Advanced Chemistry

Types of Chemical Reactions and Solution Stoichiometry

The Common

Solvent

Water

Water as a Solvent Bent (V-shaped) molecule

with 105° bond angle.

Covalent bonds – electron sharing between oxygen and hydrogen.

Unequal sharing of electrons due to electronegativity differences cause water to be a polar molecule.

Water as a Solvent Partial charges (δ) form on

the water molecule due to the unequal sharing of electrons.

Water’s polarity causes it to be a solvent for ionic compounds.

Polar Water Molecules Interact with the Positive and Negative Ions of a Salt Assisting in the Dissolving Process

Solubility

Solubility of ionic substances in water varies greatly.

Solubility depends on the relative attractions of the ions and water molecules for each other.

Once dissolved, an ionic compound becomes hydrated, therefore ions disperse independent of one another.

Solubility

Water will dissolve non-ionic substances depending upon their structure.

If a polar bond exists within the structure, the molecule can be subject to being water soluble.

In general, polar substances dissolve in polar solvents and nonpolar substances dissolve in nonpolar solvents.

An Ethanol Molecule Contains a Polar O-H Bond Similar to Those in the Water

Molecule

The Polar Water Molecule Interacts Strongly with the Polar-O-H bond in

Ethanol

Strong and Weak

Electrolytes

Nature of Aqueous Solutions

Strong and Weak Electrolytes

A solution in which a substance is dissolved in water; the substance is the solute and the solvent is the water.

A common property for a solution is electrical conductivity. Its ability to conduct an electric

current.

Strong and Weak Electrolytes

Pure water is not an electrical conductor.

Strong electrolytes conduct current very efficiently.

Weak electrolytes conduct only a small current.

Nonelectrolytes do not allow current to flow.

Electrical Conductivity of Aqueous Solutions

Svante Arrhenius

1859-1927

Studied the nature of solutions and theorized that conductivity of solutions arose from the presence of ions.

Proved that the strength of conductivity is directly related to the number of ions present in solution.

Strong Electrolytes

Strong electrolytes are substances that are completely ionized when they are dissolved in water. Soluble salts Strong acids Strong bases

NaCl Dissolves

Acids

Arrhenius discovered in his studies of solutions that when acids were dissolved in water they behaved as strong electrolytes. This result was directly related to an

acid’s ability to ionize in water.

An Acid is a substance that produces H+ ions when it is dissolved in water.

HCl is Completely Ionized

Acids

H2SO4

H2Ou xuuuu → H +(aq) + HSO−

4(aq)

An Acid is a substance that produces H+ ions when it is dissolved in water.

Acids

H2SO4

H2Ou xuuuu → H +(aq) + HSO−

4(aq)

In conductivity studies, virtually every molecule ionizes. Therefore, strong electrolytes are strong

acids.

Strong Acids Sulfuric acid, nitric acid and hydrochloric

acid are aqueous solutions and should be written in chemical equations as such.

A strong acid is one that completely dissociates into its ions. In aqueous solutions, the HCl molecule does not exist.

Sulfuric acid can produce two H+ ions per molecule. Only the first H+ ion completely dissociates. The anion HSO4

- remains partially intact.

Strong Bases

Bases are soluble ionic compounds containing the hydroxide ion (OH-).

Strong bases are strong electrolytes and these compounds ionize completely in water.

Strong Bases

NaOH (s )H2O → Na(aq)

+ +OH(aq)−

KOH(s)H2O → K(aq)

+ +OH(aq)−

An Aqueous Solution of Sodium Hydroxide

Weak Electrolytes

Weak electrolytes are substances that exhibit a small degree of ionization in water. They produce relatively few ions

when dissolved in water Most common weak electrolytes are

weak acids and weak bases.

Weak Acids

Formulas for acids are often written with the acidic hydrogen atom or atoms (the hydrogen atoms that will produce H+ ions in solution) listed first. If any nonacidic hydrogens are present they are written later in the formula.

HC2H 3O2(aq) + H 2O(l ) →← H3O(aq)

+ +C2H3O2(aq)−

Weak Acids

In Acetic acid, only 1% of its molecules ionize.

The double arrow indicates that the reaction can occur in either direction.

Acetic acid is a weak electrolyte and therefore a weak acid because it dissociates (ionizes) only to a slight extent in aqueous solutions.

HC2H 3O2(aq) + H 2O(l ) →← H3O(aq)

+ +C2H3O2(aq)−

Acetic Acid (HC2H3O2)

Weak Bases

The most common weak base is NH3.

