Chem 121 Chapter 4 Notes: Stoichiometry, Electrolytes, Acids, Bases, and Redox Reactions
Chem 121 Chapter 4: Stoichiometry of Chemical Reactions
Chemical Symbols on Different Levels and Balancing Chemical Equations
Microscopic Level Chemistry: Deals with individual atoms and molecules.
Macroscopic Level Chemistry: Deals with observable quantities of substances.
Chemical Equation: A symbolic representation of a chemical reaction.
Example: C3H8(g) + 5O2(g) \rightarrow 3CO2(g) + 4H_2O(g)
Components of a Chemical Reaction:
Correct chemical formula for each reactant and product.
Symbol to designate state of matter for each: (s) (solid), (l) (liquid), (g) (gas), (aq) (aqueous solution).
Coefficient to balance the equation.
Law of Conservation of Mass: Chemical equations must be balanced to ensure that no atoms are created or destroyed during a reaction.
Reactants (left side) yield products (right side).
A balanced chemical equation has the same number and types of atoms on both the reactants and products side.
Example balancing: C3H8(g) + 5O2(g) \rightarrow 3CO2(g) + 4H_2O(g)
Reactants:
C = 3
H = 8
O = (5×2)=10
Products:
C = 3
H = (4×2)=8
O = (3×2)+(4×1)=6+4=10
Balancing Chemical Equations (Example 1):
Equation to balance: Na3PO4(aq) + CaCl2(aq) \rightarrow NaCl(aq) + Ca3(PO4)2(s)
Initial Atom Count (Reactants vs. Products):
Na: 3 vs. 1
P: 1 vs. 2
O: 4 vs. 8
Ca: 1 vs. 3
Cl: 2 vs. 1
Balanced Equation (step-by-step mental process/example shown in transcript):
Need 3 Ca on reactants: Na3PO4(aq) + 3CaCl_2(aq)
Need 2 P on reactants: 2Na3PO4(aq) + 3CaCl_2(aq)
Now balance Na and Cl on products side: 2Na3PO4(aq) + 3CaCl2(aq) \rightarrow 6NaCl(aq) + Ca3(PO4)2(s)
Final Atom Count (Balanced):
Na: (2×3)=6 vs. 6
P: (2×1)=2 vs. 2
O: (2×4)=8 vs. (2×4)=8
Ca: (3×1)=3 vs. 3
Cl: (3×2)=6 vs. 6
Smallest Whole-Number Coefficients: Coefficients in a balanced equation should be the smallest possible whole numbers.
Incorrectly balanced example: 6Al(s) + 3Fe2O3(s) \rightarrow 3Al2O3(s) + 6Fe(l)
All coefficients are divisible by 3.
Al = 6 vs. (3×2)=6
Fe = (3×2)=6 vs. 6
O = (3×3)=9 vs. (3×3)=9
Correctly balanced (simplified): 2Al(s) + Fe2O3(s) \rightarrow Al2O3(s) + 2Fe(l)
Al = 2 vs. 2
Fe = 2 vs. 2
O = 3 vs. 3
Electrolytes and Nonelectrolytes
What happens when a compound dissolves in water?
In aqueous solutions, water is always the solvent.
Soluble ionic compounds behave as strong electrolytes.
They dissolve to give solutions of hydrated ions, meaning the ions are surrounded by water molecules.
This process of ionic compounds dissolving to form ions in solution is called dissociation.
Example: NaCl(s)→H2ONa+(aq)+Cl−(aq).
Many molecular compounds behave as nonelectrolytes.
They dissolve to give hydrated molecules, meaning the molecules remain intact but are surrounded by water.
Example: C{12}H{22}O{11}(s) \xrightarrow{H2O} C{12}H{22}O_{11}(aq).
Acids are an exception among molecular compounds, as they can behave as electrolytes.
Example: HCl(g)→H2OH+(aq)+Cl−(aq).
Definitions:
Electrolyte: A substance that dissolves in water, giving a solution that can conduct electricity.
