Chemistry 11
Solution Chemistry Unit Notes
Unit 9 Solution Chemistry
Solution: ________________________________________________________________
Solvent:__________________________________________________________________
Solute:__________________________________________________________________
 A solute is soluble in a solution if _______________________________________
 A solvent is saturated with a solute if ____________________________________
 A solvent is unsaturated with a solute if __________________________________
the solubility of a solute is ______________________________________________
_________________________________________________________________________
Solvation: ________________________________________________________________
Ionic solid:_______________________________________________________________
Molecular solid:___________________________________________________________
Electrolytic Solution: ______________________________________________________
Does the amount of solid in a saturated solution increase, decrease or stay the same if…
and explain why.
Increase the temperature
Increase the temperature in a
closed container then allow it to
cool back to room temperature
Decrease the temperature
Leave the container uncovered
overnight
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Chemistry 11
Solution Chemistry Unit Notes
1. CONDUCTIVITY
 Solids :________________________________________________________
 Pure Liquids:___________________________________________________
______________________________________________________________
 ___________________________________________________________________
___________________________________________________________________
Which compounds will dissolve to from ionic solutions?
1)
2)
3)
Need water for conductivity – otherwise ions not dissociated
How To determine if a substance will conduct electricity
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Chemistry 11
Solution Chemistry Unit Notes
2. MOLECULAR POLARITY
Dipole:___________________________________________________________________
Electropositive: atoms with ______________________tend to form ___________ ions
Electronegative: atoms with ______________________tend to form ___________ ions
London Forces: ___________________________________________________________
________________________________________________________________________
A. Non-polar molecules
 ___________________________
eg) BF3 , CH4
F
H
F
H
C
B
F
H
H
Have a __________________
B. Polar Molecules
to be polar, molecules:
 must be _____________________ (dissimilar ends)
 one end _______________________ than other end
δO
dipole moment
H
H
δ+
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Chemistry 11
Solution Chemistry Unit Notes
C. Dipole-Dipole Forces
 ___________________________________________________________________
eg) water:
δ-
δ-
O
O
H
H
H
H
δ+
δ+
D. Hydrogen bonding
 ___________________________________________________________________
___________________________________________________________________
 ______________________________________________( between different NH3 ’s)
 ___________________________________________________________________
 ___________________________________________________________________
(Explains high BP of water)
The electronegative atom must have one or more _____________
electron pairs as in the case of oxygen, flurine and nitrogen,
and has a negative partial charge.
The hydrogen, which has a partial _____________
charge tries to find another atom of oxygen or
nitrogen with excess electrons to share and is
________________ to the partial negative charge.
This forms the basis for the hydrogen bond.
E. Physical Properties of Water Related with Intermolecular Forces
 For most substances, solids are more dense than liquids. This is not true for
water. Water is _______ dense as a solid
 ice floats on liquid water! Strong hydrogen bonds formed at freezing lock
water molecules away from each other.
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Chemistry 11
Solution Chemistry Unit Notes
 When ice melts, the structure collapses and molecules move closer together.
This property plays an important role in lake and ocean ecosystems.
 Floating ice often ______________ and ________________ animals and plants
living in the water below.
Using Diagrams Compare the Structure of Ice Water to Liquid Water
F. Polarity and Boiling Point:
Review
 The polarity of the molecules determines the forces of attraction between the
molecules in the liquid state.
 Polar molecules are attracted by the opposite charge effect (the positive end of
one molecule is attracted to the negative end of another molecule.
 Molecules have different degrees of polarity as determined by the functional
group present.
The evidence for hydrogen bonding
 Many elements form compounds with hydrogen - referred to as ____________.
 If you plot the boiling points of the hydrides of the Group 14 elements, you find
that the boiling points _______________ as you go down the group.
Explain Why this happens:___________________________________________
__________________________________________________________________
__________________________________________________________________
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Chemistry 11
Solution Chemistry Unit Notes
 Although for the most part the trend is exactly the same as in group 4 (for
exactly the same reasons), the boiling point of the hydride of the first element in
each group is abnormally high.
 In the cases of NH3, H2O and HF there must be some additional
_________________________ forces of attraction, requiring significantly more
______________________________ to break.
 These relatively powerful intermolecular forces are described as
_______________________ ________________________
The origin of hydrogen bonding
 The hydrogen is attached directly to one of the __________ _________________
elements, causing the hydrogen to acquire a significant amount of
_____________ ___________________
 Each of the elements to which the hydrogen is attached is not only significantly
negative, but also has at least one "active" lone pair.
