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Chemistry 1AChapter 5
Water, H2O
Water Attractions
Liquid Water
Solutions
• A solution, also called a homogeneous mixture, is a mixture whose particles are so evenly distributed that the relative concentrations of the components are the same throughout.
• Water solutions are called aqueous solutions.
Solution (Homogeneous Mixture)
Solute and Solvent
• In solutions of solids dissolved in liquids, we call the solid the soluteand the liquid the solvent.
• In solutions of gases in liquids, we call the gas the solute and the liquid the solvent.
• In other solutions, we call the minor component the solute and the major component the solvent.
Solution of an Ionic Compound
Solution of an Ionic Compound (cont.)
Liquid-Liquid Solution
Precipitation Reactions
• In a precipitation reaction, one product is insoluble in water.
• As that product forms, it emerges, or precipitates, from the solution as a solid.
• The solid is called a precipitate. • For example, Ca(NO3)2(aq) + Na2CO3(aq)
→ CaCO3(s) + 2NaNO3(aq)
Precipitation Questions
• Describe the solution formed at the instant water solutions of two ionic compounds are mixed (before the reaction takes place).
• Describe the reaction that takes place in this mixture.
• Describe the final mixture. • Write the complete equation for
the reaction.
Solution of Ca(NO3)2
Solution of Ca(NO3)2 and Na2CO3 at the time of mixing, before the reaction
Product Mixture for the reaction of Ca(NO3)2 and Na2CO3
Complete Ionic Equation
Spectator Ions
• Ions that are important for delivering other ions into solution but that are not actively involved in the reaction are called spectator ions.
• Spectator ions can be recognized because they are separate and surrounded by water molecules both before and after the reaction.
Net Ionic Equations
• An equation written without spectator ions is called a net ionic equation.
Ca2+(aq) + CO32−(aq) → CaCO3(s)
Writing Precipitation Equations
• Step 1: Determine the formulas for the possible products using the general double-displacement equation.
AB + CD → AD + CB• Step 2: Predict whether either of the
possible products is water insoluble. If either possible product is insoluble, a precipitation reaction takes place, and you may continue with step 3. If neither is insoluble, write “No reaction”.
Water Solubility• Ionic compounds with the following ions are
soluble. – NH4
+, group 1 metal ions, NO3−, and C2H3O2
−
• Ionic compounds with the following ions are usually soluble.– Cl−, Br−, I− except with Ag+ and Pb2+
– SO42− except with Ba2+ and Pb2+
• Ionic compounds with the following ions are insoluble.– CO3
2−, PO43−, and OH− except with NH4
+ and group 1 metal cations
– S2− except with NH4+ and group 1 and 2 metal
cations
Writing Precipitation Equations (cont)
• Step 3: Follow these steps to write the complete equation.– Write the formulas for the reactants
separated by a “+”. – Separate the formulas for the reactants and
products with a single arrow. – Write the formulas for the products separated
by a “+”. – Write the physical state for each formula.
• The insoluble product will be followed by (s).• Water-soluble ionic compounds will be followed by
(aq).– Balance the equation.
Writing Precipitation Equations (cont)
• Write the complete ionic equation.– Describe aqueous ionic compounds
as ions.– Describe the solid with a complete
formula. – Make sure that you have correctly
done the following• Balanced the equation• Included charges on ions• Included states
Writing Precipitation Equations (cont)
• Write the net ionic equation.– Eliminate ions that are in the
same form on each side of the complete ionic equation.
– Rewrite what’s left and balance.
Skills to Master (1)
• Convert between names and symbols for the common elements.
• Identify whether an element is a metal or a nonmetal.
• Determine the charges on many of the monatomic ions.
• Convert between the name and formula for polyatomic ions.
Skills to Master (2)
• Convert between the name and formula for ionic compounds.
• Balance chemical equations. • Predict the products of double
displacement reactions. • Predict ionic solubility.
Arrhenius Acid Definition
• An acid is a substance that generates hydronium ions, H3O+
(often described as H+), when added to water.
