Chapter 4 - Types of Chemical Reactions and Solution Stoichiometry … · · 2016-10-1610/16/2016...

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Chapter 4

Types of Chemical Reactions and Solution Stoichiometry

Section 4.1Water, the Common Solvent

Aqueous Solutions

W t i th di l iWater is the dissolving medium, or solvent.


Some Properties of Water

Water is able to dissolve so many

substances because:

Water is “bent” or V shaped

105

H

‐ Water is “bent” or V‐shaped.

‐ The O‐H bonds are covalent.

‐ Water is a polar molecule.

‐ Hydration occurs when salts dissolve in water.

H

O2

105

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A Solute

‐ dissolves in water (or other “solvent”)

‐ changes phase (if different from the solvent)so e )

‐ is present in lesser amount (if the same phase as the solvent)


A Solvent

‐ retains its phase (if different from the solute))

‐ is present in greater amount (if the same phase as the solute)

Section 4.1Water, the Common SolventHow Ionic solids dissolve

H

H

Click here for Animation

H

HH

Polar water molecules interact with the positiveand negative ions of a salt, assisting in the dissolving process. This process is called hydration.

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One of the most important substances on Earth.

Can dissolve many

Copyright © Cengage Learning. All rights reserved 7

Can dissolve many different substances.

A polar molecule because of its unequal charge distribution.


Dissolution of a solid in a liquid


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Solubility

The general rule for solubility is:

“Like dissolves like.”

Polar water molecules can dissolve other l l l h l h l dpolar molecules such as alcohol and,

also, ionic substances such as NaCl.

Nonpolar molecules can dissolve other nonpolar molecules but not polar or ionic substances. Gasoline can dissolve grease.

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Section 4.1Water, the Common SolventMiscibility

Miscible ‐‐ two substances that will mix together in any proportion to make a solution. Alcohol and water are miscible because they are both polarmiscible because they are both polar and form hydrogen bonds.

Immiscible ‐‐ two substances that will not dissolve in each other. Oil and vinegar are immiscible because oil is nonpolar and vinegar is polar.


Solubility

How does the rule “Like dissolves like.” apply to cleaning paint brushes used for l i d h dlatex paint as opposed to those used with oil‐based paint?

Section 4.2The Nature of Aqueous Solutions: Strong and Weak Electrolytes

Nature of Aqueous Solutions

Solute – substance being dissolved.

Solvent – liquid water.

Electrolyte – substance that when dissolved in water d l i h d l i iproduces a solution that can conduct electricity.

A nonelectrolyte is a substance which, when dissolved in water, gives a nonconducting solution.


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Electrolytes

Strong Electrolytes – conduct current very efficiently (bulb shines brightly). Completely ionized in water.

soluble salts, strong acids, and strong bases.

NaCl, KNO3, HNO3, NaOH

Weak Electrolytes – conduct only a small current (bulb glows dimly). A small degree of ionization in water.

weak acids and weak bases.

HC2H3O2, aq. NH3, tap H2O



Electrolytes

Nonelectrolytes – no current flows (bulb remains unlit). Dissolves but does not produce any ions.

Molecular substances

pure H2O, sugar solution, glycerol



Electrolyte behavior




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04_1529

BaCl2(s)

Ionic Substance

= Ba2+

= Cl

2( )dissolves

When BaCl2 dissolves, the Ba2+ and Cl-

ions are randomly dispersed in the water. BaCl2 is a strong electrolyte.


Acids

Strong acids ‐ dissociate completely (~100 %) to produce H+ in solution

HCl H SO HNO HB HI & HClOHCl, H2SO4, HNO3, HBr, HI, & HClO4

Weak acids ‐ dissociate to a slight extent (~ 1 %) to give H+ in solution

HC2H3O2, HCOOH, HNO2, & H2SO3


04_1530

+

+

+

+ +

= H+

+

+

+ +

+

+

= Cl

+

HCl is completely ionized and is a strong electrolyte.

