Ch 4 Types of Chemical Reactions and Solution Stoichiometry Student.ppt

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Chapter 4

Types of Chemical Reactions and Solution

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Chapter 4

Table of Contents

4.1 Water, the Common Solvent

4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes

4.3 The Composition of Solutions 4.4 Types of Chemical Reactions 4.5 Precipitation Reactions

4.6 Describing Reactions in Solution

4.7 Stoichiometry of Precipitation Reactions 4.8 Acid–Base Reactions

4.9 Oxidation–Reduction Reactions

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Section 4.1

Water, the Common Solvent

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•

One of the most important substances on

Earth.

•

Can dissolve many different substances.

(4)

Section 4.1

•

A polar molecule because

of its unequal charge

distribution.

– More negative charge on

the oxygen

– More positive charge on

the hydrogen

– Positive pole and a

negative pole

(5)

Section 4.1

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Dissolution of an Ionic Solid in a Liquid

Water molecules “pull” ionic solids apart

Negative end (oxygen) surrounds (hydrates) cations Positive end (hydrogens) surrounds (hydrates) anions

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Section 4.1

Dissolution of an Non Ionic Solid in a Liquid

Polar water molecules interact with polar bonds in non ionic compounds

(7)

Section 4.2

The Nature of Aqueous Solutions: Strong and Weak Electrolytes

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•

Solution = homogeneous mixture

– Same throughout

•

First cup of coffee from a pot is the same as

the last cup

– Composition of a solution can be varied by

varying the amount of the dissolved

substance

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Section 4.2

•

Solutions

– Solute – substance being dissolved. – Solvent – liquid water.

• _{One useful property for characterizing solutions is} electrical conductivity

– Whether or not the solution can conduct an electrical current

(9)

Section 4.2

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Copyright © Cengage Learning. All rights reserved 9 • Strong Electrolytes – conduct current very efficiently (bulb shines brightly). • Weak Electrolytes – conduct only a small current (bulb glows dimly).

• Nonelectrolytes – no current flows (bulb remains unlit).

Electrolytes

Fig 4.4

Electrolyte – substance that when dissolved in water produces a

solution that can conduct electricity.

Place a light bulb between two electrodes immersed in an

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Section 4.2

•

Basis for conductivity properties of solutions

postulated by Arrhenius

•

Proposed that conductivity arose from the

presence of ions

– Solutions of strong electrolytes contain lots of

ions

– Solutions of weak electrolytes contain very few

ions

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Section 4.2

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Strong Electrolytes

• Dissociate 100% in water, producing positive and negative ions

• In solution NaCl comes apart into individual Na+ and

Cl – ions

– Pulled apart by the water molecules

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Section 4.2

• Dissociate 100% in water, producing positive and negative ions

• In equations show the formation of 100% ions in aqueous (aq) solutions. Arrow only to the right!

• Soluble salts

• Strong acids

• Strong bases

NaCl (s) H2O l  

Na+(aq) + Cl-(aq)

HCl (s) H2O l  

H+(aq) + Cl-(aq)

NaOH (s) H2O l  

Na+(aq) + OH-(aq) H2SO4 (s)

H₂O l 

 

(13)

Section 4.2

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• Arrhenius also helped to define the nature of acids

• An Arrhenius acid is a substance that produces H+ ions

(protons) when it is dissolved in water.

• _{HCl, H}₂_SO₄

– Water has an essential role in this process

• _{HA (aq) +H}₂_{O (l)}_→__H₃_O+ (aq) +A – (aq)

• An Arrhenius base is a substance that produces OH –

ions when dissolved in water

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Section 4.2

Learning Check

Complete each for strong electrolytes in water.

1.

2.

3.

CaCl₂ (s) H2O l

        

K₃PO₄ (s) H2O l

        

HNO₃ (s) H2O l

        

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Section 4.2

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Weak Electrolytes

• dissociates only slightly in water

• in water forms a solution of a few ions and mostly undissociated molecules

• Larger arrow pointing to the left indicates that this reaction does not proceed very far

• Most of the acid stays in the form of HF

– We will see this later

• in equations shows equilibrium between undissociated form and ions in aqueous (aq) solutions. Arrows go both ways!

