Chapter 4Types of Chemical Reactions and Solution

Stoichiometry

Water is the dissolving medium of the common solvent: Some properties

• Water is “bent” or v-shaped

• The O—H bonds are covalent

• Water is a polar molecule

• Hydration occurs when salts dissolve in water

The Water Molecule is Polar

Polar Water Molecules Interact with the Positive and Negative Ions of a Salt

A solute (substance being dissolved):• Dissolves in water or other solvent• Changes phase if different from the solvent • Present in lesser amount (if the same phase as

the solvent)A solvent: • Retains its phase (if different from the solute)• Present in greater amount (if the same phase as

the solute)• Rule for predicting solubility:

Like dissolves Like

The Nature of Aqueous Solutions: Strong and Weak Electrolytes

• Strong electrolytes: Conduct current very efficiently. Strong electrolytes are completely ionized when they are dissolved in water (soluble salts, strong acids, strong bases).

• Weak electrolytes: Conduct only a small current. Weak electrolytes are substances that exhibit a small degree of ionization in water (weak acids, weak bases).

• Non electrolytes: Permit no current to flow. Non electrolytes are substances that dissolve in water but do not produce any ions. (pure water, sugar solution).

BaCI2 Dissolving

HCI (aq) is Completely Ionized

Acetic Acid in Water

The Composition of Solutions

To perform stoichiometric calculation in solutionwe need-• The nature of the reaction• The amounts of chemicals present in the

solutions (concentration)• The concentration of a solution can be expressed

by Molarity (M).• Molarity (M) = moles of solute per volume of

solution in liters:M = molarity = (moles of solute)/(liters of solution)

Example: If you weigh out 5.84 g of NaCl and dissolved it in 500. mL of water, what is the molarity of the solution?

5.84 g NaCl x (1 mol NaCl)/(58.44 g NaCl) = 0.100 mol

500. mL x 1 L/1000 mL = 0.500 L

Molarity = (0.100 moles)/(0.500 L) = 0.200 M

Concentration of IonsExample: Give the concentration of each type of ion in thefollowing solution:

a. 0.50 M Co(NO3)2 b. 1 M Fe(ClO4)3 c. 1.75 M Al2(SO4)3

a. Solution reaction: Co(NO3)2(s) H2O Co2+(aq) + 2NO3

-(aq)

Each mole of Co(NO3)2 that is dissolved, the solution contains 1

mol Co2+ ions and 2 mol NO3- ions. Thus a solution of 0.50 M

Co(NO3)2 contains 0.50 M Co2+ and (2 x 0.50) M or 1.0 M NO3-.

b. Fe(ClO4)3(s) H2O Fe3+(aq) + 3ClO4

-(aq)

1 M Fe(ClO4)3 contains 1 M Fe3+ ions and 3 M ClO4- ions.

c. Al2(SO4)3(s) H2O 2Al3+(aq) + 3SO4

2-(aq)

1.75 M Al2(SO4)3 contains (2 x 1.75) M or 3.50 M Al3+ ions and (3

x 1.75) M or 5.25 M SO42- ions.

Standard Solution

• A standard solution is a solution whose concentration is accurately known.

Finding volume or weight:

• How many grams of K2Cr2O7 are required to prepare 1.00 L of a 0.20 M solution?

M = moles/L

Moles = M x L = 0.200 M x 1.00 L = 0.200 mol K2Cr2O7.

0.200 mol K2Cr2O7 x (294.20 g)/(1 mol) = 58.8 g

• What volume is needed to prepare a 1.20 M solution from 10.0 g CaCl2 ? (mm = 111.0 g)

M = moles/L ==> L = moles/M

10.0 g CaCl2 x (1 mole CaCl2 )/111.0 g CaCl2) = 0.090 moles

L = (0.090 moles)/(1.20 M) = 0.0750 L

= 75.0 mL

Preparation of a Standard Solution

Dilution

• The desired molarity solutions are often prepared from concentrated stock solutions (routinely used solutions prepared in concentrated form) by adding water. This process is called dilution.

• Dilution with water does not alter the numbers of moles of solute present.

