1. Lab #4: Solubility of Salts
Larissa Guillen and Abigail Delgado
Course 1411, Section 520
November 10, 2014
2. Abstract
This paper talks about several points in a conducted lab experiment known as the
“Solubility of Salts”. First we will talk about the purpose of the experiment as well as the
methods being conducted. Next, the materials and methods that we used for the experiment will
be discussed. We will then talk about the results of the experiment which include the calculations
and the raw data (table and chart). Lastly we will discuss the importance of the results that we
got and the results that others got when they conducted the experiment.
Introduction
3. The purpose of the experiment is to determine the identity and solubility of an unknown
salt. The solubility of a substance is an amount at which the substance can be dissolved in any
given amount of solvent. The theory behind this experiment is to see how much of a substance,
in this case potassium chloride, can be dissolved into another substance, such as water. In a lab
report made by a chemistry teacher, Dr. Breinan, he writes “If you add just a tiny bit of almost
any ionic compound to water, it will dissolve. But if you keep adding more and more salt
crystals, eventually you will reach a point where no more will dissolve, no matter how long you
stir” (Breinan, n.d. pg. 1). If you end up with too much salt left over in your solution, you might
end up with not enough to salt solution and not an accurate amount of salt solution.
When you mix the water and salt together, you want to make sure that you dissolve as
much salt as possible to get a more accurate result to actual result. When you transfer the salt
solution to the evaporating dish, you want to make sure to not get any solid salt in the dish. Then
you want to dry the salt solution until all the water is gone and the salt is completely dry. Then
you want to weigh the salt and evaporating dish until you get a constant weight.
Materials and Methods
Materials Needed:
4. 125 mL Erlenmeyer Flask
Evaporating Dish
Thermometer
Stirring Rod
400 mL Beaker
250 mL Beaker
Hot Plate
Salt Sample
Ring Stand
Ring Support
Bunsen Burner
Method:
A saturated salt solution will be prepared at an assigned temperature. In this case, a saturated
solution is formed when it is equal to the undissolved solute.
5. Experimental Procedure
Obtain a clean, dry evaporating dish and weigh it to the nearest 0.01 gram. Get an
unknown sample of salt from the instructor. The instructor will assign the students a temperature
which will be needed for the saturated solution to be prepared by the student. Get a 125 mL
Erlenmeyer flask, add in the unknown salt sample and 40 mL of distilled water. Heat the water
and salt slowly near boiling or until the salt has dissolved. Get a 400 mL beaker and fill it up
halfway with water. Depending on the temperature assigned, heat or cool down the water to a
temperature about 10 degrees lower or higher than the assigned temperature. Put the lower half
of the Erlenmeyer flask with the salt solution in the water. Put the thermometer in the solution in
the flask and softly stir the flask. Do not use the thermometer to stir the solution because it might
break.
While the solution is cooling down, check to see if the salt starts to crystallize. If there are
no crystals, get more salt and repeat the solution process. When the solution has reached the
temperature assigned, let the salt settle completely. Very carefully pour out 4-5 mL of the
solution into the evaporating dish. Make sure that none of the solid salt gets in the dish. Weigh
the dish and salt solution. Get a 250 mL and fill it up halfway with water to make a hot water
bath. Put the beaker on a ring on the ring stand and heat the water. Using the hot water bath,
carefully evaporate the solution. When the salt looks dry, let the dish cool down to room
temperature and weigh it. Reheat salt for a few minutes, cool down, and weigh it again until it’s
within 0.02 g of the first weight.
Find the weight of water in the solution by subtracting the weight of the salt from the mas
of the solution. To calculate the solubility of the salt by solving for “x” use: (x/100gH20) =
(weight of salt/weight of H2O). Record solubility on the data sheet and turn into the instructor.
6. Results
Assigned Temperature: 20°C
Larissa Abigail
1. Weight of dish and solution: 53.20g 68.2g
2. Weight of dish: 48.86g 40.7g
3. Weight of solution: 4.34g 4.5g
4. Weight of dish and dry salt: 50.59g 42.9g
5. Weight of Salt: 1.73g 2.2g
6. Weight of water: 2.61g 2.3g
7. Solubility of salt: 66.28g/100g H2O 95.6g/100g H2O
Calculation of solubility:
(x/100g H2O) = (weight of salt/weight of H2O)
(x/100g H2O) = (1.73g/2.61g)
(1.73g • 100g H2O) = (x • 2.61g)
173g = 2.61g • x
(173g/2.61g) = (2.61g/2.61g)
x = 66.2835249 or 66.28g/100g H2O
The salt solubility of the unknown salt was 66.28g/100g H2O.
