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1 Chapter 4 Type of Chemical Reactions and Solution Stoichiometric Water, Nature of aqueous solutions, types of electrolytes, dilution. Types of chemical reactions: precipitation, acid-base and oxidation reactions. Stoichiometry of reactions and balancing the chemical equations.
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1 Chapter 4 Type of Chemical Reactions and Solution Stoichiometric Water, Nature of aqueous solutions, types of electrolytes, dilution. Types of chemical.

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Page 1: 1 Chapter 4 Type of Chemical Reactions and Solution Stoichiometric Water, Nature of aqueous solutions, types of electrolytes, dilution. Types of chemical.

1

Chapter 4Type of Chemical Reactions and Solution

Stoichiometric

Water, Nature of aqueous solutions, types of electrolytes, dilution.

Types of chemical reactions: precipitation, acid-base and oxidation reactions.

Stoichiometry of reactions and balancing the chemical equations.

Page 2: 1 Chapter 4 Type of Chemical Reactions and Solution Stoichiometric Water, Nature of aqueous solutions, types of electrolytes, dilution. Types of chemical.

2

Aqueous Solutions

Water is the dissolving medium, or solvent.

Page 3: 1 Chapter 4 Type of Chemical Reactions and Solution Stoichiometric Water, Nature of aqueous solutions, types of electrolytes, dilution. Types of chemical.

3

Figure 4.1: (Left) The water molecule is polar. (Right) A space-filling model of the water

molecule.

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4

Figure 4.2: Polar water molecules interact with the positive and negative ions of a salt assisting

in the dissolving process.

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5

Some Properties of Water

Water is “bent” or V-shaped. The O-H bonds are covalent. Water is a polar molecule. Hydration occurs when salts dissolve in

water.

Page 6: 1 Chapter 4 Type of Chemical Reactions and Solution Stoichiometric Water, Nature of aqueous solutions, types of electrolytes, dilution. Types of chemical.

6

Figure 4.3: (a) The ethanol molecule contains a polar O—H bond similar to those in the water molecule. (b) The polar water molecule interacts strongly with the polar O—H bond in ethanol. This is a case of "like dissolving like."

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A Solute

dissolves in water (or other “solvent”)

changes phase (if different from the solvent)

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

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A Solvent

retains its phase (if different from the solute)

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

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9

General Rule for dissolution

Like dissolve likePolar dissolve polar (water dissolve

ethanol)Non-polar dissolve nonpolar (benzene

dissolve fat)

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Figure 4.5: When solid NaCl dissolves, the Na+ and Cl- ions are randomly dispersed in the water.

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Electrolytes

Strong - conduct current efficiently

NaCl, HNO3

Weak - conduct only a small current

vinegar, tap water

Non - no current flows

pure water, sugar solution

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Figure 4.4: Electrical conductivity of aqueous solutions.

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13

Acids

Strong acids - dissociate completely to produce H+ in solution

hydrochloric and sulfuric acid

HCl , H2SO4

Weak acids - dissociate to a slight extent to give H+ in solution

acetic and formic acid

CH3COOH, CH2O

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14

Bases

Strong bases - react completely with water to give OH ions.

sodium hydroxide

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

ammonia

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15

Figure 4.6: HCl(aq) is completely ionized.

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16

Figure 4.7: An aqueous solution of sodium hydroxide.

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Figure 4.8: Acetic acid (HC2H3O2) exists in water mostly as undissociated molecules. Only a small percentage of the molecules are ionized.

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18

Molarity

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

M

M

molaritymoles of soluteliters of solution

HClmoles of HCl

liters of solution3

62

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Common Terms of Solution Concentration

Stock - routinely used solutions prepared in concentrated form.

Concentrated - relatively large ratio of solute to solvent. (5.0 M NaCl)

Dilute - relatively small ratio of solute to solvent. (0.01 M NaCl): (MV)initial=(MV)Final

Page 20: 1 Chapter 4 Type of Chemical Reactions and Solution Stoichiometric Water, Nature of aqueous solutions, types of electrolytes, dilution. Types of chemical.

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Figure 4.10: Steps involved in the preparation of a standard aqueous solution.

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Figure 4.12: Dilution Procedure (a) A measuring pipet is used to transfer 28.7mL of 17.4 M acetic acid solution to a volumetric flask. (b) Water is added to the flask to the calibration mark. (c) The resulting solution is 1.00 M acetic acid.

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Practice Example

How many moles are in 18.2 g of CO2?

