1. solution

3. COMPONENTS OF A SOLUTION
SOLUTE – substance being dissolved; present in lower proportion
SOLVENT – the dissolving medium; present in greater proportion

4. TYPES OF SOLUTIONS SOLUTE SOLVENT EXAMPLE GAS O2 in N2 (air) LIQUID
CO2 in soda
SOLID
H2 in palladium
H2O(g) in air
Alcohol in H2O
Hg in Ag
(dental amalgam)
Sugar solution
Steel (C in Fe)

5. SOLVATION – dissolving of solids accompanied by heat of solution
ENDOTHERMIC – dissolution process absorbs heat from the surrounding
EXOTHERMIC – dissolution process releases heat into the surrounding

6. FACTORS AFFECTING SOLUBILITY
1. Nature of solute and solvent – the solute-solvent interaction must be greater than the solute-solute interactions and solvent –solvent interactions (or like dissolves like – polar solutes combine with polar solvent; non-polar solutes combine with non-polar solvents

7. 2. Pressure – The greater the pressure above a liquid, the greater is the solubility of the gas in the liquid.
3. Temperature – The lower the temperature of a liquid, the greater is the solubility of the gas in the liquid.

8. RELATIVE (or comparative) WAYS OF EXPRESSING CONCENTRATION
UNSATURATED or DILUTE – contains a relatively small amount of solute
SATURATED – contains the maximum amount of solute in that a given quantity of solvent can dissolve in a given temperature
SUPERSATURATED – contains more solute than saturated solutions
- prepared by adding more solute to a saturated solution at a higher temperature

9. of solubility on temperature
Solubility vs. Temperature for Solids
140 KI
130
120
gases
solids
NaNO3
110
Solubility Table
100
KNO3 90 80
HCl
NH4Cl
shows the dependence
of solubility on temperature 70
Solubility (grams of solute/100 g H2O) 60
NH3
KCl 50
“Solubility Curves for Selected Solutes”
Description: This slide is a graph of solubility curves for 10 solutes. It shows the number of grams of solute that will dissolve in 100 grams of water over a temperature range of 0cC to 10 cC.
Basic Concepts
The maximum amount of solute that will dissolve at a given temperature in 100 grams of water is given by the solubility curve for that substance.
When the temperature of a saturated solution decreases, a precipitate forms.
Most solids become more soluble in water as temperature increases, whereas gases become less soluble as temperature increases.
Teaching Suggestions
Use this slide to teach students how to use solubility curves to determine the solubilities of various substances at different temperatures. Direct their attention to the dashed lines; these can be used to find the solubility of KClO3 at 50 cC (about 21 g per 100 g of H2O). Make sure students understand that a point on a solubility curve represents the maximum quantity of a particular solute that can be dissolved in a specified quantity of solvent or solution at a particular temperature. Point out that the solubility curve for a particular solute does not depend on whether other solutes also are present in the solution (unless there is a common-ion effect; this subject usually is covered at a later stage in a chemistry course).
Questions
Determine the solubilities (in water) of the following substance at the indicated temperatures: NH3 at 50 oC; KCl at 90 oC; and NaNO3 at 0 oC.
Which of the substances shown on the graph is most soluble in water at 20 oC? Which is lease soluble at that temperature? For which substance is the solubility lease affected by changes in temperature?
Why do you think solubilities are only shown between 0 oC and 100 oC?
In a flask, you heat a mixture of 120 grams of KClO3 and 300 grams of water until all of the KClO3 has just been dissolved. At what temperature does this occur? You then allow the flask to cool. When you examine it later, the temperature is 64 oC and you notice a white powder in the solution. What has happened? What is the mass of the white powder?
Compare the solubility curves for the gases HCl, NH3, and SO2) with the solubility curves for the solid solutes. What generalizations(s) can you make about the relationship between solubility and temperature?
According to an article in an engineering journal, there is a salt whose solubility in water increases as the water temperature increases from 0 oC to 65 oC. The salt’s solubility then decreases at temperatures above 65 oC, the article states. In your opinion, is such a salt likely to exist? Explain your answer. What could you do to verify the claims of the article? 40 30
NaCl
KClO3 20 10
SO2
LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 517 9

10. How to determine the solubility of a given substance?
Find out the mass of solute needed to make a saturated solution in 100 cm3 of water for a specific temperature(referred to as the solubility).
This is repeated for each of the temperatures from 0ºC to 100ºC. The data is then plotted on a temperature/solubility graph,and the points are connected. These connected points are called a solubility curve. 10

11. 11

12. How to use a solubility graph?
A. IDENTIFYING A SUBSTANCE ( given the solubility in g/100 cm3 of water and the temperature)
Look for the intersection of the solubility and temperature. 12

