Alp Solutions Equivalent Concept & Titration Eng Jf

8/18/2019 Alp Solutions Equivalent Concept & Titration Eng Jf

1/8

Equivalent Concept & Titration - 1

1. 85.5 = Emetal

+ EOH

–

or 85.5 = Emetal

+ 17or E

metal = 68.5

2. (a) EP =

3

31 = 10.33 (b) E

Al =

3

27 = 9 (c) E

Fe =

2

56 = 28 (d) E

S =

6

32 = 5.33

3. 40 g, O ! 60 g metal " 8 g, O ! 12 g metal (E)

4. v.f. of Na2S2O3 = 2 (2.5 – 2) = 1 " Eq. wt. = M/1

5. v.f. of KBrO3 = 1 (5 – ( – 1)) = 6

" Eq. wt. =6

M

6. v.f. of oxalic acid = 2 (4 – 3) = 2 " Eq. wt. =2

M

7. meqH2SO4 = meqNaOH

0.2 × V = 0.02 × 20V = 2 mL

8. m eq. of HCl reacted with alkalline earth metal carbonate = (25 × 1) – (50 × 0.1) = 20

" m eq. of alkalline earth metal carbonate = 20

" .wt.Eq

1 × 1000 = 20

" Eq. wt. of metal carbonate =20

1000 = 50

Eq. wt. of Metal = Eq. wt. of metal carbonate – Eq. wt. of carbonate

= 50 – 2

60 = 20

At wt. of metal = (Eq. wt. of metal) × Valency of metal= 20 × 2 = 40

9. 2NH3 + H2SO4 #$ # (NH4)2SO4

2NaOH + H2SO

4 #$ # Na

2SO

4 + 2H

2O

" m.eq of H2SO

4 neutralized by NH

3 = (50 × 1) – (24.5 × 1) = 25.5

" m.moles of NH3 present in double sulphate = 25.5

" Mass of NH3 present in double sulphate = %

&

'()

* +17

1000

5.25 = 0.4335 g

" % of NH3 in double sulphate =

5

4335.0 × 100 = 8.67%

Target : JEE (IITs)

CHEMISTRY

COURSE : VISHWAAS(JF)

TOPIC : EQUIVALENT CONCEPT & TITRATIONS

SOLUTION OF ADVANCED LEVEL PROBLEMS


2/8


10. KMnO4 + Fe+2 #$ # Fe+3 + Mn+2

milli equivalent of KMnO4 = milli equivalent of Fe2+

1 × 5 × M = 1×56

140

M =556

1140

+

+ = 0.5

KHC2O

4.H

2C

2O

4 = 2 C

2O

42 – + KMnO

4 ##$ Mn2+ + CO

2

milli eq. of KMnO4 = milli eq. of KH2 C2O4 . H2C2O4100 × 5 × 0.5 = 1 × 2 × 2 × M

M = 0.0625

KHC2O

4.H

2C

2O

4 + 3NaOH $ 3H

2O + Na

2C

2O

4 + KNaC

2O

4

meq of KHC2O

4.KHC

2O

4 = meq of NaOH

3 × 1 × 0.0625 = 0.20 × V

V =20.0

0625.03+ = 0.9375mL or

16

15mL.

11. milli eq. of FeSO4 = milli eq. of KMnO

4

N1V

1 = N

2V

2

N1 × 25 =20

10

1+

N1 =

2510

20

+ = M

1 (V.f. = 1)

Weight of FeSO4.7H

2O, W =

2510

20

+ × 278 = 22.24g

% of FeSO4 .7 H

2O = 100

25

24.22+ = 88.96 %

12. Redox changes are :

Fe $ # Fe+2 + 2e – (in H2SO4)

Fe+2 $ # Fe+3 + e – (with K2Cr2O7)