In an aqueous solution, ammonia results in a basic solution.

NH 3(aq) + H2O(l ) →← NH4(aq)

+ +OH(aq)−

The Reaction of NH3 in Water

Nonelectrolytes

Nonelectrolytes are substances that dissolve in water but do not produce any ions.

The Composition of

Solutions

Concentration:

Solutions Most chemical reactions take place in the

environment of solutions. In order to perform stoichiometric calculations in solutions, one must know two things. The nature of the reaction; which depends

on the exact forms the chemicals take when dissolved.

The amounts of the chemicals present in the solutions, usually expressed as concentrations.

Concentration

Molarity (M) – is moles of solute per volume of solution in liters:

Example: 1.0M= 1.0 molar = 1.0moles solute/1liter of solution

M =molarity=moles of soluteliters of solution

Example Calculate the molarity of a solution

prepared by dissolving 11.5g of solid NaOH in enough water to make 1.50L of solution.

.192 M NaOH

Example

Calculate the molarity of a solution prepared by dissolving 1.56g of gaseous HCl in enough water to make 26.8 ml of solution.

1.60M HCl

Example

Give the concentration of each type of ion in 0.50M Co(NO3)2.

Co2+ = 0.50 M Co2+

NO3- = 1.0 M NO3

-

Example

Calculate the number of moles of Cl- ions in 1.75L of 1.0 x 10-3 M ZnCl2.

3.5 x 10-3 mol Cl-

Dilution

The process of changing the molarity of a solution from a more concentrated solution to a lesser concentrated solution. Moles of solute after dilution = moles

of solute before dilution. M x V= moles M1V1=M2V2

Example

What volume of 16M sulfuric acid must be used to prepare 1.5L of 0.10 M H2SO4 solution?

9.4 ml solution

Steps Involved in the Preparation of a Standard Aqueous Solution

Types of Chemical

Reactions

Types of Solution Reactions

Most solution reactions can be put into three types of reactions: Precipitation Reactions Acid-Base Reactions Oxidation-Reduction Reactions

Precipitation Reactions

When two solutions are mixed, an insoluble substance sometimes forms; that is, a solid forms and separates from the solution. The solid that forms is called a

precipitate.

Precipitation Reaction Example

K2CrO4(aq) + Ba(NO3)2(aq) →

2K(aq)+ +CrO4(aq)

2− + Ba(aq)2+ + 2NO3(aq)

− →

A more accurate representation is:

Reactant Solutions

Solution Post-Reaction

Precipitation Reaction Example

2K(aq)+ +CrO4(aq)

2− + Ba(aq)2+ + 2NO3(aq)

− →

We look at all the possible combinations of the ions to check for compounds that form solids.

•K2CrO4

•KNO3

•BaCrO4

•Ba(NO3)2

Precipitation Reaction Example

2K(aq)+ +CrO4(aq)

2− + Ba(aq)2+ + 2NO3(aq)

− →

Two of these combinations are the reactants and can be ruled out:

•K2CrO4

•KNO3

•BaCrO4

•Ba(NO3)2

Solubility

Predicting the identity of a solid product in a precipitation reaction requires knowledge of the solubilities of common ionic substances. Slightly soluble – the tiny amount of solid

that dissolves is not noticeable. The solid appears insoluble to the naked eye.

Insoluble and slightly soluble are often used interchangeably.

Simple Rules for the Solubility of Salts in Water

Precipitation Reaction Example

2K(aq)+ +CrO4(aq)

2− + Ba(aq)2+ + 2NO3(aq)

− →

Two of these combinations are the reactants and can be ruled out:

•K2CrO4

•KNO3

•BaCrO4

•Ba(NO3)2

Precipitation Reactions

Precipitation reactions move forward due to the decrease in energy state of the compound. Bonds forming in the compound increase stability and push the reaction forward.

Solutions

Describing Reactions in

Formula Equation

Although the formula equation shows the reactants and products of the reaction, it does not give a correct picture of what actually occurs in solution. Gives the overall reaction

stoichiometry but not necessarily the actual forms of the reactants and products.

Complete Ionic Equation

The complete ionic equation better represents the actual forms of the reactants and products in solution. All substances that are strong electrolytes are represented as ions. Complete ionic equation shows all

ions in a reaction, even those that do not participate in the reaction. These ions are called spectator ions.

Net Ionic Equation

The net ionic equation includes only those solution components that are directly involved in the reaction. Commonly used because it gives

the actual forms of the reactants and products and includes only the species that undergo a change. Spectator ions are not included.