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Bases

Strong bases ‐ react completely with water to give OH ions. sodium hydroxide

NaOH(s) ‐‐‐> Na+(aq) + OH

‐(aq)

Weak bases ‐ react only slightly with water to give OH ions. ammonia

NH3(aq) + HOH(l) <‐‐‐> NH4+(aq) + OH

‐(aq)


04_1531

+

+

+

+

+

+

+

++

+

+

+

- = OH

= Na+

An aqueous solution of sodium hydroxide which isa strong bases dissociating almost 100 %.


04_1532

Acetic acid(CH3COOH) exists in water mostly asexists in water mostly as undissociatedmolecules. Only a small percent of the molecules are ionized.

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Section 4.3The Composition of Solutions

Chemical Reactions of Solutions

We must know:

The nature of the reaction.

The amounts of chemicals present in the solutions The amounts of chemicals present in the solutions.



Molarity

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

moles of soluteM l itM


moles of solute = Molarity = liters of solution

M

6 moles of HCl3 HCl = 2 liters of solution

M


A 500.0‐g sample of potassium phosphate is dissolved in enough water to make 1.50 L of solution. What is

the molarity of the solution?

EXERCISE!EXERCISE!

y

1.57 M


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Concentration of Ions

For a 0.25 M CaCl2 solution:

CaCl2 → Ca2+ + 2Cl–

Ca2+: 1 × 0.25 M = 0.25 M Ca2+

Cl–: 2 × 0.25 M = 0.50 M Cl–.



Which of the following solutions containsthe greatest number of ions?

CONCEPT CHECK!CONCEPT CHECK!

a) 400.0 mL of 0.10 M NaCl.

b) 300.0 mL of 0.10 M CaCl2.

c) 200.0 mL of 0.10 M FeCl3.

d) 800.0 mL of 0.10 M sucrose.



Let’s Think About It

Where are we going?

To find the solution that contains the greatest number of moles of ions.number of moles of ions.

How do we get there?

Draw molecular level pictures showing each solution. Think about relative numbers of ions.

How many moles of each ion are in each solution?


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Notice

The solution with the greatest number of ions is not necessarily the one in which:

the volume of the solution is the largest.the volume of the solution is the largest.

the formula unit has the greatest number of ions.



Dilution

The process of adding water to a concentrated or stock solution to achieve the molarity desired for a particular solution.solution.

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 0.50 M solution of sodium chloride in an open beaker sits on a lab bench. Which of the following would decrease the concentration of the salt solution?


a) Add water to the solution.

b) Pour some of the solution down the sink drain.

c) Add more sodium chloride to the solution.

d) Let the solution sit out in the open air for a couple of days.

e) At least two of the above would decrease the concentration of the salt solution.


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What is the minimum volume of a 2.00 M NaOH solution needed to make 150.0 mL of a 0.800 MNaOH solution?

EXERCISE!EXERCISE!

60.0 mL


Section 4.4Types of Chemical Reactions

Precipitation Reactions

Acid–Base Reactions

Often called a neutralization reaction Because the acid neutralizes the baseacid neutralizes the base.

Oxidation–Reduction Reactions

Ionic compounds are formed through the transfer of electrons.

The reaction involves the transfer of electrons.


Section 4.5Precipitation Reactions

Precipitation Reaction

A double displacement reaction in which a solid forms and separates from the solution.

When ionic compounds dissolve in water, theWhen ionic compounds dissolve in water, the resulting solution contains the separated ions.

Precipitate – the solid that forms.


If you’re not a part of the solution, your part of the precipitate

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

Ba2+(aq) + CrO42–(aq) → BaCrO4(s)



Precipitation of Silver Chloride





Precipitates

Soluble – solid dissolves in solution; (aq) is used in reaction equation.

Insoluble – solid does not dissolve in solution; (s) is usedInsoluble solid does not dissolve in solution; (s) is used in reaction equation.

Insoluble and slightly soluble are often used interchangeably.


