Titration Experement

Experiment No. 1

Exercise: Find out the concentration of potassium permanganate in gram per litre. For this purpose you are provided with a standard solution of crystalline ferrous ammonium sulphate of molarity M/30

Theory: Titration between potassium permanganate and ferrous ammonium sulphate is a redox titration. KMnO₄ in presence of dil. H₂SO₄ oxidises ferrous ammonium sulphate to ferric sulphate [Fe₂(SO₄)₃] and itself is reduced to manganese sulphate (MnSO₄).

(1) Molecular equation

2KMnO₄+ 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 3H₂O+5[0]

[2FeSO₄(NH₄)₂SO₄.6H₂O + H₂SO₄+ [O] → Fe₂(SO₄)₃ + 2(NH₄)₂SO₄ + 13H₂0] × 5

2KMnO₄+10FeSO₄ (NH₄)₂SO₄.6H₂O+8H₂SO₄→K₂SO₄ + 2MnSO₄+5Fe₂(SO₄)₃ + 10(NH₄)₂SO₄ +68H₂O

(ii) Ionic Equation :

$\displaystyle MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{{2+}}}+4{{H}_{2}}O$

$\displaystyle \left[ {F{{e}^{{2+}}}\to F{{e}^{{3+}}}+{{e}^{-}}} \right]\times 5$

$\displaystyle MnO_{4}^{-}+5F{{e}^{{2+}}}+8{{H}^{+}}\to M{{n}^{{2+}}}+5F{{e}^{{3+}}}+4{{H}_{2}}O$

Indicator :- KMnO4 is self indicator.

Observation table:

Calculations: (A) Calculation of Molarity of KMnO₄

It is clear from the molecular equation of the reaction between KMnO₄ and ferrous ammonium sulphate that 2 moles of KMnO₄ oxidise 10 moles of ferrous ammonium sulphate. i.e. 10 moles of Ferrous ammonium sulphate = 2 mol KMnO₄

$\displaystyle \frac{{{{M}_{1}}{{V}_{1}}}}{{10}}=\frac{{{{M}_{2}}{{V}_{2}}}}{2}$

Or M₁V₁ = 5 M₂V₂

M₁ = Molarity of ferrous ammonium sulphate solution = M/30

V₁ = Volume of Ferrous ammonium sulphate solution = 20 mL

M₂ = Molarity of KMnO₄ solution = ?

V₂ = Volume of KMnO₄ solution = = 19.8 mL

M₂ = $\displaystyle \frac{{{{M}_{1}}{{V}_{1}}}}{{5{{V}_{2}}}}$

M₂ = $\displaystyle \frac{{\frac{M}{{30}}\times 20}}{{5\times 19.8}}$

M₂ = $\displaystyle \frac{2}{{3\times 5\times 19.8}}$ M

(B) Calculation of Strength is gms/litre of KMnO₄ solution:

Strength is gm/litre = Molarity × Molar mass

$\displaystyle \frac{2}{{3\times 5\times 19.8}}\times 158$

= 1.0640 gm per litre

Result: The strength of the given KMnO₄ solution is 1.0640 gL^-1

Experiment No. 2

Exercise: Find out the molarity of potassium permanganate solution. For this purpose you are provided with a standard solution of crystalline ferrous ammonium sulphate containing 13.0666 grams of the salt per litre.

Theory: Titration between potassium permanganate and ferrous ammonium sulphate is a redox titration. KMnO₄ in presence of dil. H₂SO₄ oxidises ferrous ammonium sulphate to ferric sulphate [Fe₂(SO₄)₃] and itself is reduced to manganese sulphate (MnSO₄).

(1) Molecular equation

2KMnO₄+ 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 3H₂O+5[0]

[2FeSO₄(NH₄)₂SO₄.6H₂O + H₂SO₄+ [O] → Fe₂(SO₄)₃ + 2(NH₄)₂SO₄ + 13H₂0] × 5

2KMnO₄+10FeSO₄ (NH₄)₂SO₄.6H₂O+8H₂SO₄→K₂SO₄ + 2MnSO₄+5Fe₂(SO₄)₃ + 10(NH₄)₂SO₄ +68H₂O

(ii) Ionic Equation :

$\displaystyle MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{{2+}}}+4{{H}_{2}}O$

$\displaystyle \left[ {F{{e}^{{2+}}}\to F{{e}^{{3+}}}+{{e}^{-}}} \right]\times 5$

$\displaystyle MnO_{4}^{-}+5F{{e}^{{2+}}}+8{{H}^{+}}\to M{{n}^{{2+}}}+5F{{e}^{{3+}}}+4{{H}_{2}}O$

Indicator :- KMnO₄ is self indicator.

