Titration - Gyanpoints

TITRATION

EXPERIMENT No. 1

Object – Find out the concentration of given potassium permanganate solution in gram per

litre. For this purpose, you are provided a standard solution of crystalline ferrous ammonium

sulphate of $\displaystyle \frac{M}{{30}}$ molarity.

Principle / Theory-– Titration between ferrous ammonium sulphate and potassium permanganate is a redox titration. In this reaction, KMnO₄ oxides ferrous ammonium sulphate in ferric sulphate Fe₂(SO₄)₃ in the presence of H₂SO₄ whenever it reduced itself in MnSO₄.

Chemical Equation :

Molecular Equation

Chemical reacting takes place

oxn half reaction

FeSO₄(NH₄)₂SO₄.6H₂O + H₂SO₄ +[O]Fe₂(SO₄)₃ + (NH₄)₂ SO₄ + 7H₂O

Redn half reaction

KMnO₄+3H₂SO₄K₂(SO₄) + MnSO₄ + 3H2O +5[O]

Over all reaction

2KMnO4+8H2SO4 + 10FeSO4(NH4)2SO4.6H2O K2(SO4) + 2Mn SO4 + 5Fe2(SO4)3 + 10(NH4)2 SO4 +68H2O

Ionic Equation

MnO₄^–+ 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

Observation Table:

Titration between intermediate KMnO4 solution with known FAS solution.

S.No.	Volume of Known ferrous ammonium sulphate solution taken by pipette V₁ mL	Burette Reading		Volume of KMnO₄ solution (b-a) mL	Concordant Reading V₂mL
S.No.		Initial (a)	Final (b)	Volume of KMnO₄ solution (b-a) mL	Concordant Reading V₂mL
1	20	0.0	19.8	19.8	19.6
2	20	0.0	19.6	19.6
3	20	0.0	19.6	19.6

Calculation:

To calculate the molarity. of unknown solution of KMnO₄

It is clear from the reaction that 1 mole of KMnO₄ reacts with 5 mole of ferrous ammonium sulphate

Solution of FeSO₄(NH₄)₂SO₄.6H₂O = Solution of KMnO₄

M₁V₁ = 5M₂V₂

(here, M₁ = M/30, V₁ = 20 ml, V₂ = 19.6 ml)

$\displaystyle {{M}_{2}}=\frac{{{{M}_{1}}{{V}_{1}}}}{{5{{V}_{2}}}}$

$\displaystyle {{M}_{2}}=\frac{{M\times 20}}{{30\times 5\times 19.6}}$

= 0.0068 M

To calculate the concentration of unknown solution of KMnO₄

Molarity x Molecular Weight = 0.0068. x 158 = 1.0744 g/l

Result- Concentration of unknown solution, of KMnO₄ 1.0744 gl.

EXPERIMENT No, 2

Objeot–Find out the molarity of potassium permanganate solution. For this purpose, you are provided a standard solution of crystalline ferrous ammonium sulphate containing 13.0666 grams of salt per liter.

Chemical Equation :

Molecular Equation

Chemical reacting takes place

oxn half reaction

FeSO4(NH4)2SO4.6H2O + H2SO4 +[O]Fe2(SO4)3 + (NH4)2 SO4 + 7H2O

Redn half reaction

KMnO4+3H2SO4K2(SO4) + Mn SO4 + 3H2O +5[O]

Over all reaction

2KMnO4+8H2SO4 + 10FeSO4(NH4)2SO4.6H2O K2(SO4) + 2Mn SO4 + 5Fe2(SO4)3 + 10(NH4)2 SO4 +68H2O

Ionic Equation

MnO₄^–+ 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

Observation Table:

Titration between intermediate KMnO4 solution with known FAS solution.

S.No.	Volume of Known ferrous ammonium sulphate solution taken by pipette V₁ mL	Burette Reading		Volume of KMnO₄ solution (b-a) mL	Concordant Reading V₂mL
S.No.		Initial (a)	Final (b)	Volume of KMnO₄ solution (b-a) mL	Concordant Reading V₂mL
1	20	0.0	19.8	19.8	19.6
2	20	0.0	19.6	19.6
3	20	0.0	19.6	19.6

Calculation

Molarity of std. Ferrous ammonium sulphate solution

Molarity = $\displaystyle \frac{{\text{Conc}\text{. (in g/l)}}}{{\text{Molar mass of Ferrous ammonium sulphate}}}$

= $\displaystyle \frac{{13.0666}}{{392.12}}$ M = 0.0333M

Molarity of KMnO4 solution

M1V1 = 5M2V2

where,

M1 = Molarity of known Ferrous ammonium sulphate solution.= $\displaystyle \frac{{13.0666}}{{392.12}}$ M

M2 = Molarity of KMnO4 solution.= ?

V1 = Volume of known FAS solution.= 20 ml

V2 = Volume of KMnO4 solution.= 19.6 ml

$\displaystyle {{M}_{2}}=\frac{{{{M}_{1}}{{V}_{1}}}}{{5{{V}_{2}}}}$

$\displaystyle {{M}_{2}}=\frac{{0.0333M\times 20}}{{5\times 19.6}}$

= 0.0068 M

Result – Molarity of unknown solution of KMnO₄ is 0.0068 M.

