Determine the Equivalent Weight of a Metal Volumetrically

THEORY:

Suppose, the given metal is magnesium. A definite weight of the metal is taken in the chemical balance. It is allowed to react with a known excess volume of fairly strong H₂SO₄ which has been already standardised. The metal consumes part of the acid. The excess unused acid remaining after reaction is calculated by back titration against the standard alkali supplied. Then, by knowing the volume of the acid used by the metal, the equivalent wt. of the metal can be calculated.

We have,

Equivalent wt of the acid = Eq. wt. of the metal

or 1000 ml of (N) acid = Eq wt of the metal

APPARATUS:

1) Burette 

2) Pipette 

3) Conical flask 

4) Burette stand

5) Clamps 

6) Measuring flask

CHEMICAL:

1) Methyl orange or phenolphthalein

PROCEDURE :

i) The given piece of magnesium metal is cleaned by using a sandpaper, and is accurately weighed in the chemical balance.

ii) Introduce the metal in a 250 ml measuring flask, and add a known volume of fairly strong H₂SO₄ (say 25 ml). which has already been standardised.

iii) After the completion of reaction, dilute the acid with distilled water, and make up to the volume. Shake well.

iv) Rinse a clean burette with this diluted acid, and fill it up.

v) Titrate this diluted acid against the standard alkali supplied.

RESULTS:

Wt. of magnesium metal taken = 0.18 gm

Vol. of strong H₂SO₄ taken = 25 ml of 2N H₂SO₄

Solution taken in the burette = diluted acid

Alkali taken for each titration = 10 ml

Indicator used = methyl orange

End point = from yellow to light pink

Normality of standard alkali = N / 10 (f = 1.148)

BURETTE READINGS:

No. of obs	Initial reading	Final reading	Vol. of acid (ml)	Mean
1			8.5 ——->	(Approx)
2			8.2	8.2
3			8.3
4			8.2

CALCULATIONS :

Volume of strong acid taken = 25 ml of (2 N) H₂SO₄

          = 50 ml of (N) H₂SO₄.

From titration.

8.2 ml of diluted acid = 10 ml of N / 10 (f = 1.148) alkali

=(10 x 1 x 1.148) / 10  ml (N) alkali

= 1.148 ml (N) alkali.

250 ml of diluted acid= 1.148 × 250 / 8.2  ml (N) alkali

= 35.0 ml (N) alkali

= 35.0 ml (N) H₂SO₄

= Vol. of acid left.

 Vol. of acid used by 0.18 gm metal

= (50 – 35) ml of (N) H₂SO₄

= 15 ml of (N) H₂SO₄

Hence, 15 ml of (N) H₂SO₄ = 0.18 gm Mg

1000 ml of (N) H₂SO₄ =0.18 × 1000 / 15   gm Mg = 12 gm

Equivalent wt. of Mg = 12.