6e + Cr2+6

$ # 2Cr+3

m.eq. of Fe+2 in 20 mL = m.eq. of K2Cr2O7 = 30 × 30

1 = 1

" m.eq. of Fe+2 in 100 mL =20

1001+= 5

" m.moles of Fe+2 =.f.v

meq =

1

5= 5 = m.moles of Fe

" Mass of pure Fe in wire = 5 × 10 –3 × 56 = 0.28 g

% of Fe in wire =2828.0

28.0× 100 = 99%

13. On balancing the reaction,

MnO4 – + Fe+2 + H+ #$ # Mn+2 + Fe3+

MnO4 – + 5Fe2+ + 8H+ #$ # Mn2+ + 5Fe3+ + 4H

2O

or

KMnO4 + 5FeCl

2 + 8HCl #$ # MnCl

2 + 5FeCl

3 + 4H

2O + KCI

moles158

101.07

1000

500 x 3 = 1.5

So, KMnO4 is limiting reagent.

"1

KMnOofmole 4 =

5

FeClofMole 3

mole of FeCl3 =

158

10 x 5 = 0.316


3/8


14. Let V mL of reducing agent be used for KMnO4 in different medium which act as oxidant

Acid medium, Mn7+ + n1 e – $ Mna+

" n1 = 7 – a

Neutral medium, Mn7+ + n2 e – $ Mnb+

" n2 =7 – b

Alkaline medium, Mn7+ + n3 e – $ Mnc+

" n3 = 7 – c

" meq of reducing agent = meq. of KMnO4 in acid

= meq. of KMnO4 in nautral= meq. of KMnO4 in alkali

= 1 × n1 × 20 = 1 × n

2 × 33.3 = 1 × n

3 × 100

! n1, n

2,n

3 are integers and are - 7,

" n1 = 5, n

2 = 3 and n

3 =1

Therefore, different oxidation state of Mn are :Acid media, Mn7+ + 5e – $ Mna+

" a = + 2Netrual media, Mn7+ + 3e – $ Mnb+

" b = + 4Alkaline media, Mn7+ + 1e – $ Mnc+

" c = + 6Now, same volume of reducing agent is treated with K

2Cr

2O

7 and therefore,

m.eq. of reducing agent = m.eq. of K2Cr2O7But, m.eq. of reducing agent = m.eq. of KMnO

4 in acid

" m.eq. of KMnO4 in acid = m.eq. of K

2Cr

2O

7

20 × 5 = 1 × 6 × V (v.f. for Cr2O

72 – = 6)

V =6

100 = 16.67 mL

It is important to note that the conditions are valid only when Mn in each medium exist as monomeric atom,i.e. not as Mn

2.

Balanced equations for three half reactions :

MnO4 – + 8H+ + 5e – #$ # Mn2+ + 4H

2O (Acidic medium)

MnO4

– + 2H2

O + 3e – #$ # MnO2

+ 4OH – (Neutral medium)

MnO4 – + e – #$ # MnO

42 – (Basic medium)

15. Sn2+ #$ # Sn4+ +

2e

6e + Cr2

+6 #$ #

2Cr+3

Since Sn+2 is oxidized by K2Cr

2O

7

" m.eq. of Sn+2 = m.eq. of K2Cr

2O

7 used for oxidation of tin = N × V

in mL

=50.0

6

294

5.2

+ × 10 = 1.02

%%%%

&

'

((((

)

*

+.

5.06

294

5.2N!

"2 / 119

w 2Sn/ × 1000 = 1.02

2SnW / = 0.0607 g

" % Sn =4.0

0607.0 × 100 = 15.17 %


4/8


16. KIO3

+ 5KI # # $ # HCl

3I2

v.f. = 5 v.f. = 1

214

214.0 mole (excess) 3×10 –3mole

I2

+ 2Na2S

2O

3 #$ # v.f. = 2 v.f. = 1, 50 ml, M(say)

3×10 –3mole 6×10 –3= 50 × M × 10 –3 0 M = 0.12 M.

17. Used millimoles of I2= 0.05 – 0.002x10/2

= 0.04 = millimoles of Sn2+

wt of tin = 0.04 × 119 = 4.76 mg.