 Lone pairs at the 2-level have the electrons contained in a relatively small
volume of space which therefore has a high density of negative charge. Lone
pairs at higher levels are more diffuse and not so attractive to positive things.
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Chemistry 11
Solution Chemistry Unit Notes
3. Nature of IONS IN SOLUTION
A. Polar and non Polar solvents
Polar solvents:____________________________________________________________
 ______________________________________________________________________
Non – polar solvents:______________________________________________________
 ______________________________________________________________________
B. Choosing a solvent
Like-dissolves-like
 _______________ solvents tend to dissolve ___________ or _________ solutes.
 __________________ solvents tend to dissolve __________________ molecules.
Polar solvents
 ___________________________________________________________________
____________ ___________ are attracted by _______________ ____________.
 _____________________________________________.
 __________________________________________________________________.
are not attracted by _______________________________.
 __________________________________________________________.
Polar solvents will dissolve:
 _____________________________________
 _____________________________________
 _____________________________________
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Chemistry 11
Solution Chemistry Unit Notes
Non-Polar solvents
Tend not to
have no _____________________________
dissolve
do not attract __________________________
attract non-polar solutes ____________________________________________.
Do question 17 pg 205 and then draw and name two polar and two non-polar solvents
Polar solvents
Non Polar solvents
1.
1.
2.
2.
4. The nature of SOLUTIONS OF IONS
Dissociation: _____________________________________________________________
________________________________________________________________________
Dissociation Equation:
 _________________________________________________________________
 _________________________________________________________________
 _________________________________________________________________
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Chemistry 11
Solution Chemistry Unit Notes
A. Dissolving salts
__________________________________________________.
eg) NaCl





_____________________________________________________________
______________________________________________________________
______________________________________________________________
______________________________________________________________
______________________________________________________________
B. Dissolving organic acid/alcohol
Ionization:
 __________________________________________________________________
eg) CH3COOH(l)  _________________ + ____________________
Writing dissociation and ionization equations:
 _________________________________________________________________
_________________________________________________________________
 _________________________________________________________________
 _________________________________________________________________
Homework: pg. 207 # 18 – 22
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Chemistry 11
Solution Chemistry Unit Notes
5. Calculating Ion Concentrations in Soluiton
Concentration = ________________ =
if there no Volume is given…..assume 1 L.
eg1) What is the concentration of the Cl- in 0.75 M AlCl3(aq)?
1) ________________________________________________________________
2) ________________________________________________________________
AlCl3(aq) 
(aq)
+
(aq)
[Cl-] =
=
eg2) What is concentration of each ion produced by mixing 200 mL of 0.355 M AlBr3 and
250 mL of 0.300 M CaBr2
1) ________________________________________________________________
2) ________________________________________________________________
3) ________________________________________________________________
a) for AlBr3
AlBr3 
+
[AlBr3]f =
[Al3+] f =
[Br-] f =
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Chemistry 11
Solution Chemistry Unit Notes
CaBr2 
b) for CaBr2
+
[CaBr2] f =
[Ca2+] f =
[Br-] f =
Add up the concentrations of similar ions
[Al3+] = ___________________
[Ca2+] = ___________________
[Br-] = __________________________________
6. pH
A. Self- ionization of water
H2O ionizes into H+ and OH- ions.
H3O+
H2O(l) + H2O(l) 
(aq) +
…..molecular representation:
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(aq)
Chemistry 11
Solution Chemistry Unit Notes
Only a very small amount of water self-ionizes
at 25 °C in pure water:
[H3O+] = _________________M
[OH-] = __________________ M
Ion Product of water (Kw)
Kw = [
][
]
][
]
Or
Kw = [
In pure water at 25 °C
[
]=[
]=
+
= [H3O ] [OH-] =
 Kw
=
(Kw has no units)
B. Calculating [H3O+] and [OH-]
 ___________________________________________________________________
___________________________________________________________________
Example: If [H3O+] in a solution is 1.0 · 10-2 M, then [OH-] will be 1.0 · 10-12 M.
