Download Chapter 14 - Ions in Aqueous Solutions

Chemistry 30
Solutions
Chemistry 20 Review
Objectives:
1.
List the important base and derived units of the metric system (2-2).
2.
List the important prefixes in the metric system (2-2).
3.
Perform dimensional analysis (2-3).
4.
Use and perform calculations in scientific notation (2-3).
5.
Compare and contrast between accuracy and precision (2-3).
6.
Use significant digits in calculations (2-3).
7.
Be able to write and balance chemical equations (chapter 8).
8.
Be able to identify types of chemical reactions and complete equations based on activity series (chapter 8).
9.
Be able to complete molar mass calculations.
10.
Be able to complete calculations involving number of atoms, moles, mass and molar mass.
Chapter 13 - Solutions
Objectives:
1.
Distinguish between heterogeneous and homogeneous mixtures. (13-1).
2.
Distinguish between electrolytes and nonelectrolytes. (13-1)
3.
Compare the properties of suspensions, colloids, and solutions. (13-1)
4.
List and explain three factors that influence the rate of dissolving of a solid in a liquid. (13-2)
5.
Explain terms related to solutions and solubility. (13-2)
6.
Explain how solutions form and use principles of polarity to determine whether a compound will be miscible or
immiscible in a particular solvent. (13-2)
7.
List the three interactions that contribute to the heat of solution, and explain what causes dissolving to be
exothermic or endothermic. (13-2)
8.
Compare the effects of temperature and pressure on solubility. (13-2)
9.
Calculate solution concentration in terms of moles of solute per litre of solution, both of compounds and individual
ions. Molality and percent by mass will also be calculated. (13-3)
10.
Use the relationship which links the mass of solute, volume of solution and concentration of solution so that given
two, the other can be determined. (13-3)
11.
Use the relationship which links original concentration, volume of dilutent and concentration of diluted solution so
that given two, the other may be determined. (13-3)
12.
Relate concentration expressed as ppm or ppb to those expressed as mol/L. (in-class notes)
13.
Perform dilution calculations. (in-class notes)
2
Chapter 14 - Ions in Aqueous Solutions
Objectives
1.
Write equations for the dissolution of soluble ionic compounds in water. (14-1)
2.
Predict whether a precipitate will form when solutions of soluble ionic compounds are combined. (14-1)
3.
Be able to write chemical equations, overall ionic equations and net ionic equations (14-1).
4.
Compare dissociation of ionic compounds with ionization of molecular compounds. (14-1)
5.
Calculate concentrations of ions when solutions are mixed. (in-class notes)
6.
Use a solubility table to produce a method by which a mixture of ions in solution may be separated (in-class
notes)
7.
List four colligative properties and explain why they are classified as colligative properties. (14-2)
8.
Make calculations based on freezing point depression and boiling point elevation. (14-2)
Vocabulary:
colloid
suspension
homogeneous
unsaturated
immiscible
heat of solution
fractional distillation
part per million
spectator ion
net ionic equation
osmotic pressure
electrolyte
emulsion
heterogenous
saturated
solvation
colligative properties
precipitate
part per billion
hydration
strong electrolyte
semipermeable membrane
nonelectrolyte
soluble
heat of solution
Henry’s Law
solvent
solute
solubility
miscible
dissociation
supersaturated
precipitation
boiling point elevation
freezing point depression
molarity
molality
hydronium ion
ionization
weak electrolyte osmosis
Formulae:
Molar concentration =
c
moles of solute
volume of solution
=
n
V
Units of molar concentration = mol/L
remember you still have to use
Number of moles =
n =
3
mass
molar mass
m
M
Dilution Equation
(Stock concentration)(Stock volume) = (Diluted concentration)(Diluted volume)
csVs = cdVd
Mathematics in Science
In science, most numbers are measurements and as such have both a value (a number) and a unit which gives
meaning to the number. In this class all measurements must include both a numeral and a unit.
Measurements are based on the metric system and all operations reflect the uncertainty of the measurements.
Metric System
SI BASE UNITS
Quantity
Base Unit
length
mass
volume
temperature
time
amount of matter
electric current
Symbol
metre
gram
litre
kelvin
second
mole
ampere
m
g
L
K
s
mol
A
SI DERIVED UNITS
Quantity
Name of Unit
Symbol
density
kilogram per
cubic metre
kg · m-3
force
4
Newton
N
pressure
Pascal
Pa
heat energy
Joule
J
Expressed in Terms
of SI Base Units
kg · m-3
(kg/m3)
kg · m · s-2 (kg · m / s2)
N · m-2
(kg · s-2 · m-1 , N / m2)
N · m (kg · m2 · s-2 )
METRIC PREFIXES - This table lists the metric prefixes which are significant:
Prefix
Symbol
Factor by which Unit is Multiplied
Exponential Notation
1012
tera
T
1 000 000 000 000
giga
G
1 000 000 000
109
mega
M
1 000 000
106
kilo
k
1 000
103
hecto
h
100
102
deca
da
10
101
1
100
THE BASE UNIT
deci
d
0.1
10-1
centi
c
0.01
10-2
milli
m
0.001
10-3
micro
μ
0.000 001
10-6
nano
n
0.000 000 001
10-9
pico
p
0.000 000 000 001
10-12
SCIENTIFIC NOTATION
1)
For numbers larger than 1, scientific notation is determined by counting the number of times the decimal
place must be moved to the left, leaving just one number to the left of the decimal. The number of times the
decimal must be moved is the power of 10 (the exponent).