• An acidic solution is a solution with a significant concentration of H3O+ ions.
Characteristics of Acids
• Acids have a sour taste. • Acids turn litmus from blue to red.• Acids react with bases.
Strong and Weak Acids
• Strong Acid = due to a completion reaction with water, generates close to one H3O+ for each acid molecule added to water.
• Weak Acid = due to a reversible reaction with water, generates significantly less than one H3O+ for each molecule of acid added to water.
Strong Acid and Water
When HCl dissolves in water, hydronium ions, H3O+, and chloride ions, Cl−, ions form.
Solution of a Strong Acid
Acetic Acid
Weak Acid and Water
Acetic acid reacts with water in a reversible reaction, which forms hydronium and acetate ions.
Solution of Weak Acid
Strong and
Weak Acids
Recognizing Acids
• From name– (something) acid– (metal) hydrogen sulfate– (metal) dihydrogen phosphate
• From formula– HX(aq), HaXbOc, HC2H3O2, RCO2H
(RCOOH)– (symbol for metal)HSO4
– (symbol for metal)H2PO4
Strong and Weak Acids
• Strong Acids– Monoprotic – HCl(aq), HBr(aq),
HI(aq), HNO3, HClO4
– H2SO4
• Weak Acid– The rest
Arrhenius Base Definitions
• A base is a substance that generates OH− when added to water.
• A basic solution is a solution with a significant concentration of OH−
ions.
Characteristics of Bases
• Bases have a bitter taste. • Bases feel slippery on your fingers. • Bases turn litmus from red to blue.• Bases react with acids.
Strong Bases
• Anions in ionic compounds except– Neutral anions - Cl−, Br−, I−, NO3
−, ClO4
−
– Acidic anions – HSO4−, H2PO4
−
• Uncharged, molecular bases –NH3
Strong Bases• Strong Base = due to a completion reaction with
water, generates close to one (or more) OH− for each formula unit of base added to water.
– Metal hydroxides (e.g. NaOH)NaOH(s) → Na+(aq) + OH −(aq)
– Metal hydrides (e.g. LiH) LiH(s) → Li+(aq) + H−(aq)
H−(aq) + H2O(l) → H2(aq) + OH−(aq)
– Metal amides (e.g. NaNH2) NaNH2(s) → Na+(aq) + NH2
−(aq)
NH2−(aq) + H2O(l) → NH3(aq) + OH−(aq)
Weak Base
• Weak Base = due to a reversible reaction with water, generates significantly less than one OH− for each formula unit of base added to water.
– All other bases
Ammonia and WaterAmmonia reacts with water in a reversible reaction, which forms ammonium and hydroxide ions.
Ammonia Solution
pH
• Acidic solutions have pH values less than 7, and the more acidic the solution is, the lower its pH.
• Basic solutions have pH values greater than 7, and the more basic the solution is, the higher its pH.
pH Range
Neutralization Reactions
• Reactions between Arrhenius acids and Arrhenius bases are called neutralization reactions.
HNO3(aq) + NaOH(aq)
→ H2O(l) + NaNO3(aq)
Aqueous Nitric Acid
Mixture of HNO3 and NaOH Before Reaction.
Strong Acid and Strong Base Reaction
The hydronium ion, H3O+, from the strong acid reacts with the hydroxide ion, OH−, from the strong base to form water, H2O.
Mixture of HNO3 and NaOH After the Reaction
Electrolytes• Strong electrolytes ionize (strong
acids) or dissociate (water-soluble ionic compounds) completely when added to water, causing the water to conduct electric currents strongly.
• Weak electrolytes ionize (weak acids and ammonia) incompletely when added to water, causing the water to conduct electric currents weakly.
• Nonelectrolytes (such as alcohols and sugars) do not form ions in solution, and therefore, do not cause water to conduct electric currents.
Steps to Neutralization Equations
• Do you have an acid and a base?• If you have a strong acid or a strong
base or if both are strong, write a single arrow. If both are weak, write a double arrow.