Nonelectrolyte: A substance that also dissolves in water, but gives a solution that does not conduct electricity.
Strong vs. Weak Electrolytes:
Strong electrolytes: Dissociate completely (approaching 100%) in aqueous solution.
Represented by a full forward arrow in the dissociation equation.
Example: KCl(s)→H2OK+(aq)+Cl−(aq).
Weak electrolytes: Dissociate incompletely (less than 100%) in aqueous solution.
Represented by forward and reverse arrows (equilibrium arrows) in the dissociation equation.
Example (acetic acid): CH3COOH(l) \xrightleftharpoons{H2O} CH_3COO^-(aq) + H^+(aq).
Classification of Substances as Electrolytes or Nonelectrolytes:
Strong Electrolytes:
Strong Acids: HCl, HBr, HI, HClO4,HNO3,H2SO4
Soluble Ionic Compounds: KBr, NaCl, NaOH, KOH, and other soluble ionic compounds
Weak Electrolytes:
Weak Acids: CH3CO2H (acetic acid), HF, HCN
Weak Bases: NH3 (ammonia)
Nonelectrolytes:
H2O
CH3OH (methyl alcohol)
C2H5OH (ethyl alcohol)
C{12}H{22}O_{11} (sucrose)
Most compounds of carbon (organic compounds) that are not acids.
Molecular, Ionic, and Net Ionic Equations
Molecular Equation: A chemical equation where all compounds are written in their undissociated molecular form.
Example: Pb(NO3)2(aq) + 2KI(aq) \rightarrow 2KNO3(aq) + PbI2(s).
Ionic Equation: Obtained by splitting aqueous ionic compounds into their constituent ions.
Solids, liquids, and gases CANNOT be split up into ions; they remain in their molecular or formula unit form.
Example for above reaction: Pb^{2+}(aq) + 2NO3^-(aq) + 2K^+(aq) + 2I^-(aq) \rightarrow 2K^+(aq) + 2NO3^-(aq) + PbI_2(s).
Net Ionic Equation: An ionic equation written without spectator ions.
Spectator Ions: Ions that appear the same on both the reactants and products side of the ionic equation (they do not participate in the reaction).
Example (identifying spectator ions from above): The NO3−
and K+
are spectator ions.Net Ionic Equation: Pb2+(aq)+2I−(aq)→PbI2(s).
Precipitation Reactions and Solubility Rules
Precipitation Reactions: Reactions where two aqueous solutions combine to form a solid precipitate.
The precipitate is the insoluble product that forms.
Example: Pb(NO3)2(aq) + 2KI(aq) \rightarrow 2KNO3(aq) + PbI2(s).Here,\text{PbI}_2 is the precipitate.
Solubility Guidelines (Rules) to Predict Precipitate Formation:
Compounds containing these ions will ALWAYS be soluble (aq):
Ammonium (NH4+)
Nitrate (NO3−)
Acetate (CH3COO−)
Perchlorate (ClO4−)
Group 1A Cations: Li+,K+,Na+,Rb+,Cs+
Halides (Cl−,Br−,I−): Always soluble EXCEPT when with Ag+
, Hg22+
, or Pb2+
.Sulfate (SO4^{2-} ): Always soluble EXCEPT when with Sr2+
, Ba2+
, \text{Hg}2^{2+}
, or Pb2+
.Insoluble Compounds (except when with Group 1A cations or NH4+ ):
Carbonate (CO3^{2-} ): Always insoluble EXCEPT with Group 1A cations, \text{NH}4^+
.Sulfide (S2− ): Always insoluble EXCEPT with Group 1A cations, NH4+
, Sr2+
, Ba2+
.Phosphate (PO4^{3-} ): Always insoluble EXCEPT with Group 1A cations, \text{NH}4^+
.Hydroxide (OH− ): Always insoluble EXCEPT with Group 1A cations, NH4+
, Sr2+
, Ba2+
.