Example:
3000 = 3000.0 = 3 x 103
454 000 = 454 000.0 = 4.54 x 105
3 860 000 = 3 860 000.0 = 3.86 x 106
602 000 000 000 000 000 000 000 = 6.02 x 10 23
2)
For numbers smaller than 1, scientific notation is determined by counting the number of times the decimal
place must be moved to the right, leaving just one number to the left of the decimal. The number of times the
decimal must be moved is the exponent, but it is a negative number.
Example:
0.068 = 6.8 x 10-2
0.000 049 3 = 4.93 x 10-5
0.000 000 002 41 = 2.41 x 10-9
If the decimal does not have to be moved, the exponent is zero. For example, the number 1.23 written in
scientific notation is
1.23 x 100.
5
IF A NUMBER IS LARGER THAN 9999 OR SMALLER THAN 0.001 IT MUST BE WRITTEN IN SCIENTIFIC
NOTATION. Between these extremes you may use either decimal or scientific notation.
Complete questions 54 and 58 in the Problem Bank
UNCERTAINTY
Every measurement done in science has some amount of uncertainty. For instance, if you take the mass of a
substance the scale may read 46.58 g. The last digit is rounded off, so the mass could be as low as 46.575 g and
as high as 46.584 g. A more sensitive scale could be used to reduce the uncertainty, but it will always be there.
Two terms associated with uncertainty are accuracy and precision. They are defined as follows:
Accuracy - how close a measured or calculated value is to a known or real value.
Precision - how close many repeated measurements are to each other.
Scientists repeat measurements as a way to reduce uncertainty. If a number of measurements are very close to
one another, they have good precision and the scientist can be assured that the average of the measurements is
likely close to the actual value (accurate)
SIGNIFICANT DIGITS
This is related to issues of uncertainty; the results of a calculation can be no more precise than the least
precise measurement which goes into it.
To determine the number of significant digits you must be able to handle zeros and their relationship to the nonzero digits in a number. Note the following rules:
1.
All non-zero numbers are considered significant; that is, they are counted
123 has 3 significant digits; 1267 has 4 s.d.
2.
There are two situations where zeros are significant:
i)
Zeros between two non-zero numbers are considered significant.
102 has 3 s.d.; 10203 has 5 s.d.; 1002 has 4 s.d.
ii)
A zero at the end of a decimal number is significant.
12.00 has 4 s.d.; 0.010 has 2 s.d.; 1200.000 has 7 s.d.
Note: in the last example, all the zeros are significant because they are between s.d.
3.
In any other situation zeros are not considered significant:
i)
For a number larger than 1, a zero between the decimal and the first non-zero number is not
significant.
120 has 2 s.d.; 10200 has 3 s.d.; 130 000 000 has 2 s.d.
6
ii).
For a number smaller than 1, a zero between the decimal and the first non-zero number is not
significant.
0.0012 has 2 s.d.; 0.02102 has 4 s.d.; 0.000 000 001 has 1 s.d.
Exact numbers - are numbers that are defined (conversion factors in the metric system) or numbers which result
from counting objects (like the coefficients used to balance chemical equations). Exact numbers are said to have
an infinite number of significant digits for rounding purposes.
Rounding off - is necessary when a number from a calculation must have the number of significant digits
reduced. The rules for rounding are as follows:
o
if the digit following the last digit to be kept is greater than 5, the last digit is increased by 1
e.g.
o
123.46
rounded to 4 s.d. is now 123.5
if the digit following the last digit to be kept is less than 5, the last digit stays the same
e.g.
123.44
rounded to 4 s.d. is now 123.4
o if the digit following the last digit to be kept is equal to 5, followed by a nonzero digit, the last digit is
increased by 1
e.g.
o
123.452
rounded to 4 s.d. is now 123.5
if the digit following the last digit to be kept is equal to 5, and not followed by a nonzero digit , the
last digit is increased by 1 only if it produces an even number
123.45
123.55
rounded to 4 s.d. is now 123.4
rounded to 4 s.d. is now 123.6
Complete questions 38 and 39 in the Problem Bank
Rule for addition and subtraction
Add or subtract and then round-off so that the answer is no more precise than the least precise number in the
calculation.
Rule for multiplication and division
Multiply or divide and then round-off so that the answer has no more significant digits than the number with the
fewest significant digits in the calculation. Remember that any exact numbers do not enter into the determination
of least significant digits.
For long, multi-step calculations:
Do not round off the number at each step in your calculator; keep the entire number, with all its
decimal places and use it in the next step. The danger is that you introduce ROUNDING ERROR.
Record the examples in class.
7
Complete questions 40 and 41 in the Problem Bank
Multiplication in Scientific Notation
The rule is that the integers (the first number in each scientific notation) in each number are multiplied, with the
resulting number becoming the integer in the answer. The exponents (the powers of 10) are added; the answer
becomes the exponent in the answer. For example:
Problem:
Solution:
(3.4 x 102)(2.0 x 103)
1. Multiply the integers 3.4 x 2.0 = 6.8
2. Add the exponents
102 x 103 = 102+3 = 105
3. Combine
6.8 x 105
If the integer in the solution has more than one digit to the left of the decimal point the scientific notation must be
corrected:
Problem:
(6.0 x 103)(2.5 x 107) = 15 x 1010 = 1.5 x 1011
Division in Scientific Notation
Division in s/n is similar to multiplication, except that in the case of division the integers are divided, and the
exponent of the second number is subtracted from the first. For example:
Problem:
Solution:
(3.4 x 102) / (2.0 x 103)
1. Divide the integers
3.4 / 2.0 = 1.7
2. Subtract the exponents
3. Combine
102 / 103 = 102-3 = 10-1
1.7 x 10-1
If the integer in the solution has less than one digit to the left of the decimal point the scientific notation must
be corrected:
Problem:
(6.0 x 103) / (8.0 x 107) = 0.75 x 10-4 = 7.5 x 10-5
Addition and Subtraction in Scientific Notation
These two operations are slightly different from multiplication and division. Numbers that are expressed in s/n can
only be added or subtracted if the exponents are the same. If the exponents are the same the integers are added
or subtracted and the exponent is carried into the solution. For example:
Addition:
Subtraction:
(2.07 x 106) + (1.30 x 106) = 3.37 x 106
(2.07 x 106) - (1.30 x 106) = 0.77 x 106 = 7.70 x 105
If the exponents are different there are two methods which may be followed to solve the problems:
1)
Move the decimal of one number to change the exponent.
