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Chapter 7
Reactions and Solutions
Denniston Topping Caret
4th Edition
Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
7.1 Writing Chemical Reactions
• We will learn to identify the following patterns of chemical reactions:– combination– decomposition– single replacement– double replacement
• Recognizing the pattern will help you write and understand reactions.
1
7.1
Wri
tin
g C
hem
ical
R
eact
ion
sCombination Reactions
• The joining of two or more elements or compounds.
A + B AB
• Examples:
2Na(s) + Cl2(g) 2NaCl(s)
MgO(s) + CO2(g) MgCO3(s)
7.1
Wri
tin
g C
hem
ical
R
eact
ion
sDecomposition Reactions
• produce two or more products from a single reactant
AB A + B
• Examples:
2HgO(s) 2Hg(l) + O2(g)
CaCO3(s) CaO(s) + CO2(g)
7.1
Wri
tin
g C
hem
ical
R
eact
ion
sReplacement Reactions (2 types)
1. Single replacement
A + BC B + AC
• Examples:
Cu(s)+2AgNO3(aq) 2Ag(s)+Cu(NO3)2(aq)
2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)
7.1
Wri
tin
g C
hem
ical
R
eact
ion
s2. Double Replacement
AB + CD AD + CB
• Examples:
Pb(NO3)2(aq) + 2NaCl(aq)
PbCl2(s) + 2NaNO3(aq)
HCl(aq)+NaOH(aq) NaCl(aq)+H2O(l)
7.2 Types of Chemical Reactions
Precipitation Reactions
• Chemical change in a solution that results in one or more insoluble products.
• To predict if a precipitation reaction can occur it is helpful to know the solubilities of ionic compounds (Table 7.1)
2
7.2
Typ
es o
f C
hem
ical
Rea
ctio
ns
Table 7.1 Solubilities of Common Ionic Cpds.Solubility Predictions
Sodium, potassium, and ammonium compounds are generallysoluble.
Nitrates and acetates are generally soluble
Chlorides, bromides, and iodides (halides) are generally soluble. Exceptions: halides containing Pb2+, Ag+, Hg2
2+
Carbonates and phosphates are generally insoluble. Exceptions: Na+, K+, Ca2+, NH4
+ combined with carbonatesand phosphates are soluble.
Hydroxides and sulfides are generally insoluble. Exceptions:Na+, K+, Ca2+, NH4
+ combined with OH- are soluble.
7.2
Typ
es o
f C
hem
ical
Rea
ctio
ns
Predicting Whether Precipitation Will Occur
• Have the ionic compounds exchange partners.
• Look at the new compounds found and determine if any are insoluble according to the rules in Table 7.1
• The insoluble salt will be the precipitate.
• Example:
Pb(NO3)2(aq) + NaCl(aq)
PbCl2 (?) + NaNO3 ( ?)(s) (aq)
7.2
Typ
es o
f C
hem
ical
Rea
ctio
ns
Write a balanced equation for each precipitation reaction:
1. sodium hydroxide and aluminum chloride
2. sodium carbonate and nickel(II) bromide
7.2
Typ
es o
f C
hem
ical
Rea
ctio
ns
Reactions with Oxygen• Reactions with oxygen generally
release energy.
• Example: combustion of natural gas
• Organic compounds (carbon containing cpds.) CO2 and H2O usually the products.
• Example: CH4+2O2CO2+2H2O
• Rusting (corrosion of iron) is an inorganic example.
4Fe + 3O2 2Fe2O3
7.2
Typ
es o
f C
hem
ical
Rea
ctio
ns
Acid-Base Reactions
• These reactions involve the transfer of a hydrogen ion (H+) from one reactant to another.
• Example:
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
The hydrogen ion on HCl was transferred to the oxygen in OH-, giving H2O
More to come in 9.1-9.4
7.2
Typ
es o
f C
hem
ical
Rea
ctio
ns
Oxidation-Reduction Reactions
• Reaction involves the transfer of one or more electrons from one reactant to another.
• Example:
Zn(s) + Cu2+(aq) Cu(s) + Zn2+(aq)
Two electrons are transferred from Zn to Cu2+ More to come in 9.5
7.3 Properties of Solutions
• Solution - homogeneous mixture
• Solute - the substance in the mixture present in lesser quantity
• Solvent - the substance present in the largest quantity
• Aqueous Solution - solution where the solvent is water
• Solutions can be liquids as well as solids and gases.
3
7.3
Pro
per
ties
of
Sol
uti
ons
General Properties of Liquid Solutions
• Clear
• May have color
• Remember that electrolytes are compounds that dissociate when they dissolve in water.
)(-Cl )(Na )NaCl( OH2 aqaqs
• Nonelectrolytes do not dissociate.
4
7.3
Pro
per
ties
of
Sol
uti
ons
• Volumes are not additive.
– 1 L ethanol + 1 L water does not give 2 L of solution.
Solutions and Colloids
• Colloidal suspension - contains solute particles which are not uniformly distributed.
– Due to larger size of particles (1nm - 200 nm)
– Smaller than 1 nm, have solution.
– Larger than 1 nm, have a precipitate.
