Download 2(aq)

Classifying Chemical Reactions by
What Atoms Do
Classification of Reactions
H2 (g)
O2 (g) --------->
2 H(g)
O (g)
2
2 2HgO
(s) +--------->
2 Hg (l) + O
2
Synthesis reaction
C2CaCO
H4 (g) (s)
+ H2
O2 (aq) CaO (s) + CO
C2H(g)
O2 (l)
6
--------->
3
2
Cu (s) + 2 AgNO3 (aq) ---------> 2 Ag (s) + Cu(NO3)2 (aq)
Decomposition reaction
2 NaCl (s) --------->
Cl2 (g) + 2 Na (l)
2 Al (s) + Fe2O3 (s) ---------> Al2O3 (s) + 2 Fe (l)
Single displacement reaction
Mg (s) + 2 HCl (aq) ---------> H2 (g) + MgCl2 (aq)
Ba(NO 3)2 (aq) + Na2SO4 (aq) ---------> BaSO4 (s) + 2 NaNO 3 (aq)
Double displacement reaction
PCl3 (l) + 3 AgF (s) ---------> PF3 (g) + 3 AgCl (s)
Chemical Reactions
Classified by
Reaction Type
Precipitation
Reactions
Precipitation Reactions
Precipitation reactions are reactions in which a
solid forms when we mix two solutions.
!
1) reactions between aqueous solutions of ionic compounds
2) produce an ionic compound that is insoluble in water
3) The insoluble product is called a precipitate.
Precipitation Reactions
2 KI(aq) + Pb(NO3)2(aq) ➜ PbI2(s) + 2 KNO3(aq)
No Precipitate Formation = No Reaction
KI(aq) + NaCl(aq) ➜ KCl(aq) + NaI(aq)
No precipitate forms, therefore, no reaction.
KI(aq)
KCl(aq) + NaI(aq)
NaCl(aq)
Process for Predicting the Products of
a Precipitation Reaction
1.
!
2.
!
3.
4.
!
5.
!
6.
Determine which ions are present in each aqueous reactant.
Determine formulas of possible products.
Determine solubility of each potential product in water.
!
If neither product will precipitate, write no reaction after the
arrow.
If any of the possible products are insoluble, write their formulas as
the products of the reaction using (s) after the formula to indicate
solid. Write any soluble products with (aq) after the formula to
indicate aqueous.
Balance the equation.
Remember to only change coefficients, not subscripts
Practice – Predict the products and
balance the equation
K2CO3(aq) + NiCl2(aq) ➜
K2CO3(aq) + NiCl2(aq) ➜
KCl (?) + NiCO3(?)
K2CO3(aq) + NiCl2(aq) ➜ 2 KCl (?) + NiCO3(?)
K2CO3(aq) + NiCl2(aq) ➜ 2 KCl (aq) + NiCO3(s)
Practice – Predict the products and
balance the equation
KCl(aq) + AgNO3(aq) ➜
KCl(aq) + AgNO3(aq) ➜ KNO3(?) + AgCl(?)
KCl(aq) + AgNO3(aq) ➜ KNO3(aq) + AgCl(s)
Practice – Predict the products and
balance the equation
Na2S(aq) + CaCl2(aq) ➜
Na2S(aq) + CaCl2(aq) ➜ NaCl(?) + CaS(?)
Na2S(aq) + CaCl2(aq) ➜ 2 NaCl(?) + CaS(?)
Na2S(aq) + CaCl2(aq) ➜ 2 NaCl(aq) + CaS(aq)
No Reaction !!!!!
Practice – Predict the products and
balance the equation
(NH4)2SO4(aq) + Pb(C2H3O2)2(aq) ➜
(NH4)2SO4(aq) + Pb(C2H3O2)2(aq) ➜ NH4C2H3O2(?) + PbSO4(?)
(NH4)2SO4(aq) + Pb(C2H3O2)2(aq) ➜ 2 NH4C2H3O2(?) + PbSO4(?)
(NH4)2SO4(aq) + Pb(C2H3O2)2(aq) ➜ 2 NH4C2H3O2(aq) + PbSO4(s)
Ionic Equations
Equations that describe the chemicals put into the water and
the product molecules are called molecular equations.
!
