To estimate the amount of oxalic acid dissolved in 500 ml of the given solution volumetrically.

**Estimation of
oxalic acid**

To
estimate the amount of oxalic acid dissolved in 500 ml of the given solution
volumetrically. For this you are given with a standard solution of ferrous
ammonium sulphate (FAS) of normality 0.1 N and potassium permanganate solution
as link solution.

During
these titrations, oxalic acid is oxidized to CO_{2} and MnO_{4}-
ions (from KMnO_{4} ) is reduced to Mn^{2+} ion.

**Oxidation : **MnO ^{−} + 8H^{+}
+ 5e^{−} → Mn^{2+} + 4H O

**Reduction : **

**Overall reaction**

5(COOH)_{2}
+ 2MnO_{4 }^{−}
+ 6H^{+} → 10CO_{2} + 2Mn^{2+} + 8H_{2}O

Since one
mole oxalic acid releases 2 moles of electrons, the equivalent weight of oxalic
acid = 106/2 = 63 (oxalic acid is dihydrated)

(Link
KMnO_{4})Vs **(**Standard FAS )

Burette
is washed with water, rinsed with KMnO_{4} solution and filled with
same KMnO_{4} solution up to the zero mark. Exactly 20 ml of standard
FAS solution is pipetted out into the clean, washed conical flask. To this FAS
solution, approximately 20ml of 2N sulphuric acid is added. This mixture is
titrated against KMnO_{4} Link solution from the burette. KMnO_{4}
is added drop wise till the appearance of permanent pale pink colour. Burette
reading is noted and the same procedure is repeated to get concordant values.

**(**Link KMnO_{4} )Vs **(**Standard FAS solution)

Volume of
KMnO_{4} (link) solution V_{1} = ml

Normality
KMnO_{4} (link) solution N_{1} = ? N

Volume of
standard FAS solution V_{2} = 20 ml

Normality
of standard FAS solution N_{2} = 0.1 N

According
to normality equation:

V_{1}
x N_{1} = V_{2} x N_{2}

N_{1}
= V_{2} x N_{2} / V_{1}
=

Normality
KMnO_{4} (link) solution N_{1} = ____________ N

**(**Unknown oxalic acid ) Vs (Link** **KMnO_{4}** **)

Burette
is washed with water, rinsed with KMnO_{4} solution and filled with
same KMnO_{4} solution up to the zero mark. Exactly 20 ml of unknown
oxalic acid solution is pipetted out into the clean, washed conical flask. To
this oxalic acid solution approximately 20ml of 2N sulphuric acid is added.
This mixture is heated to 60 – 700C using Bunsen burner and that hot solution
is titrated against KMnO_{4} Link solution from the burette. KMnO_{4}
is added drop wise till the appearance of permanent pale pink colour. Burette
reading are noted, the same procedure is repeated to get concordant values.

(Link
KMnO_{4} )Vs **(**Unknown oxalic
acid)

Volume of
Unknown oxalic acid solution V_{1} = 20 ml

Normality
of Unknown oxalic acid solution N_{1} = ? N

Volume of
KMnO_{4} (link) solution V_{2} = ml

Normality
KMnO_{4} (link) solution N_{2} = N

According
to normality equation:

V_{1} x N_{1} = V_{2}
x N_{2}

N_{1}
= V_{2} x N_{2} / V_{1}

Normality
of Unknown oxalic acid solution N1 = ________ Y _______ N

The
amount of oxalic acid dissolved in 1 lit of the solution =(Normality) x (equivalent weight)

The
amount of oxalic acid dissolved in 500 ml of the solution = [Y × 63 × 500] / 1000

= [ ? x
63 x 500] / 1000

= ____ g

The
amount of oxalic acid dissolved in 500 ml of given the solution = g

Tags : Volumetric Analysis | Chemistry Practical Laboratory Experiment , 12th Chemistry : Practicals

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12th Chemistry : Practicals : Estimation of oxalic acid | Volumetric Analysis | Chemistry Practical Laboratory Experiment

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