• Write the formulas and states for the products. – If the base is ammonia,
NH3(aq) + HX(aq) → NH4X(aq)– Otherwise,
AB + CD → AD + CB• Balance the complete equation.
Steps to Neutralization Equations (2)
• Write the complete ionic equation.– Describe strong electrolytes as ions.– Describe everything else with a
complete formula. – Make sure that you have correctly
done the following• Balanced the equation• Included charges on ions• Included states
Steps to Neutralization Equations (3)
• Write the net ionic equation.– Eliminate ions that are in the same
form on each side of the complete ionic equation.
– Rewrite what’s left and balance.
Steps to Neutralization Equations (4)
• Check to be sure the following is true.– Strong acids described as H+. – Weak acids described with a complete
formula. – H2SO4 described as H+ and HSO4
−. – Pure (s), (l), or (g) described with
complete formula. – Ions in strong electrolytes on both
sides of the equation are eliminated.
Reaction between an Acid and a Hydroxide Base.
• The reaction has the double displacement form.
AB + CD → AD + CB– The positive part of the acid is H+.
• The hydroxide base can be soluble or insoluble.
• The products are water and a water-soluble ionic compound.
Reaction between an Acid and a Carbonate Base.
• The reaction has the double displacement form.
AB + CD → AD + CB– The positive part of the acid is H+.
• The products are water, carbon dioxide, and a water-soluble ionic compound. The H2O and the CO2come from the decomposition of the initial product H2CO3.
NH3(aq) + HF(aq) NH4+(aq) + F−(aq)
base acidH2O(l) + HF(aq) H3O+(aq) + F−(aq) neutral acidNH3(aq) + H2O(l) NH4
+(aq) + OH−(aq) base neutralH2PO4
−(aq) + HF(aq) H3PO4(aq) + F−(aq) acid acid
Arrhenius Acid-Base Reactions?
Acid and Base Definitions
• Acid– Arrhenius: a substance that generates
H3O+ in water– Brønsted-Lowry: a proton, H+, donor
• Base– Arrhenius: a substance that generates OH-
in water– Brønsted-Lowry: a proton, H+, acceptor
• Acid-Base Reaction– Arrhenius: between an Arrhenius acid and
base– Brønsted-Lowry: a proton (H+) transfer
Brønsted-Lowry Acids and Bases
NH3(aq) + HF(aq) NH4+(aq) + F−(aq)
base acidH2O(l) + HF(aq) H3O+(aq) + F−(aq) base acidNH3(aq) + H2O(l) NH4
+(aq) + OH−(aq) base acidH2PO4
−(aq) + HF(aq) H3PO4(aq) + F−(aq) base acid
Why Two Definitions for Acids and Bases? (1)
• Positive Aspects of Arrhenius Definitions– All isolated substances can be classified as
acids (generate H3O+ in water), bases (generate OH- in water), or neither.
– Allows predictions, including (1) whether substances will react with a base or acid, (2) whether the pH of a solution of the substance will be less than 7 or greater than 7, and (3) whether a solution of the substance will be sour.
• Negative Aspects of Arrhenius Definitions– Does not include similar reactions (H+ transfer
reactions) as acid-base reactions.
Why Two Definitions for Acids and Bases? (2)
• Positive Aspects of Brønsted-LowryDefinitions– Includes similar reactions (H+ transfer reactions) as
acid-base reactions.
• Negative Aspects of Brønsted-LowryDefinitions– Cannot classify isolated substances as acids (generate
H3O+ in water), bases (generate OH− in water), or neither. The same substance can sometimes be an acid and sometimes a base.
– Does not allow predictions of (1) whether substances will react with a base or acid, (2) whether the pH of a solution of the substance will be less than 7 or greater than 7, and (3) whether a solution of the substance will be sour.