13. Example: What substance has a solubility of 90 g/100 cm3 of water at a temperature of 25ºC ?13

14. 14

15. Example:
What substance has a solubility of 200 g/100 cm3 of water at a temperature of 90ºC ? 15

16. 16

17. B. Look for the temperature or solubility
Locate the solubility curve needed and see for a given temperature, which solubility it lines up with and visa versa. 17

18. What is the solubility of potassium nitrate at 80ºC ?18

19. At what temperature will sodium nitrate have a solubility of 95 g/100 cm3 ?19

20. At what temperature will potassium iodide have a solubility of 230 g/100 cm3 ?20

21. What is the solubility of sodium chloride at 25ºC in 150 cm3 of water ?
From the solubility graph we see that sodium chlorides solubility is 36 g. 21

22. Place this in the proportion below and solve for the unknown solubility. Solve for the unknown quantity by cross multiplying.
Solubility in grams =
unknown solubility in grams
100 cm3 of water
other volume of water
___36 grams____
150 cm3 water
The unknown solubility is 54 grams. You can use this proportion to solve for the other volume of water if you're given the other solubility. 22

23. If it lies above the solubility curve, then it's supersaturated,
C. Determine if a solution is saturated, unsaturated,or supersaturated.
If the solubility for a given substance places it anywhere on it's solubility curve it is saturated.
If it lies above the solubility curve, then it's supersaturated,
If it lies below the solubility curve it's an unsaturated solution. Remember though, if the volume of water isn't 100 cm3 to use a proportion first as shown above. 23

24. Solubility  how much solute dissolves in a given amount of solvent at a given temp.
CURVE
Temp. (oC)
Solubility
(g/100 g H2O)
KNO3 (s)
KCl (s)
HCl (g)
unsaturated: solution could hold more solute; below line
saturated: solution has “just right” amt. of solute; on line
supersaturated: solution has “too much” solute dissolved in it; above the line 24

25. Solids dissolved in liquids Gases dissolved in liquidsTo
Sol. To
Sol.
As To , solubility
As To , solubility 25

26. Sometimes you'll need to determine how much additional solute needs to be added to a unsaturated solution in order to make it saturated.
For example,30 grams of potassium nitrate has been added to 100 cm3 of water at a temperature of 50ºC. 26

27. How many additional grams of solute must be added in order to make it saturated?
From the graph you can see that the solubility for potassium nitrate at 50ºC is 84 grams 27

28. If there are already 30 grams of solute in the solution, all you need to get to 84 grams is 54 more grams ( 84g-30g ). 28

29. of solubility on temperature
Solubility vs. Temperature for Solids
Solubility (grams of solute/100 g H2O) KI
KCl 20 10 30 40 50 60 70 80 90
110
120
130
140
100
NaNO3
KNO3
HCl
NH4Cl
NH3
NaCl
KClO3
SO2
gases
solids
Solubility Table
shows the dependence
of solubility on temperature
“Solubility Curves for Selected Solutes”
Description: This slide is a graph of solubility curves for 10 solutes. It shows the number of grams of solute that will dissolve in 100 grams of water over a temperature range of 0cC to 10 cC.
Basic Concepts
The maximum amount of solute that will dissolve at a given temperature in 100 grams of water is given by the solubility curve for that substance.
When the temperature of a saturated solution decreases, a precipitate forms.
Most solids become more soluble in water as temperature increases, whereas gases become less soluble as temperature increases.
Teaching Suggestions
Use this slide to teach students how to use solubility curves to determine the solubilities of various substances at different temperatures. Direct their attention to the dashed lines; these can be used to find the solubility of KClO3 at 50 cC (about 21 g per 100 g of H2O). Make sure students understand that a point on a solubility curve represents the maximum quantity of a particular solute that can be dissolved in a specified quantity of solvent or solution at a particular temperature. Point out that the solubility curve for a particular solute does not depend on whether other solutes also are present in the solution (unless there is a common-ion effect; this subject usually is covered at a later stage in a chemistry course).
Questions
Determine the solubilities (in water) of the following substance at the indicated temperatures: NH3 at 50 oC; KCl at 90 oC; and NaNO3 at 0 oC.
Which of the substances shown on the graph is most soluble in water at 20 oC? Which is lease soluble at that temperature? For which substance is the solubility lease affected by changes in temperature?
Why do you think solubilities are only shown between 0 oC and 100 oC?
In a flask, you heat a mixture of 120 grams of KClO3 and 300 grams of water until all of the KClO3 has just been dissolved. At what temperature does this occur? You then allow the flask to cool. When you examine it later, the temperature is 64 oC and you notice a white powder in the solution. What has happened? What is the mass of the white powder?
Compare the solubility curves for the gases HCl, NH3, and SO2) with the solubility curves for the solid solutes. What generalizations(s) can you make about the relationship between solubility and temperature?
According to an article in an engineering journal, there is a salt whose solubility in water increases as the water temperature increases from 0 oC to 65 oC. The salt’s solubility then decreases at temperatures above 65 oC, the article states. In your opinion, is such a salt likely to exist? Explain your answer. What could you do to verify the claims of the article?
LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 517 29