18. moles of iodine = moles of chlorine =2

2.080 + × 10 –3 = 8 × 10 –3

so required % =1.7

10718 3"" × 100% = 8 %

19. excess of Sn2+ = 30 × 0.1 × 5 meq.= 15 meqSn2+ consumed = [(35 × 0.5 × 2) – 15] meq

= 20 meq.Now Cl2 + Sn

2+ $ # Sn4+ + 2Cl – (balanced equation)

So molesof Cl2 = moles of Sn2+ =

2

1eq. of Sn2+ =

2

1 × 20 m moles = 10 mmoles

= 10 × 10 –3 × 71 g = 0.71 g.

So, required % =10

71.0× 100 = 7.1%

20. Bleaching powder # # # # # # # $ # / COOHCHKI 3

I2 # # # # $ # 322

OSNa I – + Na

2S

4O

6

The redox changes are 2e – + I2

#$ # 2 I –

2S2

+2 #$ # S4+5/2 + 2e –

m.eq. of available Cl2 = m.eq. of I

2 liberated = m.eq. of Na

2S

2O

3 used.

" m.eq. of available Cl2 in 20 mL bleaching powder solution

= m.eq. of Na2S

2O

3 used

= 20 ×10

1

" m.eq. of available Cl2 in 500 mL bleaching powder solution = 20 ×

10

1 ×

20

500 = 50

"2 / 71

w × 1000 = 50 (w = weight of available Cl

2)

" w = 100050 ×

271 = 1.775 g

" % of available Cl2 in bleaching powder =

5

775.1 × 100 = 35.5%


5/8


21. H+ + IO3 – + I – #$ # I

2

On balancing the reaction,

6H+ + IO3 – + 5I – #$ # 3I

2 + 3H

2O

Let normality of HCl is X.

I2 + 2S2O3 –2 #$ # 2I – + S4O6

2 –

m.eq. of I2 = m.eq. of Na2S2O3" (millimoles of I2) × 2 = M × v.f. × V

milli moles of I2 = 2124x021.0 +

= 0.252

Now,6

HClofmole.m =

3

ofmole.m 210

6

1 / X25 + =

3

252.0(v.f. for HCl = 1)

" X = 0.02 N

1

IOofmole.m 32

=3

ofmole.m 210 0.2 × V =

3

252.0

V = 0.42 mL

22. mole of KMnO4 = 20 × 10 –3 ×

50

1 =

5

2× 10 –3 ; so, mole of Fe2+ = 5 ×

5

2 × 10 –3 = 2 × 10 –3

so, mole of N2H4 = 41

× 2 × 10 –3 =2

1 ×10 –3 ; Now mole of N2H6SO4 = mole of N2H4

so, mass of N2H

6SO

4 =

2

1 × 10 –3×130 = 65 × 10 –3 g

so, in 10 mL solution, quantity of N2H

6SO

4 = 65 × 10 –3 g

" in 1 liter solution, quantity of N2H

6SO

4 =

10

1065 32+×1000 g = 6.5 g

23. Fe3O4 , Fe2O3 , Impurity 3 3 3x mole y mole

In first reaction,

Fe3O4 + I – # # $ #

/

H Fe+2v.f. = 2x mole 3x mole

Fe2O3 + I – # # $ #

/H Fe+2

v.f. = 2y mole 2y mole

(Eq. of Fe3O4 + Eq. of Fe2O3) in 10 mL solution = Eq. of Hypo

2(x + y) ×50

10 = 4.8 × 1 × 10 –3

2x + 2y = 4.8 × 10 –3 × 5x + y = 12 × 10 –3 .....(1)

In IInd

reaction,Fe+2 + KMnO4 # # $ #

/H Fe+3 + Mn+2

v.f. = 1 v.f. = 5moles 3x + 2y 3.2 × 10 –3 × 1

eq. of Fe+2 in 25 mL solution = eq. of KMnO4

(3x + 2y) ×20

10 = 3.2 × 1 × 5 × 10 –3

3x + 2y = 32 × 10 –3 .....(2)On solving eq. (1) and (2),

x = 8 × 10 –3 ; y = 4 × 10 –3

wt. of Fe3O4 = 8 × 10 –3 × 232 = 1.856 g

wt. of Fe2O3 = 160 × 4 × 10 –3 = 0.64 g

% Fe3O4 = 6856.1 × 100 = 30.93 % Ans. ; % Fe2O3 = 6

64.0 × 100 = 10.67 % Ans.