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Practice Example

Consider the reaction

N2 + 3H2 = 2NH3

How many moles of H2 are needed to completely react 56 g of N2?

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24

Practice Example

How many grams are in 0.0150 mole of caffeine C8H10N4O2

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Practice Example

A solution containing Ni2+ is prepared by dissolving 1.485 g of pure nickel in nitric acid and diluting to 1.00 L. A 10.00 mL aliquot is then diluted to 500.0 mL. What is the molarity of the final solution?(Atomic weight: Ni = 58.70).

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Practice Example

Calculate the number of molecules of vitamin A, C20H30O in 1.5 mg of this compound.

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Practice Example

What is the mass percent of hydrogen in acetic acid HC2H3O2

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Types of Solution Reactions

Precipitation reactionsAgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

Acid-base reactionsNaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)

Oxidation-reduction reactionsFe2O3(s) + Al(s) Fe(l) + Al2O3(s)

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Simple Rules for Solubility

1. Most nitrate (NO3) salts are soluble.

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

3. Most Cl, Br, and I salts are soluble (NOT Ag+, Pb2+, Hg22+)

4. Most sulfate salts are soluble (NOT BaSO4, PbSO4, HgSO4, CaSO4)

5. Most OH salts 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.

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Figure 4.13: When yellow aqueous potassium chromate is added to a colorless barium nitrate solution, yellow barium chromate precipitates.

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Describing Reactions in SolutionPrecipitation

1. Molecular equation (reactants and products as compounds)

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

2. Complete ionic equation (all strong electrolytes shown as ions)

Ag+(aq) + NO3- (aq) + Na+ (aq) + Cl(aq) AgCl(s)

+ Na+ (aq) + NO3- (aq)

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Describing Reactions in Solution (continued)

3. Net ionic equation (show only components that actually react)

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

Na+ and NO3 are spectator ions.

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Performing Calculations for Acid-Base Reactions

1. List initial species and predict reaction.

2. Write balanced net ionic reaction.

3. Calculate moles of reactants.

4. Determine limiting reactant.

5. Calculate moles of required reactant/product.

6. Convert to grams or volume, as required.Remember: n H+ = n OH- (MV) H+ = (MV) OH-

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Neutralization Reaction

acid + base salt + water

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

H+ + Cl- + Na+ + OH-- Na+ + Cl- + H2O

H+ + OH- H2O

4.3

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Key Titration Terms

Titrant - solution of known concentration used in titration

Analyte - substance being analyzed

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.

movie

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Oxidation-Reduction Reactions(electron transfer reactions)

2Mg (s) + O2 (g) 2MgO (s)

2Mg 2Mg2+ + 4e-

O2 + 4e- 2O2-

Oxidation half-reaction (lose e-)

Reduction half-reaction (gain e-)

2Mg + O2 + 4e- 2Mg2+ + 2O2- + 4e-

2Mg + O2 2MgO

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Redox Reactions• Many practical or everyday examples of

redox reactions:– Corrosion of iron (rust formation)– Forest fire– Charcoal grill– Natural gas burning– Batteries– Production of Al metal from Al2O3 (alumina)– Metabolic processes

combustion

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Rules for Assigning Oxidation States

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

2. Oxidation state of monatomic element = charge

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

4. H = +1 in covalent compounds

5. Fluorine = 1 in compounds

6. Sum of oxidation states = 0 in compounds Sum of oxidation states = charge of the ion

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Page 41: 1 Chapter 4 Type of Chemical Reactions and Solution Stoichiometric Water, Nature of aqueous solutions, types of electrolytes, dilution. Types of chemical.

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Zn (s) + CuSO4 (aq) ZnSO4 (aq) + Cu (s)

Zn is oxidizedZn Zn2+ + 2e-

Cu2+ is reducedCu2+ + 2e- Cu

Zn is the reducing agent

Cu2+ is the oxidizing agent

Copper wire reacts with silver nitrate to form silver metal.What is the oxidizing agent in the reaction?

Cu (s) + 2AgNO3 (aq) Cu(NO3)2 (aq) + 2Ag (s)

Cu Cu2+ + 2e-

Ag+ + 1e- Ag Ag+ is reduced Ag+ is the oxidizing agent

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NaIO3

Na = +1 O = -2

3x(-2) + 1 + ? = 0

I = +5

IF7

F = -1

7x(-1) + ? = 0

I = +7

K2Cr2O7

O = -2 K = +1

7x(-2) + 2x(+1) + 2x(?) = 0

Cr = +6

Oxidation numbers of all the elements in the following ?