Conjugate Acid-Base Pairs
NH3(aq) + HF(aq) NH4+(aq) + F−(aq)
base acid acid baseH2O(l) + HF(aq) H3O+(aq) + F−(aq) base acid acid baseNH3(aq) + H2O(l) NH4
+(aq) + OH−(aq) base acid acid baseH2PO4
−(aq) + HF(aq) H3PO4(aq) + F−(aq)base acid acid base
Brønsted-Lowry Acids and Bases
Amphoteric Substances
HCO3−(aq) + HF(aq) CO2(g) + H2O(l) + F−(aq)
base acidHCO3
−(aq) + OH−(aq) CO32−(aq) + H2O(l)
acid baseH2PO4
−(aq) + HF(aq) H3PO4(aq) + F−(aq) base acidH2PO4
−(aq) + 2OH−(aq) → PO43−(aq) + 2H2O(l)
acid base
Can be a Brønsted-Lowry acid in one reaction and a Brønsted-Lowry base in another?
Oxidation
• Historically, oxidation meant reacting with oxygen.
2Zn(s) + O2(g) → 2ZnO(s)
Zn → Zn2+ + 2e−
or 2Zn → 2Zn2+ + 4e−
O + 2e− → O2−
or O2 + 4e− → 2O2−
Oxidation Redefined (1)
• Many reactions that are similar to the reaction between zinc and oxygen were not considered oxidation.
• For example, both the zinc-oxygen reaction and the reaction between sodium metal and chlorine gas (described on the next slide) involve the transfer of electrons.
Oxidation and Formation of Binary Ionic Compounds
Similar to Oxidation of Zinc
2Na(s) + Cl2(g) → 2NaCl(s)
Na → Na+ + e−
or 2Na → 2Na+ + 2e−
Cl + e− → Cl−
or Cl2 + 2e− → 2Cl−
Oxidation = Loss of Electrons
Oxidation Redefined (2)
• To include the similar reactions in the same category, oxidationwas redefined as any chemical change in which at least one element loses electrons.
Zinc Oxide Reduction• The following equation describes one of the
steps in the production of metallic zinc.ZnO(s) + C(g) → Zn(s) + CO(g)
• Because zinc is reducing the number of bonds to oxygen atoms, historically, zinc was said to be reduced.
• When we analyze the changes taking place, we see that zinc ions are gaining two electrons to form zinc atoms.
Zn2+ + 2e− → Zn• The definition of reduction was broadened to
coincide with the definition of oxidation. According to the modern definition, when something gains electrons, it is reduced.
Reduction
• The loss of electrons (oxidation) by one substance is accompanied by the gain of electrons by another (reduction). Reduction is any chemical change in which at least one element gains electrons.
Memory Aid
Oxidizing and Reducing Agents
• A reducing agent is a substance that loses electrons, making it possible for another substance to gain electrons and be reduced. The oxidized substance is always the reducing agent.
• An oxidizing agent is a substance that gains electrons, making it possible for another substance to lose electrons and be oxidized. The reduced substance is always the oxidizing agent.
Identifying Oxidizing and Reducing Agents
2Zn(s) + O2(g) → 2ZnO(s)
Zn → Zn2+ + 2e−
O + 2e− → O2−
• Zinc atoms lose electrons, making it possible for oxygen atoms to gain electrons and be reduced, so zinc is the reducing agent.
• Oxygen atoms gain electrons, making it possible for zinc atoms to lose electrons and be oxidized, so O2 is the oxidizing agent.
Partial Loss and Gain of Electrons
N2(g) + O2(g) → 2NO(g)• The N-O bond is a polar covalent bond
in which the oxygen atom attracts electrons more than the nitrogen atom.
• Thus the oxygen atoms gain electrons partially and are reduced.
• The nitrogen atoms lose electrons partially and are oxidized.
• N2 is the reducing agent. • O2 is the oxidizing agent.
Redox Terms (1)
Redox Terms (2)
• Oxidation-Reduction Reaction– an electron transfer reaction
• Oxidation– complete or partial loss of electrons
• Reduction– complete or partial gain of electrons
• Oxidizing Agent– the substance reduced; gains electrons,
making it possible for something to lose them.
• Reducing Agent– the substance oxidized; loses electrons,
making it possible for something to gain them.
Questions Answered by Oxidation Numbers
Is the reaction redox?
