Chapter 5: Jespersen - JU Medicine · Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved Your Turn In a cup of coffee that has milk

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved

Chapter 4Molecular View of Reactions in

Aqueous Solutions

Chemistry, 7th Edition

International Student Version

Brady/Jespersen/Hyslop


Chapter in Context

▪ Describe solutions qualitatively and quantitatively

▪ Distinguish electrolytes from non-electrolytes

▪ Write balanced molecular, ionic, and net ionic equations

▪ Identify acids and bases and learn names and formulas

▪ Use metathesis reactions to plan chemical syntheses

▪ Define and use molarity in calculations

▪ Understand titrations and chemical analysis2


Importance of Water

▪ One of the most common compounds on earth

▪ Dissolves many different substances

▪ Responsible, in part, for evolution of life

▪ 60% of the human body is water

3

Distinct Properties

▪ Dissolves ionic compounds

▪ Acid-base reactions occur in water


Reactions in Solution

▪ For a reaction to occur

▪ Reactants needs to come into physical contact

▪ Happens best in gas or liquid phase

▪ Movement occurs

Solution

▪ Homogeneous mixture

▪ Two or more components mix freely

▪ Molecules or ions completely intermingled

▪ Contains at least two substances

4


Definitions

Solvent

▪ Medium that dissolves solutes

▪ Component present in largest amount

▪ Can be gas, liquid, or solid

▪ Aqueous solution—water is solvent

Solute

▪ Substance dissolved in solvent

▪ Solution is named by solute

▪ Can be gas—CO2 in soda

▪ Liquid—ethylene glycol in antifreeze

▪ Solid—sugar in syrup

5


Iodine Molecules in Ethanol

6

Crystal of solute placed in solvent

Solute molecules dispersed throughout solvent


Your TurnIn a cup of coffee that has milk and sugar in it,

which are the solutes and which are/is the solvent(s)?

A. Solutes: caffeine, sugar, and milk proteins

Solvent: water

C. Solute: water

Solvents: caffeine, sugar and milk proteins

B. Solutes: sugar and milk proteins

Solvents: water and caffeine

D. Solute: milk protein only

Solvent: water and milk7


Solutions

▪ May be characterized using

Concentration

▪ Solute-to-solvent ratio

▪ Percent concentration

8

g solute

g solvent

g solute

g solution

percentage concentration =g solute

100 g solution

or


Your Turn

What is the percent concentration by mass of NaCl in a 556 g of solution that contains 23 grams of NaCl?

A. 24.2%

B. 0.242%

C. 0.0414%

D. 0.414 %

E. 4.14%

9


Your Turn

A typical blood glucose level is 90.0 mg/dL. What is the percent concentration by mass of glucose in blood assuming a density for blood of 1.06 g/mL?

A. 95.4%

B. 0.0954%

C. 8.49%

D. 0.0849%

E. 1.18%

blood mL 1000

blood L 1

blood L0.1000

glucose 0.0900g

%100blood g

glucose g1049.8

blood g 1.06

blood mL 1 4 = −

10


Relative ConcentrationDilute solution

▪ Small solute to solvent ratio

e.g., Eye drops

Concentrated solution

▪ Large solute to solvent ratio

e.g., Pickle brine

▪ Dilute solution contains less solute per unit

volume than more concentrated solution

▪ ‘Dilute’ and ‘concentrated’ are relative terms

11


Concentration

Solubility

▪ Temperature dependent

Saturated solution

▪ Solution in which no more solute can be dissolved at a given temperature

Unsaturated solution

▪ Solution containing less solute than maximum amount

▪ Able to dissolve more solute

12

Solubility =g solute needed to make saturated solution

100 g solvent


Solubilities of Some Common Substances

13

Substance Formula

Solubility

(g/100 g water)

Sodium chloride NaCl 35.7 at 0 oC

39.1 at 100 oC

Sodium hydroxide

NaOH 42 at 0 oC

347 at 100 oC

Calcium carbonate

CaCO3 0.0015 at 25 oC


Concentrations

Supersaturated Solutions

▪ Contains more solute than required for saturation at a given temperature

▪ Formed by careful cooling of saturated solutions

▪ Unstable

▪ Crystallize out when add seed crystal – results in formation of solid or precipitate (ppt.)

14


Precipitates

Precipitate

▪ Solid product formed when reaction carried out in solutions and one product has low solubility

▪ Insoluble product

▪ Separates out of solution

15


Precipitation Reaction

Reaction that produces precipitatePb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq)

Precipitates

PbI2(s)Pb2+ NO3– K+ I–

Solid precipitate

16


Your Turn

What is the precipitate in the following reaction?