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Balancing by Half-Reaction Method

1. Write separate reduction, oxidation reactions.

2. For each half-reaction:

Balance elements (except H, O)

Balance O using H2O

Balance H using H+

Balance charge using electrons

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Balancing by Half-Reaction Method (continued)

3. If necessary, multiply by integer to equalize electron count.

4. Add half-reactions.

5. Check that elements and charges are balanced.

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Half-Reaction Method - Balancing in Base

1. Balance as in acid.

2. Add OH that equals H+ ions (both sides!)

3. Form water by combining H+, OH.

4. Check elements and charges for balance.

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Balancing Redox Equations

•Example: Balance the following redox reaction:

•Cr2O72- + Fe2+ Cr3+ + Fe3+ (acidic soln)

1) Break into half reactions:

Cr2O72- Cr3+

Fe2+ Fe3+

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Balancing Redox Equations

2) Balance each half reaction:

Cr2O72- Cr3+

Cr2O72- 2 Cr3+

Cr2O72- 2 Cr3+ + 7 H2O

Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O

6 e- + Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O

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Balancing Redox Equations

2) Balance each half reaction (cont)

Fe2+ Fe3+

Fe2+ Fe3+ + 1 e-

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Balancing Redox Reactions

3) Multiply by integer so e- lost = e- gained

6 e- + Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O

Fe2+ Fe3+ + 1 e- x 6

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Balancing Redox Reactions

3) Multiply by integer so e- lost = e- gained

6 Fe2+ 6 Fe3+ + 6 e-

6 e- + Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O

4) Add both half reactions

Cr2O72- + 6 Fe2+ + 14 H+ 2 Cr3+ + 6 Fe3+ + 7 H2O

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Balancing Redox Reactions

Cr2O72- + 6 Fe2+ + 14 H+ 2 Cr3+ + 6 Fe3+ + 7 H2O

5) Check the equation

2 Cr 2 Cr

7 O 7 O

6 Fe 6 Fe

14 H 14 H

+24 + 24

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Balancing Redox Reactions

• Procedure for Basic Solutions:– Divide the equation into 2 incomplete half

reactions• one for oxidation

• one for reduction

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Balancing Redox Reactions– Balance each half-reaction:

• balance elements except H and O

• balance O atoms by adding H2O

• balance H atoms by adding H+

• add 1 OH- to both sides for every H+ added

• combine H+ and OH- on same side to make H2O

• cancel the same # of H2O from each side

• balance charge by adding e- to side with greater overall + charge

different

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Balancing Redox Equations– Multiply each half reaction by an integer so that

• # e- lost = # e- gained

– Add the half reactions together.• Simply where possible by canceling species

appearing on both sides of equation

– Check the equation• # of atoms

• total charge on each side

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Balancing Redox Reactions

Example: Balance the following redox reaction.

NH3 + ClO- Cl2 + N2H4 (basic soln)

NH3 N2H4

ClO- Cl2

1) Break into half reactions:

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Balancing Redox Reactions

NH3 N2H4

2) Balance each half reaction:

2 NH3 N2H4

2 NH3 N2H4 + 2 H+

2 NH3 + 2 OH- N2H4 + 2 H2O

+ 2 OH- + 2 OH-

2 H2O

2 NH3 + 2 OH- N2H4 + 2 H2O + 2 e-

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Balancing Redox Reactions

2 ClO- Cl2

2) Balance each half reaction:

2 ClO- Cl2 + 2 H2O2 ClO- + 4 H+ Cl2 + 2 H2O

+ 4 OH- + 4 OH-

2 ClO- + 4 H2O Cl2 + 2 H2O + 4 OH-

2 ClO- + 2 H2O Cl2 + 4 OH-

2 e- + 2 ClO- + 2 H2O Cl2 + 4 OH-

ClO- Cl2

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Balancing Redox Reactions

3) Multiply by integer so # e- lost = # e- gained

2 NH3 + 2 OH- N2H4 + 2 H2O + 2 e-

2 e- + 2 ClO- + 2 H2O Cl2 + 4 OH-

4) Add both half reactions

2 NH3 + 2 OH- + 2ClO- + 2 H2O N2H4 + 2 H2O + Cl2 + 4 OH-

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Balancing Redox Reactions5) Cancel out common species