Example:
10
8
3
(2.75 x 103) + (3.20 x 102) = (2.75 x 103) + (0.320 x 103) = 3.07 x
2)
Convert s/n numbers back to normal notation. Add or subtract as required, then convert the answer back
to s/n format.
Example:
(5.75 x 104) - (2.37 x 103) = 57 500 - 2 370 = 55 130 = 5.51 x 104
Complete questions 55, 56 and 57 in the Problem Bank
Dimensional Analysis
-
this method involves multiplication using conversion factors.
eg.
-
15 mm =
m
to solve, set up a ratio such that the units of the unknown are the units desired:
15 mm x
mm
m
=
m
(m/mm is the conversion factor; when the units are cancelled out only 'm' is left)
-
next fill in the units of the conversion factor. These units are unitary rates; they have a value of 1. In
this case 1 m is equal to 1000 mm. You must be very familiar with the value of the prefixes to make
this work.
15 mm x
-
1 m
1000 mm
=
m
now finish the calculation:
(15 x 1 )( mm x m ) = 0.015 m
100
mm
Conversion factors are not measurements. They are defined terms. As such they are considered to have infinite
significant digits. This means that the number of significant digits you start with will be the number
you end with.
Make sure you record the examples in class.
Complete questions 1 and 2 in the Problem Bank
9
Chemical Formulas and Equations
A chemical equation is a form of shorthand which gives an outline of the progress of a chemical reaction:
H2O  H2 + O2
REACTANT
PRODUCT
One very important principle of chemistry affects chemical equations. That principle is the law of conservation
of mass. It states that matter is neither gained nor lost in a chemical reaction. In a chemical reaction you end up
with the same number and type of each atom that you started with. This applies to mass and to atoms. This
principle must be reflected in chemical equations.
Balancing equations is similar to the method used to determine the chemical formulas for ionic compounds. Using
the equation for the decomposition of water, first tally up the number of each kind of atom on each side of the
reaction equation:
1 H2O

H=2
O=1
1 H 2 + 1 O2
H=2
O=2
An equation is balanced if there are the same number of atoms on each side of the reaction equation. To do this,
it is necessary to determine if there are fewer atoms on one side of equation, compared with the other side. In
this case the product side has one more oxygen than the reactant side. In order to balance the equation for
oxygen, the number of oxygen atoms on the reactant side must be increased; the only way to do that is to
increase the number of water molecules on the reactant side:
2 H2O

H=4
O=2
1 H 2 + 1 O2
H=2
O=2
As you can see, increasing the number of water molecules balances the equation in terms of oxygen atoms, but
now the reactant side has too many hydrogen atoms; we must now balance the equation in terms of hydrogen.
The only way to do this is to increase the number of hydrogen atoms on the product side, by increasing the
number of hydrogen molecules:
2 H2O
H=4
O=2

2 H 2 + 1 O2
H=4
O=2
Now the equation is balanced. The total number of each type of atom on the reactant side is equal to that on the
product side, and the law of conservation of mass is satisfied.
For another example we use a reaction between lead nitrate (Pb(NO3)2) and sodium iodide (NaI), both of which
produce clear solutions. When these solution react a solid yellow compound is produced called lead iodide (PbI 2),
as well as another clear solution which contains sodium nitrate (NaNO 3). The reaction equation follows:
10
Pb(NO3)2 + NaI  PbI2 + NaNO3
Step 1: Tally up the number of each atom:
Pb(NO3)2 + NaI  PbI2 + NaNO3
Pb = 1
Na = 1
I=1
N=2
O=6
Pb = 1
Na = 1
I=2
N=1
O=3
Step 2: Begin balancing for each atom:
This equation is already balanced for lead and sodium; the next on the list is iodine.
Pb(NO3)2 + NaI  PbI2 + NaNO3
Pb = 1
Na = 2
I=2
N=2
O=6
Pb = 1
Na = 1
I=2
N=1
O=3
Note that the only way to increase the number of iodine atoms on the reactant side is to add another NaI
molecule, thus increasing both the number of sodium and iodine atoms. This causes the number of sodium atoms
to become unbalanced, so the number of sodium atoms on the product side must be increased, by adding another
sodium nitrate molecule:
Pb(NO3)2 + 2 NaI  PbI2 + 2 NaNO3
Pb = 1
Na = 2
I=2
N=2
O=6
Pb = 1
Na = 2
I=2
N=2
O=6
Other information can also be gained or given from a reaction. It is sometimes useful to know the state a
compound is in for a chemical reaction. By "state" we mean whether the compound is a liquid, solid, or a gas.
This information is given by the use of subscripts following each compound:
(s) (l) (g) (aq) (ppt) -
solid
liquid
gas
aqueous (meaning that the compound is dissolved in water.)
precipitate (meaning that the reaction produces a solid which falls out of solution.)