If any atoms change their oxidation number, yes.
What’s oxidized? The element that increases its oxidation number
What’s reduced? The element that decreases its oxidation number
What’s the reducing agent?
The substance with the element oxidized
What’s the oxidizing agent?
The substance with the element reduced
Steps for Determination of Oxidation Numbers
• Step 1: Assign oxidation numbers to as many atoms as you can using the guidelines described on the next slide.
• Step 2: To determine oxidation numbers for atoms not described on the pervious slide, use the following guideline.– The sum of the oxidation numbers for each
atom in the formula is equal to the overall charge on the formula. (This includes uncharged formulas where the sum of the oxidation numbers is zero.)
Oxidation Numbers
uncharged element 0 no exceptions
monatomic ions charge on ion
no exceptions
combined fluorine -1 no exceptions
combined oxygen -2 -1 in peroxides
covalently bonded hydrogen
+1 no exceptions
Single Displacement
Single Displacement Reaction
Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
oxidation: Zn(s) → Zn2+(aq) + 2e−
reduction: Cu2+(aq) + 2e− → Cu(s)
Single Displacement Reaction Example
Voltaic Cell
• The system in which two half-reactions for a redox reaction are separated allowing the electrons transferred in the reaction to be passed between them through a wire is called voltaic cell.
Voltaic Cell
Electrodes
• The electrical conductors placed in the half-cells are called electrodes.
• They can be active electrodes, which participate in the reaction, or passive electrodes, which transfer the electrons into or out of a half-cell but do not participate in the reaction.
Anode
• The anode is the site of oxidation.• Because oxidation involves loss of
electrons, the anode is the source of electrons. For this reason, it is described as the negative electrode.
• Because electrons are lost forming more positive (or less negative) species at the anode, the surroundings tend to become more positive. Thus anions are attracted to the anode.
Cathode
• The cathode is the site of reduction.• By convention, the cathode is the
positive electrode.• Because electrons come to the cathode
and substances gain these electrons to become more negative (or less positive), the surroundings tend to become more negative. Thus cations are attracted to the cathode.
Other Cell Components
• A device called a salt bridge can be used to keep the charges balanced.
• The portion of the electrochemical cell that allows ions to flow is called the electrolyte.
Leclanché Cell or Dry Cell
Anode oxidation:Zn(s) → Zn2+(aq) + 2e−
Cathode reduction:2MnO2(s) + 2NH4
+(aq) + 2e−
→ Mn2O3(s) + 2NH3(aq) + H2O(l)Overall reaction:
Zn(s) + 2MnO2(s) + 2NH4+(aq)
→ Zn2+(aq) + Mn2O3(s) + 2NH3(aq) + H2O(l)
Dry Cell Image
Alkaline Batteries
Anode oxidation:Zn(s) + 2OH−(aq)
→ ZnO(s) + H2O(l) + 2e−
Cathode reduction:2MnO2(s) + H2O(l) + 2e−
→ Mn2O3(s) + 2OH−(aq)Overall reaction:
Zn(s) + 2MnO2(s) → ZnO(s) + Mn2O3(s)
Electrolysis• Voltage, a measure of the strength of an
electric current, represents the force that moves electrons from the anode to the cathode in a voltaic cell.
• When a greater force (voltage) is applied in the opposite direction, electrons can be pushed from what would normally be the cathode toward the voltaic cell’s anode. This process is called electrolysis.
• In a broader sense, electrolysis is the process by which a redox reaction is made to occur in the nonspontaneous direction.
2NaCl(l) → 2Na(l) + Cl2(g)
Primary and Secondary Batteries
• Batteries that are not rechargeable are called primary batteries.
• A rechargeable battery is often called a secondary battery or a storage battery.