2AgNO3(aq) + (NH4)2CO3(aq) →

2NH4NO3(aq) + Ag2CO3(s)

A. AgNO3

B. (NH4)2CO3

C. NH4NO3

D. Ag2CO3

E. There is no precipitate

17


Electrolytes in Aqueous Solution

▪ Ionic compounds conduct electricity

▪ Molecular compounds don’t conduct electricity

Why?

18

Bright light

No light

MolecularIons present

CuSO4 and water Sugar and water


Ionic Compounds (Salts) in Water

▪ Water molecules arrange themselves around ions and remove them from lattice.

Dissociation▪ Salts break apart into

ions when entering solution

Separated ions ▪ Hydrated

▪ Conduct electricity

▪ Note: Polyatomic ions remain intact▪ e.g., KIO3 → K+ + IO3

–

19

NaCl(s) → Na+(aq) + Cl–(aq)


Molecular Compounds In Water

▪ When molecules dissolve in water

▪ Solute particles are surrounded by water

▪ Molecules do not dissociate

20


Electrical Conductivity

Electrolyte

▪ Solutes that yield electrically conducting solutions

▪ Separate into ions when enter into solution

Strong electrolyte

▪ Electrolyte that dissociates 100% in water

▪ Yields aqueous solution that conducts electricity

▪ Good electrical conduction

▪ Ionic compounds, e.g., NaCl, KNO3

▪ Strong acids and bases, e.g., HClO4, HCl

21



Non-electrolyte

▪ Aqueous solution that doesn’t conduct electricity

▪ Molecules remain intact in solution

e.g., Sugar, alcohol

22


Your Turn

How many ions form on the dissociation of Na3PO4?

A. 1

B. 2

C. 3

D. 4

E. 8

23


Your Turn

How many ions form on the dissociation of

Al2(SO4)3?

A. 2

B. 3

C. 5

D. 9

E. 14

24


Weak electrolyte▪ When dissolved in water only a small

percentage of molecules ionize

▪ Common examples are weak acids and bases

▪ Solutions weakly conduct electricity

▪ e.g., Acetic acid (CH3COOH), ammonia (NH3)


25


Strong vs. Weak Electrolyte

26

HCl(aq) CH3COOH(aq) NH3(aq)


Dissociation Reactions

▪ Ionic compounds dissolve to form hydrated ions

▪ Hydrated = surrounded by water molecules

▪ In chemical equations, hydrated ions are indicated by

▪ Symbol (aq) after each ions

▪ Ions are written separately

KBr(s) ⎯→ K+(aq) + Br–(aq)

Mg(HCO3)2(s) ⎯→ Mg2+(aq) + 2HCO3–(aq)

27


Learning Check

Write the equations that illustrate the dissociation of the following salts:

▪Na3PO4(aq) →

▪Al2(SO4)3(aq) →

▪CaCl2(aq) →

▪Ca(MnO4)2(aq) →

28

2Al3+(aq) + 3SO42–(aq)

3Na+(aq) + PO43–(aq)

Ca2+(aq) + 2Cl–(aq)

Ca2+(aq) + 2MnO4–(aq)


Equations of Ionic Reactions

▪ Consider the reaction of Pb(NO3)2 with KI

29

PbI2(s)Pb2+ NO3– K+ I–



▪ When two soluble ionic solutions are mixed, sometimes an insoluble solid forms.

▪ Three types of equations used to describe

1. Molecular equation

▪ Substances listed as complete formulas

2. Ionic equation

▪ All soluble substances broken into ions

3. Net ionic equation

▪ Only lists substances that actually take part in reaction

30



1. Molecular Equation

▪ Complete formulas for all reactants and products

▪ Formulas written with ions together

▪ Does not indicate presence of ions (no charges)

▪ Gives identities of all compounds

▪ Good for planning experiments

e.g.,

Pb(NO3)2(aq) + 2KI(aq) ⎯→ PbI2(s) + 2KNO3(aq)

31



2. Ionic Equation

▪ Emphasizes the reaction between ions

▪ All strong electrolytes dissociate into ions

▪ Used to visualize what is actually occurring in solution

▪ Insoluble solids written together as they don’t dissociate to any appreciable extent

e.g.,

Pb2+(aq) + 2NO3–(aq) + 2K+(aq) + 2I–(aq) ⎯→

PbI2(s) + 2K+(aq) + 2NO3–(aq)

32



Spectator Ions

▪ Ions that don’t take part in reaction

▪ They hang around and watch

▪ K+ and NO3– in our example

3. Net Ionic Equation

▪ Eliminate all spectator ions

▪ Emphasizes the actual reaction

▪ Focus on chemical change that occurs

e.g., Pb2+(aq) + 2I–(aq) ⎯→ PbI2(s)

33



Criteria for ionic and net ion equations

▪ Material balance▪ The same number of each kind of atom must be

present on both sides of the arrow.

