

KMnO4 = Molarity x Change in Oxidation state of Mn

= 0.01m x 5

=0.05 N

Using Normality equation 

N1V1 = N2V2   

\(\Rightarrow\) 0.05 N x 100ml = \(N_{{H_2}{O_2}}\,\times\,100 ml\)

\(\Rightarrow\) \(N_{{H_2}{O_2}}\) = 0.05 N

Therefore the Normality of H2O2 = 0.05 N

In basic medium

2KMn+7O4 + 3H\(O_2^-1\) \(\rightarrow\) 3\(O^0_2\) +2Mn+4O2 + 2KOH

There is no change in the Normality of H2O2. because there is same change in the oxidation state of oxygen ( In acidic medium as well as basic medium )

Therefore Normality of H2O2 = 0.05 N

but, There is change in the Normality of KMnO4 .

because , there is different change in oxidation state of Mn atom ( In acidic = 5 , In basic = 3 )

Therefore Normality of KMnO4 = molarity x n

= 0.01 \(\times\) 3 

= 0.03 N

Using Normality equation 

\((N_1V_1)_{KMnO_4} = (N_2V_2)_{H_2O_2}\)

\(\Rightarrow\) 0.03 N \(\times\,\)V1 = 0.05  \(\times \) 100 ml

\(\Rightarrow\) v1\(\frac{500}{3} ml\)

Hence, \(\frac{500}{3}\) ,of same KMnO4 required in alkaline medium to oxidize 0.1 L of same H2O2.

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