EXPERIMENT No. 3

Objeet – Find out the percentage purity of impure potassium permanganate KMnO₄sample 2.0 grams of which have been dissolved in one litre solution. For this purpose you are provided a standard solution of crystalline ferrous ammonium sulphate of $\displaystyle \frac{M}{{30}}$ molarity.

Chemical Equation :

Molecular Equation

Chemical reacting takes place

oxn half reaction

FeSO4(NH4)2SO4.6H2O + H2SO4 +[O]Fe2(SO4)3 + (NH4)2 SO4 + 7H2O

Redn half reaction

KMnO4+3H2SO4K2(SO4) + Mn SO4 + 3H2O +5[O]

Over all reaction

2KMnO4+8H2SO4 + 10FeSO4(NH4)2SO4.6H2O K2(SO4) + 2Mn SO4 + 5Fe2(SO4)3 + 10(NH4)2 SO4 +68H2O

Ionic Equation

MnO₄^–+ 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

Observation Table:

Titration between intermediate KMnO4 solution with known FAS solution.

S.No.	Volume of Known ferrous ammonium sulphate solution taken by pipette V₁ mL	Burette Reading		Volume of KMnO₄ solution (b-a) mL	Concordant Reading V₂mL
S.No.		Initial (a)	Final (b)	Volume of KMnO₄ solution (b-a) mL	Concordant Reading V₂mL
1	20	0.0	20.3	20.3	20.2
2	20	0.0	20.2	20.2
3	20	0.0	20.2	20.2

Calculation

Molarity of KMnO4 solution

M1V1 = 5M2V2

where,

M1 = Molarity of known Ferrous ammonium sulphate solution.= $\displaystyle \frac{M}{{30}}$

M2 = Molarity of KMnO4 solution.= ?

V1 = Volume of known FAS solution.= 20 ml

V2 = Volume of KMnO4 solution.= 20.2 ml

$\displaystyle {{M}_{2}}=\frac{{{{M}_{1}}{{V}_{1}}}}{{5{{V}_{2}}}}$

$\displaystyle {{M}_{2}}=\frac{{M\times 20}}{{30\times 5\times 20.2}}$

= 0.0066 M

To calculate the concentration of unknown solution of KMnO₄

Molarity x Molecular Weight = 0.0066. x 158 = 1.0428 g/l

Percentage purity of sample of given impure potassium permanganate

= $\displaystyle \frac{{Concentration\text{ }of\text{ }given\text{ }KMn{{O}_{4}}Solution}}{{Concentration\text{ }of\text{ }impure\text{ }KMn{{O}_{4}}Solution}}\times 100$

= $\displaystyle \frac{{1.0428}}{2}\times 100$ = 52.14 %

Result – Percentage purity of sample of given impure potassium permanganate is 52.14 %.

EXPERIMENT No. 4

Object – Find out the molarity of potassium permanganate solution. For this purpose you are provided a standard solution of crystalline ferrous ammonium sulphate containing 19.6000 grams of salt per litre.

Chemical Equation :

Molecular Equation

Chemical reacting takes place

oxn half reaction

FeSO4(NH4)2SO4.6H2O + H2SO4 +[O]Fe2(SO4)3 + (NH4)2 SO4 + 7H2O

Redn half reaction

KMnO4+3H2SO4K2(SO4) + Mn SO4 + 3H2O +5[O]

Over all reaction

2KMnO4+8H2SO4 + 10FeSO4(NH4)2SO4.6H2O K2(SO4) + 2Mn SO4 + 5Fe2(SO4)3 + 10(NH4)2 SO4 +68H2O

Ionic Equation

MnO₄^–+ 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

Observation Table:

Titration between intermediate KMnO4 solution with known FAS solution.

S.No.	Volume of Known ferrous ammonium sulphate solution taken by pipette V₁ mL	Burette Reading		Volume of KMnO₄ solution (b-a) mL	Concordant Reading V₂mL
S.No.		Initial (a)	Final (b)	Volume of KMnO₄ solution (b-a) mL	Concordant Reading V₂mL
1	20	0.0	19.8	19.8	19.7
2	20	0.0	19.7	19.7
3	20	0.0	19.7	19.7

Calculation –

To calculate the molarity of standard solution of ferrous ammonium sulphate

Molarity of std. Ferrous ammonium sulphate solution

Molarity = $\displaystyle \frac{{\text{Conc}\text{. (in g/l)}}}{{\text{Molar mass of Ferrous ammonium sulphate}}}$

= $\displaystyle \frac{{19.6000}}{{392.12}}$ M = 0.05M

Molarity of KMnO4 solution

M1V1 = 5M2V2

where,

M1 = Molarity of known Ferrous ammonium sulphate solution.=0.05 M

M2 = Molarity of KMnO4 solution.= ?

