100 ml of NaHC2O4 required 50 ml of 0.1 M KMnO4 solution in acidic medium.. Volume of 0.1 M NaOH required by 100 ml of NaHC2O4 is......?????

Dear student!

Let us take the reaction as:

NaHC_{2}O_{4} + KMnO_{4}---> Mn^{+2} + CO2

So, equivalent of NaHC_{2}O_{4} = equivalent of KMnO_{4}

Since, we know , n_{a}M_{a}V_{a} =n_{b}M_{b}V_{b}

By, applying the law we get, molarity of NaHC_{2}O_{4}_{ }as

0.1 x M x 2 = 0.05x 0.1 x 5

or, M = 0.25/2 = 0.125

Since, molarity = moles/Volume

Hence, no. of moles = molarity x volume

Further, we have, moles of NaHC_{2}O_{4} = moles of NaOH

so, 0.125 x 0.1 = 0.1 x V

Hence, V = 0.125 L =125 ml.

Thus, 125 ml of the given NaOH would be required.

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