1. Unit 7: Solutions

2. I. What is a solution?
Heterogeneous Mixture –
Homogenous Mixture –
Solutions are examples of _______________ mixtures
Solution:
Solute:
Solvent:
Substances in which particles are NOT uniform (example - soil)
Substances in which particles are uniform (example: aqueous (aq) solutions)
HOMOGENOUS
Homogenous mixture of substances in the same physical state
Substance that is being dissolved
Salt
Water
(solute)
Substance that dissolves the solute
(solvent)

3. I. What is a solution? POLAR IONIC + and - the polar solvent
Dissolving and Dissociation: Salt (NaCl) and Water
NaCl (s)  Na+ (aq) + Cl- (aq)
Dissolving occurs when the solvent is ______________________
Dissociation is when ____________ compounds’ ions get pulled apart by the ________________ end of _____________:
POLAR
IONIC
+ and -
the polar solvent
DISSOLVING ANIMATION

4. I. What is a solution? Atmosphere (O2, N2, CO2) Antifreeze
Types of Solutions:
- Gas with gas:
Liquid with liquid:
Gas with Liquid:
Solid with liquid:
Sold with Solid:
Atmosphere (O2, N2, CO2)
Antifreeze
Soda (CO2 (g) in H2O (l))
Salt water (NaCl (s) in H2O (l))
Alloys: Mixture of metals (brass – copper and zinc)

5. II. Solubility Factors Amount of solute that will dissolve in solvent
Nature of Solute and Solvent (video demo): -
Amount of solute that will dissolve in solvent
“like dissolves like”
Solution Type
Nonpolar Solvent
Polar Solvent
Nonpolar
Polar
Ionic
Soluble
insoluble
insoluble
soluble
insoluble
soluble

6. II. Solubility Factors 2. Temperature:
Solids: ________ temperature, _________ solubility (more solid will dissolve)
Gases: ________ temperature, _________ solubility
3. Pressure:
Only effects gases in liquids
Gases: ________ pressure, ___________ solubility -
increase
increase
decrease
increase
increase
increase

7. III. Factors affecting rate of solution
increase
increase
1. Size of particles: (_______ surface area, ________ rate of solution) 2.
3. Amount already dissolved: (_______ dissolved, ________ rate)
4. Temperature:
Solids and Liquids: _________ temperature, _______ rate
- Gases: __________ temperature, ___________ rate -
Stirring – more surface area brought into contact
more
slower
increase
increase
increase
decrease

8. IV. Expressing concentrations
A. Saturated: -
Solution that contains the maximum amount of solute at a given temperature
Saturated solutions are in equilibrium
(rate of dissolving = rate of recrystalization)

9. IV. Expressing concentrations
B. Unsaturated:
Solution is holding ______________________
If more solute is added, ____________
C. Supersaturated:
Solution is holding ____________________
Unstable – ____________________________________
LESS solute than maximum
it will dissolve
more solute than maximum
Addition of more solute will bring it back to a saturated solution
Supersaturated solution video

10. V. Solubility Curves (Found on Table ___ )G
Table G Shows: mass of solute that will dissolve in 100 g (or 100 mL) of water at different temperatures
Dilute: __________ amount of solute ______________
Concentrated: __________ amount of solute ____________
What is a salt?
SMALL
DISSOLVED
LARGE
DISSOLVED
AN IONIC COMPOUND

11. V. Solubility Curves (Found on Table ___ )G
How to use the chart:
Saturated will be ______________
Unsaturated will be ____________
Supersaturated will be ___________
ON THE LINE
BELOW THE LINE
ABOVE THE LINE

12. V. Solubility Curves (Found on Table ___ )G
Table G Practice Questions
1. A solution contains 14 g of KCl in 100. g of water at 40 oC. What is the minimum amount of KCl that must be added to make this saturation solution?
2. How many grams of the compound KCl must be dissolved in 200. g of water to make a saturated solution at 60oC?
3. Identify the following as being saturated, unsaturated, or supersaturated:
A) 20 oC and 20 g of KNO3
B) 40 oC and 20 g of KClO3
C) 90 oC and 10 g of NH3
D) 50 oC and 55 g of NH4Cl
4. Ninety grams of NaNO3 is added to 100. G of H2O at 0oC. With constant stirring, to what temperature must the solution be raised to produce a saturated solution with no solid remaining?
39 g – 14 g - = 25 g
45 g X 2 = 90 g
unsaturated
Supersaturated
saturated
supersaturated
Approx. 22oC

