Molecular weight of Na2C2O4= 134
Moles of Na2C2O4= 0.24/134=0.001791
From the reaction, 5 moles of Na2C2O4 requires 2 moles of KMnO4.
0.001791 moles requires 0.001791*2/5 =0.000716 moles
Molarity= moles/ Volume (L)= 0.000716*1000/30 =.0238M
2. TThe reaction is Na3PO4 + NaH2PO4 ----> 2 Na2HPO4
moles of Na3PO4= 0.1*100/1000 =0.01 and moles of NaH2PO4= 0.095*50/1000=0.00475
limting reactant is Na3PO4. Moles of Na3PO4remaining = 0.01-0.00475=0.00525
pH = pKa + log [Na3PO4]/[Na2HPO4)= 12.15+log (0.00475/0.00525)=12.10 ( close answer is 12.19)
KMnO_4 titrant is used to titrate 0.2400 g of pure sodium oxalate, Na_2C_2O_4. according to the...