Example Problem For the following reaction, write the formula

equation, the complete ionic equations, and the net ionic equation. Aqueous potassium chloride is added to aqueous silver nitrate to form a silver chloride precipitate plus aqueous potassium nitrate.

K(aq)+ +Cl

(aq)

− + Ag(aq)+ + NO

3(aq)

− → AgCl(s) + K(aq)+ + NO3(aq)

−

Cl(aq )

− + Ag(aq)+ → AgCl(s)

of Precipitation

Reactions

Stoichiometry

Solution Stoichiometry The rules of stoichiometry and limiting

reactant apply to chemical reactions in solutions. But two rules need special emphasis. Always write a balanced equation of the

reaction and give special attention to the products that are formed and the true form of the ions in solution.

Moles must still be calculated, but molarity and volume must be used in the calculation.

Determining the Mass of Product Formed

Example Problem Determine the mass of solid NaCl that must be

added to 1.50L of a 0.100 M AgNO3 solution to precipitate all the Ag+ ions in the form of AgCl.

Ag(aq )+ +Cl (aq)

− → AgCl (s)

Determine the amount of Ag+ ions in solution.

Determine the amount of Cl- needed to react with Ag+

.15molesAg+

x1Cl−

1Ag+ x1NaCl1Cl− x

58.44g1mole

=8.77gNaCl

Steps to Solution Stoichiometry

1. Identify the species present in the combined solution, and determine what reaction occurs.

2. Write the balanced net ionic equation for the reaction.

3. Calculate the moles of reactants.

4. Determine which reactant is limiting.

5. Calculate the moles of product or products, as required.

6. Convert to grams or other units, as required.

Reactions

Acid-Base

Acids

Arrhenius’s concept of acids and bases: An acid is a substance that produces H+ ions when dissolved in water and a base is a substance that produces OH- ions. Fundamentally correct but doesn’t

include all bases.

Brønsted-Lowry Acids

Johannes BrØnsted (1879-1947) and Lowry (1874-1936) gives a more general definition of a base that includes substances that do not contain OH-. An acid is a proton donor. A base is a proton acceptor.

Acid-Base Reactions An Acid-Base reaction often forms two

things: A salt (sometimes soluble) Water

Since water is a nonelectrolyte, large quantities of H+ and OH- cannot coexist in solution.

The net ionic equations is:

H (aq)+ +OH(aq)

− → H2O(l )

Acid-Base Reactions

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l )

H(aq)+ +Cl(aq)

− + Na(aq)+ +Cl(aq)

− → Na(aq)+ +Cl(aq)

− + H2O(l )

H(aq)+ +OH(aq)

− → H2O(l )

When a strong acid and a strong base react, we expect both substances to completely ionize. We then check to see what will form that is soluble.

In this case, the salt is soluble and remains as ions. But water, a nonelectrolyte, will form since H+ and OH- have a

strong attraction for each other and do not ionize.

Acid-Base Reactions

HC2H 3O2(aq) + KOH(aq) →

OH(aq)− + HC2H3O2(aq) → H2O(l ) +C2H3O2 (aq)

−

When a weak acid and a strong base react, the weak acid usually doesn’t ionize. However, the hydroxide ion is such a strong base that for the purposes of stoichiometric calculations it can be assumed to react completely with any weak acid.

Performing Calculations for Acid-Base Reactions

1. List the substances present in the combined solution before any reaction occurs and decide what reaction will occur.

2. Write the balanced net ionic equation for this reaction.

3. Calculate the moles of reactants. For reactions in solution, use the volumes of the original solutions and their molarities.

Performing Calculations for Acid-Base Reactions

4. Determine the limiting reactant where appropriate.

5. Calculate the moles of the required reactant or product.

6. Convert to grams or volume (of solution), as required.

Acid-Base Reaction

An acid-base reaction is often called a neutralization reaction.

When just enough base is added to react exactly with the acid in a solution, we say the acid has been neutralized.

Step by Step Example

What volume of a .100 M HCl solution is needed to neutralize 25.0ml of .350 M NaOH?

What ions are present? What are the possible reactions?

Na(aq)+ +Cl(aq)

− → NaCl(s)H(aq)

+ +OH(aq)− → H2O(l )

Example

What volume of a .100 M HCl solution is needed to neutralize 25.0ml of .350 M NaOH?

NaCl is soluble. Na+ and Cl- are spectators.

Write a balanced net ionic equation.

H (aq)+ +OH(aq)

− → H2O(l )

Example

What volume of a .100 M HCl solution is needed to neutralize 25.0ml of .350 M NaOH?

