Download Solutes

Solutions Notes
Words to Know
• Solution – homogenous mixture
• Solvent – substance present in the largest
amount
• Solutes – substance present in the smallest
amount
• Aqueous solution – solutions with water as the
solvent
• Concentration – the amount of solute in a given
volume of solution
• Concentrated – large amount of solute dissolved
in solvent
• Dilute – small amount of solute dissolved in
solvent
• Saturated – a solution that contains as much
solute as will dissolve at that temperature
• Unsaturated – a solution that hasn’t reached that
limit of solute that will dissolve
• Supersaturated - a solution that contains more
solute than should dissolve at that temperature
Effect of Temperature on Solubility
• Increasing the
temperature of a
solution, increases
the amount of
solute that can be
dissolved
• Decreasing the
temperature of a
solution, causes
the solute to
recrystallize
Learning Check
1. How many grams of
NaCl will dissolve in
100 g of H2O at 90°C?
2. 50 g of KCl is
dissolved in 100 g of
water at 50°C. Is the
solution saturated,
unsaturated or
supersaturated?
Effect of Pressure on Solubility
• Pressure has a
major effect on the
solubility of gasliquid systems
• An increase in
pressure increases
the solubility of a
gas in the liquid
“Like dissolves like” – a
solvent usually dissolves
solutes that have polarities
similar to itself
•
•
•
Polar molecules dissolve
other polar molecules and
ionic compounds. and alcohols
Nonpolar molecules
dissolve other nonpolar
molecules. and alcohols
Alcohols, which have
characteristics of both
polar & nonpolar, tend to
dissolve in both types of
solvents, but will not
dissolve ionic solids. and other
alcohols
SOLUTES
NaCl
ionic
I2
non polar
C3H7OH
alcohol
benzene
(nonpolar)
Br2
non polar
KNO3
ionic
toluene
(polar)
Ca(OH)2
ionic
methanol
alcohol
Alcohols are organic, covalent
molecules with an –OH group.
Alcohol names end with “-ol.”
NH3
polar
CO2
non polar
SOLVENTS
Water polar
CCl4non polar Alcohol alcohol
Colligative properties
Colligative properties - the physical
changes that result from adding solute to a
solvent. Colligative Properties depend on
how many solute particles are present as
well as the solvent amount, but they do NOT
depend on the type of solute particles.
•
•
•
•
•
Boiling Point Elevation
Freezing Point Depression
Osmotic Pressure Increasing
Vapor Pressure Lowering
Conductivity Increasing
More particles/ions =
greater change
Learning Check
1. Which substance will provide the greatest change in
freezing point of water?
A. NaCl
B. CaCl2
C. C6H12O6 D. H2O
2 ions
3 ions
1 particle
no change
in H2O
2. Which of the following reflect colligative properties?
(I) A 0.5 m NaBr solution has a higher vapor pressure than a 0.5 m BaCl2
3 ions
2 ions
solution.
2 ions
= vapor pressure lowering
(II) A 0.5 m NaOH solution freezes at a lower temperature than pure water
0 ions
(III) Pure water freezes at a higher temperature than pure methanol.
no solutions – colligative properties compare impact of solute
on property of a solvent in a solution
A. only I
B. only II C. only III
D. I and II E. I and III
= freezing
point
depression
3. A student measured the conductivity in water, of
unlabeled liquids, after each added drop. The
following graph was produced...
a. Identify the line that represents:
aluminum chloride AlCl3, 4 ions
water H2O, no change in H2O
magnesium chloride MgCl2, 3 ions
sugar C6H12O6, 1 particle
sodium chloride NaCl, 2 ions
b. Which line could
also represent
potassium iodide?
AlCl3
MgCl2
Conductivity (µs/cm)
–
–
–
–
–
NaCl
C6H12O6
H2O
# of Drops
Solution Composition - Mass
Percent
Mass percent – describes a solution’s composition expresses the mass
of solute present in a given mass of solution
Mass Percent
=
mass of solute x 100%
mass of solution*
* mass of solution = mass of solute + mass of solvent
Example – A solution is prepared by mixing 1.00g of C2H5OH, with
100.0g of H2O. Calculate the mass percent of ethanol.
