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......?????
Let us take the reaction as:
NaHC2O4 + KMnO4---> Mn+2 + CO2
So, equivalent of NaHC2O4 = equivalent of KMnO4
Since, we know , naMaVa =nbMbVb
By, applying the law we get, molarity of NaHC2O4 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 NaHC2O4 = 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.