Classification of Reactions
4 Al (s) + 3 O2 (g) 2 Al2O3 (s)
2 H2 (g) + O2 (g) ---------> 2 H2O (g)
C2H4 (g) + H2O2 (aq) C2H6O2 (l)
Synthesis reaction 2 HgO (s) ---------> 2 Hg (l) + O2 (g)
CaCO3 (s) ---------> CaO (s) + CO2 (g)
2 NaCl (s) ---------> Cl2 (g) + 2 Na (l)
Decomposition reaction Cu (s) + 2 AgNO3 (aq) ---------> 2 Ag (s) + Cu(NO3)2 (aq)
2 Al (s) + Fe2O3 (s) ---------> Al2O3 (s) + 2 Fe (l)
Mg (s) + 2 HCl (aq) ---------> H2 (g) + MgCl2 (aq)
Single displacement reaction
Ba(NO3)2 (aq) + Na2SO4 (aq) ---------> BaSO4 (s) + 2 NaNO3 (aq)
PCl3 (l) + 3 AgF (s) ---------> PF3 (g) + 3 AgCl (s)
HCl (aq) + NaOH (aq) ---------> H2O(l) + NaCl (aq)
Double displacement reaction
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.
No Precipitate Formation = No Reaction
KI(aq) + NaCl(aq) ➜ KCl(aq) + NaI(aq)
KI(aq)
NaCl(aq)
KCl(aq) + NaI(aq)
No precipitate forms, therefore, no reaction.
Process for Predicting the Products ofa Precipitation Reaction
1. Determine which ions are present in each aqueous reactant. !2. Determine formulas of possible products. !3. Determine solubility of each potential product in water.
!4. If neither product will precipitate, write no reaction after the
arrow. !5. 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.
!6. 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) ➜ 2 KCl (?) + NiCO3(?)
K2CO3(aq) + NiCl2(aq) ➜ 2 KCl (aq) + NiCO3(s)
K2CO3(aq) + NiCl2(aq) ➜ KCl (?) + NiCO3(?)
Practice – Predict the products and balance the equation
KCl(aq) + AgNO3(aq) ➜ KNO3(?) + AgCl(?)
KCl(aq) + AgNO3(aq) ➜ KNO3(aq) + AgCl(s)
KCl(aq) + AgNO3(aq) ➜
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 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.
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)
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)
2 Ag+(aq) + SO42−(aq) ➜ Ag2SO4(s)
2K+ (aq) + SO42-(aq) + 2Ag+ (aq) + 2NO3-(aq) ➜ 2K+ (aq) + 2NO3-(aq) + Ag2SO4(s)
Na2CO3(aq) + 2 HCl(aq) ➜ 2 NaCl(aq) + CO2(g) + H2O(l)
CO32−(aq) + 2 H+(aq) ➜ CO2(g) + H2O(l)
2Na+ (aq) + CO32-(aq) + 2H+ (aq) + 2Cl-(aq) ➜ 2Na+ (aq) + 2Cl-(aq) + CO2(g) + H2O(l)
Write the ionic and net ionic equation
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
Acid-Base ReactionsAlso 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.
H+(aq) + OH-(aq) ➜ H2O(l)
(as long as the salt that forms is soluble in water)
The net ionic equation for an acid-base reaction is
Common AcidsName Formula Uses Strength
Perchloric HClO explosives, catalysts Strong
Nitric HNO explosives, fertilizers, dyes, glues Strong
Sulfuric H explosives, fertilizers, dyes, glue, batteries Strong
Hydrochloric HCl metal cleaning, food prep, ore refining, stomach acid
Strong
Phosphoric H fertilizers, plastics, food preservation Moderate
Chloric HClO explosives Moderate
Acetic HC plastics, food preservation, vinegar
Weak
Hydrofluoric HF metal cleaning, glass etching Weak
Carbonic H soda water, blood buffer Weak
Hypochlorous HClO sanitizer Weak
Boric H eye wash Weak
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
Ammonium Hydroxide
NH Ammonia Waterfertilizers, detergents,
explosives Weak
Common Bases
Write the molecular, ionic, and net-ionic equation for the acid-base reaction
HNO3(aq) + Ca(OH)2(aq) ➜
2H+ (aq) + 2NO3-(aq) + Ca2+ (aq) + 2OH-(aq) ➜ Ca2+ (aq) + 2NO3-(aq) + 2H2O(l)
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) + 2OH-(aq) ➜ 2H2O(l)
HCl(aq) + Ba(OH)2(aq) ➜
Write the molecular, ionic, and net-ionic equation for the acid-base reaction
HCl(aq) + Ba(OH)2(aq) ➜ BaCl2(aq) + H2O(l)
2H+(aq) + 2OH-(aq) ➜ 2H2O(l)
2H+ (aq) + 2Cl-(aq) + Ba2+ (aq) + 2OH-(aq) ➜ Ba2+ (aq) + 2Cl-(aq) + 2H2O(l)
2HCl(aq) + Ba(OH)2(aq) ➜ BaCl2(aq) + 2H2O(l)
H2SO4(aq) + Sr(OH)2(aq) ➜
2H+ (aq) + SO42-(aq) + Sr2+ (aq) + 2OH-(aq) ➜ SrSO4 (s) + 2H2O(l)
2H+(aq) + SO42-(aq) + Sr2+ (aq) + 2OH-(aq) ➜ SrSO4 (s) + 2H2O(l)
H2SO4(aq) + Sr(OH)2(aq) ➜ SrSO4(s) + 2 H2O(l)
Write the molecular, ionic, and net-ionic equation for the acid-base reaction
TitrationA 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).