Given
Mass Percent =
mass of solute x 100%
mass of solution
mass of solute = 1.00 g
mass of solution = 100.0 g + 1.00 g = 101.0 g
Mass % = 1.00 g x 100 %
101.0 g
Mass % = 0.990 %
Solution Composition – Molarity
Molarity – measure of concentration
- number of moles of solute per volume of solution in liters
Molarity =
moles of solute = mol = M
L of solution
L
Example – Calculate the molarity of a solution prepared by
dissolving 11.5 g NaOH in enough water to make 1.50L
solution.
1 mol NaOH x __________
1
11.5 g NaOH x ___________
=
40.00 g NaOH 1.50 L NaOH
0.192 M
Ex: Calculate the mass of solid AgCl formed
when 1.50L of a 0.100M AgNO3 solution
is reacted with excess NaCl.
NaCl + AgNO3  AgCl + NaNO3
1.50 L
?g
0.100 M
0.100 mol AgNO3 x ____________
1 mol AgCl x143.32
g AgCl =
1.50 L AgNO3 x ______________
____________
L AgNO3
1
1 mol AgNO3
1 mol AgCl
no grams?
start with liters
use M as conversion
factor to conver to mol
# M = # mol
1L
mole ratio
convert to desired unit
21.5 g
Example – How many moles of Ag+ ions are
present in 25mL of a 0.75M Ag2SO4
solution?
Ag2SO4  2 Ag+1 + SO4-2
1 L Ag2SO4
0.75 mol Ag2SO4 x ______________
2 mol Ag+1
25 mL Ag2SO4 x _____________
x ______________
=
1000 mL Ag2SO4
1 L Ag2SO4
1 mol Ag2SO4
0.038 mol Ag+1
Learning check
Calculate the molarity of a solution prepared
by dissolving 25.6 g NaC2H3O2 in enough
water to make 200.0 mL solution.
Standard Solution
• Standard Solution – a solution whose
concentration is accurately known
Example – A chemist needs 1.0 L of a
0.200M K2Cr2O7 solution. How much solid
K2Cr2O7 must be weighed out to make
this solution?
1.0 L K2Cr2O7 x ________________
0.200 mol K2Cr2O7
1
L K2Cr2O7
g K2Cr2O7
x 294.20
______________
=
1 mol K2Cr2O7
59 g K2Cr2O7
Dilution
Dilution – process of adding more solvent to a solution
Moles of solute before dilution = Moles of solute after dilution
M1V1 = M2V2
Example: What volume of 16M H2SO4 must be used to
prepare 1.5L of a 0.10M H2SO4 solution?
Given
V1 = ?
M1 = 16 M
V1 = _____
M2V2
M1
V1 = (0.10
M)(1.5 L)
____________
16 M
V2 = 1.5 L
M2 = 0.10 M
V1 = 0.0094 L
Learning Check
Example: Prepare 500.0mL of 1.00 M
HC2H3O2 from a 17.5 M stock solution.
What volume of the stock solution is
required?
Given
V1 = 500.0 mL
M1 = 1.00 M
V2 = _____
M1V1
M2
V2 = (1.00
M)(500.0 mL)
_______________
17.5 M
M2 = 17.5 M
V2 = ?
V2 = 28.6 mL
NotesAcids and Bases
Acids and Bases
Arrhenius ACIDS – produces hydrogen ions in
aqueous solutions, sour taste, low pH, and the
fact that they turn litmus paper red
HCl (aq)  H+ (aq) + Cl- (aq)
Arrhenius BASES – produces hydroxide ions in
aqueous solutions, bitter taste, slippery feel, high
pH, and the fact that they turn litmus paper blue
NaOH (aq)  Na+ (aq) + OH- (aq)
Arrhenius definition – limits the concept of a base
Bronsted – Lowry definition – gives a broader definition of
a base
Bronsted – Lowry ACID – a proton (H+) donor
Bronsted – Lowry BASE – a proton (H+) acceptor
General Reaction –
proton donor
HA (aq) + H2O (l) 
H3O+ (aq) +
A- (aq)
Acid
Base
Conjugate
Conjugate
Acid
Base
proton acceptor
Conjugate Base – everything that remains of the acid
molecule after a proton is lost
Conjugate Acid – the base with the transferred proton (H+)
Conjugate Acid – Base Pair – two substances related to
each other by the donating and accepting of a single
proton
Examples: Finish each equation and
identify each member of the conjugate
acid –base pair.