Using the two previous examples, these subscripts can be quite valuable:
2 H2O(l)

2 H2 (g) + 1 O2 (g)
Pb(NO3)2 (aq) + 2 NaI
11
(aq)
 PbI2 (s) + 2 NaNO3 (aq)
Types of Chemical Reactions
There are five basic types of chemical reactions:
1) Synthesis -
two or more simple substances combine to form a more complex compound
A + B  AB
another way to look at it is:
element + element  compound
2) Decomposition - One substance breaks down to form two or more simpler substances.
AB  A + B
compound  element + element
3) Combustion - involves the burning of a hydrocarbon in the presence of oxygen to form carbon dioxide and
water.
CxHy + O2  CO2 + H2O
4) Single Replacement - reactions occur when one element is replaced by another in a compound
A + BC  AC + B
element + compound  element + compound
5) Double Replacement - reactions occur when the elements in a solution of reacting compounds exchange
places, or replace each other.
AB + XY 
AY + XB
compound + compound  compound + compound
Complete page 1 of Balancing Chemical Equations Assignment
12
Completing Chemical Reactions
You can now predict the products of chemical reactions. For instance, if you are given this partial reaction:
Pb(NO3)4 (aq) + Mg(s) 
you can tell what kind of reaction this is. Compare this reaction with the list on the previous page. Pb(NO 3)4 is a
compound; Mg is an element. There is only one type of reaction that has a compound and an element on the
reactant side, single replacement. Thus we know that the Mg will replace the Pb in the compound (Mg forms a
cation and so will replace the cation in the compound). Check the ion chart. Mg form a 2+ ion so it will form
Mg(NO3)2 with the 1- nitrate ion.
The finished equation looks like this:
Pb(NO3)4 (aq) + 2 Mg(s)  Pb + 2 Mg(NO3)2
- remember to balance the equation
- at this stage you can’t predict the phase of the products
Likewise, if you get
C3H8 (g) + O2 (g) 
You know it is a combustion reaction because the reactants are a hydrocarbon and oxygen. The products are thus
carbon dioxide and water:
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O(g)
Write down the rest of the examples in class.
Complete the first 4 questions on the back side of the Balancing Chemical Equations Assignment
Activity Series
This is a measure of the ability of one substance in its elemental state to react with another substance and form a
compound (like in a synthesis reaction) or to force another substance in a compound to revert to its elemental state
(like in a single replacement reaction).
It turns out there is a pecking order of metals and non-metals (see the activity series attached to this package). The
elements higher on the list are more reactive; they can replace elements below them on the Periodic Table. Thus if
we take the reaction:
Mg(s) + FeCl3 (aq) 
we need to determine if the Mg can replace the Fe in FeCl 3 (since Mg forms a cation it will replace the cation in the
compound). Look at the activity series. Find Mg and Fe. Note that Mg is higher than Fe on the table. Mg is
therefore able to replace Fe in the reaction, so it will go to completion:
3 Mg(s) + 2 FeCl3 (aq)  2 Fe(s) + 3 MgCl2 (aq)
13
Take another reaction:
NaNO3 (aq) + Ni(s) 
Since Na is higher on the table than Ni, the Ni is unable to replace Na in the compound. No reaction will occur:
NaNO3 (aq) + Ni(s) 
NO REACTION
The same principles apply to the list of halogens. Remember that halogens form anions in compounds.
Note: The table also includes reactivity with oxygen, water and acids. The bottom group are unreactive. The group
second from the bottom (H2, Cu and Hg) will form oxides in the presence of oxygen. Please note that reactivity
applies to all metals higher on the table as well; they all react with oxygen to form oxides. Likewise the third group
(Ni, Sn and Pb) will react with acids to form hydrogen. This applies to all the metals above on the chart as well (but
not below).
Complete the last 6 questions on Balancing Chemical Equations
Molar Calculations
Record examples from class
Complete questions 101 to 107 in the Problem Bank
Chapter 13 - Solutions
The first 8 objectives are covered well in your text. You are responsible for making notes.
Water
-
is an unusual substance, primarily because of polarity and hydrogen bonding.
-
exists as a liquid at room temperature. It’s melting and boiling points are very high compared to
substances of similar molecular weight:
14
Substance
Molar Mass
Boiling Point
CH4
16.04 g/mol
- 164 ºC
C3H8
44.11 g/mol
- 42.1 ºC
C4H10
58.14 g/mol
- 0.5 ºC
O2
32.00 g/mol
- 183 ºC
CO2
44.01 g/mol
- 78.5 ºC
N2
28.02 g/mol
- 196 ºC
H2O
18.02 g/mol
+ 100 ºC
-
occurs in three phases normally on Earth (solid, liquid, vapour).
-
solid is less dense than the liquid.
-
has a very high heat of fusion and vapourization.
-
has a very high specific heat capacity.
-
has a high surface tension.
-
is the universal solvent:
Soluble in water
polar solutes ionic solutes
-
sugars
alcohols
-
salts
low molecular weight, non-polar molecules
-
O2 N 2
CO2
Not soluble in water
most hydrocarbons
-
fats
oils
gasoline
Concentration
-
is a measure of the amount of solute, per unit of solution or solvent.
-
is central to quantitative chemistry, since most chemical reactions happen in solution.
Molar Concentration
-
is the most common measure of concentration used in the lab.
-
is a measure of the number of moles of solute, per litre of solution (mol/L, mol∙L-1).
-
is also called Molarity (M).
-
for instance, 0.12 mol/L ≡ 0.12 M
15
Examples:
1)
Calculate the molar concentration when 0.256 mol of CuSO4 is dissolved in 1.1 L of solution.
molar concentration
c
2)
=
moles of solute
volume of solution
=
n
v
=
0.256 mol
1.1 L
0.23 mol/L (2 s.d.)
Calculate the molar concentration when 28.2 g of KNO 3 is dissolved in 750. mL of solution.