TitrationThe 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.
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)
L
mL
L of !NaOH soln
mol of NaOH
L
mol
mol of !HCl
mol
mol
mL of NaOH soln
mL of HCl soln
L of !HCl soln Molarity =
mol of HCl
L of HCl soln
12.54 mL NaOH solution x x !! x = mol HCl in the sample
0.100 mol NaOH!1.000 L NaOH soln
0.001 L NaOH soln!1.000 mL NaOH 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)
1.00 mol HCl!1.00 mol NaOH 1.25 x 10-3
10.00 mL HCl solution x = L HCl soln0.001 L HCl soln!1.000 mL HCl soln
0.0100
Molarity of HCl solution = = 1.25 x 10-3 mol HCl!0.0100 L HCl soln 0.125 M
What is the concentration of NaOH solution that requires !27.5 mL to titrate 50.0 mL of 0.1015 M H2SO4 ?
L
m
L of !H2SO4 soln
mol of H2SO4
L
mol
mol of !NaOH
mol
mol
mL of H2SO4 soln
mL of NaOH soln
H2SO4 (aq) + 2 NaOH (aq) ➜ Na2SO4 (aq) + 2 H2O (l)
L of !NaOH soln
Molarity =mol of NaOH
L of NaOH soln
27.50 mL NaOH soln x = L NaOH soln
2.00 mol NaOH!1.00 mol H2SO4
50.00 mL H2SO4 soln x x !! x = mol NaOH in the sample
0.1015 mol H2SO4!1.000 L H2SO4 soln
0.001 L H2SO4 soln!1.000 mL H2SO4 soln
0.1015
0.001 L NaOH soln!1.000 mL NaOH soln
0.02750
Molarity of NaOH soln = = 0.1015 mol NaOH!0.02750 L NaOH soln 0.3691 M
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)
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)
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)
Reactant Reactant Exchange Product
Gas Formed
Example
Metal sulfide Metal hydrogensulfide Acid H H K2
H
Metal carbonate Metal hydrogencarbonate Acid H CO K2
(g)
Metal sulfite Metal hydrogensulfite Acid H SO K2
SO
Ammonium salt Base NH NH KOH (aq) + NHNH
Compounds that UndergoGas-Evolving Reactions
Practice – Predict the products and balance the equations
Na2CO3(aq) + 2 HNO3(aq) ➜
2 HCl(aq) + Na2SO3(aq) ➜
H2SO4(aq) + CaS(aq) ➜
2 NaNO3(aq) + H2O (l) + CO2(g)
2 NaCl (aq) + H2O (l) + SO2 (g)
CaSO4(aq) + H2S(aq)
“2 NaNO3(aq) + H2CO3”
“2 NaCl(aq) + H2SO3”
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)
Redox Reactions
Atoms in Elements-------> Ions in Compound
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 BookkeepingFor 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. (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] !4. (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 *
reduction
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.
CH4 + 2 O2 → CO2 + 2 H2O-4 +4 0 -2
oxidation
Assign oxidation states, determine the element oxidized and reduced, and determine the oxidizing agent and
reducing agent in the following reactions:
Sn4+ + Ca → Sn2+ + Ca2+
2 F2 + S → SF4
Sn4+ is being reduced; Sn4+ is the oxidizing agent. Ca is being oxidized; Ca is the reducing agent.
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.
0
0 0 S 4+
F -