Download AP - 04 - Reactions in Aqueous Solutions

AP / College Chemistry – Chapter 4 Notes
Reactions in Aqueous Solution
- General Properties of Aqueous Solutions
o Aqueous solution—solution in which water is the dissolving medium
o Solution is a HOMOGENEOUS mixture
o Solvent – the dissolving medium (what the solute is dissolved in)
o Solute – substance that is dissolved in the solvent
o Solution = Solute + Solvent
o Electrolytic Properties
 Water is a very poor conductor of electrical current
 Conductivity of water results from what is dissolved in it
 Conductivity indicator uses a bulb that will not light unless a
solution conducts electricity
 Electrolyte—a substance that conducts electrical current in
aqueous solution because it forms a solution containing IONS
 Nonelectrolyte—a substance that does not conduct electrical
current in aqueous solution because it does NOT form ions
o Ionic Compounds in Water
 Ionic compounds DISSOCIATE when they dissolve in water (ions
separate and become solvated or surrounded with water
molecules)
 Polar water molecules attract to the ions and surround them in a
process called SOLVATION which helps to stabilize the ions in
solution and prevents recombination
 Ions become equally dispersed throughout the solution
o Molecular Compounds in Water
 Molecules typically dissolve as intact molecules like C12H22O11
(sucrose) or CH3OH (methanol)
 Since molecules like these do not form ions they will not conduct
electric current (NONELECTROLYTES)
 Some molecules (like ACIDS) will IONIZE in aqueous solution and
will conduct electric current (ELECTROLYTES)
 STRONG ELECTROLYTES – completely or nearly completely form
ions in solution and conduct well
o Ionic compounds (NaCl) and STRONG acids or bases (HCl)
o Forward reaction only (does not reform reactants)
 WEAK ELECTROLYTES – solutes that exist MOSTLY as neutral
molecules with some ions formed
o Polar molecules and WEAK acids or molecular bases
o Reversible reaction (EQUILIBRIUM) which usually FAVORS
the reactants and only forms some products
 Water soluble ionic compounds are strong electrolytes (except
NH4+ compounds!)
- Precipitation Reactions:
o Some reactions of ionic compounds in aqueous solutions form insoluble
ionic products that fall out of solution (PRECIPITATE)
o These reactions are called PRECIPITATION REACTIONS
o Precipitation reactions occur because pairs of oppositely charged ions
attract each other so strongly, water cannot compensate for that so the
product does NOT dissolve and become solvated
o Solubility Guidelines for Ionic Compounds
 Solubility—the amount of a substance that can be dissolved in a
given amount of solvent at a particular temperature
 Substances with solubilities of 0.01 mol/L or LESS will be
considered INSOLUBLE (although a small amount DOES dissolve!)
 Nitrates (NO3-) and Acetates (CH3COO-) are always soluble as
compounds
 Compounds of Group IA (Li+, Na+, K+, Rb+, Cs+, Fr+) and Ammonium
(NH4+) are always soluble as well
 Use the solubility rules to determine which species (reactants and
products) will be soluble or insoluble in an aqueous reaction
 Reacting magnesium nitrate with sodium hydroxide:
 This type of reaction is called a Metathesis Reaction (exchange
reaction) in which cations and anions exchange partners (double
replacement)
 To balance a metathesis reaction, follow these steps:
 Use the chemical formulas of the reactants to determine
which ions are present
 Write the chemical formulas for the products by combining
the cation from one reactant with the anion from the other
(BE SURE to use the CORRECT formula based on the charges
of the ions!!)
 Use the solubility rules to decide what is soluble and
insoluble
 Balance the equation
 Write the balanced equation for the reaction of barium chloride
and potassium sulfate
BaCl2(aq) + K2SO4(aq) → BaSO4(s) + 2 KCl(aq)
o Ionic Equations
 Molecular equation—shows the complete chemical formulas of
reactants and products without indicating ionic character
Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq)
 Complete Ionic equation—shows all ions as dissociated in
aqueous solution
 Spectator ions—ions that appear on BOTH the reactants and
products side and do NOT really change chemically in the reaction
 2 K+(aq) and 2 NO3-(aq) are exactly the same on each side so
they cancel
 HINT: THE SPECTATOR IONS ARE ALWAYS THE IONS IN
THE SOLUBLE PRODUCT!!! (KNO3)
 Net Ionic equation—shows the overall chemical reaction with the
spectator ions removed
 If every ion in a complete ionic equation is a SPECTATOR, then they
will all cancel out and NO REACTION occurs (NOTHING CHANGES!)
