Chapter 4 Chemical Quantities and Aqueous Reactions Part2

Chapter 4Chemical

Quantities and Aqueous Reactions

Part2

2008, Prentice Hall

Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro

Roy KennedyMassachusetts Bay Community College

Wellesley Hills, MA

Tro, Chemistry: A Molecular Approach

2

Ionic Equations• equations which describe the chemicals put into the water

and the product molecules are called molecular equations2 KOH(aq) + Mg(NO3)2(aq) ® 2 KNO3(aq) + Mg(OH)2(s)

• equations which describe the actual dissolved species are called complete ionic equations – aqueous strong electrolytes are written as ions

• soluble salts, strong acids, strong bases– insoluble substances, weak electrolytes, and nonelectrolytes

written in molecule form• solids, liquids, and gases are not dissolved, therefore molecule form

2K+1(aq) + 2OH-1

(aq) + Mg+2(aq) + 2NO3

-1(aq) ® 2K+1

(aq) + 2NO3-1

(aq) + Mg(OH)2(s)


3

Ionic Equations• ions that are both reactants and products are called

spectator ions

2K+1(aq) + 2OH-1

(aq) + Mg+2(aq) + 2NO3

-1(aq) ® 2K+1

(aq) + 2NO3-1

(aq) + Mg(OH)2(s)

• an ionic equation in which the spectator ions are removed is called a net ionic equation

2OH-1(aq) + Mg+2

(aq) ® Mg(OH)2(s)


4

Acid-Base Reactions• also called neutralization reactions because the

acid and base neutralize each other’s properties2 HNO3(aq) + Ca(OH)2(aq) ® Ca(NO3)2(aq) + 2 H2O(l)

• the net ionic equation for an acid-base reaction isH+(aq) + OH(aq) ® H2O(l)

– as long as the salt that forms is soluble in water


5

Acids and Bases in Solution• acids ionize in water to form H+ ions– more precisely, the H from the acid molecule is donated to a

water molecule to form hydronium ion, H3O+

• most chemists use H+ and H3O+ interchangeably

• bases dissociate in water to form OH ions– bases, like NH3, that do not contain OH ions, produce OH by

pulling H off water molecules• in the reaction of an acid with a base, the H+ from the

acid combines with the OH from the base to make water• the cation from the base combines with the anion from

the acid to make the saltacid + base ® salt + water

Common AcidsChemical Name Formula Uses Strength Perchloric Acid HClO4 explosives, catalyst Strong

Nitric Acid HNO3 explosives, fertilizer, dye, glue Strong

Sulfuric Acid H2SO4 explosives, fertilizer, dye, glue,

batteries Strong

Hydrochloric Acid HCl metal cleaning, food prep, ore refining, stomach acid Strong

Phosphoric Acid H3PO4 fertilizer, plastics & rubber,

food preservation Moderate

Chloric Acid HClO3 explosives Moderate

Acetic Acid HC2H3O2 plastics & rubber, food preservation, vinegar Weak

Hydrofluoric Acid HF metal cleaning, glass etching Weak Carbonic Acid H2CO3 soda water Weak

Hypochlorous Acid HClO sanitizer Weak Boric Acid H3BO3 eye wash Weak


7

Common BasesChemical

Name Formula Common Name Uses Strength

sodium hydroxide

NaOH lye,

caustic soda soap, plastic,

petrol refining Strong

potassium hydroxide

KOH caustic potash soap, cotton, electroplating

Strong

calcium hydroxide

Ca(OH)2 slaked lime cement Strong

sodium bicarbonate

NaHCO3 baking soda cooking, antacid Weak

magnesium hydroxide

Mg(OH)2 milk of

magnesia antacid Weak

ammonium hydroxide

NH4OH, {NH3(aq)}

ammonia water

detergent, fertilizer,

explosives, fibers Weak


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


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Example - Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide

1. Write the formulas of the reactantsHNO3(aq) + Ca(OH)2(aq) ®

2. Determine the possible productsa) Determine the ions present when each reactant dissociates

(H+ + NO3-) + (Ca+2 + OH-) ®

b) Exchange the ions, H+1 combines with OH-1 to make H2O(l)(H+ + NO3

-) + (Ca+2 + OH-) ® (Ca+2 + NO3-) + H2O(l)

c) Write the formula of the salt cross the charges

(H+ + NO3-) + (Ca+2 + OH-) ® Ca(NO3)2 + H2O(l)