H2SO4 (aq) + H2O (l) 
Acid
Base
HSO4-1(aq) + H3O+ (aq)
Conjugate
Base
Conjugate
Acid
CO32- (aq) + H2O (l) 
HCO3-1(aq) + OH- (aq)
Base
Conjugate
Acid
Acid
Conjugate
Base
The hydronium ion, H3O+, forms when water behaves as a base. This
happens when the two unshared pairs of electrons on O bond covalently with
the H+.
Learning check
Write the conjugate ACID
a. NH3
b. HCO3-1
Write the conjugate BASE
a. H3PO4
b. HBr
Finish each equation and identify each member of
the conjugate acid –base pair.
a. H2SO3 (aq) + H2O (l) 
b. SO4-2 (aq) + H2O (l) 
Water as an Acid and a Base
Amphoteric – a substance that can behave
as either an acid or a base
- water is the most common amphoteric
substance
Ionization of Water –
H2O (l) + H2O (l)  H3O+ (aq) + OH- (aq)
In the shorthand form:
H2O (l)  H+ (aq) + OH- (aq)
Ion-product constant – Kw refers to the
ionization of water
[ ] = concentration
Kw = [H+][OH-]
[H+] = hydrogen ion
concentration in M
[OH-] = hydroxide ion
concentration in M
At 25C, Kw = [H+][OH-] = [1.0 x 10-7] [1.0 x 10-7]
= 1.0 x 10-14
If [H+] increases, the [OH-] decreases, so the
products of the two is still 1.0 x 10-14.
There are three possible situations –
1. A neutral solution, where [H+] = [OH-]
2. An acidic solution, where [H+]  [OH-]
3. A basic solution, where [H+]  [OH-]
Example: Calculate [H+] or [OH-] as
required for each of the following solutions
at 25C, for each solution state whether it
is neutral, acidic, or basic.
a. 1.0 x 10-5 M OHKw = [H+][OH-]
1 x 10-14 = [H+][1.0 x 10-5 M]
[H+] = 1.0 x 10-9 M
BASIC
b. 10.0 M H+
Kw = [H+][OH-]
1 x 10-14 = [10.0 M][OH-]
[OH-] = 1.00 x 10-15 M
ACIDIC
pH scale
pH scale – because the [H+] in an aqueous solution is
typically small, logarithms are used to express solution
acidity
pH = -log [H+]
Graphing calculator
1. Press the +/- key
2. Press the log key
3. Enter the [H+]
pOH = -log [OH-]
Non graphing calculator
1. Enter the [H+]
2. Press the log key
3. Press the +/- key
Significant Figure Rule – The number of places to the right
of the decimal for a log must be equal to the number of
significant figures in the original number.
pH
Scale
Example – Calculate the pH or pOH
a. [H+] = 5.9 x 10-9 M
pH = - log [H+]
pH = - log (5.9 x 10-9 M)
pH = 8.23
b. [OH-] = 2.4 x 10-6 M
pOH = - log [OH-]
pOH = - log (2.4 x 10-6 M)
pOH = 5.62
Since Kw = [H+][OH-] = 1.0 x 10-14 ,
pH + pOH = 14.00
Example - The pH of blood is about 7.4.
What is the pOH of blood?
pH + pOH =14.00
7.4 + pOH = 14.00
pOH = 6.6
In order to calculate the concentration from the pH
or pOH,
[H+] = 10-pH
Graphing calculator
1. Press the 2nd
function, then log
2. Press the +/- key
3. Enter the pH
[OH-] = 10-pOH
Non-graphing calculator
1. Enter the pH
2. Press the +/- key
3. Press the inverse
log key
Example - The pH of a human blood
sample was measured to be 7.41. What is
the [H+] in blood?
[H+] = 10-pH
[H+] = 10-7.41
[H+] = 3.9 x 10-8 M
Example – The pOH of the water in a fish
tank is found to be 6.59. What is the [H+]
for this water?