 Write a balanced molecular equation for the reaction and
indicate which species are soluble and insoluble
 Rewrite the equation as a complete ionic equation and split
apart any aqueous substances into the ions they form
 Identify and cancel spectator ions
 Write the balanced net ionic equation
- Acids, Bases & Neutralization Reactions:
o Acid—substances that ionize in aqueous solution to form H+ (PROTON
DONORS)
Only ONE ionizable proton in
CH3COOH because only ONE
carboxylic acid group (COOH)
o Acids have different numbers of ionizable protons
o Monoprotic acids – yield one H+ per molecule (HCl, HNO3, CH3COOH)
o Diprotic acids – yields two H+ per molecule (H2SO4, H2CO3) because the
ionization happens in 2 steps
o Triprotic acids – yields three H+ per molecule (H3PO4)
Citric acid has 3 ionizable
protons because it has 3
carboxylic acid groups
(COOH)
o Bases—substances that accept or react with H+ and produce OH- when
they dissolve in water
o Ionic bases contain the OH- ion (NaOH, KOH, Ca(OH)2)
o Molecular bases like NH3 forms OH- by reacting with water!
o Ammonia is a weak electrolyte because only about 1% of NH3 forms
NH4+ and OH- ions
o Strong and Weak Acids and Bases
 Strong acids and strong bases are strong electrolytes (completely
ionized in solution)
 Weak acids and weak bases are weak electrolytes (partially
ionized in solution)
 The more H+ ionized in solution, the stronger the acid
 The more OH- produced in solution, the stronger the base
o Identifying Strong and Weak Electrolytes
 Determine first if it is a soluble ionic or molecular substance
 If it is soluble and ionic, it is a strong electrolyte
 If it is a molecule, determine if it is an acid or base
 If it is an acid, determine if it is strong or weak
 If it is a strong acid, it is a strong electrolyte
 If it is a weak acid or weak base, it is a weak electrolyte
 All other compounds are probably nonelectrolytes
o Neutralization Reactions and Salts
 Acidic solutions do not have the same properties as basic
solutions
 Acids have a sour taste and turn litmus paper RED and bases have
a bitter taste and turn litmus paper BLUE
 Neutralization reaction—an acid reacts with a base and forms
WATER and a SALT
 Salt—ionic compound whose cation comes from a base and anion
from an acid
For the reaction between magnesium hydroxide and hydrochloric acid
o Neutralization Reactions with Gas Formation
 Many bases besides OH- can react with H+ to form molecular
compounds
 Sulfides (S2-) react to form hydrogen sulfide gas (H2S(g))
 Carbonates (CO32-) and bicarbonates (HCO3-) react to form
carbonic acid (H2CO3) which is unstable and when present in
solution breaks down into H2O and CO2(g).
 Carbonates and bicarbonates are used to neutralize acid spills and
bicarbonates are used as antacids
o Driving Forces for Reactions in Aqueous Solution:
 Formation of a precipitate
 Formation of a molecule like WATER
 Formation of an insoluble gas
 If one of these things does not happen then there is NO NET
IONIC EQUATION and nothing reacts CHEMICALLY!!!
- Oxidation-Reduction Reactions
o Oxidation—Reduction Reactions involve a TRANSFER of ELECTRONS
from one reactant to another
o Also called REDOX reactions
o Oxidation and Reduction
 One example is CORROSION of metal into a metal compound by a
reaction with a substance in its environment
 Green coating forms when copper is oxidized, rust forms when
iron corrodes, silver corrodes to form a black tarnish
 Oxidation—the loss of electrons
 Reduction—the gain of electrons
The REDUCING AGENT is OXIDIZED
The OXIDIZING AGENT is REDUCED
 Oxidation and reduction MUST occur together
o Oxidation Numbers
 Oxidation Numbers (Oxidation States)—assigning electrons to
each atom in a neutral substance or ion
 Rules for assigning oxidation numbers
 Any atom in its ELEMENTAL FORM has an oxidation number
of ZERO (0).