6/8


24. Let , weight of H2C

2O

4 = 'a' g and weight of NaHC

2O

4 = 'b' g

for acid base reaction

(meq. of H2C

2O

4 + meq . of NaHC

2O

4 ) in 10 mL = 3 × 0.1

" meq. of H2C

2O

4 + meq. of NaHC

2O

4 in one litre = 3 × 0.1 × 100 = 30

"2 / 90

a × 1000 +

1 / 112

b × 1000 = 30

"

45

a1000 +

112

b1000 = 30 ......(1)

For redox change :

C2

3+ #$ # 2 C4+ + 2e –

Mn7+ + 5e – #$ # Mn2+

(meq. of H2C

2O

4 + meq. of NaHC

2O

4) in 10 mL = 4 × 0.1

" meq. of H2C

2O

4 + meq. of NaHC

2O

4 in 1 litre = 4 × 0.1 × 100 = 40

"2 / 90

a × 1000 +

2 / 112

b × 1000 = 40

(! eq. wt. of H2C

2O

4 =

2

M and eq. wt. of NaHC

2O

4 =

2

M as reductant)

"45

a1000 +112

b2000 = 40 ......(2)

Solving equation (1) and (2), we get :

a = 0.9 g and b = 1.12 g

(Also given : a + b = 2.02 and thus equation (1) or (2) can be used to find a and b by using a + b = 2.02)

25. Let 'a' mole of Cu+2 and 'b' mole of C2O

42 – be present in solution.

Case I :The solution is oxidized by KMnO4 which reacts with only C

2O

42 –.

5e + Mn+7 #$ # Mn+2

C2+3 #$ # 2C+4 + 2e

" m. eq. of C2O

4 –2 = m. eq. of KMnO

4

" b × 2 × 1000 = 0.02 × 5 × 22.6

" b = 1.13 × 10 –3

Case II : After oxidation of C2O

4 –2 , the resulting solution is neutralized by Na

2CO

3, acidified with dilute

CH3COOH and then treated with excess of KI. The liberated I

2 required Na

2S

2O

3 for its titration :

Cu+2 #$ # KI

Cu+ + I2 # # # # $ # 322

OSNaNa

2S

4O

6 + I –

" m. eq. of Cu+2 = m. eq. of I2 liberated = m. eq. of Na2S2O3 used

" m. eq. of Cu+2 = m. eq. of Na2S2O3 used

a × 1 × 1000 = 11.3 × 0.05 × 1

" a = 5.65 × 10 –4

" Molar ratio = 242

2

OCCu 2

/

=ba = 3

4

1013.11065.5

2

2

++ =

21

26. Number of millimoles of KIO3 in 30 mL of solution = Molarity × Volume in mL

=10

1 × 30 = 3

Given equation : K1O3 + 2K1 + 6HCl #$ # 31CI + 3KCl + 3H

2O

According to the equation of the reaction given, 1 mole of KIO3 is equivalent to 2 moles of KI

! No. of millimoles of K1 in 20 mL of stock solution = 2 × 3 = 6

" No of millimoles of K1 in 50 mL of the same solution = 6 ×20

50 = 15


7/8


No . of millimoles of K1O3 in 50 mL of solution =

10

1 × 50 = 5

" No . of millimoles of K1 used with 50 mL of K1O3 solution = 2 × 5 = 10 (from reaction)

" No. of millimoles of K1 used with AgNO3 = 15 – 10 = 5

Ag NO3 + K1 #$ # AgI + KNO

3

1 mole of AgNO3 reacts with 1 mole of KI. Therefore,

" No. of millimoles of AgNO3 is equal to 5 .

