Investigating Factors Affecting the Solubility of an Unknown Powder in Water

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Investigating Factors Affecting the Solubility of an Unknown Powder in Water

Aim: To investigate the effect of temperature on an unknown powder's

solubility.

Prediction: As the temperature increases so will the solubility of the

powder. I think that the temperature and solubility will be directly

proportional - so that at temperature doubles so will the solubility

of the powder.

Hypothesis: I think this because my scientific theory says that:

Molecular solids, like sugar, and ionic solids, salts, both dissolve

in water. However, they both dissolve in different ways. The

intermolecular forces holding molecules of sugar together are quite

weak so when sugar is placed in water these bonds are broken and

individual molecules are released into solution. It takes energy to

break bonds between the molecules and it also takes energy to break

the hydrogen bonds in water. These hydrogen bonds have to be disrupted

in order to insert a sugar molecule into the substance. The energy

needed for this is produced by the forming of bonds between slightly

polar sucrose molecules and polar water molecules. This process works

so well between sugar and water that up to 800g of sugar can dissolve

in one litre of water.

The positive and negative ions in ionic solids, or salts, are held

together by the strong force of attraction between particles with

opposite charges. When a salt dissolves in water the ions are released

and become associated with the polar solvent molecules. Salts separate

from their ions when they dissolve in water.

There are several factors that affect solubility between different

compounds. These are:

* Temperature- If the solution process absorbs energy, endothermic,

then the solubility will be increased if there is a temperature

increase. If the solution

releases energy, exothermic, then solubility will decrease.

* Molecular Size- If the size or weight of the individual molecules is

large then

solubility will be low because larger molecules are difficult to

surround with solvent molecules.

* Polarity- Generally only polar solute molecules will dissolve in

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