-
given 28.2 g ; need to convert mass to moles
molar mass:
KNO3
1xK
1xN
3xO
moles of substance
-
=
=
=
=
1 x 39.10 g/mol
1 x 14.01 g/mol
3 x 16.00 g/mol
101.11 g/mol
28.2 g
101.11 g/mol
=
0.279 mol
volume must be reported in L:
750. mL x
-
=
1L
1000 mL
= 0.750 L
molar concentration:
c
=
n
V
=
=
3)
0.372 mol/L
Calculate the volume of solution made with 0.455 mol of sodium phosphate in a 2.00 M solution.
c
=
n
V
V
=
n
c
=
0.455 mol
2.00 mol/L
=
16
0.279 mol
0.750 L
0.228 L
4)
How many moles are present in 4.50 L of a 0.025 mol/L solution of magnesium nitrate?
c
=
n
V
n
=
cV
=
(0.025 mol/L)(4.50 L)
=
0.11 mol
Complete questions 332 to 337, 345, 346, and 348 in the Problem Bank
Molal Concentration
is used in industry and for measurements of colligative properties (see Ch. 14); it is useful because it does
not depend on the final volume of solution.
-
is a measure of the number of moles of solute, per kilogram of solvent (mol/kg, mol∙kg -1).
-
is also called Molality (M).
-
for instance, 0.12 mol/kg ≡ 0.12 M
Examples:
1)
Calculate the molal concentration when 2.56 mol of CuSO4 is dissolved in 12.0 kg of water.
molal concentration
M
=
moles of solute
mass of solvent
=
n
ms
=
2.56 mol
12.0 kg
=
2)
0.213 mol/kg
Calculate the molality when 19 g of Mg(NO3)2 is dissolved in 1200 g of water.
Mg(NO3)2
1 x Mg
2xN
6xO
1200 g x
17
1 kg
1000 g
=
=
=
1 x 24.30 g/mol
2 x 14.01 g/mol
6 x 16.00 g/mol
148.32 g/mol
=
1.2 kg
M
=
n
ms
M
=
m
msM
M
=
n
=
=
m
M
19 g
(1.2 kg)(148.32 g/mol)
0.11 mol/kg
Note: It is perfectly acceptable to do the calculations in 2 steps; moles first, then molal concentration.
3)
Calculate the mass of solvent used with 61.2 g of sodium phosphate in a 2.00 M solution.
sodium phosphate = Na3PO4
4)
3 x Na
1xP
4xO
=
=
=
M
=
n
ms
ms
=
n
M
ms
=
m
MM
ms
=
ms
=
3 x 22.99 g/mol
1 x 30.97 g/mol
4 x 16.00 g/mol
163.94 g/mol
n
=
m
M
61.2 g
(2.00 mol/kg)(163.94 g/mol)
0.187 kg
What mass of lithium hydroxide is present in a 0.025 mol/kg solution with 360. kg of water?
lithium hydroxide = LiOH
1 x Li
1xO
1xH
18
=
=
=
1 x 6.94 g/mol
1 x 16.00 g/mol
1 x 1.01 g/mol
23.95 g/mol
M
=
M
=
m
=
M msM
=
(0.025 mol/kg)(360. kg)(23.95 g/mol)
=
220 g
n
ms
n
m
msM
=
m
M
Complete questions 338 to 340, 347, 350, and 352 in the Problem Bank
Other Measures of Concentration (adapted from Wikipedia)
“Parts per” are based on molal concentrations:






Parts per hundred (denoted by "%") — denotes one particle of a given substance for every 99 other
particles. This is the common percent. One part in 102. This is equal to 1 g of a solute in 100 g of solution,
or 100 mL of solution.
Parts per thousand (denoted occasionally "ppt") denotes one particle of a given substance for every 999
other particles. This is roughly equivalent to one drop of ink in a cup of water, or one second per 17
minutes. "Parts per thousand" is often used to record the salinity of seawater. One part in 103.
Parts per million ("ppm") denotes one particle of a given substance for every 999,999 other particles. This
is roughly equivalent to one drop of ink in a 150 litre (40 gallon) drum of water, or 1 g of solute in 1000 kg
of solution. . One part in 106 — a precision of 0.0001%.
Parts per billion ("ppb") denotes one particle of a given substance for every 999,999,999 other particles.
This is roughly equivalent to one drop of ink in a lane of a public swimming pool, or 1 gram of solute in 106
kg of solution (roughly Riversdale pool). One part in 109.
Parts per trillion ("ppt") denotes one particle of a given substance for every 999,999,999,999 other
particles. This is roughly equivalent to one drop of ink in a shipping canal lock full of water , or 1 g of
solute in Blackstrap lake. One part in 1012.
Parts per quadrillion ("ppq") denotes one particle of a given substance for every 999,999,999,999,999
other particles. This is roughly equivalent to a drop of ink in a medium-sized lake. Very few analytical
techniques can measure with this degree of accuracy; nevertheless, it is still used in some mathematical
models of toxicology and epidemiology. One part in 1015.
Dilution
-
many substances, especially acids, are received in a concentrated form (hydrochloric acid is 12.4
mol/L). This is called the stock solution.
-
to use the chemicals in the lab they are usually diluted to a concentration much less than they are
received.
-
the problem is to decide how much of the stock solution you need to make the solution you want.
For instance; you have a 12.4 mol/L stock solution of HCl and you want to make 2.00 L of 0.100 mol/L
solution.