15

HF (s)

_H_₂O

 

H

+

(aq) + F

-

(aq)

HF (s)

dissociation recombination



(16)

Section 4.2

Nonelectrolytes

• _{dissolve as molecules in water}

• Does not dissociate

• DO NOT produce ions in water

• do not conduct an electric current

C₁₂H₂₂O₁₁(s) CH₂O(l) ₁₂H₂₂O₁₁(aq) sucrose solution of sucrose

Sucrose s( ) H2O

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Section 4.2

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Classification of Solutes in Aqueous Solutions

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Section 4.3

The Composition of Solutions

•

We must know:

 The nature of the reaction.

 The amounts of chemicals present in the solutions.

 Concentration



There are lots of ways to express

concentration

 g/L

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Section 4.3

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•

Molarity (

M

) = moles of solute per

volume of solution in liters:

Molarity

M

= Molarity = moles of solute

liters of solution

6 moles of HCl

3 HCl =

2 liters of solution

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Section 4.3

Example

A 250.0-g sample of potassium phosphate

is dissolved in enough water to make 1.50 L

of solution. What is the

molarity

of the

solution?

250.0 g K₃PO₄

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Section 4.3

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Learning Check

Consider a solution made by dissolving 200.0

g of KOH in a total 350.0 mL of solution.

(22)

Section 4.3

• When we dissolve an ionic compound the molarity is not as straightforward as it seems • 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–.

 This is especially important for applications like electrochemistry or electrophysiology

where the ion concentration is what conducts the electrical current

(23)

Section 4.3

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• What is the concentration of Na+ ions in a 0.75 M solution of

Na₂CO₃?

Determining Concentration of Ions

0.75 moles Na₂CO₃

L x

2 moles Na+

mole Na₂CO₃ =1.5 M Na

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Section 4.3

• What is the concentration of Cl- ions in a 0.30 M solution of FeCl 2?

(25)

Section 4.3

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How many moles of K+ ions in 25.0 mL a 0.475 M solution of K

2CO3?

Determining Moles of Ions

25.0 mL x 1L_{1000 mLx}0.475 moles K2CO3

L x 2 moles K

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Section 4.3

How many moles of OH - ions in 125 mL a 0.255 M solution of Ca(OH) 2?

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Section 4.3

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•

The solution with the greatest number of

ions is not necessarily the one in which:



the volume of the solution is the

largest.



the formula unit has the greatest

number of ions.

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Section 4.3

Concept Check

Which of the following solutions contains

the

greatest

number of ions?

(29)

Section 4.3

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Determining Mass to make a certain Molarity

So far we have done two types of Molarity calculations

1.You are given moles and volume and told to calculate molarity 2.You are given mass and volume and told to calculate molarity

These are rarely the types of calculation you actually do in the lab.

A more typical calculation is where you are told to make a certain amount of a solution of a certain molarity. So you need to

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Section 4.3

•

A solution whose concentration is

accurately known.

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Section 4.3

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Determining Mass to make a certain Molarity

How many grams of Na₃PO₄ do you need to weigh out to make 500.0 mL of a 0.250M solution of Na₃PO₄? This is pretty typical.

500.0 mL x 1 L

1000 mL x 0.250 molL x163.97 g Na3 PO₄

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Section 4.3

Learning Check

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Section 4.3

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•

The process of adding water to a

concentrated or

stock solution to achieve

the molarity desired for a particular

solution.

•

Dilution with water does not alter the

numbers of moles of solute present.

•

Moles of solute before dilution = moles of

solute after dilution

M

₁

V

₁

=

M

₂

V

₂

(34)

Section 4.3

Exercise

What is the

volume

of a 2.00

M

NaOH

solution needed to make 150.0 mL of a 0.800

M

NaOH solution?

M₁V₁ = M₂V₂

M₁ = 2.00M V₁ = ?

M₂= 0.800 M



V₁ = M2V2

M₁

V₁ = 0.800M150.0 mL 2.00 M

(35)

Section 4.3

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Learning Check

What is the

volume

of a 12.0

M

HCl solution

needed to make 750.0 mL of a 1.00

M

HCl

solution?

(36)

Section 4.4

Types of Chemical Reactions

•

Classification of Chemical Reactions

– Precipitation Reactions

– Acid–Base Reactions

(37)

Section 4.5

Precipitation Reactions

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•

A reaction that occurs when two

solutions are mixed and a solid product

is formed



Precipitate – the solid that forms.