[OH-] = 10-pOH
[OH-] = 10-6.59
[OH-] = 2.6 x 10-7 M
Kw = [H+][OH-]
1 x 10-14 = [H+][2.6 x 10-7 M]
[H+] = 3.8 x 10-8 M
Learning check
• Determine the pH of a solution with a hydrogen ion concentration of
3.2 x10-12 M.
• What is the [OH-] concentration of a solution with a hydrogen ion
concentration of 8.9x10-4M?
• What is the pH of a solution with a hydroxide ion concentration of
5.7x10-10 M?
How Do We Measure pH?
• For less accurate
measurements, one
can use
– Litmus paper
• “Red” paper turns
blue above ~pH = 8
• “Blue” paper turns
red below ~pH = 5
– An indicator
How Do We Measure pH?
For more accurate
measurements, one
uses a pH meter,
which measures the
voltage in the
solution.
Strong Acids
• seven strong acids
are HCl, HBr, HI,
HNO3, H2SO4,
HClO3, and HClO4.
• These are, by
definition, strong
electrolytes and
exist totally as ions
in aqueous
solution.
Strong Bases
• Strong bases are the soluble hydroxides, which
are the alkali metal and heavier alkaline earth
metal hydroxides (Ca2+, Sr2+, and Ba2+).
• Again, these substances dissociate completely
in aqueous solution, strong electrolytes
Strong, Weak, or Nonelectrolyte
•
•
•
•
Electrolytes are substances which, when dissolved in water, break up into
cations (plus-charged ions) and anions (minus-charged ions). We say they
ionize. Strong electrolytes ionize completely (100%), while weak electrolytes
ionize only partially (usually on the order of 1–10%). The ions in an
electrolyte can be used to complete an electric circuit and power a bulb.
Strong electrolytes fall into three categories: strong acids, strong bases, and
soluble salts.
The weak electrolytes include weak acids, weak bases and insoluble salts.
Molecules are nonelectrolytes.
Substance
Classification - Strong acid, weak
acid, strong base, weak base, soluble
salt, insoluble salt, molecule
Strong electrolyte,
weak electrolyte,
nonelectrolyte
sodium hydroxide
strong base
acetic acid
weak acid
weak electrolyte
potassium nitrate
soluble salt
strong electrolyte
hydrobromic acid
weak acid
weak electrolyte
silver chloride
insoluble salt
weak electrolyte
Carbon dioxide
molecule
nonelectrolyte
strong electrolyte
Learning check
Substance
chloric acid
barium carbonate
nitric acid
sulfurous acid
strontium sulfate
ethanol
octane (gasoline)
Classification - Strong
Strong electrolyte,
acid, weak acid, strong
weak electrolyte,
base, weak base, soluble
nonelectrolyte
salt, insoluble salt, molecule
Titration
A known concentration of base (or acid) is slowly
added to a solution of acid (or base).
Titration
A pH meter or
indicators are used to
determine when the
solution has reached
the equivalence point,
at which the
stoichiometric amount
of acid equals that of
base.
Titration of a Strong Acid with
a Strong Base
From the start of the
titration to near the
equivalence point,
the pH goes up
slowly.
Titration of a Strong Acid with
a Strong Base
Just before and after
the equivalence point,
the pH increases
rapidly.
Titration of a Strong Acid with
a Strong Base
At the equivalence
point, moles acid =
moles base, and the
solution contains only
water and the salt from
the cation of the base
and the anion of the
acid.
Titration of a Strong Acid with
a Strong Base
As more base is
added, the increase
in pH again levels
off.
Neutralization
Neutralization Reaction =
Acid + Base  Salt + Water
Salt – ionic compound containing a positive
ion other than H+ and a negative ion other
than OH-
Buffered solutions – resists a change in its
pH even when a strong acid or base is
added to it
- A solution is buffered in the presence of a
weak acid and its conjugate base
pH and pOH Calculations
+
H
[OH-] = 1 x 10-14
[H+]
-
OH
pH
pOH = 14 - pH
pH = 14 - pOH
pOH = -log[OH -]
[OH-] = 10 -pOH
pH = -log[H +]
[H+] = 10 -pH
[H+] = 1 x 10-14
[OH-]
pOH