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Simple Rules for Solubility

1. Most nitrate (NO3) salts are soluble.

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

3. Most Cl, Br, and I salts are soluble (except Ag+, Pb2+, Hg22+).

4. Most sulfate salts are soluble (except BaSO4, PbSO4, Hg2SO4, CaSO4).

5. Most OH 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,

except for those containing the cations in Rule 2.



Which of the following ions form compounds with Pb2+ that are generally soluble in water?


a) S2–

b) Cl–

c) NO3–

d) SO42–

e) Na+


Section 4.6Describing Reactions in Solution

Formula Equation (Molecular Equation)

Gives the overall reaction stoichiometry but not necessarily the actual forms of the reactants and products in solution.p

Reactants and products generally shown as compounds.

Use solubility rules to determine which compounds are aqueous and which compounds are solids.

AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)


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Complete Ionic Equation

All substances that are strong electrolytes are represented as ions.

Ag+(aq) + NO (aq) + Na+(aq) + Cl(aq)Ag (aq) + NO3 (aq) + Na (aq) + Cl (aq)

AgCl(s) + Na+(aq) + NO3(aq)



Net Ionic Equation

Includes only those solution components undergoing a change.

Show only components that actually react.

Ag+(aq) + Cl(aq) AgCl(s)

Spectator ions are not included (ions that do not participate directly in the reaction).

Na+ and NO3 are spectator ions.



Write the correct formula equation, complete ionic equation, and net ionic equation for the reaction between cobalt(II) chloride and sodium hydroxide.

Formula Equation:


q

CoCl2(aq) + 2NaOH(aq)

Co(OH)2(s) + 2NaCl(aq)

Complete Ionic Equation:

Co2+(aq) + 2Cl(aq) + 2Na+(aq) + 2OH(aq)

Co(OH)2(s) + 2Na+(aq) + 2Cl(aq)

Net Ionic Equation:

Co2+(aq) + 2OH(aq)

Co(OH)2(s)Copyright © Cengage Learning. All rights reserved 42

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Section 4.7Stoichiometry of Precipitation Reactions

Solving Stoichiometry Problems for Reactions in Solution

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

2 Write the balanced net ionic equation for the2. 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(s), as required.

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



10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume no volume change).

(Part I)CONCEPT CHECK!CONCEPT CHECK!

g

What precipitate will form?

lead(II) phosphate, Pb3(PO4)2

What mass of precipitate will form?

1.1 g Pb3(PO4)2




Where are we going? To find the mass of solid Pb3(PO4)2 formed.

How do we get there?How do we get there? What are the ions present in the combined solution?

What is the balanced net ionic equation for the reaction?

What are the moles of reactants present in the solution?

Which reactant is limiting?

What moles of Pb3(PO4)2 will be formed?

What mass of Pb3(PO4)2 will be formed?


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(Part II)CONCEPT CHECK!CONCEPT CHECK!

g

What is the concentration of nitrate ions left in solution after the reaction is complete?

0.27 M




Where are we going?

To find the concentration of nitrate ions left in solution after the reaction is complete.


What are the moles of nitrate ions present in the combined solution?

What is the total volume of the combined solution?




(Part III)CONCEPT CHECK!CONCEPT CHECK!

g

What is the concentration of phosphate ionsleft in solution after the reaction is complete?

0.011 M


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Where are we going?

To find the concentration of phosphate ions left in solution after the reaction is complete.

Ho do e get there? How do we get there?

What are the moles of phosphate ions present in the solution at the start of the reaction?

How many moles of phosphate ions were used up in the reaction to make the solid Pb3(PO4)2?

How many moles of phosphate ions are left over after the reaction is complete?

What is the total volume of the combined solution?Copyright © Cengage Learning. All rights reserved 49

Section 4.8Acid‐Base Reactions

Acid–Base Reactions (Brønsted–Lowry)

Acid—proton donor

Base—proton acceptor

What is the net ionic equation for the reaction of HCl(aq)What is the net ionic equation for the reaction of HCl(aq) and KOH(aq)?