7.3
Pro
per
ties
of
Sol
uti
ons
• Tyndall effect - the ability of a colloidal suspension to scatter light.
– See this as a haze when shining light through the mixture.
– Solutions: light passes right through without scattering.
• Suspension:
– mixture contains particles much larger than a colloidal suspension
– particles do not easily settle out and form a precipitate
7.3
Pro
per
ties
of
Sol
uti
ons Degree of Solubility
• Solubility - how much solute can dissolve in a solvent.
• Factors which affect solubility:1 Polarity of solute and solvent
• The more they are different, the less soluble.
2 Temperature
• increase in temp. usually increases solubility.
3 Pressure
• for gas solutes, increased pressure of gas increases solubility
7.3
Pro
per
ties
of
Sol
uti
ons • Saturated solution - contains all the
solute that can be dissolved at a particular temperature.
• Supersaturated solution - contains more solute than can be dissolved at that temperature.
• How is this done? • Heat solvent, saturate it with solute then
cool slowly.
• Sometimes the excess will precipitate out.
• If it doesn’t precipitate, the solution will be supersaturated.
7.3
Pro
per
ties
of
Sol
uti
ons
Solubility and Equilibrium
• A saturated solution is an example of a dynamic equilibrium.
5
7.3
Pro
per
ties
of
Sol
uti
ons Solubility of Gases: Henry’s Law
• Henry’s Law - the number of moles of a gas dissolved in a liquid is proportional to the partial pressure of the gas above the liquid. (At constant temperature)
• Carbonated Beverages
• Respiration
7.4 Concentration of Solutions: Percentage
• Concentration - amount of solute dissolved in a given amount of solution.
6
solution ofamount
solute ofamount ion concentrat
• We will learn to express concentration in 3 ways:
Weight/Volume Percent, Weight/Weight Percent and Molarity
7.4
Per
cen
tage
1. Calculate the %W/V of 2.00 x 102 mL containing 20.0 g sodium chloride.
(ans:10.0%)
2. How many grams is required to make 500.0 mL of a 1.50% glucose solution.
(ans: 7.50 g glucose)
3. What volume is required to make a 15.0% glucose solution containing 10.0 g glucose
(ans: 66.7 mL solution)
7.4
Per
cen
tage
Weight/Weight Percent
%100solutions grams
solute grams
W
W%
Used most often with solid solutions.
Note: there is an error with the above formula on page 186 of your book.
7.4
Per
cen
tage
1. Calculate the %(W/W) of copper in an alloy containing 15.0 g copper and 10.0 g nickel.
(ans: 60.0% Cu)
2. A sample of silver and gold is 35%(W/W) of silver. If the sample has a mass of 13.5 g, what is the mass of silver?
(ans: 4.7 g Ag)
7.5
Mol
es a
nd
Eq
uiv
alen
ts
1. Calculate the molarity of a 150.0 mL solution contain 10.0 g NaOH.
(ans: 1.67 M)
2. How many grams of NaCl is required to make 200.0 mL of a 0.125 M solution?
(ans: 1.46 g NaCl)
7.5
Mol
es a
nd
Eq
uiv
alen
ts Dilution
• A dilution is required if..
– you have a solution in stock and
– wish to make a solution of less concentration.
– Simply add the correct amount of water.
• Since:
solution L
solute molesM
8
• Then: M x V = moles of solute
• In a dilution will the moles of solute change?
M1V1 = M2V2
This is the dilution formula, DO NOT USE IT FOR ANYTHING ELSE!7.
5 M
oles
an
d E
qu
ival
ents
• You may use %W/V concentration units as well
•No
•So, since M x V=moles
7.5
Mol
es a
nd
Eq
uiv
alen
ts1. Calculate the molarity of a solution made by
diluting 25.0 mL of 6.0 M HCl with enough water to make 500 mL of solution.
(ans: 0.30 M)
2. Calculate amount of 6.0 M HCl needed to make 1.0 L of 0.10 M HCl.
(ans: 16.7 mL)
3. How many mL of water must be added to 15.0 mL of 6.0 M HCl to produce a 1.0 M solution? Assume volumes are additive.
(ans: 90.0 mL - 15.0 mL = 75.0 mL water)
7.5
Mol
es a
nd
Eq
uiv
alen
tsRepresentation of Concentration of Ions in Solution 9
• Two common ways of expressing concentration of ions in solution:
1. moles per liter (molarity)
2. equivalents per liter (emphasizes charge) We will not cover this.
7.5
Mol
es a
nd
Eq
uiv
alen
ts• Let’s look at Na3PO4.
• If we dissolved 1 mol of sodium phosphate into 1 L of water, what would be the concentration (M) of the solution?
• 1 M
• Now let’s consider the individual ions. What would the concentration of Na+ be?
• 3 M
7.5
Mol
es a
nd
Eq
uiv
alen
ts 1M Na3PO4
• What would the concentration of PO4
3- ions be?
• 1 M
• Equivalent is defined by the charge.