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3. Determine the solubility of the saltCa(NO3)2 is soluble

4. Write an (s) after the insoluble products and a (aq) after the soluble productsHNO3(aq) + Ca(OH)2(aq) ® Ca(NO3)2(aq) + H2O(l)

5. Balance the equation2 HNO3(aq) + Ca(OH)2(aq) ® Ca(NO3)2(aq) + 2 H2O(l)



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6. Dissociate all aqueous strong electrolytes to get complete ionic equation– not H2O

2 H+(aq) + 2 NO3-(aq) + Ca+2(aq) + 2 OH-(aq) ® Ca+2(aq) +

2 NO3-(aq) + H2O(l)

7. Eliminate spectator ions to get net-ionic equation

2 H+1(aq) + 2 OH-1(aq) ® 2 H2O(l)H+1(aq) + OH-1(aq) ® H2O(l)


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Titration• often in the lab, a solution’s concentration is

determined by reacting it with another material and using stoichiometry – this process is called titration

• in the titration, the unknown solution is added to a known amount of another reactant until the reaction is just completed, at this point, called the endpoint, the reactants are in their stoichiometric ratio– the unknown solution is added slowly from an

instrument called a burette• a long glass tube with precise volume markings that

allows small additions of solution


13

Acid-Base Titrations• the difficulty is determining when there has been just

enough titrant added to complete the reaction– the titrant is the solution in the burette

• in acid-base titrations, because both the reactant and product solutions are colorless, a chemical is added that changes color when the solution undergoes large changes in acidity/alkalinity– the chemical is called an indicator

• at the endpoint of an acid-base titration, the number of moles of H+ equals the number of moles of OH – aka the equivalence point


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Titration

http://wps.prenhall.com/wps/media/objects/165/169763/Acid-BaseTitrationMovie.html


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TitrationThe base solution is thetitrant in the burette.

As the base is added tothe acid, the H+ reacts withthe OH– to form water. But there is still excess acid present so the colordoes not change.

At the titration’s endpoint,just enough base has been added to neutralize all theacid. At this point the indicator changes color.


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Example 4.14:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?

• Write down the given quantity and its units. Given: 10.00 mL HCl

12.54 mL of 0.100 M NaOH


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• Write down the quantity to find, and/or its units.

Find: concentration HCl, M

InformationGiven: 10.00 mL HCl

12.54 mL of 0.100 M NaOH



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• Collect Needed Equations and Conversion Factors:

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) 1 mole HCl = 1 mole NaOH

0.100 M NaOH 0.100 mol NaOH 1 L sol’n


12.54 mL of 0.100 M NaOHFind: M HCl


solution literssolute molesMolarity


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• Write a Concept Plan:

mLNaOH

LNaOH

molNaOH

NaOH L 1NaOH mol 1000.

mL 1L 0010.

molHCl

NaOH mol 1HCl mol 1


12.54 mL of 0.100 M NaOHFind: M HClCF: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L

M = mol/L


mLHCl

LHCl

mL 1L 0010. HCl liters

HCl molesMolarity


20

NaOH mole 1HCl mol 1

L 1NaOH mol 1000

mL 1L 0.001NaOH mL 2.541

.• Apply the Solution Map:

= 1.25 x 10-3 mol HCl


12.54 mL of 0.100 M NaOHFind: M HClCF: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L

M = mol/LCP: mL NaOH → L NaOH →

mol NaOH → mol HCl; mL HCl → L HCl & mol M

Example:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?


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HCl L 010000mL 1

L 0.001NaOH mL 0.001 .

• Apply the Concept Plan:


12.54 mL NaOHFind: M HClCF: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L




M 1250HCl L 0.01000

HCl moles 10 x 1.25Molarity-3

.


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• Check the Solution:HCl solution = 0.125 M The units of the answer, M, are correct.

The magnitude of the answer makes sense since the neutralization takes less HCl solution than

NaOH solution, so the HCl should be more concentrated.


