Chapter 4
[Brown/LeMay]
[Silberberg]
A solution is a Homogeneous mixture (a mixture that has uniform properties through out the material)
A solution is composed of a solute dissolved in a solvent.
Normally, the solute is the lesser quantity and the solvent
is the greater quantity.
There is one solvent, there can be one or more solutes.
Are solutions with water as the solvent
An expression of concentration
Molar Concentration, also Molarity, symbol (M), is defined as
Molarity = moles of solute/ liters of solution
M = n/V or n = MV
Molarity is often more useful since chemicals react on a mole stoichiometry basis and not a weight basis.
The unit factor from molarity is moles/liter or it can also be written moles/1,000 mL
The concept of molarity allows you to associate moles of solute with volume of solution. Using molar mass also allows a connection of mass of solute with volume of solution.
How many grams of NaCl is required to make 1.000 L of 1 M solution?
1.000 L x 1 mole/L x 58.44 g/mole = 58.44 grams NaCl
149.1 grams of KCl was added to enough water to make 1.000 L of solution. What is the Molarity?
(149.1 g KCl/74.55 g/mol)/ 1.000L = 2.000 moles/L = 2.000 M KCl
What volume of 8.00 M KCl solution is required to make 500 mL of 2.00 M KCl solution?
(2.00 M x 0.500 L) = (8.00 M x ? L)
(2.00 M x 0.500 L)/8.00 M = 0.125 L = 125 mL
When an ionic compound dissolves, the Molarity represents the concentration of the compound in solution. It does not necessarily represent the concentration of ions in solution. The concentration of ions will depend on the number of ions produced by the compound.
Since concentration (molarity) times volume gives moles [MV = n], this can be used to calculate dilution volumes using
M1V1 = n = M2V2
Titration is commonly used to determine the concentration of an unknown compound. A solution of known concentration is reacted with a solution of unknown concentration, with the amount added is controlled until the reaction is just complete. This provides enough information to calculate the concentration of the unknown.
M1V1 (n2/n1) = M2V2
or
(M1V1/n1) = (M2V2/n2)
The process of mixing a solute into a solution.
Ionic compounds, acids, and bases dissociate or ionize in the
process of dissolving in water,
Covalent (molecular) compounds generally (except for acids and bases) do not
dissociate in water.
The process where an ionic compound, acid, or base separates into or reacts to form individual cations and anions.
Example (ionic compound): Table salt is a solid consisting
of sodium and chlorine; NaCl(s)
As table salt is dissolved in water, the solid separates into individual ions.
NaCl(s) à
Na+(aq) + Cl-(aq)
Example (acid): Vinegar is a solution consisting of acetic
acid in water. Acetic acid is a liquid
that readily mixes with water. A
minority of the acetic acid molecules will ionize in water producing the
hydrogen ion and the acetate ion.
CH3COOH(aq) à CH3COO-(aq) + H+(aq)
Example (base): Ammonia, NH3, is a weak base
that produces hydroxide ions by reaction with water.
NH3(aq) + H2O(l) à NH4+(aq)
+ OH-(aq)
Example (water): Water self-ionizes to a small degree
(nonelectrolyte)
2H2O(l) ßà H3O+(aq) + OH-(aq)
are compounds that in solution will conduct electricity. This occurs because the compound forms ions in solution.
Electrolytes are either ionic compounds, acids or bases.
carry a large current, this is when compounds form ions readily, i.e. they completely ionize in water.
Strong Electrolytes include all ionic compounds and strong acids
carry little current, this is when compounds don't form ions readily, they dissociate only partially in water.
Weak electrolytes include weak acids and weak bases.
do not carry electrical current. These are compounds that do not form ions.
Nonelectrolytes include covalent compounds that are not acids nor bases.
Solubility is the maximum amount of solute that can dissolve in solution at a given temperature; usually given in grams solute per 100 mL of solvent.
Soluble: when a compound has an appreciable solubility in water.
Insoluble: when a compound has a minor or negligible solubility in water.
Solubility is temperature dependent: solubility of solids and liquids tend to increase with increasing temperature; solubility of gases decrease with increasing temperature.
Equilibrium Condition
Characteristics - Forward/Reverse
Many reactions are reversible, such that the products can react together to form the reactants.
Equilibrium is dynamic with forward and reverse reactions continuing to occur but are balanced.
Dynamic Equilibrium: The rate of the forward reaction is equal to the rate of the reverse reaction.