What are the moles of reactant present in solution?

25.0ml NaOHx

1L

1000mlx0.350mol OH −

LNaOH=8.75x10−3molOH−

Example

What volume of a .100 M HCl solution is needed to neutralize 25.0ml of .350 M NaOH?

How many moles of H+ are needed?

=8.75x10−3molH +

The mole ratio is 1:1

Example

What volume of a .100 M HCl solution is needed to neutralize 25.0ml of .350 M NaOH?

What volume of HCl is required?

Vx0.100molH +

L=8.75x10−3molH +

V =8.75x10−2 L

Example In a certain experiment, 28.0ml of 0.250 M

HNO3 and 53.0ml of .320 M KOH are mixed. Calculate the amount of water formed in the resulting reaction. What is the concentration of H+ or OH- ions in excess after the reaction goes to completion?

H+ is limiting7.00x10-3mol

H2O.123 M OH-

excess

Acid-Base Titrations

Volumetric analysis is a technique for determining the amount of a certain substance by doing a titration.

A titration involves delivery (from a buret) of a measured volume of a solution of known concentration (the titrant) into a solution containing the substance being analyzed (the analyte).

Acid-Base Titrations

The point in the titration where enough titrant has been added to react exactly with the analyte is called the equivalence point or the stoichiometric point.

This point is often marked with an indicator, a substance added at the beginning of the titration that changes color at the equivalence point.

Acid-Base Titrations

The point at which the indicator actually changes color is called the endpoint of the titration.

The procedure for determining accurately the concentration of a solution is called standardizing the solution.

Indicators A common indicator for acid-base

titrations is phenolphthalein, which is colorless in an acidic solution and pink in a basic solution.

Example A student carries out an experiment to standardize a

sodium hydroxide solution. To do this, the student weighs out a 1.3009g sample of potassium hydrogen phthalate (KHC8H4O4 or KHP). KHP (molar mass 204.22g/mol) has one acidic hydrogen. The student dissolves the KHP is distilled water, add phenolphthalein as an indicator, and titrates the resulting solution with the sodium hydroxide solution to the phenolphthalein endpoint. The difference between the final and initial buret readings indicate that 41.20 ml of the sodium hydroxide solution is required to react exactly with the 1.3009g KHP. Calculate the concentration of the sodium hydroxide solution.

Example 1.3009g KHP, molar mass = 204.22g/mol

41.20 ml NaOH solution to neutralize KHP

Calculate concentration of NaOH

KHP has 1 acidic hydrogen so it should react in a 1 to 1 ratio:

KHC8H 4O4(aq) + NaOH(aq) → H2O(l ) +C8H4O4(aq)2− + K(aq)

+ + Na(aq)+

.1546 M

Reactions

Oxidation-Reduction

Oxidation and Reduction

Oxidation Reduction reactions transfer one or more electrons from one species to another. Called Redox reactions. Common and important type of

reaction: photosynthesis, energy production and combustion

Oxidation States

Also called oxidation numbers provides a way to keep track of electrons in oxidation-reduction reactions. In a covalent compound when electrons

are shared, oxidation numbers are based on the relative electron affinity of the elements involved.

Oxidation States

In a covalent compound: If the bond is between two

identical atoms, the electrons are divided equally.

If the bond is between different atoms, the electrons are divided based on electron attraction.

Oxidation State Example

H2O Total electrons: 4 O has more electronegativity and

maintains a δ-. Therefore O is assumed to have

taken both electrons, one from each of the Hydrogens.

H has an oxidation state of +1 (each) O has an oxidation state of -2

Oxidation States

The sum of the oxidation states must be zero for an electrically neutral compound.

The sum of the oxidation states must equal the charge of the ion. Ion charges are written as n+ or

n-, while oxidation numbers are written +n or -n

Examples

CO2

SF6

NO3-

C +4, O -2

S +6, F -1

N +5, O -2

Non- Integer Example

Fe3O4 Oxygen is assigned first, -2 Giving Fe a +8/3 state Acceptable because all Fe

is assumed to have the same charge within the compound. But the compound actually contains two Fe3+ ions and one Fe2+ ion.

Redox Reactions

Characterized by a transfer of electrons

2Na(s ) +Cl2(g) → 2NaCl(s)

Redox Reactions

C -4 C +4 Carbon loses 8 electrons Increase in oxidation state is Oxidation

O 0 O -2 Oxygen gains 8 electrons: 4 (-2) = -8 Decrease in oxidation state is

Reduction

Redox Reactions

C -4 C +4 Carbon is Oxidized Oxygen gas is the oxidizing agent

O 0 O -2 Oxygen is Reduced Methane is the reducing agent.