Moles of solute before dilution

= moles of solute after dilution

M1V1 = M2V2

(a) A Measuring Pipet (b) A Volumetric (transfer) Pipet

Example: What volume of 16 M sulfuric acid

must be used to prepare 1.5 L of 0.10 M H2SO4

solution?

V1 = Volume before dilution = ?

M1 = Concentration before dilution = 16 M

V2 = Volume after dilution = 1.5 L

M2 = Concentration after dilution = 0.10 M

M1V1 = M2V2

V1 =M2V2 / M1 = (0.10 M x 1.5 L)/ 16 M = 0.0094 L = 9.4 mL

Types of Chemical Reactions

Solution Reactions are classified as follows:• Precipitation reactions• Acid – base reactions• Oxidation – reduction reactionsOther classification:• Formation reaction• Decomposition reaction• Single replacement• Double replacement

Precipitation Reactions

When two solutions are mixed, an insoluble substance (solid) sometimes forms and separates from the solution. Such a reaction is called a precipitation reaction, and the solid that forms is called a precipitate.

Example: K2CrO4(aq) + Ba(NO3)2(aq) Products (Soluble) (Soluble)

Contains the ions: K+, CrO42-, Ba2+, NO3

-

• Product must contain both anions and cations

• Most ionic products contain one type of cation and one type of anion

… Precipitation Reactions continued…

The possible combinations are-

K2CrO4, KNO3, BaCrO4, Ba(NO3)2

(reactant) (reactant)

The only real possibility for the solid yellow

products are-KNO3 or BaCrO4

Both K+ and NO3- ions are colorless but CrO4

2-

ion is yellow, so the yellow solid is BaCrO4.

K2CrO4(aq) + Ba(NO3)2(aq) BaCrO4(s) + 2KNO3(aq)

The Reaction of K2CrO4(aq) and Ba(NO3)2(aq)

Describing Reactions in Solutions

• Regular balanced equation or molecular equation: (gives the overall reaction stoichiometry but not the actual forms of reactants and products)

K2CrO4(aq) + Ba(NO3)2(aq) BaCrO4(s) + 2KNO3(aq)

• Complete ionic equation: (represents as ions all reactants and products that are strong electrolyte)

2K+(aq) + CrO4

2-(aq) + Ba2+

(aq) + 2NO3-(aq)

BaCrO4(s) + 2K+(aq) + 2NO3

-(aq)

• Net ionic equation: (Ions participate in reaction)

Ba2+(aq) + CrO4

2-(aq) BaCrO4(s)

• Spectator ions: The ions that do not participate directly in the reaction, present in solution both before and after the reaction. K+ and NO3

- ions.

2K+(aq) + CrO4

2-(aq) + Ba2+

(aq) + 2NO3-(aq)

BaCrO4(s) + 2K+(aq) + 2NO3

-(aq)

Simple Rules for Solubility1.Most nitrate (NO3

) salts are soluble.

2.Most alkali (group 1A) salts and NH4+ are

soluble.3.Most Cl, Br, and I salts are soluble (NOT Ag+,

Pb2+, Hg22+)

4.Most sulfate salts are soluble (NOT BaSO4, PbSO4, HgSO4, CaSO4)

5.Most OH salts are only slightly soluble (NaOH, KOH are soluble, Ba(OH)2, Ca(OH)2 are marginally soluble)

6.Most S2, CO32, CrO4

2, PO43 salts are only

slightly soluble.

Stoichiometry of Precipitation ReactionsStep 1: Identify the species present in the combined solution, and determine what reaction occurs.Step 2: Write the balanced net ionic equation for the reaction.Step 3: Calculate the moles of the reactants (use volume and molarity).Step 4: Determine which reactant is limiting.Step 5: Calculate the moles of product or products, as required.Step 6: Convert to grams or other units, as required.

Example: Calculate the mass of solid NaCl that must be added to 1.50 L of a 0.100 M AgNO3 solution to precipitate all the Ag+ ions in the form of AgCl.