13. V. Solubility Curves (Found on Table ___ )G
Table G Practice Questions
1. A solution contains 14 g of KCl in 100. g of water at 40 oC. What is the minimum amount of KCl that must be added to make this saturation solution?
2. How many grams of the compound KCl must be dissolved in 200. g of water to make a saturated solution at 60oC?
3. Identify the following as being saturated, unsaturated, or supersaturated:
A) 20 oC and 20 g of KNO3
B) 40 oC and 20 g of KClO3
C) 90 oC and 10 g of NH3
D) 50 oC and 55 g of NH4Cl
4. Ninety grams of NaNO3 is added to 100. G of H2O at 0oC. With constant stirring, to what temperature must the solution be raised to produce a saturated solution with no solid remaining?
39 g – 14 g - = 25 g
45 g X 2 = 90 g
unsaturated
Supersaturated
saturated
supersaturated
Approx. 22oC

14. VI. How to determine if a compound is soluble (Found on Table ___ )
Ionic compounds can or cannot be soluble (dissolve in water).
If it is soluble the phase is ___________, if it is insoluble the phase is ___________
Use Table _____ to identify the phase:
Examples:
1) PbCO3_____ ) KNO3 _______ 3) Li2S ______
aqueous (aq)
solid (s) F
soluble
Insoluble
soluble

15. VII. Concentration
Molarity:
Equation (Table T)
Tells how much solute is dissolved in a given amount of solvent
# of moles of solute in 1 L of solution
Molarity = moles of solute
liters of solution
M = mol L

16. VII. Concentration
Examples:
What is molarity of a solution that contains 4.0 mol of NaOH in L of solution?
2. Calculate the molarity of 2.0 moles of HCl dissolved in 500. mL solution.
mol = 4.0 mol NaOH
V = 0.50 L
M = ?
mol = L
4.0 mol NaOH
= 8.0 mol/L (M)
Molarity (M) =
0.50 L
mol = 2.0 mol HCl
V = 500. mL = L
M = ?
mol = L
2.0 mol HCl
M =
= M
0.500 L

17. VII. Concentration Step 1: GFM = 1(40g/mol) + 2(14g/mol) +
Molarity from Grams: If you are given grams instead of mole, follow the steps:
Step 1: Convert grams to moles by finding the GFM
Step 2: Use the molarity equation
Example 1: What is the molarity of a solution containing 82.0 g of Ca(NO3)2 in 2.0 L of solution?
Step 1: GFM =
1(40g/mol) +
2(14g/mol) +
6(16g/mol) =
164 g/mol
1 mol
82.0 g x
= mol
164 g
0.500 mol
Step 2: Molarity =
mol = L
= 0.25 M
2.0 L

18. VII. Concentration Step 1: GFM = 2(23g/mol) + 1(12g/mol) +
Molarity from Grams: If you are given grams instead of mole, follow the steps:
Step 1: Convert grams to moles by finding the GFM
Step 2: Use the molarity equation
 Example 2: There is a L solution with 53 g of Na2CO3 completely dissolved. What is the molarity of the solution?
Step 1: GFM =
2(23g/mol) +
1(12g/mol) +
3(16g/mol) =
106 g/mol
1 mol
= 0.50 mol
53 g x
106 g
0.50 mol
Step 2: Molarity =
mol = L
= 2.0 M
0.25 L

19. VII. Concentration M = mol/L  mol = (M)(L) = (0.250 M) (2.00L)
How to make a solution:
What mass of sodium carbonate is required to prepare 2.00 L of M Na2CO3 solution?
Step 1: Find out how many moles are needed
Step 2: Convert moles to grams
M = mol/L
 mol = (M)(L)
= (0.250 M) (2.00L)
= mol Na2CO3
GFM =
2(23g/mol) +
1(12g/mol) +
3(16g/mol)
= 106 g/mol
106 g Na2CO3
0.500 mol Na2CO3 x
= 53.0 g Na2CO3
1 mol Na2CO3

20. VII. Concentration
Parts Per Million (PPM):
Equation (See Ref. Tabs.)
Ratio between mass of solute and total mass of solution
PPM = mass of solute* 1,000,000
mass of solution