"

Weight of AgNO3 = 5 × 10 –3 × 170 g = 0.85 g

" % of AgNO3 =

1

10085.0 + = 85 %

27. Assume 10 mL of titrant, so there is 0.1 × 10 = 1 m.mole NaOH or KMnO4. The acidity is due of

KHC2O

4.H

2C

2O

4(KH

3A

2)

m.eq. of KH3A

2 = m.eq. of NaOH

3 × m.mole of KH3A

2 = m.mole of NaOH x 1

" m.moles of KH3A

2 = 1 ×

3

1 = 0.33 ....... (1)

m.eq. of KMnO4 = m.eq. of Na

2A + m.eq. of KH

3A

2

5 × m.mole of KMnO4 = m.mole of Na

2A × 2 + m.mole of KH

3A

2 × 4 (C

2O

42 – $ 2CO2, v.f. = 2)

" 5 × 1 = m.mole of Na2A × 2 + 0.33 × 4 (from (1))

" m.mole of Na2A = 1.83 ....... (2)

From (1) and (2),

23

2

AKH

ANa

m

m

=2181033.0

1341083.1

3

3

++

++2

2

= 3.38

" Mixing proportion by mass = 3.38 : 1.

28. The redox changes are :

for reducing of Fe2O

3 by zinc dust

Fe23+ + 2e – #$ # 2Fe2+

Fe2+ #$ # Fe3+ + e –

oxidant + ne – #$ # product

meq. of Fe2O

3 in 25 mL

= meq. of Fe3+ in Fe2O

3

= meq. of Fe2+ formed

= meq. of oxidant used to oxidize Fe2+ again

" meq. of Fe2O

3 in 25 mL = meq of oxidant

= 17 × 0.0167 × n

Where, n is the number of electron gained by 1 mole of oxidant

" meq. of Fe2O3 in 100 mL = 17 × 0.0167 × n × 25

100

"

2

M100

10002.551

+

++ = 17 × 0.0167 × n × 4

! molecuar wt. of Fe2O

3 = 160

" n = 40167.017160100

210002.551

++++

+++ = 6

Hence, number of moles of electrons gained by one mole of oxidant = 6 .


8/8


29. Let total mole of Mg used for MgO and Mg3N

2 be 'a' and 'b' respectively.

2Mg + O2 #$ # 2MgO

Before reaction a 0

After reaction 0 a

3Mg + N2 #$ # Mg3N2

Before reaction b 0

After reaction 0 b/3

Now (a + b/3) mole of MgO and Mg3N

2 are present in the mixture.

MgO + 2HCl #$ # MgCl2 + H2O ; Mg3N2 + 8HCl #$ # 3MgCl2 + 2NH4Cl

or the solution contains 'a' mole of MgCl2 from MgO and 'b' mole of MgCl

2 from Mg

3N

2 and

3

'b2' mole of NH

4Cl.

Also mole of HCl used for this purpose = 2a +3

b8

(for MgO) (for Mg3N

2 )

Now mole of HCl reacted with MgO and Mg3N

2 =

1000

1260 2 = 0.048

2a +3

b8 = 0.048 ... (1)

Further, mole of NH4Cl formed = mole of NH

3 liberated

= mole of HCl used for absorbing NH3

= Total equivalent of acid – equivalent of base used for back titration

=1000

610 2 = 4 × 10 –3

"3

b2= 4 × 10 –3 or b = 6 × 10 –3 ... (2)

From (1) 2a +31068 3

2++ = 0.048 or a = 16 × 10 –3

Thus, % of Mg burnt to Mg3N

2 =

)166(

6

/ × 100 = 27.27 %

30. In presence of Hph indicator

m. eq. of HCl = m. eq. of Na2CO

3 + m.eq. of NaOH

30 + 0.1 = (a + 1) + (a + 1)

" a = 1.5

In presence of MeOH indicator

m. eq. of H2SO

4 = m. eq. of Na

2CO

3 + m. eq. of NaOH

0.15 + 2 + V = (a + 2) + (a + 1)V = 15 ml

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Alp Solutions Equivalent Concept & Titration Eng Jf