2 KOH(aq) + Mg(NO3)2(aq) ➜ 2 KNO3(aq) + Mg(OH)2(s)
Equations that describe the material’s structure when
dissolved are called complete ionic equations.
Aqueous strong electrolytes are written as ions.
Insoluble substances, weak electrolytes, and nonelectrolytes are
written as molecules.
2K+(aq) + 2OH−(aq) + Mg2+(aq) + 2NO3−(aq) ➜ 2K+(aq) + 2NO3−(aq) + Mg(OH)2(s)
Ionic Equations
Ions that are both reactants and products are called spectator ions.
!
2 K+(aq) + 2 OH−(aq) + Mg2+(aq) + 2 NO3−(aq) ➜ 2 K+(aq) + 2 NO3−(aq) + Mg(OH)2(s)
An ionic equation in which the spectator ions are removed is called a
net ionic equation.
2 OH−(aq) + Mg2+(aq) ➜ Mg(OH)2(s)
Write the ionic and net ionic equation
K2SO4(aq) + 2 AgNO3(aq) ➜ 2 KNO3(aq) + Ag2SO4(s)
2K+ (aq) + SO42-(aq) + 2Ag+ (aq) + 2NO3-(aq) ➜ 2K+ (aq) + 2NO3-(aq) + Ag2SO4(s)
2 Ag+(aq) + SO42−(aq) ➜ Ag2SO4(s)
Write the ionic and net ionic equation
Na2CO3(aq) + 2 HCl(aq) ➜ 2 NaCl(aq) + CO2(g) + H2O(l)
2Na+ (aq) + CO32-(aq) + 2H+ (aq) + 2Cl-(aq) ➜ 2Na+ (aq) + 2Cl-(aq) + CO2(g) + H2O(l)
CO32−(aq) + 2 H+(aq) ➜ CO2(g) + H2O(l)
Acids and Bases
Acids and Bases in Solution
Acids ionize in water to form H+ ions.
(More precisely, the H+ from the acid molecule is donated
to a water molecule to form hydronium ion, H3O+)
Bases dissociate in water to form OH- ions.
(Bases, such as NH3, that do not contain OH- ions,
produce OH- by pulling H+ off water molecules.)
!
In the reaction of an acid with a base, the H+ from the acid
combines with the OH- from the base to make water.
!
The cation from the base combines with the anion from the
acid to make a salt.
acid + base ➜ salt + water
Molecular Models of Selected Acids
Acid-Base Reactions
Also called neutralization reactions because the
acid and base neutralize each other’s properties
2 HNO3(aq) + Ca(OH)2(aq) ➜ Ca(NO3)2(aq) + 2 H2O(l)
Note that the cation from the base combines with the
anion from the acid to make the water soluble salt.
The net ionic equation for an acid-base reaction is
H+(aq) + OH-(aq) ➜ H2O(l)
(as long as the salt that forms is soluble in water)
Common Acids
Name
Formula
Uses
Strength
Perchloric
HClO
explosives, catalysts
Strong
Nitric
HNO
explosives, fertilizers, dyes, glues
Strong
Sulfuric
H
explosives, fertilizers, dyes, glue,
batteries
metal cleaning, food prep, ore
refining, stomach acid
fertilizers, plastics, food
preservation
Strong
Hydrochloric
Phosphoric
Chloric
Acetic
Hydrofluoric
Carbonic
Hypochlorous
Boric
HCl
H
HClO
HC
HF
H
HClO
H
Strong
Moderate
explosives
Moderate
plastics, food preservation,
vinegar
Weak
metal cleaning, glass etching
Weak
soda water, blood buffer
Weak
sanitizer
Weak
eye wash
Weak
Common Bases
Name
Formula
Common Name
Uses
Strength
Sodium
Hydroxide
NaOH
Lye, Caustic Soda
soap, plastic production,
petroleum refining
Strong
Potassium
Hydroxide
KOH
Caustic Potash
soap, cotton processing,
electroplating
Strong
Calcium
Hydroxide
Ca(OH)
Slaked Lime
cement
Strong
Sodium
Bicarbonate
NaHCO
Baking Soda
food preparation, antacids
Weak
Magnesium
Hydroxide
Mg(OH)
Milk of Magnesia
antacids
Weak
Ammonia Water
fertilizers, detergents,
explosives
Weak
Ammonium
Hydroxide
NH
HCl(aq) + NaOH(aq) ➜ NaCl(aq) +
H2O(l)
HCl(aq)
NaOH(aq)
NaCl(aq) +!