Precipitation Reactions
Acid-Base Reactions
Oxidation-Reduction Reactions
Precipitation reactions are double replacement reactions in which one of the products is insoluble.
Double replacement reaction: Two compounds exchange anions, or cations. (two cations exchange anions, or two anions exchange cations) [Reactants and products consist of two ionic compounds with the ions exchanged with each other]
AX + BY à AY + BX
Example: NaI(aq) + AgNO3(aq) à AgI(s) + NaNO3(aq)
One product will be listed as (s), which means that it is solid (a precipitate).
Solubility rules are used to predict if a reaction will occur.
The products are predictable since the ionic charge of the
metals will not change.
The precipitation product is neutral.
Generally, the precipitate will contain one type of cation and one type of
anion.
General form: AX + BZ à AZ + BX
NaBr(aq) + AgC2H3O2 à NaC2H3O2(aq) + AgBr(s) [This reaction occurs since one product is insoluble]
NaCl(aq) + NH4NO3 à NR {NaNO3(aq) + NH4Cl(aq)} [This reaction does not occur since both products are soluble]
2NaCl(aq) + Pb(C2H3O2)2(aq) à NaC2H3O2(aq) + PbCl2(s) [This reaction occurs because one product is insoluble]
NaCl(aq) + KNO3(aq) à NR (NaNO3(aq) + KCl(aq)) [This reaction does not occur because both products are soluble]
BaBr2(aq) + Ag2SO4(aq) à 2AgBr(s) + BaSO4(s) [This reaction occurs because both products are insoluble]
Compounds containing the following ions are generally soluble.
·
Nitrate ion, NO3-
·
Most halide ions, Cl-,
Br-, I-, [Halides
of Ag+, Hg22+, and Pb2+ are
insoluble]
·
Sulfate ion, SO42-,
[Sulfates of Sr, Ca, Ba, and Pb are marginally soluble]
Compounds containing
the following ions are generally insoluble in water.
·
Hydroxide ion, OH-
[Hydroxides of alkali metal ions and ammonium ion are soluble, Ca, Sr, Ba are
slightly soluble]
·
Carbonate ion, CO32-
[alkali metal ions and ammonium
ion are soluble]
·
Phosphate ion, PO43-
[alkali metal ions and ammonium
ion are soluble]
·
Sulfide ion, S2-
[alkali
metal ions and ammonium ion are soluble]
The molecular equation shows the molecules or formula's of compounds that react and are formed.
Na2CrO4(aq) + CaS(aq) à CaCrO4(s) + Na2S(aq)
The Complete Ionic Equation shows the form of all species in solution. All aqueous ionic compounds get separated into ions.
2Na+(aq) + CrO42-(aq) + Ca2+(aq) + S2-(aq) à CaCrO4(s) + 2Na+(aq) + S2-(aq)
Ions that do not participate in the reaction are spectator ions. These ions show up as both reactants and products in the complete ionic equation.
The Net Ionic Equation shows only the species that
directly participate in the reaction.
Write the complete ionic equation without the spectator ions to get the net
ionic equation.
CrO42-(aq) + Ca2+(aq) à CaCrO4(s)
Ch. 16 Sections 1-5 [Brown/LeMay]
Arrhenius Acid is a substance that ionizes in water to produce hydrogen ions. These compounds consist of molecules with a hydrogen attached by a polar bond.
Examples: HCl, HF, H2SO4, H3PO4, CH3COOH (HC2H3O2)
HCl(g) + H2O(l) à H+(aq) + Cl-(aq) + H2O(l) à H3O+(aq) + Cl-(aq)
In water, the hydrogen ion, H+, combines with water to form the hydronium ion, H3O+. The use of either form is acceptable.
Arrhenius Base is a substance that ionizes in water to produce hydroxide ions.
Examples: NaOH, Ca(OH)2, KOH, Mg(OH)2, NH3, amines
Acids can classified by the number of protons they can release
in solution.
Acids with 2 or more acidic protons are called polyprotic acids.
Usually only the first dissociation step contributes to the concentration of H+.
Monoprotic acids have one ionizable proton (HCl)
Diprotic acids have two ionizable protons (H2SO4)
Triprotic acids have three ionizable protons (H3PO4)
Hydrogen with a nonmetal element (most but not all).