Ag+ NO-3 Na+ Cl-

Ag+(aq) + Cl-

(aq) AgCl(s)

1.50 L x (0.100 mol Ag+) = 0.150 mol Ag+

Ag+ and Cl- react in a 1:1 ratio, 0.150 mol Cl- ions and 0.150 mol NaCl x (58.45 g NaCl)/(1 mol NaCl) = 8.77 g NaCl

Acid-Base ReactionsArrhenius:• Acid is a substance that produces H+ ions when

dissolved in water• Base is substances that produces OH- ionsBronsted-Lowry:• An acid is a proton donor• A base is a proton acceptor

Balanced equation: HCl + NaOH NaCl + H2OComplete ionic equation: H+ + Cl- + Na+ + OH-

Na+ + Cl- + H2O

Net ionic equation: H+ + OH- H2O

Neutralization ReactionExample: What volume of a 0.100 M HCl solution is needed to neutralize 25.0 mL of 0.350 M NaOH?

The species available for reaction are – H+, Cl-, Na+, OH-

NaCl is soluble, so Na+, Cl- are spectator ions, net ionic equation, H+

(aq)+ OH-(aq) H2O(l)

Moles of OH- ions: 25.0 mL x (1L)/(1000 mL) x (0.350 mol OH- )/(L NaOH) = 8.75 x 10-3 mol OH-

H+and OH- ions react in a 1:1 ratio, 8.75 X 10-3 mol H+ ions is required to neutralize OH- ions present.

V x (0.100 mol H+)/L = 8.75 x 10-3 mol H+

V = [(8.75 x 10-3 mol H+)]/[(0.100 mol H+)/L] =

8.75 x 10-2 L x (1000 mL)/1L = 87.5 mL

Acid-Base Titration• Acid-base reaction is called a neutralization reaction (the

acid and base are destroyed to leave a “neutral” solution).• Volumetric analysis is a technique for determining the

amount of a certain substance by doing a titration.• A titration involves delivery of a measured volume of a

solution of known concentration (the titrant) into a solution containing the substance being analyzed (the analyte).

• Equivalence point or stoichiometric point: enough titrant added to react exactly with the analyte.

• Indicator: A substance added at the beginning of the titration that changes color at or near the equivalence point.

• End point: The point where the indicator actually changes color.

Oxidation-Reduction Reactions• In oxidation-reduction (redox) reaction, one

species loses electrons while another species gains electrons.

• Both processed must occur simultaneously. We can’t have a net ionic reaction that contain only oxidation or only reduction.

• The number of electron lost by the oxidation process must equal the number gained by the reduction process. We cannot create or destroy electrons.Cu Cu2+ + 2e- (oxidation)2Ag+ + 2e- 2Ag (reduction)Cu + 2Ag+ Cu2+ +2Ag (net ionic reaction)

Oxidation States

Oxidation states or oxidation numbers provides a way to keep track of electrons (accounting for electrons) in redox reactions.

• Rules for Assigning Oxidation States:1. The oxidation state of an atom in an element is

O (element in its elemental form is O).2. The oxidation state of a monatomic ion is the

same as its charge.3. In its compound, fluorine is always assigned

an oxidation state of –1.

…Oxidation States continued…

4. Oxygen is assigned an oxidation state –2 in covalent compound (except in peroxides where it is –1)

5. Hydrogen is assigned an oxidation state +1 in its covalent compounds with non metals.

6. Sum of the oxidation states must be zero for an electrically neutral compound, for an ion equals the charge of the ion.

Assigning Oxidation States

Example: Assign oxidation states to all atoms in thefollowing.

a. CO2 O = -2 (for each oxygen); C = +4

b. SF6 F = -1 (for each fluorine); S = +6

c. NO3- O = -2 (each oxygen); N = +5

d. MnO4- O = -2 (each oxygen); Mn = +7

e. HSO3- O = -2 (each oxygen); H = +1, S = +4

f. H2O H = +1 (each); O = -2

g. Li3N Li = +1 (each); N = -3

Oxidation-Reduction• Oxidation is an increase in oxidation state or a loss of

electron.• Reduction is a decrease in oxidation state or a gain of

electron.• Oxidizing Agent is the electron acceptor (reduced).• Reducing Agent is the electron donor (oxidized)