• One Equivalent of an ion is the number of grams of the ion corresponding to Avogadro’s number of electrical charges.
ionon charges ofnumber
(g)ion of massmolar ionan of equivalent One
7.5
Mol
es a
nd
Eq
uiv
alen
ts
• 1 mol PO43- = 3 equivalents PO4
3-
• O.K. back to the 1 M Na3PO4 solution. What are the equivalents of Na+?
• 3 mol Na+ = 3 equivalents of Na+.
• What are the equivalents of PO43-?
• 1 mol PO43- = 3 equivalents of PO4
3-.
• Do you see where the word equivalent might come form?
• Another way of looking at it: it is the mole of charges:
• 1 mol Na+ = 1 equivalent Na+
7.5
Mol
es a
nd
Eq
uiv
alen
ts • So the following are things we can say about a 1 M solution of Na3PO4.
3 mol/L Na+
1 mol/L PO43-
3 equivalents/L Na+
3 equivalents/L PO43-
7.6 Concentration-Dependent Solution Properties
• Colligative properties- properties of solutions that depend on the concentration of the solute particles, rather than the identity of the solute.
• There are four:– vapor pressure lowering– boiling point elevation– freezing point depression– osmotic pressure
10
7.6
Col
liga
tive
Pro
per
ties Vapor Pressure Lowering
• Raoult’s Law - when a nonvolatile solute is added to a solvent, the vapor pressure of the solvent decreases in proportion to the concentration of the solute.
7.6
Col
liga
tive
Pro
per
ties Freezing Point Depression and
Boiling Point Elevation
• Before we can discuss the above colligative properties we must learn a new concentration unit.
• Molality (m) = moles of solute per kg of solvent.
solvent kg
solute moles Molality
• Note: the denominator is in kg solvent, not in kg solution.
7.6
Col
liga
tive
Pro
per
ties Freezing point depression (Tf) is
proportional to the number of solute particles Tf=kf m
freezing point depression constantEach solvent has it’s own value.
• Note, it is solute particles not just solute.• If the solute is an electrolyte, what will
it do?• It will break apart into ions.• Therefore the same concentration of NaCl
will affect the freezing point twice as much as glucose (a nonelectrolyte)
7.6
Col
liga
tive
Pro
per
ties Boiling point elevation (Tb) is
proportional to the number of solute particles
Tb=kb m
boiling point elevation constantEach solvent has it’s own value.
• Again, an electrolyte will affect the boiling point to a greater degree than a nonelectrolyte with the same concentration.
7.6
Col
liga
tive
Pro
per
ties Osmotic Pressure
• Semipermeable membranes- allow solvent but not solute to diffuse from one side to another.
• Osmosis - the movement of solvent from a dilute solution to a more concentrated solution through the membrane.
• Osmotic pressure ( - the amount of pressure required to stop the flow.
7.6
Col
liga
tive
Pro
per
ties
=MRT
Again, M is molarity of particles. Called osmolarity.
7.6
Col
liga
tive
Pro
per
ties
Calculate the osmolarity and osmotic pressure of a 2.5 x 10-2 M solution of CaCl2 at 25oC.
7.6
Col
liga
tive
Pro
per
ties
Cell remains unchanged when surrounded by an isotonic solution. (water entering = water leaving)
Cell rupture occurs when cells are surrounded by hypotonic solution (water entering > water leaving)
Crenation occurs when cells are surrounded by a hypertonic solution (water leaving > water entering)
7.7 Water as a Solvent
• Water often referred to as the “universal solvent”
• Excellent solvent for polar molecules.
• Most abundant liquid on earth.
• 60% of the human body is water.– transports ions, nutrients and waste into and out of
cells.– solvent for biochemical reactions in cells and
digestive tract– reactant or product in some biochemical processes.
11
7.8 Electrolytes in Body Fluids
CATIONS IN THE BLOOD and CELLS
• Na+ and K+ two most important cations
12
Na+
K+
Blood Cells
135 meq/L 3.5 – 5.0 meq/L
10 meq/L 125 meq/L
• Active transport: the transporting of the above ions across the cell membrane.
• Cellular energy must be expended to make concentration of ions different on each side of the cell membrane.
• This is accomplished via large protein molecules embedded in cell membranes.
7.8
Ele
ctro
lyte
s in
Bod
y F
luid
s
7.8
Ele
ctro
lyte
s in
Bod
y F
luid
s• Danger to body occurs when Na+ and
K+ both in blood and in cells becomes too high or low.
- Na+ too low: - Decrease of urine output
- Drymouth
- flushed skin
- fever
- Na+ too high:- confusion, stupor or coma
7.8
Ele
ctro
lyte
s in
Bod
y F
luid
s- K+ too high:
- death by heart failure
- K+ too low:- death by heart failure
ANIONS IN THE BLOOD
• Cl- - acid/base balance- maintenance of osmotic pressure- oxygen transport by hemoglobin
7.8
Ele
ctro
lyte
s in
Bod
y F
luid
s• HCO3
-
- Form in which most waste CO2 is carried out of the body.
PROTEINS IN THE BLOOD
• Blood clotting factors
• antibodies
• albumins (carriers of non-polar substances which cannot dissolve in water)
• Proteins are transported as a colloidal suspension.
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