This experiment is to investigate if CaCl2 and Na2CO3 in Distilled Water, Hydrochloric Acid, Sodium Hydroxide, Butanol and Ethanol at room temperature.

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Sagar Manilal

15-Nov-2003

Chemistry Lab: Solubility of Solids

Aim

        This experiment is to investigate if CaCl2 and Na2CO3 in Distilled Water, Hydrochloric Acid, Sodium Hydroxide, Butanol and Ethanol at room temperature.

Hypothesis

        This investigation is to see which one of the two solids dissolves in which of the liquids. These solids are CaCl2 and Na2CO3. I find that Na2CO3 may be more soluble in the liquids as the Na is an element that can lose only 1 electron unlike Ca that needs to lose 2. However I find that the difference will be minute since the ionization energies are both in the s-shell. Na has an s-shell missing so it is more likely to lose 1 electron unlike Ca that needs to lose 2. This is difficult for the Calcium as the number of protons remains the same and is having a greater pull on the electrons after losing 1 electron.

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Another factor that should be included is that the shielding effect is present in the Ca but not in Na since Ca has more that 3 shells unlike Na. In conclusion to my assumption, I find that Na2CO3 will be more soluble in the liquids than CaCl2.

Materials

Chemicals

Liquids        → H2O, HCl, NaOH, C4H6, CH4COOH

Solids                → CaCl2, Na2CO3

Variables

Dependent Variable        → Solubility

Independent Variable        → Mass of solid, volume of liquid, temperature (22°C)

Control Variable        → Solid and liquid compounds

Procedure

  1. Assemble 10 ml of distilled H2O in a test tube.
  2. ...

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