Acids and Bases: Identification and Naming
The Arrhenius Definition of Acids and Bases:
Acid: A molecule or polyatomic ion that produces H+ (hydrogen ions) in aqueous solution.
H+
is very reactive and typically associates with water to form the hydronium ion (H3extO+
).Thus, when discussing aqueous acids, H+=H3extO+
.Example: HCl(g)→H2OH+(aq)+Cl−(aq)
Characteristics of Acids (which hydrogens are acidic?): To be acidic, a hydrogen must be bonded to a very electronegative atom or group that is stable as an ion.
Examples of Acids: HCl, \text{HNO}3,CH3COOH(CH3CO2H),\text{H}2 ext{PO}4^-
, HCO3−
.Examples of Non-Acids: \text{CH}4,\text{C}2 ext{H}6,otherhydrocarbons,andalcohols(CH3CH_2OH
).
Base: A molecule or ion that produces OH− (hydroxide ions) in water.
Strong Base Examples: NaOH, KOH, Ba(OH)2
.Weak Base Example: NH3
(ammonia) is a common weak base.
Polyprotic Acids
Polyprotic Acids: Acids that can donate more than one proton (H+).
Monoprotic Acids: Have one dissociable H+ .
HCl →H++Cl−
(strong electrolyte / acid)\text{HNO}3 \rightarrow H^+ + NO3^-
(strong electrolyte / acid)CH3COOH \xrightleftharpoons H^+ + CH3COO^-
(weak electrolyte / acid)
Diprotic Acids: Have two dissociable H+ .
\text{H}2 ext{SO}4 \rightarrow 2H^+ + SO_4^{2-}
(strong electrolyte / acid for the first proton).
Triprotic Acids: Have three dissociable H+ .
\text{H}3 ext{PO}4 \xrightleftharpoons 3H^+ + PO_4^{3-}
(weak electrolyte / acid).
Strong Acids and Strong Bases
Strong Acid or Base: A molecule or ion that dissociates completely in water.
Weak Acids and Bases: Dissociate very little in water.
Memorize the 6 Strong Acids: HCl, HBr, HI, \text{HClO}4,\text{HNO}3,\text{H}2 ext{SO}4 (sulfuric acid).
Any other acid encountered will generally be weak.
Strong Bases: All soluble metal hydroxides.
Group 1 metal hydroxides (e.g., NaOH, KOH).
Hydroxides of Ca2+
, Ba2+
, and Sr2+
.
Weak Bases: Ammonia (NH3 ) is the most common example.
Compounds with \text{NH}2groups(e.g.,CH3NH_2
, methylamine) are also weak bases.
Neutralization Reactions
Definition: Acid-base reactions are called neutralization reactions.
Reaction between Strong Acid and Strong Base: Produces a salt and water.
Molecular Equation: KOH(aq)+HCl(aq)→KCl(aq)+H2O(l)
Base + Acid →
A Salt + Water
Ionic Equation: K+(aq)+OH−(aq)+H+(aq)+Cl−(aq)→K+(aq)+Cl−(aq)+H2O(l)
Net Ionic Equation (for strong acid + strong base): H+(aq)+OH−(aq)→H2O(l)
Neutralization with a Carbonate Base:
Example: Na2CO3(aq) + HNO3(aq) \rightarrow NaNO3(aq) + H2O(l) + CO2(g)
Neutralization with a Bicarbonate Base:
Example: NaHCO3(aq) + CH3COOH(aq) \rightarrow CH3COONa(aq) + H2O(l) + CO_2(g)
Oxidation-Reduction (Redox) Reactions
Definition: Redox reactions involve the transfer of electrons between species.
Identification: Redox reactions are identified by changes in oxidation numbers or charges of atoms and ions.
Example (Ionic Reaction): Sn2+(aq)+Fe3+(aq)→Sn4+(aq)+Fe2+(aq)
Tin (Sn): Goes from a +2 charge to a +4 charge.
This is a loss of electrons (2 electrons lost).
