Lab 13: Solubility (A Very Salty Lab)

Introduction:

The goal of this inquiry lab was to create and implement our own procedure to identify an unknown solid using experimental data and known solubility data.  We used a graph of the solubility curves of three solids (NaCl, NaNO3, and KNO3) to help identify.  We were only allowed to use the given solid, 10 mL of water, and other equipment such as hot plates, balances, thermometers, flasks, and stirring rods.

Important terms:

-solute: the substance that is dissolved
-solvent: the substance that the solute is dissolving in
-solubility: the measure of the amount of solute that will dissolve in a given amount of solvent

Procedure:

1.  We chose 45°C as our testing temperature because the solubility varied greatly at that temperature, so it was easier to differentiate between the solids based on the amount of sample dissolved.

2.  We measured 10 mL of water and poured that into a small beaker.  We knew that that was 10 g because 1 gram of water is equal to 1 mL of water.

3.  Next, we filled a larger beaker about halfway with water to act as the hot bath for the small beaker, and placed that on a hot plate.

4.  While that was heating up, we used the solubility graph to find the solubility in grams for each salt at 45°C.  We divided that mass by ten because the mass given on the graph is per 100g of water, and we only had 10g water.

5.  We measured out 4.8 grams of NaCl, which was one gram more than the mass on the solubility line, so we would know for sure if the substance was above the solubility line and unsaturated or saturated.

6.  When the thermometer in bath on the hot plate reached 45°C, we added that mass of substance to the small beaker and placed that into the bath (bigger beaker).  After stirring with a stirring rod for a couple of minutes, we checked to see if the substance had dissolved, which it had, meaning it was an unsaturated solution.

7.  We then moved on to the next salt, KNO3, and added 3.5 more g to the beaker to obtain 8.3 total g of substance, which was one more gram than the mass on the solubility line for KNO3.

8.  Because the substance dissolved for both NaCl and KNO3, we knew that the mystery substance was NaNO3.  11.7 grams of NaNO3 did not dissolve in the 10 g of water, making it a saturated solution.

Data:

Trial 1 (NaCl test):

Temperature - 45°C
Mass of solute - 4.8 grams
Mass of solvent (water) - 10 grams
Dissolved

Trial 2 (KNO3):

Temperature - 45°C
Mass of solute - 8.3 grams
Mass of solvent (water) - 10 grams
Dissolved

Trial 3 (NaNO3):

Temperature - 45°C
Mass of solute - 11.7 grams
Mass of solvent (water) - 10 grams
Not dissolved

Conclusion:

The mystery substance we had was NaNO3.  We found this out because in the first trial, we measured 4.8 grams of solute into 10 grams of solvent at 45°C and the solute dissolved.  In the second trial, we measured 8.3 grams of solvent into the same amount of solvent at the same temperature and the solute again dissolved.  This left NaNO3 as the identity, because the solution would be saturated if it was NaCl or KNO3.  The relationship between the solubility of a substance and the temperature is that as solubility increases, temperature also increases.  For NaNO3, the saturation point at 45°C is 38 grams, and the saturation point at 100°C is 242 grams.  A challenge I encountered during the trials was making sure the temperature was right at 45°C when we started adding and stirring the substance, or else the results would be off.