Kw = [H3O+] [OH-] = 1.0 · 10-14
Solve for [OH-]
 [OH-] =
Kw
[ H 3O  ]
=
[OH-] =
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Chemistry 11
Solution Chemistry Unit Notes
Note:
 ___________________________________________________________________
___________________________________________________________________
 ___________________________________________________________________
___________________________________________________________________
 ___________________________________________________________________
___________________________________________________________________
C. Acidic, Basic, or Neutral Solutions
Acidic: ________________________________________________________________
Basic: ________________________________________________________________
Neutral: ______________________________________________________________
eg) If the [H3O+] (or [H+]) in blood is 4.0 · 10-8 M, what is the [OH-] at 25 °C? Is blood
acidic, basic or neutral.
Kw = [H3O+] [OH-] = 1.0 · 10-14
[H3O+] = 4.0 · 10-8 M
Solve for [OH-]
[OH-] =
Kw
[ H 3O  ]
=
=
Since [H3O+] =
Or [OH-] =
Blood is __________________________
Practice Problems: Do on a separate piece of paper
1) What is the [OH-] in chocolate milk if [H3O+] = 4.5 · 10-7 M? Is chocolate milk
acidic or basic or neutral.
2) What is the [H3O+] in black coffee if [OH-] = 1.3 · 10-9 M. Is black coffee acidic,
basic, or neutral?
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Chemistry 11
Solution Chemistry Unit Notes
D. The pH Scale
 ___________________________________________________________________
 ___________________________________________________________________
pH =
or pH =
or pOH =
pH =
Example: In one brand of vegetable juice, [H3O+] = 7.3 · 10-5 M. What is the pH of the
juice?
pH =
=
On your calculator:
1)
2)
3)
4)
5)
6)
pH =
pH and Significant figures.
 ___________________________________________________________________.
eg)
[H3O+] =
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Chemistry 11
Solution Chemistry Unit Notes
 pH =
E. pH and pOH scale @25oC
More Acidic
pH
0
1
2
[H3O+]
1 10-1
10-2
[OH-]
pOH
3
4
Neutral
5
6
More Basic
7
8
9
10
11
12
13
14
10-3 10-4 10-5 10-6 10-7 10-8 10-9 10-10 10-11 10-12 10-13 10-14
10-14 10-13 10-12 10-11 10-10 10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 1
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Conversion Picture:
Kw =
pH =
pOH =
pH + pOH =
Practice problems:
1) Normal rainwater has a pH near 6. In rain near a coal – burning power plant, [H3O+]
= 6.23 · 10-4 M. What is the pH of this rainwater?
2) In household bleach, [OH-] = 5.0 · 10-2 M. What is the pH of household bleach?
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Chemistry 11
Solution Chemistry Unit Notes
F. Acid Base Titrations
Remember from unit 7:
In a titration the molarity of one chemical is determined by reacting it with another one
with known molarity (standard).
1.
6.50 mL of 0.100 M H2C2O4 is required to neutralize 10.0 mL of KOH solution in a
titration. Calculate the base concentration.
H2C2O4 +
2KOH

K2C2O4 + 2H2O
If you get data in a table, you need to subtract the final burette reading from the initial to
get the volume of acid or base added.
If you have multiple trials, average them but reject any volumes that are way off.
Use some common sense.
Burette Volume in mL
Initial
0.00 mL
Final
8.95 mL
Volume Added
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8.95 mL
17.41mL
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17.41 mL
25.85 mL
Chemistry 11
Solution Chemistry Unit Notes
2.
8.95 mL , 8.46 mL and 8.44 mL of 0.200 M H2SO4 was required to neutralize 25.0
mL of KOH solution in a titration. Calculate the base concentration.
3.
Calculate the mass of H2C2O4.2H2O required to make 100.0 mL of a 0.1000 M
standard solution to use in your titration lab tomorrow.
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Chemistry 11
Solution Chemistry Unit Notes
G. ‘ICE’ charts with Acid Base Reactions:
For reactions with Excess acids or bases
1.
250.0 mL of 0.100 M H2SO4 reacts with 600.0 mL of 0.0500 M NaOH. Calculate the
concentration of the excess acid or base.
Step 1: Complete the balanced reaction equation
Step 2: Build the ICE table (I = Initial, C = Change, E = Equilibrium or End)
I
C
E
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Chemistry 11
Solution Chemistry Unit Notes
2.
300.0 mL of 0.100 M H2SO4 is mixed with 100.0 mL of 0.650 M NaOH , calculate the
concentration of the excess acid or base?
I
C
E
The two solutions are mixed therefore you need to ______________________________
to get the molarity!
The new volume =
[NaOH] =
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