H2O(l)
Write the molecular, ionic, and net-ionic
equation for the acid-base reaction
HNO3(aq) + Ca(OH)2(aq) ➜
HNO3(aq) + Ca(OH)2(aq) ➜ Ca(NO3)2(aq) + H2O(l)
2HNO3(aq) + Ca(OH)2(aq) ➜ Ca(NO3)2(aq) + 2H2O(l)
2H+ (aq) + 2NO3-(aq) + Ca2+ (aq) + 2OH-(aq) ➜ Ca2+ (aq) + 2NO3-(aq) + 2H2O(l)
2H+(aq) + 2OH-(aq) ➜ 2H2O(l)
Write the molecular, ionic, and net-ionic
equation for the acid-base reaction
HCl(aq) + Ba(OH)2(aq) ➜
HCl(aq) + Ba(OH)2(aq) ➜ BaCl2(aq) + H2O(l)
2HCl(aq) + Ba(OH)2(aq) ➜ BaCl2(aq) + 2H2O(l)
2H+ (aq) + 2Cl-(aq) + Ba2+ (aq) + 2OH-(aq) ➜ Ba2+ (aq) + 2Cl-(aq) + 2H2O(l)
2H+(aq) + 2OH-(aq) ➜ 2H2O(l)
Write the molecular, ionic, and net-ionic
equation for the acid-base reaction
H2SO4(aq) + Sr(OH)2(aq) ➜
H2SO4(aq) + Sr(OH)2(aq) ➜ SrSO4(s) + 2 H2O(l)
2H+ (aq) + SO42-(aq) + Sr2+ (aq) + 2OH-(aq) ➜ SrSO4 (s) + 2H2O(l)
2H+(aq) + SO42-(aq) + Sr2+ (aq) + 2OH-(aq) ➜ SrSO4 (s) + 2H2O(l)
Titration
A solution’s concentration is determined by
reacting it with another solution and using
stoichiometry – this process is called titration.
!
In the titration, the unknown solution is added to
a known amount of another reactant until the
reaction is just completed. At this point, called
the endpoint, the reactants are in their
stoichiometric ratio.
!
The unknown solution is added slowly from an
instrument called a burette.
Acid-Base Titrations
The difficulty is determining when there has
been just enough titrant added to complete
the reaction.
!
In acid-base titrations, because both the
reactant and product solutions are colorless,
a chemical (indicator) is added that changes
color when the solution undergoes large
changes in acidity/alkalinity
!
At the endpoint of an acid-base titration, the
number of moles of H+ equals the number of
moles of OH-(equivalence point).
Titration
The titrant is the base
solution in the burette.
As the titrant is added to
the flask, the H+ reacts
with the OH– to form
water. But there is still
excess acid present so the
color does not change.
At the titration’s endpoint,
just enough base has been
added to neutralize all the
acid. At this point the
indicator changes color.
Titration
The titration of 10.00 mL of HCl solution of unknown concentration
requires 12.54 mL of 0.100 M NaOH solution to reach the end point.
What is the concentration of the unknown HCl solution?
HCl(aq) + NaOH(aq) ➜ NaCl(aq) + H2O(l)
mL of
NaOH soln
L
mL
mL of
HCl soln
mol of
NaOH
L of !
NaOH soln
L of !
HCl soln
mol of !
HCl
L
mol
mol
mol
Molarity
=
mol of HCl
L of HCl soln
The titration of 10.00 mL of HCl solution of unknown concentration
requires 12.54 mL of 0.100 M NaOH solution to reach the end point.
What is the concentration of the unknown HCl solution?
HCl(aq) + NaOH(aq) ➜ NaCl(aq) + H2O(l)
12.54 mL NaOH solution x 0.001 L NaOH soln! x 0.100 mol NaOH!
!
1.000 mL NaOH soln 1.000 L NaOH soln
!
x 1.00 mol HCl!
