Lab 13: Solubility

Introduction: Solubility is the property of one substance dissolving in another. In this lab we were given a solute and were asked to figure out what the solute was. This was based on the solubility of the salt in 10 mL of water at 50 degrees Celsius. By dissolving a certain amount of solute in the water and looking if there is any solute left undissolved we can figure out what solute it based on the graph given. Our solute was Potassium Nitrate.

Procedure:

1. Measure out 13 grams of solute between three weigh boats in 5 grams, 5 grams, and 3 grams.
2. Measure out 10 mL of distilled water into a graduated cylinder and then pour into a beaker
3. Plug in hot plate and turn it on
4. Fill up a large beaker with water and place on hot plate, check the temperature with a thermometer until the water reaches of 50 degrees Celsius.
5. Place small beaker in large beaker filled with water and heat indirectly until small beaker has a temperature of 50 degrees Celsius (take care that no water from the large beaker enters the small beaker)
6. Pour 5 grams of solute in and stir with glass stirrer keeping the water at 50 degrees Celsius (this may require taking off the heat, just check the temperature with the thermometer every so often)
7. If the solute dissolves repeat step 6 pouring in the other 5 grams of solute
8. If solute dissolves repeat step 6 again pouring in the final 3 grams

Observations (Qualitative Data):  I observed that when the first amount of solute was poured in it started as cloudy but then the solution became clear and no particles of solute were left at the bottom. When the second amount of solute was added it remained cloudy when stirred, and when the solution settled there was particles of solute at the bottom of the beaker. That meant that the solution was saturated.

Quantitative Data:

Grams of Water	10 grams
Grams of Solute (Trial 1)	5 grams
Grams of Solute (Trial 2)	10 grams
Grams of Solute (Trial 3)	13 grams
Temperature of Water	50 °C
Trial in which Water was Saturated	Trial 2

Conclusion: Since the water was saturated with 10 grams of solute of at 50 °C that means according to the graph we were given the solute was Potassium Nitrate. We knew that the solution was saturated with 10 g of solute at 50 °C because there were particle of the solute at the bottom of the beaker, meaning it had not dissolved all the way making it saturated. That differed from when 5 g of solute were dissolved in the solution at 50 °C because then the solution had now particles floating around and was completely clear, so the solute was dissolved. So, the solute could not be Sodium Chloride because sodium Chloride would have saturated the solution at 5 g and it could not be Sodium Nitrate because the solute would have dissolved all the way if there was 10 g of solute and the solute was Sodium Nitrate. This means that the solute could only be Potassium Nitrate. Certain amounts of solute dissolve in solutions at certain temperatures because the transfer of energy from breaks the intermolecular forces, and the strength of the forces depend on the molecule. For Ions the more heat the more solute can be dissolved for gases the opposite is true.