Examples: HCl, HBr, HI, HF, H2S
Not Acids: H2O, NH3
Many common acids are oxyacids, in which the acidic proton is attached to an oxygen atom [contains hydrogen, oxygen, plus another element{hydrogen with a polyatomic anion}]. Examples: H2SO4, H3PO4
Inorganic acids are the binary acids and oxyacids.
Organic acids have a carbon backbone with hydrogen and a carboxyl group (-COOH)
A strong acid is an acid that is almost all in the
dissociated/ionized form.
The dissociation equilibrium lies far to the right.
A strong acid yields a weak conjugate base. (low affinity for a
proton)
[much weaker base than water]
A weak acid has a dissociation equilibrium that lies far
to the left.
Most of the acid is in the undissociated/unionized form.
A weak acid yields a relatively strong conjugate base.
The conjugate base is stronger than water.
Strong bases dissociate essentially completely (LiOH, NaOH, KOH, RbOH, CsOH)
dissociate or forms ions only partially in water
An Arrhenius acid neutralizes an Arrhenius base to form a salt (ionic compound) and water.
HCl(aq) + NaOH(aq) à NaCl(aq) + H2O(l)
Acid + base à salt
+ water
The net ionic equation of these neutralization reactions is
generally:
H+(aq) + OH-(aq) à H2O(l)
Bronsted Acid is a proton donor (proton is a hydrogen ion)
Bronsted Base is a proton
acceptor [neutralizes the acid by
removing its proton]
[expands
the definition of base]
BAAD: Base Accepts, Acid Donates
Conjugate acid
Conjugate acid is the species formed when the base accepts the proton.
Example: NH4+ is the conjugate acid when NH3 accepts a proton.
Conjugate base
The species that remains from the Bronsted acid species after the proton is donated.
Example: F- is the conjugate base when HF donates a proton.
Conjugate acid-base pair
A Bronsted acid and its conjugate base. The acid-base pair differs by only one proton.
Example: H2S and HS- are a conjugate acid-base pair.
A Bronsted Acid-Base neutralization reaction is the reaction of an acid and base, this produces a weaker acid and base.
The hydroxide ion can be assumed to react completely with any acid (strong or weak).
These reactions produce gases; H2S and CO2
respectively.
Examples:
CaCO3(s) + 2HCl(aq) à
CaCl2(aq) + CO2(g) + H2O(l)
CaCO3(s) + H2SO4(aq) à CaSO4(s) + CO2(g)
+ H2O(l)
Fe2S3(s) + 6HCl(aq) à 2FeCl3(aq)
+ 3H2S(g)
A substance is amphoteric if it can behave as either an acid or base.
Water is the most common amphoteric substance, as shown in the autoionization
of water where water is both acid and base in one reaction.
2H2O(l) ßà H3O+(aq)
+ OH-(aq)
The equilibrium expression is
Kw = [H3O+][OH-] = [1.0 x 10-7
M][1.0 x 10-7 M] = 1.0 x 10-14
Where the equilibrium constant, Kw, is called the ion-product
constant or dissociation constant of water
MCF08
A way of expressing the concentration of the hydrogen ion is by
using the pH scale
Where
pH = -log[H+]
This also means that [H+] = 10-pH
Significant Figures: The number of decimal places in the log is equal to the number of significant figures in the original number.
Similar log scales are used for other quantities including
hydroxide.
pOH = -log[OH-]
Where pH + pOH = 14.
Strong acids are essentially completely dissociated. Use this information in identifying the major species in solution. Identifying all the major species will aid in determining the concentration in solution.
[H+] is calculated using Kw (1.0 x 10-14) = [H+][OH-]
Ch. 20 Sections 1-2 [Brown/LeMay]
Electrochemistry is the study of the interchange of chemical and electrical energy
Oxidation-Reduction Reactions (or Redox Reactions) involve the transfer of electrons between species.
Oxidation States (or oxidation numbers) provide a way to keep track of electrons in Oxidation-Reduction Reactions.
For covalent bonds with identical atoms, the electrons are shared between the atoms.
For two different atoms, the electrons are assigned to the atom that is more electronegative (has a stronger attraction for electrons).
For ionic compounds, the oxidation state equals the ionic charge for monoatomic ions.
· The oxidation state of an atom in an element is 0.
· The oxidation state of a monoatomic ion is the same as its charge.
· The sum of oxidation states for a neutral compound must be zero.
· The sum of oxidation states for a polyatomic ion equals the ionic charge.
· Group I metals, in compounds, have an oxidation state of +1.
· Group II metals, in compounds, have an oxidation state of +2.