V1 = Volume of known FAS solution.= 20 ml

V2 = Volume of KMnO4 solution.= 19.7 ml

$\displaystyle {{M}_{2}}=\frac{{{{M}_{1}}{{V}_{1}}}}{{5{{V}_{2}}}}$

$\displaystyle {{M}_{2}}=\frac{{0.05M\times 20}}{{5\times 19.7}}$

= 0.0101 M

Result – Molarity of unknown solution of KMnO₄ is 0.0101 M.

EXPEROMENT No.5

Object -Find out the concentration of given potassium permanganate solution in gram per litre. For this purpose, you are provided a standard solution of crystalline oxalic acid of $\displaystyle \frac{M}{{40}}$ molarity.

Principle – Titration between oxalic acid and potassium permanganate is a redox titration. In this reaction, KMnO₄ oxides oxalic acid in CO₂ in the presence of H₂SO₄, whenever it reduced itself in MnO₄.

Chemical Equation

Molecular Equation

2 MnO₄ +3H₂SO₄+ 5H₂C₂O₄ →K₂SO₄+ 2MnSO₄+ 8H₂O+ 10CO₂

Ionic Equation

2MnO₄^–+ 16H⁺ + 5C₂O₄^2- →Mn²⁺ + 10CO₂ + 8H₂O

Observation Table :

S.No.	Volume of Known oxalic acid solution taken by pipette V₁ mL	Burette Reading		Volume of KMnO₄ solution (b-a) mL	Concordant Reading V₂mL
S.No.	Volume of Known oxalic acid solution taken by pipette V₁ mL	Initial (a)	Final (b)	Volume of KMnO₄ solution (b-a) mL	Concordant Reading V₂mL
1	20	0.0	20.3	20.3	20.2
2	20	0.0	20.2	20.2
3	20	0.0	20.2	20.2

Calculation–

To calculate the molarity of unknown solution of KMnO₄

It is clear from the reaction that 2 mole of KMnO₄ reacts with 5 mole of oxalic acid.

Solution of (COOH)₂.2H₂O = Solution of KMnO₄

2 M₁V₁=5 M₂V₂

$\displaystyle {{M}_{2}}=\frac{{2{{M}_{1}}{{V}_{1}}}}{{5{{V}_{2}}}}$

(here, M₁ = M/40, V₁ = 20 ml, V₂ = 20.2 ml)

$\displaystyle {{M}_{2}}=\frac{{2\times 20}}{{5\times 40\times 20.2}}$ M = 0.0099 M

Concentration of KMnO₄ of Solution (in g/)

= Molarity × Molecular Weight = 0.0099 × 158 = 1.5721 g/l

Result – Concentration of unknown 8olution of given KMnO₄ is 1.5721 g/l

EXPERIMENT No,6

Object – Find out the percentage purity of impure potassium permanganate sample 2.5 grams of which have been dissolved in one litre solution of oxalic acid of $\displaystyle \frac{M}{{30}}$ molarity.

Chemical Equation

Molecular Equation

2 MnO₄ +3H₂SO₄+ 5H₂C₂O₄ →K₂SO₄+ 2MnSO₄+ 8H₂O+ 10CO₂

Ionic Equation

2MnO₄^–+ 16H⁺ + 5C₂O₄^2- →Mn²⁺ + 10CO₂ + 8H₂O

Observation Table :

S.No.	Volume of Known oxalic acid solution taken by pipette V₁ mL	Burette Reading		Volume of KMnO₄ solution (b-a) mL	Concordant Reading V₂mL
S.No.	Volume of Known oxalic acid solution taken by pipette V₁ mL	Initial (a)	Final (b)	Volume of KMnO₄ solution (b-a) mL	Concordant Reading V₂mL
1	20	0.0	19.9	19.9	19.8
2	20	0.0	19.8	19.8
3	20	0.0	19.8	19.8

Calculation-

To calculate the molarity of unknown solution of KMnO₄

It is clear from the reaction that 2 mole of KMnO4 reacts with 5 mole of oxalic acid

Solution of (COOH)₂.2H₂O = Solution of KMnO₄

2 M₁V₁=5 M₂V₂

$\displaystyle {{M}_{2}}=\frac{{2{{M}_{1}}{{V}_{1}}}}{{5{{V}_{2}}}}$

(here, M₁ = M/30, V₁ = 20 ml, V₂ = 19.8 ml)

$\displaystyle {{M}_{2}}=\frac{{2\times 20}}{{5\times 30\times 19.8}}$ M = 0.0134 M

Concentration of KMnO₄ of Solution (in g/l)

= Molarity × Molecular Weight= 0.0134 × 158 = 2.117 g/l

Percentage purity of sample of given impure potassium permanganate

=Concentration of given KMnO₄ Solution/Concentration of impure KMnO₄ Solution

= $\displaystyle \frac{{2.117}}{{2.5}}\times 100$ = 84.68%

Result – Percentage purity of sample of given impure potassium permanganate is 84.68%.

Leave a Reply Cancel reply

Leave a Reply Cancel reply