21. VII. Concentration PPM = mass of solute x 1,000,000 mass of solution
Example 1: Approximately g of oxygen can be dissolved in 100. mL of water at 20oC. Express this in terms of parts per million.
Example 2: 2.5 grams of a groundwater solution is found to contain 5.4 x 10-6 grams of the Cu+2 ion. What is the concentration of the copper ion in ppm?
100. mL H2O = 100. g H2O
PPM = mass of solute x 1,000,000
mass of solution
= g O2 x 106
(100.g g)
= 43 PPM
PPM = mass of solute x 106
mass of solution
= 5.4 x10-6 g Cu2+ ions x 106
(2.5 g)
= 2.2 PPM

22. VIII. Colligative properties of solutions
Properties of solvent that change when a solute is added
They depend on the concentration of the solute in solution:
_____ ___concentration, _____ _____change
Electrolytes:
CaCl2 (s) 
Nonelectrolytes:
C6H12O6 (s) 
greater
greater
Solutions that conduct electricity due to the free-moving ions (mostly ionic compounds)
Ca +2 (aq) + 2 Cl -1 (aq)
Substances formed from covalent bonding do not dissociate into ions upon entering water
C6H12O6(aq)

23. VIII. Colligative properties of solutions
Boiling Point Elevation:
- The Higher the concentration of a solution, the higher the boiling point of the solution
- The more ionic particles in a solution, the higher the boiling point of the solution
MgCl2 (s) Mg2+ (aq) 2Cl- (aq) (3 particles – higher BP)
NaCl (s) Na+ (aq)+ Cl- (aq) (2 particles)
In 2 cups of water

24. CHECK FOR UNDERSTANDING
1) Which of the following solutions will boil at the highest temperature?
a) 100 g NaCl in 1000 g of water b) 100 g NaCl in 500 g water
c) 100 g NaCl in 250 g of water d) 100 g NaCl in 125 g of water
The answer is D because it has the highest concentration:
As the concentration of solution increases, the boiling point increases.

25. CHECK FOR UNDERSTANDING
2) Which solution has the highest boiling point?
a) 1.0 M KNO3 b) 2.0 M KNO3
c) 3.0 M C6H12O6 d) 2.0 M Ca(NO3)2
The answer is D because it has the most amount of particles when dissolved:
1.0 M KNO3  1 particle of K - and 1 particle of NO3 - = 2 total particles
3.0 M C6H12O6  3 particles of C6H12O6 = 3 total particles
2.0 M KNO3  2 (1 particles of K - and 1 particles of NO3 - ) = 4 total particles
2.0 M Ca(NO3)2  2 (1 particles of Ca – 2 and 2 particles of NO3 - ) = 6 total particles

26. VIII. Colligative properties of solutions
Freezing Point Depression:
Example: Salt is added to roads in the winter time
- The higher the concentration of a solution, the lower the freezing point of the solution
- The more ionic particles in a solution, the lower the freezing point of the solution
MgCl2 (s) Mg2+ (aq) 2Cl - (aq) (3 particles – lower FP)
NaCl (s) Na+ (aq)+ Cl - (aq) (2 particles)

27. VIII. Colligative properties of solutions
Which type of solute will have the greatest effect on MP and BP, ionic or covalent? Why?
- Ionic, they break apart when in water releasing more particles
NaCl  Na + (aq) + Cl – (aq) (2 particles)
C6H12O6  C6H12O6 (aq) (1 particle)

28. CHECK FOR UNDERSTANDING
1) Which of the following solutions will freeze at the lowest temperature?
a) 100 g NaCl in 150 g of water b) 100 g NaCl in 600 g water
c) 100 g NaCl in 125 g of water d) 100 g NaCl in 250 g of water
The answer is C because it has the highest concentration:
. As the concentration of the solution decreases, the freezing point increases.

29. CHECK FOR UNDERSTANDING
2) Which of the following solutions will freeze at the lowest temperature?
a) 100 g C6H12O6 in 500 g of water b) 100 g AlCl3 in 500 g of water
c) 100 g KBr in 500 g of water d) 100 g MgF2 in 500 g of water
The answer is B because AlCl3 breaks into 4 particles, the most of any of the choices. The more particles that a solute ionizes into, the higher the boiling point will be.