PbO(s) + CO(g) Pb(s) + CO2(g)

+2 -2 +2 -2 0 +4 -2 (each)Pb reduced, C oxidized, PbO oxidizing agent, CO reducing agent

A Summary of an Oxidation-Reduction Process

Balancing Oxidation-Reduction EquationsRedox reactions in acidic solutionStep 1: Write separate equation for the oxidation and

reduction half reactions.Step 2: For each half reaction-a. Balance all the elements except hydrogen and oxygen

b. Balance oxygen using H2Oc. Balance hydrogen using H+

d. Balance the charge using electronsStep 3: If necessary, multiply one or both balanced half-

reactions by an integer to equalize the number of electrons transferred

Step 4: Add the half reactions and cancel identical speciesStep 5: Check that the elements and charges are balanced

Balancing Oxidation-Reduction Reactions

Example: Balance the equation for the reactionbetween permanganate and iron (II) ion in acidicsolution

MnO4-(aq) + Fe2+

(aq) Acid Fe3+(aq) + Mn2+

(aq)

Half reactions: MnO4- Mn2+ (reduction)

Fe2+ Fe3+ (oxidation)

Balance half reactions: MnO4- Mn2+

a. Manganese is balanced

b. Balance oxygen by adding 4H2O to the right side of the equation

..continued..

c. Balance hydrogen by adding 8H+ to the left side

8H+(aq) + MnO4

-(aq) Mn2+

(aq) + 4H2Od. Equalize the charge by adding electrons

5e- + 8H+(aq) + MnO4

-(aq) Mn2+

(aq) + 4H2O(l)

Fe2+(aq) Fe3+

(aq) + e- x 5

5 Fe2+(aq) 5 Fe3+

(aq) + 5e-

Add half reactions –

5Fe2+(aq) + MnO-

4(aq) + 8H+(aq)

5Fe3+(aq) + Mn2+

(aq) + 4H2O(l)

Check that elements and charges are balanced.

Balancing Oxidation-Reduction Reactions in Basic Solution

• Step 1: Use the half reaction method as

specified for acidic solution.• Step 2: Add a number of OH- ions that is equal

to the number of H+ ions (to both sides).

• Step 3: Form H2O by combining H+ and OH-.

• Step 4: Check that elements and charges are balanced.

Example:

ClO- + CrO2- Cl - + CrO4

2- (basic solution)

+1 +3 -1 +6First half reaction:

ClO- +2e- Cl-

ClO- + 2e- Cl- + H2O

ClO- + 2e- +2H+ Cl- + H2OSecond half reaction:

CrO2- CrO4

2-

CrO2- CrO4

2- + 3e-

CrO2- + 2H2O CrO4

2- + 3e-

CrO2- + 2H2O CrO4

2- + 3e- + 4H+

…Continued…

Multiply first half reaction by 3 and second half reaction

by 2 then add two half reactions.

3ClO- + 6e- + 6H+ 3Cl- + 3H2O

2CrO2- + 4H2O 2CrO4

2- + 6e- + 8H+

3ClO- + 2CrO2- + H2O 3Cl- + 2CrO4

2- + 2H+

Add OH- to both sides to remove H+

3ClO- + 2CrO2- + H2O + 2OH- 3Cl- + 2CrO4

2- + 2H+ + 2OH-

3ClO- + 2CrO2- + H2O + 2OH- 3Cl- + 2CrO4

2- + 2H2O

3ClO- + 2CrO2- + 2OH- 3Cl- + 2CrO4

2- + H2O

Summary• Like dissolves Like

• Strong, weak and non electrolytes

• Molarity: moles of solute/Liters of solution

• Standard Solution: concentration accurately known

• Dilution: M1V1 = M2V2

• Precipitation, acid-base and oxidation reduction reactions

• Spectator ions

• Acids: produces H+ or proton donor

• Bases: produces OH- and proton acceptor

• Acid-Base titration: Neutralization reaction

• Oxidation-Reduction reactions

• Oxidation States

• Balancing redox equations