Since you know the volume and concentration of the dilute solution, you can calculate the number of
moles in the solution:
n
= cv
= (0.100 mol/L)(2.00 L)
= 0.200 mol
This represents the number of moles of HCl you need from the stock solution. Now you have the
number of moles and the concentration, so you can calculate the volume of concentrated HCl you need
to add to 2.00 L of water to make the dilute solution:
v
19
=
n
c
=
0.200 mol
12.4 mol/L
= 0.0161 L (1000 mL/L)
= 16.1 mL of concentrated HCl is added to water to make 2.00 L of a
0.100 mol/L solution.
Fortunately, there is an easier way to do this. Since the moles taken from the stock solution (n s) is the
same number of moles that goes into the dilute solution (nd)
ns = nd
and
ns = csvs
nd = cdvd
so the formula for dilution is:
csvs = cdvd
Examples:
1.
What volume of concentrated sulfuric acid (18 M) is needed to make 500. mL of 2.00 M
dilute solution?
2.
What is the concentration of dilute solution if 25.0 mL of glacial acetic acid (24.0 mol/L) is
added to make 2.00 L of dilute solution?
3.
What volume of a 1.00 mol/L stock solution of aluminum chloride is needed to make 500. mL of
3.0 x 10-4 mol/L solution?
Complete questions 365, 366, 370 to 372 in the Problem Bank
20
Serial Dilution
Serial dilutions are an accurate method of making solutions of low molar concentrations. A stock solution with a
molarity greater than that which is required is accurately diluted using a suitable solvent. Since measuring small
volumes of solution is prone to error, a series of dilutions are performed in order to gradually reduce the
concentration of the solution from that of the stock solution.
Serial "ten-fold" dilutions are commonly used.
To carry out a serial "ten-fold" dilution you would do the following:





Add 9 ml of distilled water to each test tube.
To the first test tube, add 1 ml of the stock solution.
Mix or vortex.
Now add 1 ml of this solution to the second test tube and mix.
Repeat until you have reached your desired concentration.
In the diagram above, calculate the concentration in each of the 6 test tubes if the original concentration of stock
solution was 24 mol/L acetic acidl
21
Ions in Aqueous Solution
-
when a substance dissolves it is called a solvation reaction:
sugar is a non-electrolyte:
-
C12H22O11 (s) → C12H22O11 (aq)
soluble ionic compounds undergo dissociation in solution; they break up into their constituent ions:
NaCl(s) → Na1+(aq) + Cl1-(aq)
(this represents both solvation and dissociation)
Where (s) means solid
(aq) means aqueous (dissolved in water)
(l) means liquid
(g) means gas
(ppt) means precipitate
-
to write a dissociation equation for any ionic compound you must find out the identity of the cation and
anion, then write the equation, paying attention to the stoichiometry:
Al2(SO4)3 (s) contains the Al3+ and SO42- ions. When it dissociates you get 2 Al3+ and 3 SO42- ions:
Al2(SO4)3 (s) → 2 Al3+(aq) + 3 SO42-(aq)
-
this means that the concentration of the ions may be different than the calculated concentration of the
substance in solution:
If the aluminum sulfate is a 0.200 mol/L solution, then
Al2(SO4)3 (s)
0.200 mol/L
→
2 Al3+(aq)
+
2(0.200 mol/L)
= 0.400 mol/L
3 SO42-(aq)
3(0.200 mol/L)
= 0.600 mol/L
Write the equation for the dissociation of the following ionic substances. Calculate the concentration of each ion in
solution:
a)
0.35 mol/L NaOH
b)
1.12 mol/L (NH4)2CO3
c)
0.056 mol/L V3(PO4)5
22
Chemical Reactions in Solution
-
there are 4 indications that a chemical reaction has occurred:
i)
ii)
iii)
iv)
gas production
energy change
colour change
precipitate formed
a precipitate is a solid formed from the reaction of two soluble ions in solution; it is the most common
example of a chemical reaction in solution. For example:
Pb(NO3)2 (aq) + 2 KI(aq) → PbI2
(s)
+ 2 KNO3 (aq)
this is the chemical reaction equation for the combination of lead (II) nitrate and potassium iodide. If
we recognized the fact that both ionic reactants had dissociated and were in fact ions in solution we would get the
overall ionic equation:
Pb2+(aq) + 2 NO31-(aq) + 2 K1+(aq) + 2 I1-(aq)
-
→
PbI2(s) + 2 K1+(aq) + 2 NO31-(aq)
the lead (II) iodide does not break up into ions because it is a solid at the bottom of the beaker.
there are ions that are the same on both sides of this reaction equation (K 1+ and I1-). These do not
participate in the chemical reaction and are called spectator ions. If we eliminate the spectators we get the net
ionic equation:
Pb2+(aq) + 2 I1-(aq) → PbI2 (s)
-
if there is no precipitate formed, no chemical reaction occurs.
Try writing the balanced chemical equation, overall ionic equation and net ionic equation for the following:
a)
FeCl2 (aq) + K2S
b)
AlBr3 (aq) + NaOH
c)
(NH4)3PO4 (aq) + Ca(NO3)2 (aq) → NH4NO3 (aq) + Ca3(PO4)2 (s)
d)
Aqueous solutions of silver nitrate and sodium carbonate react to produce a precipitate of silver
carbonate and aqueous sodium nitrate.
e)
Aqueous solutions of barium chloride and potassium sulphate react to produce a precipitate of barium
sulphate and aqueous potassium chloride.