Iron (Fe): Goes from a +3 charge to a +2 charge.
This is a gain of electrons (1 electron gained).
Balanced Ionic Equation (balancing electron transfer): Sn2+(aq)+2Fe3+(aq)→Sn4+(aq)+2Fe2+(aq)
OIL RIG Mnemonic:
Oxidation Is Loss (of electrons).
Reduction Is Gain (of electrons).
Species Oxidized and Reduced:
An element or ion that loses electrons is OXIDIZED (Sn2+
is oxidized).An element or ion that gains electrons is REDUCED (Fe3+
is reduced).
Recognizing Redox Reactions in Complex Compounds: If a reaction is not a precipitation or acid-base reaction, it is likely a redox reaction.
Example: 2Fe2O3(s) + 3C(s) \rightarrow 4Fe(s) + 3CO_2(g)
Assigning Oxidation Numbers (Rules):
An atom in its elemental state has an oxidation number of 0.
Examples: C, \text{H}2,\text{Cl}2,\text{S}8,\text{Br}2
, Mg.
An atom in a monatomic ion has an oxidation number equal to its charge.
Examples: Na+=+1
, Fe3+=+3
, O2−=−2
.
If a compound is ionic, the oxidation numbers of the elements are equal to their charge.
Example in \text{Fe}2 ext{O}3
: Fe is +3
, O is −2
.
For elements in a molecule or polyatomic ion (common assignments):
Oxygen (O): Usually −2
.Hydrogen (H): Usually +1
.Fluorine (F): Always −1
.
For other elements in a molecule or polyatomic ion: Use algebra.
The sum of the oxidation numbers must equal the charge on the molecule or polyatomic ion.
Example calculation for CO2
: x+2(−2)=0⇒x−4=0⇒x=+4
for Carbon.
Applying Oxidation Number Rules (Example: 2Fe2O3(s) + 3C(s) \rightarrow 4Fe(s) + 3CO_2(g)):
Reactants:
Fe in \text{Fe}2 ext{O}3
: +3O in \text{Fe}2 ext{O}3
: −2C: 0
Products:
Fe: 0
C in CO2
: +4O in CO2
: −2
Analysis of Changes:
Oxygen: No change ($-2 \rightarrow -2).
Iron: +3→0 (oxidation number decreases, so Iron in \mathbf{Fe2O3} is reduced).
Carbon: 0→+4 (oxidation number increases, so Carbon in C is oxidized).
Reducing Agents and Oxidizing Agents:
Reducing Agent: The species that loses electrons and causes reduction in another species. It itself becomes oxidized.
In the example: Carbon (C) is the reducing agent.
Oxidizing Agent: The species that gains electrons and causes oxidation in another species. It itself becomes reduced.
In the example: Iron in \text{Fe}2 ext{O}3$$
is the oxidizing agent.
Balancing Simple Redox Reactions
Steps for Balancing Redox Reactions in Acidic and Basic Solution:
Write the two half-reactions representing the oxidation and reduction processes.
Balance all elements in each half-reaction except oxygen and hydrogen.
Balance oxygen atoms by adding H2extO
molecules to the side that needs more oxygen.Balance hydrogen atoms by adding H+
ions to the side that needs more hydrogen.Balance charge by adding electrons (e−
) to the more positive side of each half-reaction.Equalize electrons: If necessary, multiply each half-reaction's coefficients by the smallest possible integers to yield equal numbers of electrons transferred in both half-reactions.
Add the balanced half-reactions together and simplify by removing species (like H2extO or H+) that appear on both sides of the equation.
For reactions occurring in basic media (excess hydroxide ions), carry out these additional steps:
a. Add OH−
ions to both sides of the equation in numbers equal to the number of H+
ions.
b. On the side of the equation containing both H+
and OH−
ions, combine these ions to yield water molecules (H++OH−→H2extO).
c. Simplify the equation by removing any redundant water molecules from both sides.Final Check: Verify that both the number of atoms and the total charges are balanced on both sides of the final equation.