Chapter 4 Chemical Quantities and Aqueous Reactions Part2
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Transcript of Chapter 4 Chemical Quantities and Aqueous Reactions Part2
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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
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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)
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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)
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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
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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
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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
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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)
Example - Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide
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Example - Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide
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
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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
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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
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?
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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
InformationGiven: 10.00 mL HCl
12.54 mL of 0.100 M NaOHFind: M HCl
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?
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
InformationGiven: 10.00 mL 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/L
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?
mLHCl
LHCl
mL 1L 0010. HCl liters
HCl molesMolarity
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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
InformationGiven: 10.00 mL 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:
InformationGiven: 10.00 mL HCl
12.54 mL 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?
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.
InformationGiven: 10.00 mL HCl
12.54 mL 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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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)
Example 4.15 - When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves
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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)
Example 4.15 - When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves
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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
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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
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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
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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
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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
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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
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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
+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)
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Practice – Complete the Reactions
• combustion of C3H7OH(l)
• combustion of CH3NH2(g)
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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)