(38)

Section 4.5

• Aqueous solution of potassium chromate, K₂CrO₄(aq)

– soluble

• Aqueous solution of barium nitrate, Ba(NO₃)₂(aq)

– soluble

• When these solutions are

mixed, a yellow (insoluble) solid forms.

(39)

Section 4.5

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• To write this equation it is critical to remember that in

solution ionic compounds don’t exist as formula units they exist as ions

.

• K₂CrO₄(aq) is 2 K+ and CrO

4 2 –

• Ba(NO₃)₂(aq) is Ba2+ and 2NO 32 –

• 2K+ (aq) + CrO

42 – (aq) + Ba2+ (aq) 2NO32 – (aq) → products

(40)

Section 4.5

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Section 4.5

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• 2 K+ (aq) + CrO

42 – (aq) + Ba2+ (aq) 2NO3– (aq) → products

• How do we predict the products of this reaction

• First write the formula for the opposite combination of ions • K+ with NO

3– and Ba2+ with CrO42 –

• 2 possible products • KNO₃ and BaCrO₄

• Have a look at solubility rules (you DO NOT need to memorize these)

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Section 4.5

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+, Hg 22+).

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–, CO

32–, CrO42–, PO43– salts are only slightly soluble,

except for those containing the cations in Rule 2.

(43)

Section 4.5

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• KNO₃ is soluble (rule 2)

• BaCrO₄ is insoluble (rule 6)

• So the complete equation for the reaction looks like this

• K₂CrO₄(aq) +Ba(NO₃)₂(aq)→BaCrO₄(s) +2KNO₃(aq)

• * A precipitation reaction can also be called a double displacement reaction

.

•

(44)

Section 4.5

Precipitation Reactions

Concept Check

Use the solubility rules to predict the products

when aqueous solutions of silver nitrate

(45)

Section 4.5

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Concept Check

Which of the following ions form compounds

with Pb

2+

that are

soluble

in water?

a)

S

2–

b

Cl

–

c)

NO

₃–

d)

SO

₄2–

(46)

Section 4.6

Describing Reactions in Solution

•

Gives the overall reaction stoichiometry

but not necessarily the actual forms of

the reactants and products in solution.

•

Reactants and products generally shown

as compounds.

•

Use solubility rules to determine which

compounds are aqueous and which

compounds are solids.

(47)

Section 4.6

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•

Represents as ions all reactants and

products that are strong electrolytes.

Ag+(aq) + NO

3–(aq) + Na+(aq) + Cl–(aq) →

AgCl(s) + Na+(aq) +

NO₃–(aq)

(48)

Section 4.6

•

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 NO

3– are spectator ions.

(49)

Section 4.6

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Formula Equation:

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

Complete Ionic Equation: Ag+(aq) + NO

3–(aq) + Na+(aq) + Cl–(aq) → AgCl(s) + Na+(aq) + NO3–(aq)

Net Ionic Equation:

(50)

Section 4.5

•

Net Ionic Equation for the first example

•

Ba

2+

(

aq

) + CrO

42–

(

aq

) → BaCrO

4

(

s

)

(51)

Section 4.6

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Concept Check

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

(52)

Section 4.7

Stoichiometry of Precipitation Reactions

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

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

3. Calculate the moles of reactants.

4. Determine which reactant is limiting.

(53)

Section 4.7

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Example

Calculate the mass of solid NaCl that must be

added to 1.50 L of a 0.100 M AgNO

₃

solution to

precipitate all the Ag

+

ions in the form of AgCl.

1.50 L x 0.100 mol Ag+ ions

L = 0.150 mol Ag+ ions

0.150 mol Cl_{ions x 1mol NaCl}

(54)

Section 4.7

Concept Check (There are two parts to this)

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).



What

precipitate

will form?

(55)

Section 4.7

Return to TOC

•

Where are we going?

 To find the mass of solid Pb₃(PO₄)₂ formed.

•

How do we get there?

 What are the ions present in the combined solution?

 What is the balanced 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?

(56)

Section 4.7

Concept Check (Part I)

10.0 mL of a 0.30

M

sodium phosphate solution

M

lead(II) nitrate

(57)

Section 4.7

Return to TOC

Concept Check (Part II)

10.0 mL of a 0.30

M

sodium phosphate solution

M

lead(II) nitrate



What mass of precipitate will form?