o Ex. H2, P4, He
 Any MONATOMIC ION has an oxidation number equal to
the CHARGE of the ion
o Ex. Na+ (+1), Mg2+ (+2), Cl- (-1)
 NONMETALS usually have NEGATIVE oxidation numbers
o OXYGEN is usually (-2) except in PEROXIDES (-1) and
SUPEROXIDES (-1/2)
o HYDROGEN is (+1) when bonded to nonmetals and
(-1) when bonded to metals
o FLUORINE is (-1) in ALL COMPOUNDS
o HALOGENS other than F are (-1) in most binary
compounds unless combined with OXYGEN in
OXYANIONS in which they are some (+) number
 The SUM of all the oxidation numbers in a NEUTRAL
compound is ZERO. The sum of the oxidation numbers in a
POLYATOMIC ION is equal to the CHARGE of the ion
o H3O+ = each H is (+1) and O is (-2) so 3(+1) + 1(-2) =
+1
o Determine the oxidation numbers of everything in
H2S, S8, Na2SO3, and SO42(a) When bonded to a nonmetal, hydrogen has an oxidation number of +1 (rule 3b).
Because the H2S molecule is neutral, the sum of the oxidation numbers must equal zero
(rule 4). Letting x equal the oxidation number of S, we have 2(+1) + x = 0. Thus, S has
an oxidation number of −2.
(b) Because this is an elemental form of sulfur, the oxidation number of S is 0 (rule 1).
(c) Because this is a binary compound, we expect chlorine to have an oxidation number
of −1 (rule 3c). The sum of the oxidation numbers must equal zero (rule 4).
Letting x equal the oxidation number of S, we have x + 2(−1) = 0. Consequently, the
oxidation number of S must be +2.
(d) Sodium, an alkali metal, always has an oxidation number of +1 in its compounds
(rule 2). Oxygen has a common oxidation state of −2 (rule 3a). Letting x equal the
oxidation number of S, we have 2(+1) + x + 3(−2) = 0. Therefore, the oxidation
number of S in this compound is +4.
(e) The oxidation state of O is −2 (rule 3a). The sum of the oxidation numbers equals
−2, the net charge of the SO42− ion (rule 4). Thus, we have x + 4(−2) = −2. From this
relation we conclude that the oxidation number of S in this ion is +6.
Comment These examples illustrate that the oxidation number of a given element
depends on the compound in which it occurs. The oxidation numbers of sulfur, as seen
in these examples, range from −2 to +6.
o Oxidation of Metals by Acids and Salts
 The reaction between a metal and an acid or metal salt conforms
to the general pattern of DISPLACEMENT REACTIONS because the
ion in solution is diplaced (replaced) through oxidation of an
element
 SINGLE REPLACEMENT REACTION
 Many metals undergo displacement reactions with acids and form
H2(g) as a product
 Notice that the spectator ions do NOT change their oxidation
number in the reaction
 Metals can also be oxidized by aqueous solutions of various salts
0
+2
+2
0
2+
 The oxidation of Fe  Fe is accompanied by the reduction of
Ni2+  Ni
 Whenever one substance is oxidized, another substance must be
reduced!!
 Write a balanced molecular and net ionic equation for the
reaction between aluminum and hydrobromic acid
2 Al(s) + 6 HBr(aq) → 2 AlBr3(aq) + 3 H2(g)
Both HBr and AlBr3 are soluble strong electrolytes. Thus, the complete
ionic equation is
2 Al(s) + 6H+(aq) + 6Br−(aq) → 2 Al3+(aq) + 6 Br−(aq) + 3 H2(g)
Because Br− is a spectator ion, the net ionic equation is
2 Al(s) + 6 H+(aq) → 2 Al3+(aq) + 3 H2(g)
0
+1
+3
0
Comment The substance oxidized is the aluminum metal because its
oxidation state changes from 0 in the metal to +3 in the cation, thereby
increasing in oxidation number. The H+ is reduced because its oxidation
state changes from +1 in the acid to 0 in H2.
