Effect of Temperature on Solubility of Potassium Chloride in Water Essay

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Pages: 4

IB Chemistry Lab Report Design Example—

Effect of Temperature on Solubility of Potassium Chloride in Water

Research question

How does temperature affect the solubility of potassium chloride in water?


As the temperature of water increases, the particles of solid Potassium chloride, KCl, which are absorbing energy from its surrounding, start moving more easily between the solution and its solid state because. According to the second law of thermodynamics, the particles will shift to the more disordered, more highly dispersed solution state. I predict that as the temperature of a KCl and water mixture increases, then the solubility of the KCl will also increase.


Dependant variable

The dependant
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Weigh each of the 50ml beakers labeled 1, 2, 3, 4, 5 and 6 and record their masses.

Introduce 100ml of distilled water into beakers 1 through 6.

Beakers 1, 2, 3, 4, 5 and 6 are held at 0°C, 10°C, 20°C, 30°C, 40°C and 50°C respectively.

Dissolve 50 grams of Potassium chloride into each beaker by using a stirring rod. The stirring in all 6 beakers should last 10 minutes.

Insert the syringe at the midpoint between the solution's surface and the bottom of the beaker and extract 40ml as such from each solution.

Push the contents of each syringe containing samples from the solutions in beakers 1, 2, 3, 4, 5 and 6 into the 50ml beakers labeled 1, 2, 3, 4, 5 and 6 respectively.

Heat each of the 50 ml beakers until complete evaporation of the water occurs, and then immediately place each beaker on the balance and record the mass.
By subtracting from the combined mass of (1) the salt residue from the evaporated extracted solution and (2) the 50 ml beaker, the mass of the corresponding 50ml beaker, the mass of the salt dissolved in the extracted solution can be obtained. The number of moles can be calculated, by using the formula n= m/MM , where m is the mass of the salt and MM the molar mass, which is equal to 74.60g/mol . The number of moles is then divided by the volume of the extracted sample (40ml). The obtained concentration is then multiplied by a factor of 5, in order to get the concentration of the initial solution. As expected, the