· The following rules are hierarchical, the first listings has priority.
· The oxidation state of fluorine in compounds is always assigned a value of –1.
· The oxidation state of hydrogen in compounds with nonmetals is assigned a value of +1.
· The oxidation state of hydrogen in combination with metals is –1.
· The oxidation state of oxygen in compounds is usually assigned a value of –2, except for peroxides (O22-) where oxygen is assigned a value of –1; superoxide (O21-) where the oxygen is assigned a value of –1/2; or when combined with fluorine the oxygen value is variable.
· Group 7A nonmetals, Halogens (except fluorine) in compounds, generally are –1 except when combined with oxygen or fluorine.
· Group 6A nonmetals, in compounds, has an oxidation state of –2.
· Group 5A nonmetals, in compounds, has an oxidation state of –3.
Oxidation-reduction (redox) reactions involves transfer of electrons from the reducing agent to the oxidizing agent.
Oxidation and Reduction have to occur simultaneously.
Sometimes redox are obvious where the ionic charge of an element changes, ions are formed or removed.
Sometimes, such as combustion reactions, it is less obvious because all reactants and products are neutral covalent compounds.
Redox is a combined word from the two words:
reduction and oxidation
Involves an increase in oxidation number/state (a lose of electrons)
Involves a decrease in oxidation number/state (a gain of electrons)
OIL RIG: Oxidation Involves Loss; Reduction Involves Gain
LEO says GER: Lose Electrons, Oxidized; Gain Electrons, Reduced
Oxidizing Agent (electron acceptor) promotes oxidation in another species. The Oxidizing Agent is reduced itself.
Reducing Agent (electron donor) promotes reduction in another species. The Reducing Agent is oxidized.
Redox reactions can be broken into half reactions; one involving oxidation and one involving reduction.
A Half Reaction shows either an oxidation reaction alone or a reduction reaction alone.
The half reactions shows the electrons transferred; as a product for oxidation reactions and as a reactant for reduction reactions.
Half-reactions are balanced for elements and charge.
For any redox reaction to be balanced, the number of electrons gained and the number of electrons lost must be equal.
Redox reactions can occur in an acidic environment, an alkaline (basic) environment, or a neutral environment.
Balancing these types of reactions depends on the type of environment in which the reaction occurs.
· Assign oxidation states to each element on each side of the reaction
· Separate the overall reaction into separate half reactions
· Balance each half-reaction
· Balance all elements except hydrogen and oxygen
· Balance oxygen using H2O
· Balance hydrogen using H+
· Balance the charge using electrons
· Multiply one or both balanced half-reactions by an integer to equalize the number of electrons transferred in both half-reactions
· Add the half-reactions and cancel any species which appears as both reactant and product.
· For basic solutions add OH- to both sides of the equation to neutralize any H+ into H2O, then remove any excess water so water only appears at most on one side of the equation.
· Check that all elements and charges are balanced
Problem: Balance the following reaction in acidic
solution:
Cr2O72-(aq) + NO2-(aq) à
Cr3+(aq) + NO3-(aq)
Solution:
· Assign oxidation states to each element on each side of the reaction
Cr2O72-(aq) + NO2-(aq)
à
Cr3+(aq) + NO3-(aq)
+6 -2 +3 -2 +3 +5 -2
· Separate the overall reaction into separate half reactions
Oxidation: NO2-(aq) à NO3-(aq)
Reduction: Cr2O72-(aq) à Cr3+(aq)
· Balance each half-reaction
· Balance all elements except hydrogen and oxygen
Oxidation: NO2-(aq) à NO3-(aq)
Reduction: Cr2O72-(aq) à 2Cr3+(aq)
· Balance oxygen using H2O
Oxidation: NO2-(aq) + H2O à NO3-(aq)
Reduction: Cr2O72-(aq) à 2Cr3+(aq) + 7H2O
· Balance hydrogen using H+
Oxidation: NO2-(aq) + H2O(l) à NO3-(aq) + 2H+(aq)
Reduction: Cr2O72-(aq) + 14H+(aq) à 2Cr3+(aq) + 7H2O(l)
· Balance the charge using electrons
Oxidation: NO2-(aq) + H2O(l) à NO3-(aq) + 2H+(aq) + 2e-
Reduction: Cr2O72-(aq) + 14H+(aq) + 6e- à 2Cr3+(aq) + 7H2O(l)
· Multiply one or both balanced half-reactions by an integer to equalize the number of electrons transferred in both half-reactions
Oxidation (x3): 3NO2-(aq) + 3H2O(l) à 3NO3-(aq) + 6H+(aq) + 6e-
Reduction: Cr2O72-(aq) + 14H+(aq) + 6e- à 2Cr3+(aq) + 7H2O(l)
· Add the half-reactions and cancel any species, which appears as both reactant and product.
