SS2 Chemistry Lesson Note on Solubility

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SOLUBILITY

Definition: The Solubility of a solute (solid) in a solvent (liquid) is the concentration of the saturated solution. Solubility can be defined thus:

1. It is the maximum amount of the solid that dissolves in 1 dm³ of the solution at a given temperature. It is expressed in mol.dm^-3.
2. It is the maximum mass, in grams, of the solid that dissolve in 100 g of the solvent at a given temperature.

When some common salts is added to a beaker of water and the mixture stirred, the salt gradually disappear, and the clear colourless mixture is obtained. The salt is said to have dissolved in the water. Thus, the salt that dissolved in the water is called solute and the water that does the dissolving is known as the solvent and the product obtained by dissolving the salt in the water is called a solution.

Therefore, a solution is a homogenous mixture which is formed when a solute is completely dissolved in a solvent. A solution can be saturated or supersaturated.

Determination of solubility of KNO₃ at 30^oC

Stage 1: Preparing of Saturated Solution

1. Put 100 cm³ of distilled water in a beaker, add the salt little at a time and warm on a Bunsen burner with stirring. Continue the addition of the salt with stirring and keeping the beaker warm at about 50^oC, until the salt can no longer dissolve.
2. Allow the saturated solution to cool to 30^oC; this is the saturated solution of the salt at 30^oC. It will be noticed that as the hot saturated solution cools, the excess solid in the solution separates out, to give a saturated solution at a lower temperature.
3. Stage 2: Evaporation of Saturated Solution to Dryness.
4. Weigh accurately a clean dry evaporating dish: x g.
5. Rinse a 25 cm3 pipette with the saturated solution, in order to warm it.
6. Pipette 25 cm³ of the saturated solution at 30Oc, as quickly as possible (without pipetting any undissolved solid) into the evaporating dish and re-weigh: y g
7. Place the dish on a steam bath and evaporate to dryness. Allow to cool in a charged desiccators, and then re-weigh.
8. Repeat the process of heating and cooling, until a constant mass is obtained: w g.

Calculation of the solubility at 30^oC

The following hypothetical values will be used to illustrate how to calculate the solubility of a salt.

Data required:

• Mass of the evaporating dish:                                                x = 15.20 g
• Mass of evaporating dish + 25 cm³ of saturated solution: y = 40.70 g
• Mass of dish + anhydrous salt:                                              w = 20.70 g

Solubility in mol.dm^-3

Mass of anhydrous salt  = (w – x)   = (20.70  – 15.20) g = 5.50 g

i.e. 25 cm³ of saturated solution contain 5.50 g of salt,

Hence, 100 cm³ of saturated solution contain 5.50  x 1000/25  = 220 g of salt.

To convert 220 g to moles

Amount (mol)  = (Mass of salt)/ (Molar mass).

Molar mass of KNO₃ = 39 + 14 + 48 = 101 g.mol^-1

i.e. Amount, in moles = 220/ 101  = 2.18 moles

Therefore, Solubility of KNO₃ at 30^oC = 2.18 mol.dm^-3

Solubility in grams per 100 g of water

Mass of water used = (y – w)  = (40.70  – 20.70 ) g = 20.0 g

Mass of salt used =      (w – x)  = (20.70  – 15.20) g = 5.50 g

i.e. 20.0 g of water at 30^oC saturated 5.50 g of salt

Therefore, 100 g of water will saturate 5.50 x 100/20 g of salt

i.e.                                          5.50 x 5 g = 27.5 g of salt.

Hence, the solubility of the salt at 30^oC is 27.5 g per 100 g of water.

WORKED EXAMPLES

1. If 11.87g of potassium trioxonitrate(V) were dissolved in 43.4 g of distilled water at 60 , calculate the solubility of the solute in moldm^-3

Solution:

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