30. Classify as unsaturated, saturated, or supersaturated.
Solubility vs. Temperature for Solids
Solubility (grams of solute/100 g H2O) KI
KCl 20 10 30 40 50 60 70 80 90
110
120
130
140
100
NaNO3
KNO3
HCl
NH4Cl
NH3
NaCl
KClO3
SO2
gases
solids
Classify as unsaturated, saturated, or supersaturated.
80 g 30oC
45 g 60oC
50 g 10oC
70 g 70oC
=unsaturated
per
100 g
H2O
=saturated
=unsaturated
=supersaturated 30

31. Solubility vs. Temperature for Solids
Solubility (grams of solute/100 g H2O) KI
KCl 20 10 30 40 50 60 70 80 90
110
120
130
140
100
NaNO3
KNO3
HCl
NH4Cl
NH3
NaCl
KClO3
SO2
gases
solids
So sat. 40oC for 500 g H2O = 5 x 66 g = 330 g
120 g < 330 g unsaturated
saturation 40oC for 100 g H2O = 66 g KNO3
Per 500 g H2O, 120 g 40oC 31

32. Describe each situation below.
(A) Per 100 g H2O, 100 g Unsaturated; all solute
50oC. dissolves; clear solution.
(B) Cool solution (A) very Supersaturated; extra
slowly to 10oC solute remains in solution;
still clear.
(C) Quench solution (A) in Saturated; extra solute
an ice bath to 10oC. (20 g) can’t remain in
solution, becomes visible. 32

33. QUANTITATIVE WAYS OF EXPRESSING CONCENTRATION
PERCENTAGE COMPOSITION
PARTS PER MILLION
MOLE FRACTION
MOLARITY
MOLALITY

34. PERCENTAGE CONCENTRATION
% by mass of solute =
mass of solute X %
mass of solution
% by mass of solvent
mass of solvent X %
% by volume of solute
volume of solute X %
volume of solution
% by volume of solvent
volume of solvent X %

35. (Ex.) What is the % by mass of sugar prepared by dissolving 18 g of sugar in 54 g of water?
Solution:
Mass of solute (sugar) = 18 g
Total mass of solution = mass of sugar + mass of H2O
= 18 g + 54 g = 72 g
% sugar (by mass) = mass of sugar X 100%
mass of solution
= 18 g X 100% = 25 %
72 g

36. PARTS PER MILLION (ppm)
ppm of solute =
mass of solute X
mass of solution
ppm of solvent
mass of solvent X

37. MOLARITY (M) – number of moles (n) of solute per liter of solution (V)

38. Ex. A solution is prepared by dissolving 12 g of NaOH in water to make 80 mL of solution. What is the molarity of the solution?
Solution: Molar Mass (MM) of NaOH
= ( ) g
= 40 g/mole
Number of moles (n) = Given Weight (GW)
Molar Mass (MM)
= g = 0.3 mole
40 g/mole
Molarity (M) = 0.3 mole = 3.75 mole = 3.75 M
0.08 L L

39. MOLALITY (m) – number of moles (n) of solute kilogram of solvent (kg)

40. Ex. What is the molality of a solution if 196 g of H2SO4 is dissolved in 500 grams of water?
Solution: Molar Mass of H2SO4
= 2(1)g + 1(32)g + 4(16)g = 98 g/mole
Number of moles (n) = Given Weight (GW)
Molar Mass (MM)
= g = 2 moles
98 g/mole
m = 2 moles = 4 m
0.5 kg

41. MOLE FRACTION (mf) – number of moles of solute OR solvent divided by the total number of moles in the solution

42. Solution: Molar Mass of H2SO4 = 2(1)g + 1(32)g + 4(16)g = 98 g/mole
Ex. What is the mole fraction of the solute if 196 g of H2SO4 is dissolved in 500 grams of water?
Solution: Molar Mass of H2SO4
= 2(1)g + 1(32)g + 4(16)g = 98 g/mole
Number of moles (n) = Given Weight (GW)
Molar Mass (MM)
H2SO4 = g = 2 moles H2SO4
98 g/mole
H2O = g = 27.8 moles H2O
18 g/mole
mf H2SO4 = 2 moles H2SO4 = 0.067
2 moles H2SO moles H2O