o The Activity Series
 Different materials may vary in the ease with which they are
oxidized
 A list of metals arranged in order of decreasing ease of oxidation
is called the ACTIVITY SERIES
 Metals at the top of the table are the most easily oxidized (REACT
MOST READILY to form compounds) (called ACTIVE METALS)
 Metals at the bottom of the activity series are very stable and
form compounds less readily (LESS REACTIONS) (called NOBEL
METALS)
 Any metal on the list can be oxidized by the ions of elements
BELOW it
 Any metal HIGHER on the list can REPLACE metals below them
 Only metals ABOVE HYDROGEN on the activity series are able to
react with acids to form H2
 Cu will not replace H in an acid, but Cu DOES however react with
nitric acid but the Cu is oxidized to Cu2+ by the NITRATE ION
accompanied by the formation of brown NO2(g)
 Here the NO3- is reduced to NO2 (Nitrogen goes from +5  +4)
 Will an aqueous solution of iron(II) chloride oxidize Mg metal? If
so write the balanced molecular and net ionic equations.
Because Mg is above Fe in the table, the reaction occurs. To write the
formula for the salt produced in the reaction, we must remember the
charges on common ions. Magnesium is always present in compounds as
Mg2+; the chloride ion is Cl−. The magnesium salt formed in the reaction
is MgCl2, meaning the balanced molecular equation is
Mg(s) + FeCl2(aq) → MgCl2(aq) + Fe(s)
Both FeCl2 and MgCl2 are soluble strong electrolytes and can be written in
ionic form, which shows us that Cl− is a spectator ion in the reaction. The
net ionic equation is
Mg(s) + Fe2+(aq) → Mg2+(aq) + Fe(s)
0
+2
+2
0
The net ionic equation shows that Mg is oxidized and Fe2+ is reduced in
this reaction.
- Concentrations of Solutions
o Concentration—the amount of solute dissolved in a given quantity of
solvent or quantity of solution
o The greater the amount of solute dissolved, the more CONCENTRATED
the resulting solution will be
o Molarity
 Molarity (M) – the number of moles of solute per LITER of
solution
 Calculate the molarity of a solution made by dissolving 23.4 g
sodium sulfate in enough water to make a 125 mL solution
o Expressing The Concentration of an Electrolyte
 When an ionic compound dissolves, the relative concentration of
the ions depend on the chemical formula of the compound.
 1.0 M NaCl  [Na+] = 1.0 M and [Cl-] = 1.0 M
 Because NaCl(aq)  Na+(aq) + Cl-(aq)
 But Na2SO4(aq)  2 Na+(aq) + SO42-(aq)
 So 1.0 M Na2SO4  [Na+] = 2.0 M and [SO42-] = 1.0 M
 In general,
AxBy(aq)  x Ay+ + y Bx[Ay+] = x [AxBy]
[Bx-] = y [AxBy]
 What is the molar concentration of each ion present in 0.025 M
aqueous calcium nitrate?
Calcium nitrate is composed of calcium ions (Ca2+) and nitrate ions (NO3−), so its
chemical formula is Ca(NO3)2. Because there are two NO3− ions for each Ca2+ ion,
each mole of Ca(NO3)2 that dissolves dissociates into 1 mol of Ca2+ and 2 mol of
NO3−. Thus, a solution that is 0.025M in Ca(NO3)2 is 0.025 M in Ca2+ and 2 ×
0.025 M = 0.050 M in NO3−:
o Interconverting Molarity, Moles and Volume
 Can use the molarity equation to find any missing value
 Can also use molarity concentration as a CONVERSION FACTOR
(ratio of moles and liters!!)
For example, if we know the molarity of an HNO3 solution to be 0.200 M, which
means 0.200 mol of HNO3 per liter of solution, we can calculate the number of
moles of solute in a given volume, say 2.0 L. Molarity therefore is a conversion
factor between volume of solution and moles of solute:
To illustrate the conversion of moles to volume, let's calculate the volume of
0.30 M HNO3 solution required to supply 2.0 mol of HNO3:
 How many grams of Na2SO4 are required to make 0.350 L of
0.500 M Na2SO4?
o Dilution
 Stock Solution—lab solution that is prepared in concentrated
form
 Dilution—solutions of LOWER concentration made from the stock
solution by adding water
 Main thing to remember is in a solution, the MOLES of solute
remains UNCHANGED!!