Example Metallurgy, the process of producing a metal from its

ore, always involves oxidation-reduction reactions. In the metallurgy of galena (PbS), the principal lead-containing ore, the first step is the conversion of lead sulfide to its oxide (a process called roasting):

The oxide is then treated with carbon monoxide to produce the free metal:

For each reaction, identify the atoms that are oxidized and reduced, and specify the oxidizing and reducing agents.

PbO(s ) +CO(g) → Pb(s) +CO2(g)

Example

Pb +2 Pb +2S -2 S +4O 0 O -2

Sulfur is oxidized and oxygen is reduced.Oxygen gas is the oxidizing agent and lead

sulfide is the reducing agent.

Example

Pb +2 Pb 0

O -2 O -2

C +2 C +4

PbO(s ) +CO(g) → Pb(s) +CO2(g)

Lead is reduced and carbon is oxidized. PbO is the oxidizing agent, and CO is the

reducing agent.

Oxidation-Reduction

Equations

Balancing

Balancing Redox Reactions

Two methods are normally used:

1.Balancing of Oxidation states

2.Separation of the reaction into two half-reactions Normally used for more complex

reactions

Oxidation States Balancing Method

We know that in a redox reaction we must ultimately have equal numbers of electrons gained and lost, and we can use this principle to balance redox equations.

Balancing Redox Steps

1. Write the unbalanced equation.

2. Determine the oxidation states of all atoms.

3. Show electrons gained and lost.

4. Use coefficients to equalize the electrons gained and lost.

5. Balance the rest of the equations by inspection.

6. Add appropriate states.

Example

H (aq)+ +Cl(aq)

− + Sn(s) + NO3(aq)− → SnCl6(aq)

2− + NO2(g) + H2O(l )

+1 -1 0 +5 -2 +4 -1 +4 -2 +1 -2

Note that hydrogen, chlorine, and oxygen do not change oxidation states and are not involved in electron exchange.

0 +5 -2 +4 -1 +4 -2

Example

0 +5 -2 +4 -1 +4 -2

Tin lost 4 electrons and each Nitrogen gained 1 electron. Therefore each nitrogen must have a

coefficient of 4.

Balance the rest as usual

8H (aq)+ + 6Cl(aq)

− + Sn(s) + 4NO3(aq)− → SnCl6(aq)

2− + 4NO2(g) + 4H2O(l )

ENDThe

Figure 4.11a-b Measuring Pipets and Volumetric Pipets Measure Liquid Volume

Figure 4.12a-c A Measuring Pipet is Used to Add Acetic Solution to a

Volumetric Flask

Figure 4.15 a&b The Reaction of K2CrO4 and Ba(NO3)2

Figure 4.17 Molecular-Level Representations Illustrating the

Reaction of KCl (aq) with AgNO3 (aq) to Form AgCl (s)

Determining the Mass of Product Formed

Performing Calculations for Acid-Base Reactions

Neutralization Reactions I

Neutralization Reactions II

Neutralization Titration

Figure 4.19 The Reaction of Solid Sodium and Gaseous Chlorine to Form

Solid Sodium Chloride

Figure 4.20 A Summary of Oxidation-Reduction Process

The Half-Reaction Method (Acidic Solution)

The Half-Reaction Method (Basic Solution)

Figure 4.4a-c Electrical Conductivity of Aqueous

Solutions

An Aqueous Solution of Co(NO3)2.

Figure 4.10 Steps Involved in the Preparation of a Standard Aqueous

Solution

Yellow Aqueous Potassium

Figure 4.14a-b Reactant Solutions

Figure 4.16 Addition of Silver Nitrate to Aqueous Solution of Potassium Chloride

Figure 4.17 Reaction of KCI(aq) with AgNO3(aq) to

form AgCI(s).

Lead Sulfate

KOH and Fe(NO3)3 Mix to Create Solid Fe(OH)3.

Figure 4.18a-c The Titration of an Acid with a Base

Figure 4.19 The Reaction of Solid Sodium and Gaseous Chlorine to Form Solid

Sodium Chloride

Oxidation of Copper Metal by Nitric Acid

Magnetite

Aluminum and Iodine Mix to Form Aluminum Iodide

Chocolate

When Potassium Dichromate Reacts with Ethanol, the Solution

Contains Cr3+.

Table 4.2 Rules for Assigning Oxidation States

The End