23
(aq)
→ FeS
(aq)
(s)
+ KCl
(aq)
→ NaBr (aq) + Al(OH)3 (s)
Mixing Solutions
What happens if you mix two ionic solutions ? Using what we just learned you can calculate the concentration of
the ions in solution before mixing. Take these two solutions:
CaCl2(aq) →
0.25 mol/L
Ca2+(aq) +
2 Cl1-(aq)
1(0.25 mol/L) 2(0.25 mol/L)
= 0.25 mol/L
= 0.50 mol/L
Mg3(PO4)2(aq)
0.011 mol/L
→
3 Mg2+(aq)
+
2 PO43-(aq)
3(0.011 mol/L)
2(0.011 mol/L)
= 0.033 mol/L
= 0.022 mol/L
Now suppose we take 300. mL of the CaCl2 and 200. mL of the Mg3(PO4)2 and we mix them. Assuming that no
precipitate is formed, what will the concentration of each ion in solution be after mixing ?
To solve this we go to the dilution formula:
CsVs = CdVd
In this case, Cs and Vs is the original concentration and volume of each ion (0.25 mol/L and 300. mL for the Ca 2+,
for instance) and d2 is the total volume of the mixture (300. mL + 200. mL). We are looking for Cd, the ion
concentration after mixing:
Cd =
CsVs
Vd
For the Ca2+ ion the calculation is as follows:
[Ca2+] = (0.25 mol/L)(300. mL)
(300. mL + 200. mL)
= 0.15 mol/L
The rest look like this:
[Cl1-] = (0.50 mol/L)(300. mL)
= 0.30 mol/L
(300. mL + 200. mL)
[Mg2+] = (0.033 mol/L)(200. mL)
(300. mL + 200. mL)
= 0.013 mol/L
[PO43-] = (0.022 mol/L)(200. mL)
= 0.0088 mol/L
(300. mL + 200. mL)
This is a long problem, but at least they are simple. Calculate the concentration of each ion in each of the following
solutions in which no reaction occurs:
a)
b)
c)
24
2.0 L of 0.40 mol/L MgSO4 mixed with 2.0 L of 0.080 mol/L KI
200. mL of 0.600 mol/L AlBr3 mixed with 300. mL of 0.400 mol/L BaBr2
20.0 mL of 0.50 mol/L FeCl3 mixed with 80.0 mL of 0.15 mol/L NH4Cl
Predicting the Solubility of Ionic Compounds
substances dissolve in water when the attraction of the solute particles to the water molecules exceeds the
attraction of the undissolved solute for itself.
if the solute particles are held together very strongly, the attraction of the water molecules may not be
enough to cause it to dissolve. Such solutes will be insoluble, or have a very low solubility.
we can’t tell from looking at a molecule whether it will be soluble or insoluble in water. We can use
empirical evidence to make predictions. See the solubility table.
-
the solubility table allows one to make predictions as to the solubility of various ionic substances:
i)
Na2S
soluble, because Na is an alkali metal (group 1)
ii)
NH4Cl
soluble, because the NH41+ ion is soluble with any anion
iii)
CuNO3
soluble, because NO31- is soluble with any cation
iv)
AgCH3COO
insoluble
v)
Mg(CH3COO)2
soluble
vi)
PbCl4
soluble; in this compound the Pb is 4+
vii)
PbCl2
insoluble; in this compound the Pb is 2+
Predict the solubility of the compounds in question 2 on page 447 of the text.
Predicting the Results of Chemical Reactions
-
given any two ionic solutions you should be able to predict the products and if a precipitation reaction
occurs:
AgNO3 (aq) + NaCl(aq) → ???
-
first, rewrite the reactants as separate ions in solution:
Ag1+(aq) + NO31-(aq) + Na1+(aq) + Cl1-(aq) →
-
in a double replacement reaction the cation of the first compound combines with the anion of the second,
and vice versa:
Ag1+(aq) + Cl1-(aq) → AgCl
Na1+(aq) + NO31-(aq) → NaNO3
25
-
-
next use the solubility table to predict the solubility of the two products of this reaction:
AgCl
insoluble
NaNO3
soluble
write the chemical equation for this reaction:
AgNO3 (aq) + NaCl(aq) → AgCl(s) + NaNO3 (aq)
-
write the overall ionic equation:
Ag1+(aq) + NO31-(aq) + Na1+(aq) + Cl1-(aq) →
-
AgCl(s) + Na1+(aq) + NO31- (aq)
write the net ionic equation:
Ag1+(aq) + Cl1-(aq) → AgCl(s)
Try the following; for each:
1.
determine the identity of the possible products.
2.
find out of the products are soluble or insoluble.
3.
if no insoluble compound is formed, simply write No Reaction in the product place of the
chemical equation:
NaCl(aq) + NH4NO3 (aq) → No Reaction
4.
26
if one (or both) of the possible products is insoluble, write a chemical equation, an
overall ionic equation and a net ionic equation.
i)
NaBr and AgNO3
ii)
NiCl2 and KOH
iii)
CuSO4 and K2S
iv)
CoCl2 and (NH4)3PO4
v)
Cu(NO3)2 and KI
Selective Precipitation of Ions
We have mixed solutions for double replacement reactions, with two cations and two anions. This is not the way it
is in the real world. The real world involves reactions of many ions mixed together in complex interactions.
Industry requires that valuable ions be removed from mixtures for further refining and sale. Selective precipitation
allows ions to be removed one at a time. This is important for the chemical and mining industries.
Example. You have an aqueous mixture containing Pb2+, Al3+ and Ag1+. What anions can be added which will
precipitate these cations, one at a time ?
1. add acetate (CH3COO1-). This will precipitate the Ag1+, leaving the Pb2+ and the Al3+ in solution.
2. add Cl1-, Br1-, I1-, or SO42-. Any of these will precipitate the Pb2+, leaving the Al3+ in solution.
3. add any of the anions from the bottom half of the solubility chart (S 2-, OH1-, CO32-, ...). This will precipitate
the Al3+.
Example. You have an aqueous mixture containing Br1-, SO42- and CO32-. What cations can be added which will
precipitate these anions, one at a time ?