12.54 mL NaOHFind: M HClCF: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L





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Gas Evolving Reactions• Some reactions form a gas directly from the ion

exchangeK2S(aq) + H2SO4(aq) ® K2SO4(aq) + H2S(g)

• Other reactions form a gas by the decomposition of one of the ion exchange products into a gas and water

K2SO3(aq) + H2SO4(aq) ® K2SO4(aq) + H2SO3(aq)H2SO3 ® H2O(l) + SO2(g)


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NaHCO3(aq) + HCl(aq) ® NaCl(aq) + CO2(g) + H2O(l)


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Compounds that UndergoGas Evolving Reactions

ReactantType

ReactingWith

Ion ExchangeProduct

Decom-pose?

GasFormed

Example

metalnS,metal HS

acid H2S no H2S K2S(aq) + 2HCl(aq) ®2KCl(aq) + H2S(g)

metalnCO3,metal HCO3

acid H2CO3 yes CO2 K2CO3(aq) + 2HCl(aq) ®2KCl(aq) + CO2(g) + H2O(l)

metalnSO3

metal HSO3

acid H2SO3 yes SO2 K2SO3(aq) + 2HCl(aq) ®2KCl(aq) + SO2(g) + H2O(l)

(NH4)nanion base NH4OH yes NH3 KOH(aq) + NH4Cl(aq) ®KCl(aq) + NH3(g) + H2O(l)


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Example 4.15 - When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves

1. Write the formulas of the reactantsNa2CO3(aq) + HNO3(aq) ®

2. Determine the possible productsa) Determine the ions present when each reactant dissociates

(Na+1 + CO3-2) + (H+1 + NO3

-1) ®b) Exchange the anions(Na+1 + CO3

-2) + (H+1 + NO3-1) ® (Na+1 + NO3

-1) + (H+1 + CO3-2)

c) Write the formula of compounds cross the charges

Na2CO3(aq) + HNO3(aq) ® NaNO3 + H2CO3


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3. Check to see either product H2S - No4. Check to see if either product decomposes –

Yes– H2CO3 decomposes into CO2(g) + H2O(l)Na2CO3(aq) + HNO3(aq) ® NaNO3 + CO2(g) + H2O(l)



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5. Determine the solubility of other productNaNO3 is soluble

6. Write an (s) after the insoluble products and a (aq) after the soluble products

Na2CO3(aq) + 2 HNO3(aq) ® 2 NaNO3(aq) + CO2(g) + H2O(l)

7. Balance the equationNa2CO3(aq) + 2 HNO3(aq) ® 2 NaNO3 + CO2(g) + H2O(l)



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Other Patterns in Reactions

• the precipitation, acid-base, and gas evolving reactions all involved exchanging the ions in the solution

• other kinds of reactions involve transferring electrons from one atom to another – these are called oxidation-reduction reactions– also known as redox reactions– many involve the reaction of a substance with O2(g)

4 Fe(s) + 3 O2(g) ® 2 Fe2O3(s)


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Combustion as Redox2 H2(g) + O2(g) ® 2 H2O(g)


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Redox without Combustion2 Na(s) + Cl2(g) ® 2 NaCl(s)

2 Na ® 2 Na+ + 2 e

Cl2 + 2 e ® 2 Cl


32

Reactions of Metals with Nonmetals

• consider the following reactions:4 Na(s) + O2(g) → 2 Na2O(s)2 Na(s) + Cl2(g) → 2 NaCl(s)

• the reaction involves a metal reacting with a nonmetal• in addition, both reactions involve the conversion of

free elements into ions 4 Na(s) + O2(g) → 2 Na+

2O– (s)2 Na(s) + Cl2(g) → 2 Na+Cl–(s)


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Oxidation and Reduction• in order to convert a free element into an ion, the

atoms must gain or lose electrons– of course, if one atom loses electrons, another must accept

them• reactions where electrons are transferred from one

atom to another are redox reactions• atoms that lose electrons are being oxidized, atoms

that gain electrons are being reduced

2 Na(s) + Cl2(g) → 2 Na+Cl–(s)Na → Na+ + 1 e– oxidationCl2 + 2 e– → 2 Cl– reduction

Leo

Ger


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Electron Bookkeeping• for reactions that are not metal + nonmetal, or do

not involve O2, we need a method for determining how the electrons are transferred