2 NH3 + 2 OH- + 2 ClO- + 2 H2O N2H4 + 2 H2O + Cl2 + 4 OH-

2

2 NH3 + 2 ClO- N2H4 + Cl2 + 2 OH-

6) Check final equation:

2 N 2 N

6 H 6 H

2 Cl 2 Cl

2 O 2 O

-2 -2

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Practice ExampleIn the following the oxidizing agent is:

5H2O2 + 2MnO4- + 6H+ 2Mn2+ + 8H2O + 5O2

a. MnO4-

b. H2O2

c. H+

d. Mn2+

e. O2

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Practice ExampleDetermine the coefficient of Sn in acidic solution

Sn + HNO3 SnO2 + NO2 + H2O

1

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Practice ExampleThe sum of the coefficients when they are

whole numbers in basic solution:

Bi(OH)3 + SnO22- Bi + SnO3

2-

13

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http://www.chemistrycoach.com/balancing_redox_in_acid.htm#BalancingRedoxEquationsinAcidicorBasicMedium

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http://www.chemistrycoach.com/tutorials-5.htm#Oxidation-Reduction

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• http://www.sstdt.org.sa/arc-e.htm

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QUESTIONAn unknown substance dissolves readily in water but not in benzene (a nonpolar solvent). Molecules of what type are present in the substance? 1) Neither polar nor nonpolar 2) Polar 3) Either polar or nonpolar 4) Nonpolar 5) none of these

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ANSWER2) Polar Section 4.1 Water, the Common Solvent (p. 134) The solubility rule for molecular compounds is “like dissolves like”, that is, polar molecules dissolve in polar solvents and nonpolar molecules dissolve in nonpolar solvents.

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QUESTIONHow many grams of NaCl are contained in 350. mL of a 0.250 M solution of sodium chloride? 1) 41.7 g 2) 5.11 g 3) 14.6 g 4) 87.5 g 5) None of these

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ANSWER2) 5.11 g Section 4.3 The Composition of Solutions (p. 140) Volume (L) times concentration (mol/L) gives moles. Moles are then converted to grams.

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QUESTIONWhat volume of 18.0 M sulfuric acid must be used to prepare 15.5 L of 0.195 M H2SO4? 1) 168 mL 2) 0.336 L 3) 92.3 mL 4) 226 mL 5) None of these

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ANSWER1) 168 mL Section 4.3 The Composition of Solutions (p. 140) Use the dilution formula, M1

V1 = M2 V2, M is

in mol/L and V is in L.

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QUESTIONThe net ionic equation for the reaction of aluminum sulfate and sodium hydroxide contains which of the following species? 1) 3Al

3+(aq)

2) OH–(aq)

3) 3OH–(aq)

4) 2Al3+

(aq) 5) 2Al(OH)3(s)

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ANSWER

3) 3OH–(aq)

Section 4.6 Describing Reactions in Solution (p. 154) The net ionic equation is found by canceling the spectator ions from the total ionic equation.

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QUESTIONWhich of the following is a strong acid? 1) HF 2) KOH 3) HClO4 4) HClO 5) HBrO

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ANSWER3) HClO4 Section 4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes (p. 136) Memorization of the list of strong acids will allow one to determine the difference between strong acids and weak acids.

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QUESTIONAll of the following are weak acids except: 1) HCNO. 2) HBr. 3) HF. 4) HNO2. 5) HCN.

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ANSWER2) HBr. Section 4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes (p. 136) Knowing the list of strong acids will allow one to determine which acids are strong and which are weak.

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QUESTIONWhich of the following is not a strong base? 1) Ca(OH)2 2) KOH 3) NH3 4) LiOH 5) Sr(OH)2

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ANSWER3) NH3 Section 4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes (p. 136) Knowing the list of strong bases will allow one to determine which bases are strong and which are weak.

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QUESTIONThe interaction between solute particles and water molecules, which tends to cause a salt to fall apart in water, is called: 1) hydration. 2) polarization. 3) dispersion. 4) coagulation. 5) conductivity.

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ANSWER1) hydration. Section 4.1 Water, the Common Solvent (p. 134) Hydration is the process of water molecules surrounding and stabilizing ions so that they can be pulled into solution.

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QUESTION

The concentration of a salt water solution that sits in an open beaker decreases over time.

1) True 2) False

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ANSWER2) False Section 4.3 The Composition of Solutions (p. 140) The amount of water decreases over time so the concentration (mol NaCl/volume of water) increases over time.

