To determine the solubility of inorganic salt at room temperature.

 Experiment

DETERMINATION OF SOLUBILITY OF DRUG AT ROOM TEMPERATURE

Aim:

To determine the solubility of inorganic salt at room temperature

Requirements:

• Pure inorganic salt 

• Distilled water

• Porcelain dishes (or) watch glasses.

• Beaker

• Pipette(10 ml)

Theory:

Principle:

Solubility is defined in quantitative terms as the concentration of solute in a saturated solution as a certain temperature qualitatively its defined as a spontaneous interactions of two (or) more subtances to form a homogeneous molecular dispension. 

Quantitatively solubility is expressed in molarity, Molarity and percentage (w/v (or) w/w). a saturated solution is defined as the one in which the solute is in equilibrium in with the solid phase (solute) at a certain temperature.

A solution may sometimes contain more amount of solute than in saturated one, such a solution is known as super saturated solution

The solubility of the solid depends on temperature, melting point of the solid and molar heat of 

fusion. 

Procedure:

1. Clean all glasswares using detergent solution and chronic acid solution.

2. Wash two (or) three times using purified water.

3. Take 50ml of distilled water in a beaker (100).

4. Add some amount of Nacl in distilled water and stir using glass rod (or) by an electric motor 

driven shaker.

5. Take sample using a pipette with a piece of filter paper, tie at the tip of the pipette. So that 

the insoluble salt can be removed. 

6. Remove the piece of the filter paper from the tip of the pipette and transfer 10l of the 

solution in weighed porcelain dish (or) watch glass used in direct heat (or) on the water bath 

(or) micro oven.

7. Dry the solution till constant weight. 

8. Take weight of the dish using double pan balance (or) chemical balance calculate weight of the powder.

9. Record all the parameter in the form of table are in proper way of that it becomes easy to

understand the concept of solubility.

Observation Table:

Calculation:

Solubility of the salt= wt. of residue ×100

                               Wt. of solvant 

 

                               W3-W1 ×100

                     W2-W3

Result:

Solubility of the given salt at a room temperature was found to be –