• chemists assign a number to each element in a reaction called an oxidation state that allows them to determine the electron flow in the reaction– even though they look like them, oxidation states are

not ion charges!• oxidation states are imaginary charges assigned based on

a set of rules • ion charges are real, measurable charges


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Rules for Assigning Oxidation States• rules are in order of priority1. free elements have an oxidation state = 0– Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g)

2. monatomic ions have an oxidation state equal to their charge– Na = +1 and Cl = -1 in NaCl

3. (a) the sum of the oxidation states of all the atoms in a compound is 0– Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0


36

Rules for Assigning Oxidation States3. (b) the sum of the oxidation states of all the atoms in

a polyatomic ion equals the charge on the ion– N = +5 and O = -2 in NO3

–, (+5) + 3(-2) = -1

4. (a) Group I metals have an oxidation state of +1 in all their compounds– Na = +1 in NaCl

5. (b) Group II metals have an oxidation state of +2 in all their compounds– Mg = +2 in MgCl2


37

Rules for Assigning Oxidation States5. in their compounds, nonmetals have oxidation

states according to the table below– nonmetals higher on the table take priority

Nonmetal Oxidation State ExampleF -1 CF4

H +1 CH4

O -2 CO2

Group 7A -1 CCl4

Group 6A -2 CS2

Group 5A -3 NH3


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Practice – Assign an Oxidation State to Each Element in the following

• Br2

• K+

• LiF

• CO2

• SO42-

• Na2O2


39

Practice – Assign an Oxidation State to Each Element in the following

• Br2 Br = 0, (Rule 1)

• K+ K = +1, (Rule 2)• LiF Li = +1, (Rule 4a) & F = -1, (Rule 5)

• CO2 O = -2, (Rule 5) & C = +4, (Rule 3a)

• SO42- O = -2, (Rule 5) & S = +6, (Rule 3b)

• Na2O2 Na = +1, (Rule 4a) & O = -1, (Rule 3a)


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Oxidation and ReductionAnother Definition

• oxidation occurs when an atom’s oxidation state increases during a reaction

• reduction occurs when an atom’s oxidation state decreases during a reaction

CH4 + 2 O2 → CO2 + 2 H2O-4 +1 0 +4 –2 +1 -2

oxidation

reduction

http://wps.prenhall.com/wps/media/objects/165/169763/Oxidation-ReduxII.html


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Oxidation–Reduction• oxidation and reduction must occur simultaneously – if an atom loses electrons another atom must take them

• the reactant that reduces an element in another reactant is called the reducing agent– the reducing agent contains the element that is oxidized

• the reactant that oxidizes an element in another reactant is called the oxidizing agent– the oxidizing agent contains the element that is reduced

2 Na(s) + Cl2(g) → 2 Na+Cl–(s)Na is oxidized, Cl is reduced

Na is the reducing agent, Cl2 is the oxidizing agent


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Identify the Oxidizing and Reducing Agents in Each of the Following

3 H2S + 2 NO3– + 2 H+ ® 3 S + 2 NO + 4 H2O

MnO2 + 4 HBr ® MnBr2 + Br2 + 2 H2O


43

Identify the Oxidizing and Reducing Agents in Each of the Following

3 H2S + 2 NO3– + 2 H+ ® 3 S + 2 NO + 4 H2O

MnO2 + 4 HBr ® MnBr2 + Br2 + 2 H2O

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

ox agred ag

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

oxidationreduction

oxidation

reduction

red agox ag


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Combustion Reactions• Reactions in which O2(g) is a

reactant are called combustion reactions

• Combustion reactions release lots of energy

• Combustion reactions are a subclass of oxidation-reduction reactions

2 C8H18(g) + 25 O2(g) ® 16 CO2(g) + 18 H2O(g)


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Combustion Products• to predict the products of a combustion

reaction, combine each element in the other reactant with oxygen

Reactant Combustion Product

contains C CO2(g)

contains H H2O(g)

contains S SO2(g)

contains N NO(g) or NO2(g)

contains metal M2On(s)


46

Practice – Complete the Reactions

• combustion of C3H7OH(l)

• combustion of CH3NH2(g)


47

Practice – Complete the Reactions

C3H7OH(l) + 5 O2(g) ® 3 CO2(g) + 4 H2O(g)

CH3NH2(g) + 3 O2(g) ® CO2(g) + 2 H2O(g) + NO2(g)

Documents

Chapter 4 Chemical Quantities and Aqueous Reactions Part2