Add Half-reactions
Cr2O72-(aq) + 3NO2-(aq)
+ 3H2O(l) + 14H+(aq) + 6e- à
2Cr3+(aq) + 3NO3-(aq) + 7H2O(l) +
6H+(aq) + 6e-
Cancel duplicate species
Cr2O72-(aq) + 3NO2-(aq)
+ 8H+(aq) à 2Cr3+(aq) + 3NO3-(aq)
+ 4H2O(l)
· For basic solutions add OH- to both sides of the equation to neutralize any H+ into H2O, then remove any excess water so water only appears at most on one side of the equation.
· Check that all elements and charges are balanced
Cr2O72-(aq) + 3NO2-(aq) + 8H+(aq) à 2Cr3+(aq) + 3NO3-(aq) + 4H2O(l)
In single replacement reactions, one metal replaces another metal from a compound. The metal that has a stronger tendency to become the ionic compound is more active than the metal that becomes metallic. Activity is the tendency of a metal to loose an electron to become ionic.
A combination of an oxidation reaction higher on the list combined with a reduction reaction from lower on the list is spontaneous.
Each half reaction is associated with an electrical potential (i.e. voltage). A combination of reactions that creates a positive voltage is spontaneous.
When a reaction is written in the reverse direction, it changes the sign of the voltage.
Most Easily Oxidized [Most Reactive]
Li(s) à Li+(aq) + e- ξo = 3.05 V
K(s) à K+(aq) + e- ξo = 2.93 V
Ca(s) à Ca2+(aq) + 2e- ξo = 2.87 V
Na(s) à Na+(aq) + e- ξo = 2.71 V
Mg(s) à Mg2+(aq) + 2e- ξo = 2.37 V
Al(s) à Al3+(aq) + 3e- ξo = 1.66 V
Mn(s) à Mn2+(aq) + 2e- ξo = 1.18 V
Zn(s) à Zn2+(aq) + 2e- ξo = 0.76 V
Cr(s) à Cr3+(aq) + 3e- ξo = 0.74 V
Fe(s) à Fe2+(aq) + 2e- ξo = 0.44 V
Ni(s) à Ni2+(aq) + 2e- ξo = 0.23 V
Sn(s) à Sn2+(aq) + 2e- ξo = 0.14 V
Pb(s) à Pb2+(aq) +
2e- ξo
= 0.13 V
H2(g) à 2H+(aq) + 2e- ξo = 0.0 V
Cu(s) à Cu2+(aq) + 2e- ξo = -0.34 V
Ag(s) à Ag+(aq) + e- ξo = -0.80 V
Hg(l) à Hg2+(aq) + 2e- ξo = -0.854 V
Pt(s) à Pt2+(aq) + 2e- ξo = -?
Au(s) à Au3+(aq) + 3e- ξo = -1.50 V
Least Easily Oxidized [Least Reactive]
If a metal is higher on the activity series list than Hydrogen, then it will dissolve in acid.
If the electric potential for the oxidation of the metal is positive, then it will dissolve in acid.
A titration (volumetric analysis) involves precisely adding a measured volume of a known concentration of a reactant (titrant) to a measured volume of an unknown concentration of another reactant (analyte).
This is done until a chosen endpoint is reached. The endpoint is often marked by the change of color of an indicator chemical.
The indicator is chosen to be close to the stoichiometric point or equivalence point.
For a good titration:
The reaction occurring must be known and relatively fast
The stoichiometric point must be marked
The volumes must be measured accurately.
Phenolphthalein is usually the chosen indicator for an acid-base titration.
The titrant is usually a strong acid or base.
Titration calculations are conducted identically to stoichiometry calculations.
Reactions conducted in solution use solution concentration in calculations of how much of one reactant will react with another reactant.
MV yields moles of solute.
For a reaction of the form
n1Reactant1 + n2Reactant2 à
Products
These relationships between reactant concentrations can be calculated using
M1V1/n1 = M2V2/n2
Rearranging slightly gives
M1V1(n2/n1) = M2V2
or
moles[1](reaction stoichiometry) = moles[2]