Lab 13: Solubility: A Guided Inquiry Lab

Introduction

In this lab, my lab partner Meghana and I used the "Solubility of 3 Salts" graph in order to design our own procedure to find the identity of the solid we were given. The solubility curve graph clearly illustrates the curve in which each solid (NaNO3, KNO3, and NaCl) will dissolve or not dissolve in water. The curve follows the largest mass in grams of each solid at degrees Celsius in which the solute will completely dissolve. Because of this, any amount of solid (in grams) below the line will make an unsaturated solution and completely dissolve, and any amount of solid above the line will be a saturated solution and not completely dissolve. Therefore, by making the solution one temperature and seeing if each mass dissolved or not, we were able to find which solid we had.

Procedure

1. We chose to raise the temperature of the solution to 44 degrees Celsius because at that temperature, there were large gaps in between the mass that was on the solubility line.
2. We measured out 5 grams of the solid in order to test if the solid was NaCl, as it was 3.7g at the solubility line, and if it dissolved, we knew that the solution would be unsaturated, and therefore not NaCl since it was above the solubility line.
3. We filled a larger beaker that would act as a hot bath, and filled the smaller beaker to 10 mL. Since we filled it to 10 mL, we divided all the values on the graph by 10 since the graph was based on 100g/100mL of H2O.
4. We heated the smaller beaker to 44 degrees Celsius by placing it in the larger beaker and placing that on the hot plate.
5. Once the smaller beaker was 44 degrees C (measured with a thermometer) we took it out of the larger beaker and checked if the solid was dissolved. After stirring, it dissolved completely, so we knew it couldn't be NaCl.
6. We added 4 grams to the small beaker, making the total 9g, as KNO3's solubility line value was 7g at 44 degrees C.
7. We again heated the smaller beaker to 44 degrees C and stirred the solid to see if it would dissolve. The solid dissolved, so we knew the solution was unsaturated, and therefore couldn't be KNO3, as it was above the solubility line.
8. Since the solid was dissolved for both tries and therefore was not NaCl or KNO3, we knew that it had to be NaNO3. NaNO3 was the only solid whose solubility line was above 9 grams at 44 degrees C, and the only solid in which 9 grams would make a unsaturated solution in 10 mL of water.

Data

• Trial One:
• Temperature: 44 degrees Celsius
• Mass of Solute: 5 grams
• Mass of Solvent: 10 grams
• Dissolved/Didn't Dissolve: Dissolved

• Trial Two:
• Temperature: 44 degrees Celsius
• Mass of Solute: 9 grams
• Mass of Solvent: 10 grams
• Dissolved/Didn't Dissolve: Dissolved

Conclusion

The unknown solid we had was NaNO3. We know this because in the first trial, we measured 5 grams of solute into 10 grams of solvent, and after being heated to 44 degrees, the solute dissolved. This meant that the solid could either be KNO3 or NaNO3, since if it was NaCl, it would be a saturated solution. Then in the second trial, we measured 9 grams of solute into 10 grams of solvent, and after being heated to 44 degrees, the solute dissolved again. This meant that the solid had to be NaNO3, since if it was KNO3, it would be a saturated solution and wouldn't dissolve. The relationship between temperature and solubility of a solid is that as the temperature of the solution increases, the solubility of the solid increases. For example, in NaNO3, the largest amount of grams that could be dissolved completely is about 10g at 36 degrees Celsius, and at 100 degrees Celcius, the number is 18g.

9 grams of our solid dissolved at 44 degrees Celsius