Solubility and Saturation: Common Salt

Unit: Matter and Its Behaviour

Matter is defined in the language of science; anything around us that has some mass and occupies a certain space when we keep it is called matter.

A mixture is obtained by physically intermixing two or more components in any proportion. The components of a mixture retain their individual properties.

We can have different types of mixtures depending on the nature of the components that form a mixture.

1. A mixture whose composition is uniform is called a homogeneous mixture or solution. 
2. Mixtures that contain physically distinct parts and have non-uniform compositions are called heterogeneous mixtures. 

Solution

A solution is a homogeneous mixture of two or more substances. A salt solution, lemonade or soda water are all examples of solutions.

A solute (salt, sugar, etc.) dissolved in a solvent (water, alcohol, etc.) is known as a solution. Hence, in general, there are two components in a solution: solute and solvent. One component (solvent) is present in a larger amount and dissolves another component (solute), present in a smaller amount.

Similarly, there are solid solutions (alloys) and gaseous solutions (air). A saturated solution is one in which no more solute can be dissolved in the solvent at a given temperature. As the temperature of the solvent increases, the solvent is capable of dissolving more solute. 

• Saturated solution: A saturated solution is a solution in which no more solute can be dissolved at a given temperature. 
• Solubility: Solubility is determined in grams per 100 g of solvent. 

Here, solubility can be obtained by two methods.

The first one being the density method and the other being the evaporation method. Density is mass per unit volume of a solution. In this method, the mass of the saturated solution and the density of water are used to determine the mass of the solute present in 100 g of water. In the evaporation method, 100 mL of a saturated solution is evaporated to obtain the amount of solute dissolved in the solvent to obtain its solubility. 

Let us understand this through a simulation. 

The aim of this experiment is to prepare a saturated solution of common salt in distilled water and to determine its solubility at room temperature.

Apparatus Required

The apparatus required for this experiment will be: 

• Distilled water
• Measuring cylinder
• Beakers
• Tripod stand
• Wire gauze
• Burner
• Common salt in watch glass
• Filter paper
• Funnel
• Beaker with saturated solution
• Glass rod
• Digital balance
• China dish
• Tongs
• Spatula

The procedure for this experiment will be: 

To prepare a saturated solution:

1. Take a 250 mL beaker and measure out 150 mL of water into it using a measuring cylinder. 
2. Add some salt into the beaker and stir.
3. Warm the solution slightly. 
4. Add salt into the beaker and stir until saturation is obtained, i.e., some salt remains at the bottom of the beaker.

To determine the solubility of the prepared saturated solution using the density method:

5. Take a clean, empty beaker and weigh it.
6. Tare the weight of the empty beaker. 
7. Transfer 100 mL of the prepared saturated solution using a measuring cylinder into the empty beaker.
8. Record the weight of the saturated solution.
9. Using the density of the distilled water, calculate the amount of solute present in 100 mL of water. 

To determine the solubility of the prepared saturated solution using the evaporation method: 

10. Take a clean and dried china dish and weigh it. 
11. Transfer 25 mL of the prepared saturated solution into the china dish. 
12. Heat to remove all the solvent present until the salt is obtained. 
13. Cool and weigh the china dish with salt. 
14. Record the readings and calculate the amount of solute present in 25 mL of water and 100 mL of water.

Thus, it can be concluded that, 

1. A saturated solution was prepared by dissolving salt in 150 mL of water. 
2. The solute was added with constant stirring until a small amount of solute remained at the bottom of the beaker. 
3. Filtration was performed to remove any insoluble particles to obtain a saturated solution at room temperature. 
4. The density method uses the mass of the solution and the density of water to determine the solubility of the solution. By this method, the amount of salt obtained was 36 g in 100 g of water at room temperature. 
5. The evaporation method uses direct evaporation of water from the solution to obtain the salt present in it. By this method, the amount of salt obtained was 36 g in 100 g of water at room temperature.

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