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QUESTIONThe following reactions: Pb

2+ + 2I

– PbI2

2Ce4+ + 2I

– I2 + 2Ce

3+

HOAc + NH3 NH4

+ + OAc

are examples of

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QUESTION (continued)

1) acid-base reactions. 2) unbalanced reactions. 3) precipitation, acid-base, and redox

reactions, respectively. 4) redox, acid-base, and precipitation

reactions, respectively. 5) precipitation, redox, and acid-base

reactions, respectively.

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ANSWER5) precipitation, redox, and acid-base

reactions, respectively. Section 4.5 Precipitation Reactions (p. 151) PbI2 is insoluble, Ce

4+ changes to Ce

3+ and

HOAc is an acid while NH3 is a base.

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QUESTIONWhich of the following salts is insoluble in water? 1) Na2S 2) K3PO4 3) Pb(NO3)2 4) CaCl2 5) All of these are soluble in water.

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ANSWER5) All of these are soluble in water. Section 4.5 Precipitation Reactions (p. 153) According to the solubility rules for ionic compounds, compounds containing Group IA ions or nitrate ions will always be soluble. Compounds containing halides are generally soluble, aside from silver, lead and mercury(I) halides.

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QUESTION

When NH3(aq) is added to Cu2+

(aq), a precipitate initially forms. Its formula is: 1) Cu(NH3)4

2+

2) Cu(NO3)2 3) Cu(OH)2 4) Cu(NH3)4

2+

5) CuO

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ANSWER3) Cu(OH)2 Section 4.4 Types of Chemical Reactions (p. 148) Ammonia produces hydroxide ion in water: NH3 + H2O NH4

+ + OH

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QUESTIONIn the balanced molecular equation for the neutralization of sodium hydroxide with sulfuric acid, the products are: 1) NaSO4 + H2O 2) NaSO3 + 2H2O 3) 2NaSO4 + H2O 4) Na2S + 2H2O 5) Na2SO4 + 2H2O

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ANSWER5) Na2SO4 + 2H2O Section 4.8 Acid-Base Reactions (p. 158) The salt is made from the anion of the acid and the cation of the base.

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QUESTIONWhat mass of NaOH is required to react exactly with 25.0 mL of 1.2 M H2SO4? 1) 1.2 g 2) 1.8 g 3) 2.4 g 4) 3.5 g 5) None of these

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ANSWER3) 2.4 g Section 4.8 Acid-Base Reactions (p. 158) Remember that the reaction is 2NaOH + H2SO4 Na2SO4 + 2H2O, so there are two moles of NaOH used per one mole of H2SO4.

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QUESTIONIn which of the following does nitrogen have an oxidation state of +4? 1) HNO3 2) NO2 3) N2O 4) NH4Cl 5) NaNO2

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ANSWER2) NO2 Section 4.9 Oxidation-Reduction Reactions (p. 164) Oxygen almost always has an oxidation state of –2 when part of a compound. The exception is when it is part of a peroxide. For example, hydrogen peroxide H2O2. Then it has an oxidation state of –1.

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QUESTION

In the reaction 2Cs(s) + Cl2(g) 2CsCl(s), Cl2 is 1) the reducing agent. 2) the oxidizing agent. 3) oxidized. 4) the electron donor. 5) two of these

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ANSWER2) the oxidizing agent. Section 4.9 Oxidation-Reduction Reactions (p. 164) Metals lose electrons, so they are oxidized, making the other reactant an oxidizing agent.

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QUESTIONWhich of the following statements is(are) true? Oxidation and reduction: 1) cannot occur independently of each other. 2) accompany all chemical changes. 3) describe the loss and gain of electron(s),

respectively. 4) result in a change in the oxidation states of

the species involved. 5) 1, 3, and 4 are true

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ANSWER5) 1, 3, and 4 are true. Section 4.9 Oxidation-Reduction Reactions (p. 167) (2) is false because certain reactions, such as double displacement reactions, are not redox reactions.

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QUESTIONHow many of the following are oxidation-reduction reactions? NaOH + HCl NaCl + H2O Cu + 2AgNO3

2Ag + Cu(NO3)2 Mg(OH)2

MgO + H2O N2 + 3H2

2NH3

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QUESTION (continued)

1) 0 2) 1 3) 2 4) 3 5) 4

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ANSWER3) 2 Section 4.9 Oxidation-Reduction Reactions (p. 164) If an element is found on the reactant’s side, this is almost always a redox reaction, since an element usually becomes part of a compound during a chemical reaction.