(58)

Section 4.8

Acid–Base Reactions

•

At the beginning of the chapter we saw

the Arrhenius definition of an acid and

base

– Acid (H+ donor) and Base (OH

–

donor)

•

The Br

Ø

nsted Lowry definition is more

general

– Acid - proton donor

– Base - proton acceptor

(59)

Section 4.8

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•

What are the products of a reaction between

a strong acid and a strong base*?

•

NaOH (s) + HCl (l) → NaCl (aq) + H

₂

O(l)

•

Water and salt

•

The salt stays in solution because it is soluble

•

Water forms because large quantities of the

OH

–

and H

+

ions will not stay in solution

– We know this because pure water is a non

electrolyte

(60)

Section 4.8

• Net ionic equation for the reaction of a strong acid and strong base

• Molecular Equation

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

• Complete ionic equation

Na+ (aq) + OH– (aq) + H+ (aq) + Cl– (aq) → Na+ (aq) + Cl – (aq) + H

2O (l)

• Net ionic equation

H+ _{(aq) + OH}− _{(aq) → H}

2O(l)

(61)

Section 4.8

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•

What are the products of a reaction

between a weak acid and a strong base?

• KOH (s) + HC₂H₃O₂ (l) →?

•

This one is a little trickier

– HC₂H₃O₂is a weak acid (dissociates very little). – Three species in solution are the K+ ion OH – ion

and undissociated acid (HC₂H₃O₂)

(62)

Section 4.8

• Net ionic equation for the reaction of a strong acid and strong base

• Molecular Equation

– KOH (s) + HC₂H₃O₂ (l) → KC₂H₃O₂ (aq) + H₂O(l)

• Complete ionic equation – K+_{(aq) + OH}–_{(aq) + HC}

2H3O2 (l) → K+ (aq) + C2H3O2l– (aq) + H2O (l)

– The primary reaction here is between the OH – from the KOH and the H from the HC₂H₃O₂

– Net ionic equation

(63)

Section 4.8

Acid–Base Reactions

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solution before any reaction occurs, and decide what reaction will occur.

2. Write the balanced molecular 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.

(64)

Section 4.7

Example

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

Molecular equation NaOH (s) + HCl (l) → NaCl (aq) + H₂O(l)

Calculate moles of reactant

Determine volume of HCl to neutralize NaOH.

25.0 mL x 1 L

1000 mL x 0.350 mol NaOHL = 0.00875 mol NaOH

(65)

Section 4.7

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Concept Check

What volume of a 0.200 M H

₂

SO

₄

solution is

(66)

Section 4.7

Example

When 25.0 mL of 0.250 M HNO

₃

and 55.0 mL of

0.350 M KOH are mixed

How much water is formed?

What is the concentration of H+ or OH – ions in excess after

(67)

Section 4.7

Return to TOC

Example

When 25.0 mL of 0.250 M HNO

₃

and 55.0 mL of

0.350 M KOH are mixed, how much water is formed?

Molecular equation KOH (s) + HNO₃ (l) → KNO₃(aq) + H₂O(l)

H+ is the limiting reactant and 0.00625 mol of H

2O forms

25.0 mL x 1 L

1000 mL x 0.250 mol HNOL 3 x 1 mol H

+

mol HNO3

= 0.00625 mol H+

55.0 mL x 1 L

1000 mL x 0.350 mol KOHL x1 mol OH



mol KOH = 0.0193 mol OH

(68)

Section 4.7

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Example

When 25.0 mL of 0.250 M HNO

₃

and 55.0 mL of

0.350 M KOH are mixed

How much water is formed? 0.00625 mol of H₂O forms

What is the concentration of H+ or OH – ions in excess after

the reaction goes to completion?

H+ was the limiting reactant so there is none left

(69)

Section 4.8

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•

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.

(70)

Section 4.8

•

Calculations – we have already done some

calculations of this type

– Neutralization reaction is what you get in

an acid – base titration

•

What volume to neutralize

– Lets look at some other types of

calculations that are seen with titrations

(71)

Section 4.8

Return to TOC

Concept Check

For the titration of sulfuric acid (H

₂

SO

₄

) with

sodium hydroxide (NaOH), how many

moles

of sodium hydroxide

would be required to

react with 1.00 L of 0.500

M

sulfuric acid to

reach the endpoint?