= 1.25 x 10-3 mol HCl in the sample
1.00 mol NaOH
10.00 mL HCl solution x
Molarity of HCl solution =
0.001 L HCl soln!
1.000 mL HCl soln
= 0.0100 L HCl soln
1.25 x 10-3 mol HCl!
= 0.125 M
0.0100 L HCl soln
What is the concentration of NaOH solution that requires !
27.5 mL to titrate 50.0 mL of 0.1015 M H2SO4 ?
H2SO4 (aq) + 2 NaOH (aq) ➜ Na2SO4 (aq) + 2 H2O (l)
mL of
H2SO4 soln
mol of
H2SO4
L of !
H2SO4 soln
mol of !
NaOH
L
L
mol
m
mol
mol
mol of NaOH
mL of
NaOH soln
L of !
NaOH soln
Molarity
=
L of NaOH soln
What is the concentration of NaOH solution that requires !
27.50 mL to titrate 50.00 mL of 0.1015 M H2SO4 ?
H2SO4 (aq) + 2 NaOH (aq) ➜ Na2SO4 (aq) + 2 H2O (l)
50.00 mL H2SO4 soln x 0.001 L H2SO4 soln! x 0.1015 mol !H2SO4!
1.000 mL H2SO4 soln 1.000 L H2SO4 soln
!
x 2.00 mol NaOH!
= 0.1015
mol NaOH in the sample
1.00 mol H2SO4
27.50 mL NaOH soln x 0.001 L NaOH soln! = 0.02750 L NaOH soln
1.000 mL NaOH soln
Molarity of NaOH soln =
0.1015 mol NaOH!
=
0.02750 L NaOH soln
0.3691 M
Gas-Evolving
Reactions
Gas-Evolving Reactions
Some reactions form a gas directly from the ion exchange:
K2S(aq) + H2SO4(aq) ➜ K2SO4(aq) + H2S(g)
Other reactions form a gas by the decomposition of one
of the ion exchange products into a gas and water.
K2SO3(aq) + H2SO4(aq) ➜ K2SO4(aq) + H2SO3(aq)
!
!
H2SO3 ➜ H2O(l) + SO2(g)
NaHCO3(aq) + HCl(aq) ➜ NaCl(aq) + CO2(g) + H2O(l)
NaHCO3(aq)
NaCl(aq) + CO2(g)
+ H2O(l)
HCl(aq)
Other reactions form a gas by the decomposition of one
of the ion exchange products into a gas and water.
NaHCO3(aq) + HCl(aq) ➜ NaCl(aq) + H2CO3(aq)
!
!
H2CO3 ➜ H2O(l) + CO2(g)
Compounds that Undergo
Gas-Evolving Reactions
Reactant
Reactant
Metal sulfide
Metal hydrogensulfide
Acid
Metal carbonate
Metal hydrogencarbonate
Acid
Metal sulfite
Metal hydrogensulfite
Ammonium salt
Exchange
Product
Gas
Formed
Example
H
K2
H
CO
K2
(g)
Acid
H
SO
K2
SO
Base
NH
NH
KOH (aq) + NH
NH
H
H
Practice – Predict the products and
balance the equations
H2SO4(aq) + CaS(aq) ➜
CaSO4(aq) + H2S(aq)
Na2CO3(aq) + 2 HNO3(aq) ➜
“2 NaNO3(aq) + H2CO3”
2 NaNO3(aq) + H2O (l) + CO2(g)
2 HCl(aq) + Na2SO3(aq) ➜
“2 NaCl(aq) + H2SO3”
2 NaCl (aq) + H2O (l) + SO2 (g)
Redox Reactions
Oxidation/Reduction Basic Definitions
Oxidation and Reduction - Symbolic Representation
Oxidation and Reduction at the Atomic Level
Redox Reactions
Oxidation/reduction reactions involve transferring
electrons from one atom to another.
!
Also known as redox reactions
!
Many involve the reaction of a substance with O2(g).
!
4 Fe(s) + 3 O2(g) ➜ 2 Fe2O3(s)
Atoms in Elements-------> Ions in Compound
Combustion as Redox
2 H2(g) + O2(g) ➜ 2 H2O(g)
Redox without Combustion
2 Na(s) + Cl2(g) ➜ 2 NaCl(s)
Reactions of Metals with Nonmetals
Consider the following reactions:
!