1. notice that both the Br1- and the SO42- are more soluble than the CO32-:
a. the Br1- is soluble with everything except Ag 1+, Pb2+,Cu1+ and Hg22+
b. the SO42- is soluble with everything except the group 2 ions (except Mg 2+), Ag1+ and Pb2+
2. that means that we can add any other cation to precipitate the CO32-, except the ones above, the group
1 ions and NH41+. Pick a cation off your ion chart. The first one there is Al3+. That will do. This will
precipitate the CO32-, leaving the Br1- and SO42- in solution
3. There are two acceptable procedures to remove the final two ions:
a.
i. add any group 2 ion except Mg2+. This will precipitate the SO42-, leaving the Br1- in
solution
ii. add Ag1+, Pb2+,Cu1+ or Hg22+. This will precipitate the Br1b.
i. add Cu1+ or Hg22+. This will precipitate the Br1-, leaving the SO42- in solution
ii. add Ca2+ or another group 2 element, Ag1+ or Pb2+. This will precipitate the SO42-.
Devise a procedure for separating the following mixtures by precipitating them individually:
1.
2.
3.
27
Ba2+ and Pb2+
Cl1- and OH1Cu2+, Mg2+ and Sr2+
Colligative properties are properties of solutions that have nothing to do with the identity of the solute, only
with the molal concentration of non-volatile solutes.
There are four colligative properties:
1.
2.
3.
4.
Vapour Pressure
Freezing Point Depression
Boiling Point Elevation
Osmotic Pressure
Vapour Pressure
This can be thought of as a measure of the tendency of molecules to escape from a liquid and form a vapour.
Solutes affect vapour pressure by lowering the concentration of solvent molecules at the surface of the liquid. The
greater the concentration of solute, the lower the concentration of solvent at the surface and the lower the vapour
pressure.
Freezing Point Depression
Freezing point depression means that an aqueous solution has a lower freezing temperature than pure water. The
amount by which the freezing point is depressed increases with the molal concentration.
If water turns into ice, the energy of attraction between water molecules has to exceed their kinetic energy. That
is, the energy of the bond between molecules has to make them stay put in one place, rather than allowing them
to wander off in the liquid.
If a solute is present, these molecules tend to get in the way of the water molecules getting together. The whole
system needs to be at a lower total energy for the solute to be pushed out of the way to allow freezing. This
means a lower temperature.
Applications:
1.
Road salt
2.
Antifreeze
3.
Fish, amphibians and reptiles
28
To calculate the freezing point depression of a solution, use this formula:
For water, the freezing point depression constant is negative 1.86 °C / mol/kg. In other words, for every 1 mol/kg
of solute in solution, the freezing point of an aqueous solution is decreased by 1.86°C.
Boiling Point Elevation
This is related to freezing point depression and happens for the same reasons. The attraction of the solute particles
for the water molecules prevents them from moving from the liquid to the vapour phase as easily. More energy
needs to be applied, so the boiling temperature increases.
This effect is far less pronounced than freezing point depression. The boiling point elevation constant for water is
0.51 °C / mol/kg, less than one-third of that for the freezing point. The formula, however, is the same:
Examples:
1.
Calculate the freezing point depression and boiling point elevation of a 2.40 mol/kg solution of sucrose.
Point to Remember: Sucrose is a hydrocarbon molecule, it is polar, but it does not dissociate. Thus:
C12H22O11 (s)  C12H22O11 (aq)
2.40 mol/kg
29
1(2.40 mol/kg)
= 2.40 mol/kg
Freezing point depression:
ΔT
= Kfm
= (- 1.86 ºC / mol/kg)(2.40 mol/kg)
= - 4.46 ºC
Since water freezes at 0 ºC, the freezing point would be – 4.46 ºC.
Boiling Point Elevation
ΔT
= K bm
= (0.51 ºC / mol/kg)(2.40 mol/kg)
= 1.2 ºC
Since water boils at 100 ºC, the boiling point would be 101.2 ºC.
2.
Calculate the freezing point depression and boiling point elevation of a 2.40 mol/kg solution of
ammonium phosphate.
Point to Remember: (NH4)3PO4 is an ionic compound that dissociates in solution. Thus:
(NH4)3PO4 (s) 
2.40 mol/kg
3 NH4 (aq)
+
3(2.40 mol/kg)
= 7.20 m
PO4 (aq)
1(2.40 mol/kg)
= 2.40 m
The total molal concentration is
7.20 mol/kg + 2.40 mol/kg = 9.60 mol/kg
Freezing point depression:
ΔT
= Kfm
= (- 1.86 ºC / mol/kg)(9.60 mol/kg)
= - 17.9 ºC
Since water freezes at 0 ºC, the freezing point would be – 17.9 ºC.
Boiling Point Elevation
ΔT
= K bm
= (0.51 ºC / mol/kg)(9.60 mol/kg)
= 4.9 ºC
Since water boils at 100 ºC, the boiling point would be 104.9 ºC.
Complete questions 387 to 391 in the Problem Bank
30
Osmotic Pressure
Semipermeable membranes allow the movement of some particles while blocking the movement of others. Cell
membranes are an example. Water can pass freely across the membrane, while ions pass with difficulty. This
process is called osmosis.
Osmotic pressure is the external pressure that must be applied to stop osmosis.
Like other colligative properties, the greater the solute concentration difference on either side of the membrane,
the greater the osmotic pressure.
An isotonic solution provides an osmotic pressure of zero, because it has the same solute concentration as the
cells. An intravenous solution is usually an NaCl solution that is isotonic to the cell cytoplasm. It doesn’t matter
that the solute makeup does not match the cytoplasm, just that the concentration is the same.
31