The identity of the salt is Potassium Chloride.
9. The significance of the results is to show the relationship between the temperature and
solubility of a salt when it is in water. In this experiment the end result came out to be
49.64g/100g of H2O and we found that our unknown salt was Potassium Chloride. When we
first conducted the experiment, we first got 66.28g/100g of H2O and 95.6g/100g of H2O. We got
66.28g/100g of H2O as our first result because the salt would not dissolve all the way so we had
very little solution. Our other result that we got was 95.6g/100g of H2O. We got this because we
forgot to weigh the evaporating dish at the beginning of the experiment and then again when it
had the solution in it. When we had finished with our experiment, we told the instructor what our
results were and we wrote them on the board. Another person that did the same experiment as us
had 49.64g/100g of H2O, which was the closest to correct answer and we were told to use his
answer for the end result.
The reason why we got 66.28g/100g of H2O because when we were heating the salt
sample and water, it was taking a long time to heat up and hardly any salt was dissolving. We
ended up to work with the little solution that we had. For the second result, we got 95.6g/100g of
H2O because when we first started the experiment, we forgot to weigh the evaporating dish. We
also forgot to weigh the evaporating dish and salt solution after we dried the salt solution the first
time we had to weigh it. We weighed the evaporating dish and solution three times to find the
constant weight. These errors could have been avoided if we had a properly working hot plate
and if we read the directions over more than once and carefully. Had the hot plate been working
properly, we would have obtained a better amount of salt solution to put in the evaporating dish.
This would have also changed the outcome for the solubility of salt.
The setup could have been more effective if we read the directions before we started the
experiment and highlighted the important parts of the experimental procedure. The purpose of
10. the experiment wasn’t achieved. The actual solubility of potassium chloride is 34.0 g/100 g of
H2O and we were off by 20-60 grams. Our results were off due to mistakes at the beginning of
the experiment and not being able to dissolve all of the salt in the water. Also on the website,
http://www.chem1.com/acad/webtext/solut/solut-6a.html, one section talks about how the
“solubility of usually increases with temperature but not always” (section 1).
When we were heating our salt solution, we had to heat it up to about 10 degrees more
than our assigned temperature of 20 degrees Celsius. Once we heated the solution up, we put it in
an ice water bath to cool it down to our assigned temperature. We believe that we may have left
our salt solution in the ice bath for too long and that affected our end result. The results of our
other classmates were closer to the actual solubility of potassium chloride but because they
didn’t make the mistakes we did. Also on the website
http://chemicals.etacude.com/p/more/kcl.html, it shows in a graph and a chart what the solubility
of potassium chloride. It shows that the solubility is 34.03 g/100 g of H2O, which is the closest
to the actual solubility of the salt.
Conclusion
11. It can be concluded that the point at which potassium chloride salt is soluble in water at a
temperature of 20 degrees Celsius is about 34.0g/100g of H2O. Also the temperature at which
you heat the salt solution makes a difference when you’re trying to find the solubility of any kind
of salt. We also now know that if you have a high amount of salt in your water, it can a long time
or require a higher temperature to dissolve the salt.
12. References
Dr. Breinan. (n.d). Solubility Lab. Retrieved from https://www.google.com/url?
sa=t&rct=j&q=&esrc=s&source=web&cd=6&ved=0CC8QFjAF&url=https%3A%2F
%2Fwww.glastonburyus.org%2Fstaff%2FBREINANH%2Fregular%2Ffirstlabs%2FDocuments
%2FSolubilityCurve.lab.student.doc&ei=Kw9dVPOFJM71yASR2oCIAQ&usg=AFQjCNEVwB
UXnXY8gkM5_F0LZoyO_Kylhg&sig2=jl1mdoxtlClxyIRy3c3vaw.
Solubility equilibria of salts Part 1: Solubility products and calculations. (n.d.). Retrieved
from http://www.chem1.com/acad/webtext/solut/solut-6a.html.
[Solubility of potassium chloride]. (n.d). Retrieved from
http://chemicals.etacude.com/p/more/kcl.html.