Observation table:

Calculations: (i)Molarity of Ferrous ammonium sulphate :

Weight of Ferrous Ammonium sulphate = 13.0666 grams (given)

Volume of Ferrous Ammonium sulphate solution = 1000 mL

Molar mass of Ferrous Ammonium sulphate (FAS) = 392.12

Molarity of solution M₁ = $\displaystyle \frac{{Conc.(gm/litre)}}{{Molar\text{ mass}}}$

Molarity of the solution (M₁) = $\displaystyle \frac{{13.0666}}{{392.12}}$ M

(2) Calculation of Molarity of KMnO₄

$\displaystyle \frac{{{{M}_{1}}{{V}_{1}}}}{{10}}=\frac{{{{M}_{2}}{{V}_{2}}}}{2}$

Or M₁V₁ = 5 M₂V₂

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

V₁ = Volume of Ferrous ammonium sulphate solution = 20 mL

M₂ = Molarity of KMnO₄ solution = ?

V₂ = Volume of KMnO₄ solution = 19.8 mL

M₂ = $\displaystyle \frac{{13.0666}}{{392.12}}\times \frac{{20}}{{5\times 19.8}}$ M

M₂ = 0.0067 M

Result: The molarity of the given KMnO₄ solution is 0.0067 mol L^-1

Experiment No. 3

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

(1) Molecular equation

2KMnO₄+ 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 3H₂O+5[0]

[2FeSO₄(NH₄)₂SO₄.6H₂O + H₂SO₄+ [O] → Fe₂(SO₄)₃ + 2(NH₄)₂SO₄ + 13H₂0] × 5

2KMnO₄+10FeSO₄ (NH₄)₂SO₄.6H₂O+8H₂SO₄→K₂SO₄ + 2MnSO₄+5Fe₂(SO₄)₃ + 10(NH₄)₂SO₄ +68H₂O

(ii) Ionic Equation :

$\displaystyle MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{{2+}}}+4{{H}_{2}}O$

$\displaystyle \left[ {F{{e}^{{2+}}}\to F{{e}^{{3+}}}+{{e}^{-}}} \right]\times 5$

$\displaystyle MnO_{4}^{-}+5F{{e}^{{2+}}}+8{{H}^{+}}\to M{{n}^{{2+}}}+5F{{e}^{{3+}}}+4{{H}_{2}}O$

Indicator :- KMnO₄ is self indicator.

Observation table:

Calculations: (i) Calculation of Molarity of KMnO₄

$\displaystyle \frac{{{{M}_{1}}{{V}_{1}}}}{{10}}=\frac{{{{M}_{2}}{{V}_{2}}}}{2}$

Or M₁V₁ = 5 M₂V₂

M₁ = Molarity of ferrous ammonium sulphate solution = $\displaystyle \frac{M}{{30}}$ M

V₁ = Volume of Ferrous ammonium sulphate solution = 20 mL

M₂ = Molarity of KMnO₄ solution = ?

V₂ = Volume of KMnO₄ solution = 19.6 mL

M₂ = $\displaystyle \frac{{{{M}_{1}}{{V}_{1}}}}{{5{{V}_{2}}}}$

M₂ = $\displaystyle \frac{{\frac{M}{{30}}\times 20}}{{5\times 19.6}}$

M₂ = $\displaystyle \frac{2}{{15\times 19.6}}$ M

(ii) Strength purity of KMnO₄ =

Strength (gL^-1) = molarity × molar mass

= $\displaystyle \frac{2}{{15\times 19.6}}\times 158$

(iii) Percentage purity of KMnO₄ = $\displaystyle \frac{{\text{Strength of given KMn}{{\text{O}}_{4}}\text{ in g}{{\text{L}}^{{-1}}}}}{{\text{Weight of KMn}{{\text{O}}_{4}}\text{ in g}{{\text{L}}^{{-1}}}}}\times 100$