If 1 mL of a KMnO4 solution react with 0 140 g Fe2+ and if 1 mL of KHC2O4 H2C2O4 - Chemistry - General Principles and Processes of Isolation of Elements

Question

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

Poonam Desai · Accepted Answer

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
â€‹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 clearthat10KHC2O4

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

â€‹5 × 55
845 = 1 × 158 033279 225 = 158
0330 140 gm = x?x= 0 140×158 033279
225x= 0 079 gmso 1 ml contains 0
079 gm KMnO4

therefore in 1000 ml gms of KMnO4 will be

= 0
079 × 1000 =79 gmmolarity of KMnO4 will be,=79158
033=0 499 gmoles/litr=0 499 M of KMnO4

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

1 25(1 ml × M1) = 0
1 ml × 0
499 M1 ml × M1= 0 1×0 4991
25M1=0 03991 mlM1=0 0399 mol of KHC2O4
H2C2O4

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
0
0399 ×1 ml =4(0
2 M × V2 ml)0
2 M × V2 ml = 0
0399 ×14V2=0 0099750 2V2=0
0498 ml of NaOH
Hope this information will clear your doubts about the
topic

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