# If 1 mL of a KMnO4 solution react with 0.140 g Fe2+ and if 1 mL of KHC2O4 H2C2O4 solution react with 0.1 mL of previous KMnO4 solution, how many millilitres of 0.20 M NaOH will react with 1 mL of previous KHC2O4, H2C2O4 solution in which all the protons (H+) are ionisable? (A) 15/16 mL                   (B) 13/16                   (C) 11/14                    (D) None of these

Dear Student,

To find out ml of NaOH required for neutrilising 1ml of KHC2O4.H2C2O4, we need to Calculate,
1. Concentration of KMnO4
2. Concentration of KHC2O4.H2C2O4
3. Finally ml of NaOH.

lets first see the balanced equation of all the three reactions, moles equivalent, and molecular weights.

Reaction 1.  5Fe 2+ + 1MnO 4- +8H+  → 5Fe 3+  + Mn 2+ + 4H2O

from reaction, it is clear that 5Fe reacts with 1 KmnO4
​5Fe = 1 KMnO4
molecular weights - KMnO4 = 158.033 g/mol
-Fe2+ = 55.845 g/mol

Reaction 2.  10KHC2O4.H2C2O4 + 8 KMnO4 + 17H2SO4 → 8MnSO 4 + 9K2SO4 + 40CO2 + 32H2O

from reaction, it is clear
that10KHC2O4.H2C2O4 reacts with 8 KmnO4
1.25 KHC2O4.H2C2O4 = 1 KMnO4
molecular weights - KHC2O4.H2C2O4= 218.160 g/mol

Reaction 3.  KHC2O4.H2C2O4 + 4NaOH → 2Na2C2O4 + 3H2O +KOH

from the reaction, it is clear that
1KHC2O4.H2C2O4 reacts with 4NaOH
4 NaOH = 1 KHC2O4.H2C2O4
molecular weights -NaOH = 39.99 g/mol

now let's move towards the main calculation

1. Calculation of Concentration of KMnO4
5 Fe2+ = 1 KMnO4

therefore in 1000 ml gms of KMnO4 will be

2. Concentration of KHC2O4.H2C2O4
M1V1=M2V2
M= molarity
V= volume
1= KHC2O4.H2C2O4
2= KMnO4
by considering moles equivalent the formula will be
1.25 M1V1=M2V2

Now last step calculation of NaOH ml required

M1V1=M2V2
1=KHC2O4
2=NaOH
by considering moles equivalent the formula will be

M1V1=4 M2V2