Solubility Practice a Curvy Subject

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Chemistry I
EQ: Why do some substances dissolve in water
while others do not?
Solubility Practice a
Curvy Subject
Enduring Understanding
Resource use is determined by properties and availability.
Broad Brush Knowledge Concentrations, Measurement, Colligative Properties,
Saturation
Concepts Important to Know and Understand
Chemical Quantities, Interactions of Matter
Core Objectives: Describe the nature of water and aqueous solutions:
Describe the structure of the water molecule and explain how its polarity affects
its properties such as surface tension, solubility, boiling point, freezing point, and
specific heat. Describe solutions in terms of solute, solvent, solvation, degrees of
saturation, relative concentration, and colligative properties.
TEKS 12A demonstrate and explain effects of temperature and the nature of solid solutes on the solubility of solids.
13 The student knows relationships among the concentration, electrical conductivity, and colligative properties of a solution. The student is
expected to: A: compare unsaturated, saturated, and supersaturated solutions B: interpret relationships among ionic and covalent
compounds, electrical conductivity, and colligative properties of water, 2D: organize, analyze, evaluate, make inferences, and predict trends
from data; 2E: communicate valid conclusions.
TEACHER MANAGEMENT
Estimated Time: 20-30 minutes could be used as homework in Unit I Section C.1.
Materials: Copies for all students
Teacher Prep: None
Safety Precautions: None
PRACTICE
PURPOSE: to practice reading and interpreting solubility graphs.
Solubility Graph of Solids and Gases
KI KNO 3 NaNO 3
140
130
120
110
Grams of solute / 100. g H2O
100
90
80
70
60
50
40
30
20
10
0 10 20 30 40 50 60 70 80 90 100
NH 4 Cl
KClO 3
KCl
HCl
NaCl
NH 3
SO 2

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KEY
Temperature ( 0 C)
Directions: Answer in complete sentences and show all work for calculations.
1. What relationship exists between solubility and temperature for most of the substances shown?
A direct relationship is shown for most substances, as temperature increases the
solubility of most substances increases.
2. What is the exception?
The exception is gas solubility. Gases are less soluble at higher temperature,
illustrating an indirect relationship
3. What explains why solids become more soluble as temperature increases and why gases become
less soluble? Use the following words to help justify your response: temperature, collisions and
energy.
As temperature increases both speed of the water molecules and the number of
collisions increase. A solid remains dissolved by the continual motion of the eater
molecules. The more energy and motion, the more particles that can remain in
solution. This is especially true for ionic compounds, because water is polar and the
opposite charges help stick the water molecules to the ions. Gases are often nonpolar
and they have very low boiling points, so as temperature increases they tend to “boil
away”. That is why a soda will go flat much faster if it is left in a warm place.
4. Which is more soluble NaNO 3 or KCl? Does the temperature matter? Why or why not?
NaNO 3 because the higher line indicates that more NaNO 3 can be dissolved at ALL
temperatures.
5. How does the line drawn for a particular substance relate to the saturation of a solution of that
substance? Use the following words to help justify your response: saturated, unsaturated,
supersaturated.
The line marks the saturation point of a solute in 100 grams of water. If you have less
grams dissolved than indicated by the line then your solution is unsaturated at that
temperature. If you have more, then your solution is supersaturated at that
temperature.
6. How many grams of NH 4 Cl will dissolve in 100. grams of water at 90. 0 C? 72 grams
7. How many grams of KClO 3 will dissolve in 300. grams of water at 30. 0 C?
300. g of H 2 O X 10. g KClO 3 = 30. g KClO 3
100. g H 2 O
8. How would you make a saturated solution of KNO 3 at 60. 0 C in 50. grams of water?
The graph indicates that that 106 grams of KNO 3 will dissolve in 100. grams of water,
so I need only 106g /2= 53 grams of KNO 3 to make the solution. So I would measure
out 50 grams of water and 53 grams of KNO 3 and mix them together to make the
solution.
9. If you were asked to make a saturated solution of KCl in 100. grams of water, what other piece of
information would you need to before you could start? Why?
You would need to know the temperature of the solution. The saturation point is very
different depending on the temperature
10. If you start with a saturated solution of NH 3 in 100. grams of water at 10. 0 C, how many grams of NH 3
gas would bubble out of solution if you raise the temperature to 80. 0 C?

100. grams of water at 10. 0 C can hold 70. grams of NH 3 .
100. grams of water at 80.0C can hold 14 grams of NH 3 .
So 70. g -14 g = 56 grams of NH 3 will bubble out.
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11. A saturated solution of NaNO 3 was made with 300. grams of water at 40. 0 C. How much NaNO 3
could be recovered by evaporating the solution to dryness?
So 300. grams of water can hold 3 X 106 g = 318 g NaNO 3 will be recovered.
12. A saturated solution KNO 3 in 400. grams of water at 50. 0 C is cooled to 10. 0 C. How much KNO 3 will
come out of solution as crystals?
From the graph 100. grams of water at 50. 0 C can hold 83 g of KNO 3
So 400. grams of water can hold 4 X 83 g = 332 g KNO 3
From the graph 100. grams of water at 10. 0 C can hold 21 g of KNO 3
So 400. grams of water can hold 4 X 21 g = 84 g KNO 3
As the solution is cooled 332 g- 84g = 248 g KNO 3 will form crystals.
13. You start with 200. grams of ice saturated with SO 2 at 0. 0 C. How many grams of SO 2 will bubble out
of solution if you melt the ice and raise the temperature of the water to 80. 0 C?
From the graph 100.g of ice at 0. 0 C can hold 23 g of SO 2
So 200. grams of ice can hold 2 X 24 g = 48 g SO 2
From the graph 100g. of water at 80. 0 C can hold 4 g of SO 2
So 200. grams of water can hold 2 X 4g = 8 g of SO 2
As the solution is heated 48 g – 8 g = 40 g of SO 2 will bubble out.