4 Na(s) + O2(g) → 2 Na2O(s)
2 Na(s) + Cl2(g) → 2 NaCl(s)
!
The reactions involve a metal reacting with a nonmetal.
In addition, both reactions involve the conversion of free
elements into ions. !
!
Na2O = 2 Na+ + O2!
NaCl = Na+ + Cl-
Oxidation and Reduction
To convert a free element into an ion, the atoms
must gain or lose electrons (of course, if one
atom loses electrons, another must accept
them).
!
Atoms that lose electrons are being oxidized,
atoms that gain electrons are being reduced.
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
Na → Na+ + 1 e– oxidation
Cl2 + 2 e– → 2 Cl– reduction
Electron Bookkeeping
For reactions that are not metal + nonmetal, or do not involve
O2, we need a method for determining how the electrons
are transferred.
!
Chemists assign a number to each element in a reaction called
an oxidation state that allows them to determine the
electron flow in the reaction.
!
Even though they look like them, oxidation states
are not ion charges!
Oxidation states are imaginary charges assigned
based on a set of rules.
!
Ion charges are real, measurable charges.
Rules for Assigning Oxidation States
(in order of priority)
1.
!
Free elements have an oxidation state = 0.
In Na (s), Na = 0 ; In Cl2 (g), Cl2 = 0
!
2. Monatomic ions have an oxidation state equal to
their charge.
!
In NaCl, Na = +1 and Cl = −1
!
3. (a) The sum of the oxidation states of all the atoms
in a compound is 0.
!
Na = +1 and Cl = −1 in NaCl, (+1) + (−1) = 0
Rules for Assigning Oxidation States
(in order of priority)
3.
!
!
4.
(b) The sum of the oxidation states of all the atoms in a
polyatomic ion equals the charge on the ion.
In NO3–, N = +5 and O = −2 [3 x (2-) + 1 x (5+) = -1]
(a) Group I metals have an oxidation state of +1 in all their
compounds.
!
!
(b) Group II metals have an oxidation state of +2 in all their
compounds.
!
Rules for Assigning Oxidation States
(in order of priority)
5. In their compounds, nonmetals have oxidation
states according to the table below
Assign an oxidation state to each
element in the following
Br2
!
K+
!
LiF
!
CO2
!
SO42−
!
Na2O2
Br = 0, (Rule 1)
K = +1, (Rule 2)
Li = +1, (Rule 4a) & F = −1, (Rule 5)
O = −2, (Rule 5) & C = +4, (Rule 3a)
O = −2, (Rule 5) & S = +6, (Rule 3b)
Na = +1, (Rule 4a) & O = −1 , (Rule 3a)
Determine the oxidation states of all the atoms in a
propanoate polyatomic anion, C3H5O2–
There are no free elements or free ions in propanoate, so
the first rule that applies is Rule 3b
(C3) + (H5) + (O2) = −1
!
Because all the atoms are nonmetals, the next rule we
use is Rule 5, following the elements in order:
H = +1
O = −2
(C3) + 5(+1) + 2(−2) = −1
(C3) = −2
C = −⅔
*
Oxidation and Reduction
Another Definition
Oxidation occurs when an atom’s oxidation
state increases during a reaction.
Reduction occurs when an atom’s oxidation
state decreases during a reaction.
-4
CH4
0
+
+4
-2
2 O2 → CO2 + 2 H2O
oxidation
reduction
Assign oxidation states, determine the element oxidized
and reduced, and determine the oxidizing agent and
reducing agent in the following reactions:
Sn4+ + Ca0 → Sn2+ + Ca2+
Sn4+ is being reduced; Sn4+ is the oxidizing agent.
Ca is being oxidized; Ca is the reducing agent.
2
0
F2
0
+ S → SF4
S 4+
F-
F is being reduced from F0 to F-;F2 is the oxidizing agent.
S is being oxidized from S0 to S+4;S is the reducing agent.
Assign oxidation states, determine the element oxidized
and reduced, and determine the oxidizing agent and
reducing agent in the following reactions:
0
+7
+3
+4
Fe + MnO4− + 4 H+ → Fe3+ + MnO2 + 2 H2O
oxidation
reduction
Fe is the reducing agent.
MnO4− is the oxidizing agent.