Nickel-Cadmium Battery
Anode reaction:Cd(s) + 2OH−(aq) Cd(OH)2(s) + 2e−
Cathode reaction:NiO(OH)(s) + H2O(l) + e−
Ni(OH)2(s) + OH−(aq) Net Reaction:Cd(s) + 2NiO(OH)(s) + 2H2O(l)
Cd(OH)2(s) + 2Ni(OH)2(s)
Pb(s) + HSO4−(aq) + H2O(l)
PbSO4(s) + H3O+(aq) + 2e−Cathode reaction:PbO2(s) + HSO4
−(aq) + 3H3O+(aq) + 2e−
PbSO4(s) + 5H2O(l)Net Reaction:Pb(s) + PbO2(s) + 2HSO4
−(aq) + 2H3O+(aq) 2PbSO4(s) + 4H2O(l)
Lead Acid Battery
Conversions to Moles
Molarity
• Converts between moles of solute and volume of solution
Equation Stoichiometry (1)
Equation Stoichiometry (1)
• Tip-off - The calculation calls for you to convert from amount of one substance to amount of another, both of which are involved in a chemical reaction.
• General Steps1. If you are not given it, write and balance
the chemical equation for the reaction.2. Start your dimensional analysis in the
usual way.
Equation Stoichiometry (2)
3. Convert from the units that you are given for substance 1 to moles of substance 1. – For pure solids and liquids, this means
converting mass to moles using the molar mass of the substance.
– Molarity can be used to convert from volume of solution to moles of solute.
Equation Stoichiometry (3)
4. Convert from grams of substance 1 to moles of substance 1.
5. Convert from moles of substance 2 to the desired units for substance 2.
– For pure solids and liquids, this means converting moles to mass using the molar mass of substance 2.
– Molarity can be used to convert from moles of solute to volume of solution.
6. Calculate your answer and report it with the correct significant figures (in scientific notation, if necessary) and unit.
Titration• Titration involves the addition of one
solution (solution 1) to another solution (solution 2) until a chemical reaction between the components in the solutions is complete. Solution 1 is called the titrant, and we say that it is used to titrate solution 2. The completeness of the reaction is usually shown by a change of color caused by a substance called an indicator.
Titration Apparatus
Steps for Titration (1)
• A specific volume of the solution to be titrated (solution 2) is added to an Erlenmeyer flask. For example, 25.00 mL of a phosphoric acid solution of unknown concentration might be added to a 250-mL Erlenmeyer flask.
Steps for Titration (2)• A solution of a substance that reacts with
the solute in the solution in the Erlenmeyer flask is added to a buret. This solution in the buret, which has a known concentration, is the titrant. The buret is set up over the Erlenmeyer flask so the titrant can be added to the solution to be titrated. For example, a 1.02 M NaOH solution might be added to a buret, which is set up over the Erlenmeyer flask containing the phosphoric acid solution.
Steps for Titration (3)
• An indicator is added to the solution being titrated. The indicator is a substance that changes color when the reaction is complete. In our example, phenolphthalein, which is a common an acid-base indicator, is added to the phosphoric acid solution in the Erlenmeyer flask.
Phenolphthalein
• Phenolphthalein is a substance that has two forms. In acidic conditions, it is in the acid form, which is colorless. In basic conditions, an H+ ion is removed from each phenolphthalein molecule, converting it to its base form, which is red.
Steps for Titration (4)
• The titrant is slowly added to the solution being titrated until the indicator changes color, showing that the reaction is complete. This stage in the procedure is called the endpoint. In our example, the NaOH solution is slowly added from the buret until the mixture in the Erlenmeyer flask changes from colorless to red. At this point, 34.2 mL of 1.02 M NaOH has been added.
Making Solutions
• From pure solids• From a pure or almost pure acid
(e.g. 17 M HC2H3O2 or 18 M H2SO4)
• From any other more concentrated solution
Making a Solution from Pure Solid
Making a Solution From Concentrated Acid
Add Concentrated Acid to Water
Dilution Problems
mol solute concentrated = mol solute dilute--- mol solute conc --- mol solute dil--- L conc soln = --- L dil soln
L conc soln L dil soln
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
conc conc dilut
C C
e dilute
D D
V M = V M M V = M V
Making a Solution from a More Concentrated Solution (not
concentrated acid)