Need a balanced equation

(72)

Section 4.7

Example – Standardize a NaOH solution

• Weighs out a 1.3009-g sample of the weak acid potassium hydrogen phthalate KHC₈H₄O₄ (KHP).

• Dissolve the KHP in distilled water, adds phenolphthalein as an indicator, and titrates the resulting solution with the unknown

sodium hydroxide solution.

• 41.20 mL of the sodium hydroxide solution is required to react exactly with the 1.3009 g KHP.

(73)

Section 4.7

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Example – Standardize a NaOH solution What do we need to solve this?

Molecular equation – this is a weak acid like HC₂H₃O₂

KOH (s) + KHC₈H₄O₄ (l) → KC₂H₃O₂ (aq) + H₂O(l)

Moles of acid (KHC₈H₄O₄)

Use moles of acid to get to molarity of base (NaOH)

1.3009 g KHC8H4O4 x 1 mol KHC_{204.22 g KHC}8H4O4 8H4O4

= 6.3701x10-3_{mol KHC}

8H4O4

molarity of NaOH = 6.3701x10-3mol NaOH

(74)

Section 4.9

Oxidation–Reduction Reactions

•

Reactions in which one or more electrons

are transferred.

(75)

Section 4.9

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• Way to keep track of movement of electrons in chemical reactions

• Oxidation state of an atom in its elemental form is always 0.

• Na (s), O₂(g), O₃(g), Hg (l)

• For monatomic ions the oxidation state is the same as the charge

• Na+ = +1, Ca2+ = +2, O2 – = – 2

• Oxidation numbers do not refer to real charges on the atoms, except in the case of actual ionic substances. • For covalent compounds the electrons are shared and

(76)

Section 4.9

• In covalent compounds

– If the atoms are identical the electrons are shared equally (no charge on either atom) – If the atoms are different the shared

electrons are assigned to the atom with the stronger attraction for electrons

(77)

Section 4.9

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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

(78)

Section 4.9

(79)

Section 4.9

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Concept Check

Find the

oxidation states

for each of the

elements in each of the following

compounds:

• K₂Cr₂O₇ • CO₃

(80)

Section 4.9

•

Why do we need to assign oxidation

states at all ?

– To keep track of transfer of electrons

in reactions

•

Transfer may occur to form ions

– Ionic compounds

•

Transfer maybe be more subtle

– Covalent compounds

(81)

Section 4.9

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Transfer of Electrons

Na loses an electron, Cl

₂

gains an electron

C changes its oxidation state from – 4 to +4

(82)

Section 4.9

• Oxidation

–

Increase in oxidation state – Involves loss of electrons

– The substance that is oxidized is the reducing agent

• Reduction

–

Decrease in oxidation state – Involves gain of electrons

(83)

Section 4.9

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Redox Characteristics

Na loses an electron

Increase in oxidation state Is oxidized

Is reducing agent

Cl₂gains an electron

Decrease in oxidation state Is reduced

(84)

Section 4.9

Redox Characteristics

Carbon in methane loses electrons Carbon is oxidized

Methane (CH₄) is the reducing agent (entire compound)

Oxygen gains electrons Oxygen is reduced

(85)

Section 4.9

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Concept Check

For the following oxidation-reduction reactions

Identify which atom is oxidized and which is reduced. Identify the oxidizing agent and the reducing agent.

a) Zn (s) + 2 HCl (aq) → ZnCl₂(aq) + H₂(g) b) 2 CuCl (aq) → CuCl₂(aq) + Cu (s)

(86)

Section 4.10

Balancing Oxidation–Reduction Equations

1. Write the unbalanced equation.

2. Determine the oxidation states of all atoms in

the 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.

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Section 4.10

Balancing Oxidation–Reduction Equations

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1. By oxidation states

2. Method of half reactions (Ch 18)

Why do we have to do either of these?

Normally when we balance an equation we are only concerned about the atoms

In a redox reaction we also have to be concerned with the movement of electrons.

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Section 4.10

Balancing Oxidation–Reduction Equations Unbalanced equation

Oxidation states

Electrons gained and lost

Balance for electrons

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Section 4.9

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Concept Check

• Balance the reaction between solid zinc and aqueous