Question

How could I determine experimentally the solubility of sodium bromide in water?

Answer 1

Here are two methods.

Explanation:

Method 1

Put a known amount of solvent — for example, 100 mL — in a container. Slowly add known amounts of the solute until no more will dissolve, even after vigourous and prolonged stirring.

If you do this experiment at 20 °C with water as the solvent and sodium bromide as the solute, you will find that 90.8 g of the salt dissolve in 100 mL water. The solubility of sodium bromide is 90.8 g/100 mL water at 20°C.

Method 2

This is a little more complicated.

Repeat the above process, but this time add a little extra solute, so that some solid remains undissolved.

Heat the mixture until all the solute dissolves. Then cool the solution slowly to its original temperature.

Excess solute should precipitate out of the solution. If it does not, you may have a supersaturated solution. Add a small crystal of the solute to induce crystallization.

Filter the crystals through a pre-weighed filter paper. Pour a small amount of a slightly polar solvent such as acetone through the filter paper. The acetone will dissolve the water but, we hope, not the crystals.

Let the paper dry. Determine the mass of paper + crystals, and hence the mass of crystals. Subtract the mass of crystals from the original mass of solute.

For example, assume you added 94.6 g of NaBr to 100 mL of water at 20°C. You heat to 30 °C to dissolve everything, cool to 20 °C, and recover 3.8 g of NaBr.

The mass of NaBr in the solution is (94.6 – 3.8) g = 90.8 g. The solubility is 90.8 g/100 mL of water at 20 °C.