 Since the moles of solute = Molarity (mol/L) x Volume (L) both for
before and after the dilution, we get the dilution formula:
 To prepare a dilute solution (250.0 mL of 0.100M CuSO4) from a
concentrated stock of 1.00 M CuSO4
 How many mL of 3.0 M H2SO4 are needed to make 450 mL of
0.10 M H2SO4?
So use 15.0 ml of 3.0 M stock and add to about 400 mL of water {ALWAYS ADD
ACID!!!!} and then bring the final volume up to 450 mL total!
- Solution Stoichiometry and Chemical Analysis
o If a solution’s concentration is known then MOLARITY x VOLUME (L) =
MOLES!
o If we know MOLES then we can enter into stoichiometry to find an
amount of anything else in the reaction using the balanced equation
o This is illustrated by the following flowchart for solution stoichiometry
o How many grams of Ca(OH)2 are needed to neutralize 25.0 mL of 0.100
M HNO3?
2 HNO3(aq) + Ca(OH)2(s) → 2 H2O(l) + Ca(NO3)2(aq)
o Titrations
 Titration—combining a solution of UNKNOWN concentration with
a solution of KNOWN concentration (STANDARD SOLUTION) in
order to completely react with the solute in the unknown
concentration solution
 Equivalence point—point where stoichiometrically equivalent
quantities are brought together in a reaction
 Titrations can be conducted using neutralization, precipitation or
oxidation-reduction reactions
 For acid-base neutralization reactions, an INDICATOR is added
which is a dye substance whose color is pH sensitive
 Standard solution is slowly added to the unknown solution until
the equivalence point is reached (moles of H+ = moles of OH-)
 Then the unknown concentration is calculated via stoichiometry
 45.7 mL of 0.500 M H2SO4 is required to neutralize 20.0 mL of
NaOH solution. What is [NaOH]?
H2SO4(aq) + 2 NaOH(aq) → 2 H2O(l) + Na2SO4(aq)
 The quantity of Cl- is determined by titrating the sample with
Ag+. The precipitation reaction taking place is
Ag+(aq) + Cl-(aq)  AgCl(s)
The endpoint is marked by a special type of indicator. How many
grams of Cl are in the sample of water if 20.2 mL of 0.100 M Ag+ is needed to
react with the Cl- sample? If the sample has a mass of 10.0 grams, what percent
of Cl- does it contain?
-
 A sample of 70.5 mg of potassium phosphate is added to 15.0 mL
of 0.050 M AgNO3 resulting in the formation of a precipitate.
Write the molecular equation for the reaction. What is the
limiting reactant? Calculate the theoretical yield (in grams) of
the precipitate that forms.
K3PO4(aq) + 3 AgNO3(aq) → Ag3PO4(s) + 3 KNO3(aq)
Comparing the amounts of the two reactants, we find that there are (7.5 ×
10−4)/(3.32 × 10−4) = 2.3 times as many moles of AgNO3 as there are moles of
K3PO4. According to the balanced equation, however, 1 mol K3PO4 requires 3 mol
AgNO3. Thus, there is insufficient AgNO3 to consume the K3PO4, and AgNO3 is the
limiting reactant.
Suggested Problems: 4.1, 4.2, 4.4, 4.5, 4.7, 4.8, 4.14, 4.15, 4.16, 4.17, 4.19, 4.21,
4.23, 4.24, 4.26, 4.27, 4.30, 4.35, 4.36, 4.37, 4.39, 4.43, 4.45, 4.47, 4.48, 4.51,
4.52, 4.53, 4.55, 4.57, 4.58, 4.60, 4.61, 4.63, 4.64, 4.65, 4.66, 4.67, 4.69, 4.71,
4.73, 4.75, 4.77, 4.79, 4.81, 4.82, 4.84, 4.85, 4.87, 4.89, 4.94, 4.96, 4.99, 4.103,
4.104, 4.106, 4.108, 4.109, 4.111, 4.113