Acid + Base salt + water

For a strong acid and base reaction:

H+(aq) + OH–(aq) H2O(l)



Neutralization of a Strong Acid by a Strong Base




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Performing Calculations for Acid–Base Reactions

1. List the species 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 moles of reactants.

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 Titrations

Titration – 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 – enough titrant added to react exactly with the analyte.

Endpoint – the indicator changes color so you can tell the equivalence point has been reached.



For the titration of sulfuric acid (H2SO4) with sodium hydroxide (NaOH), how many moles of sodium hydroxide would be required to react with 1.00 L of


y q0.500 M sulfuric acid to reach the endpoint?

1.00 mol NaOH


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Where are we going?

To find the moles of NaOH required for the reaction.


What are the ions present in the combined solution? What is the reaction?

What is the balanced net ionic equation for the reaction?

What are the moles of H+ present in the solution?

How much OH– is required to react with all of the H+ present?


Section 4.9Oxidation‐Reduction Reactions

Redox Reactions

Reactions in which one or more electrons are transferred.



Reaction of Sodium and Chlorine


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Rules for Assigning Oxidation States

1. Oxidation state of an atom in an element = 0

2. Oxidation state of monatomic ion = charge of the ion

3. Oxygen = 2 in covalent compounds (except in peroxides where it = 1)

4. Hydrogen = +1 in covalent compounds

5. Fluorine = 1 in compounds

6. Sum of oxidation states = 0 in compounds

7. Sum of oxidation states = charge of the ion in ions



Find the oxidation states for each of the elements in each of the following compounds:

EXERCISE!EXERCISE!

K2Cr2O7

CO32‐

MnO2

PCl5 SF4


K = +1; Cr = +6; O = –2

C = +4; O = –2

Mn = +4; O = –2

P = +5; Cl = –1

S = +4; F = –1


Redox Characteristics

Transfer of electrons

Transfer may occur to form ions

Oxidation – increase in oxidation state (loss of Oxidation – increase in oxidation state (loss of electrons); reducing agent

Reduction – decrease in oxidation state (gain of electrons); oxidizing agent


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Which of the following are oxidation‐reduction reactions? Identify the oxidizing agent and the reducing agent.


g g

a)Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

b)Cr2O72‐(aq) + 2OH‐(aq) 2CrO4

2‐(aq) + H2O(l)

c)2CuCl(aq) CuCl2(aq) + Cu(s)

Copyright © Cengage Learning. All rights reserved

Section 4.10Balancing Oxidation‐Reduction Equations

Balancing Oxidation–Reduction Reactions by Oxidation States

1. Write the unbalanced equation.

2. Determine the oxidation states of all atoms in the reactants and products.reactants and products.

3. Show electrons gained and lost using “tie lines.”

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

5. Balance the rest of the equation by inspection.

6. Add appropriate states.



Balance the reaction between solid zinc and aqueous hydrochloric acid to produce aqueous zinc(II) chloride and hydrogen gas.y g g


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1. What is the unbalanced equation?

Zn(s) + HCl(aq) Zn2+(aq) + Cl–(aq) + H2(g)



2. What are the oxidation states for each atom?

Zn(s) + HCl(aq) Zn2+(aq) + Cl–(aq) + H2(g)0 +1 –1 +2 –1 0



3. How are electrons gained and lost?

1 e– gained (each atom)


2 e– lost

The oxidation state of chlorine remains unchanged.


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4. What coefficients are needed to equalize the electrons gained and lost?

1 e– gained (each atom) × 2


2 e– lost

Zn(s) + 2HCl(aq) Zn2+(aq) + Cl–(aq) + H2(g)



5. What coefficients are needed to balance the remaining elements?

Zn(s) + 2HCl(aq) Zn2+(aq) + 2Cl–(aq) + H2(g)


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Chapter 4 - Types of Chemical Reactions and Solution Stoichiometry … · · 2016-10-1610/16/2016...

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