▪ Electrical balance ▪ The net electrical charge on the reactants must

equal the net electrical charge on the products▪ Charge does not necessarily have to be zero

34


Your Turn

Which of the following is not electrically balanced?

A. Ag+(aq) + Cl–(aq) ⎯→ AgCl(s)

B. NH4+(aq) + H2O ⎯→ NH3(aq) + H3O

+(aq)

C. NO2–(aq) + H2O ⎯→ HNO2(aq) + OH–(aq)

D. Mg+(aq) + 2OH–(aq) ⎯→ Mg(OH)2(s)

E. Cu2+(aq) + Sn(s) ⎯→ Cu(s) + Sn2+(aq)

35

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved 36

Your Turn

Which of the following is not electrically balanced?

A. PO43-(aq) + H+(aq) ⎯→ HPO4

2-(aq)

B. 2Ag+(aq) + Zn(aq) ⎯→ Zn2+(aq) + 2Ag(s)

C. H2PO4–(aq) ⎯→ 2H+(aq) + PO4(aq)

D. CO32-(aq) + 2H+(aq) ⎯→ H2CO3(aq)

E. Pb2+(aq) + S2-(aq) ⎯→ PbS(s)


Net Ionic Equations

▪ Many ways to make PbI2

1. Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq)

2. Pb(C2H3O2)2(aq) + 2NH4I(aq) →

PbI2(s) + 2NH4C2H3O2(aq)

▪ Different starting reagents

▪ Same net ionic equation

▪ Pb2+(aq) + 2I–(aq) → PbI2(s)

37


Converting Molecular Equations to Ionic Equations

Strong electrolytes exist as dissociated ions in solution

Strategy

1. Identify strong electrolytes

2. Use subscript coefficients to determine total number of each type of ion

3. Separate ions in all strong electrolytes

4. Show states as recorded in molecular equations

38


Learning Check: Convert Molecular to Ionic Equations:

Write the correct ionic equation for each:

Pb(NO3)2(aq) + 2NH4IO3(aq) →

Pb(IO3)2(s) + 2NH4NO3(aq)

2NaCl (aq) + Hg2(NO3)2 (aq) → 2NaNO3 (aq) + Hg2Cl2 (s)

39

2Na+(aq) + 2Cl–(aq) + Hg22+(aq) + 2NO3

–(aq) →2Na+(aq) + 2NO3

–(aq) + Hg2Cl2(s)

Pb2+(aq) + 2NO3–(aq) + 2NH4

+(aq) + 2IO3–(aq) →

Pb(IO3)2(s) + 2NH4+(aq) + 2NO3

–(aq)


Your TurnConsider the following reaction :

Na2SO4(aq) + BaCl2(aq) → 2NaCl(aq) + BaSO4(s)

Write the correct ionic equation.

A.2Na+(aq) + SO42–(aq) + Ba2+(aq) + Cl2

2–(aq) → 2Na+(aq) + 2Cl–(aq) + BaSO4(s)

B. 2Na+(aq) + SO42–(aq) + Ba2+(aq) + 2Cl–(aq) →

2Na+(aq) + 2Cl–(aq) + BaSO4(s)

C. 2Na+(aq) + SO42–(aq) + Ba2+(aq) + Cl2

2–(aq) →2Na+(aq)+ 2Cl–(aq) + Ba2+(s) + SO4

2–(s)

D. Ba2+(aq) + SO42–(aq) → BaSO4(s)

E. Ba2+(aq) + SO42–(aq) → Ba2+(s) + SO4

2–(s)

40


Converting Ionic Equations to Net Ionic Equations

Strategy

1. Identify spectator ions

2. Cancel from both sides

3. Rewrite equation using only substances that actually react.

4. Show states as recorded in molecular and ionic equations

41


Learning Check: Convert Ionic Equation to Net Ionic Equation

Write the correct net ionic equation for each.

Pb2+(aq) + 2NO3–(aq) + 2K+(aq) + 2IO3

–(aq) →

Pb(IO3)2(s)+ 2K+(aq) + 2NO3–(aq)

2Na+(aq) + 2Cl–(aq) + Hg22+(aq) + 2NO3

–(aq) →

2Na+(aq)+ 2NO3–(aq) + Hg2Cl2(s)

42

2Cl–(aq) + Hg22+(aq) → Hg2Cl2(s)

Pb2+(aq) + 2IO3–(aq) → Pb(IO3)2(s)


Your Turn

Consider the following molecular equation:

(NH4)2SO4(aq) + Ba(CH3CO2)2(aq) →

2NH4CH3CO2(aq) + BaSO4(s)

Write the correct net ionic equation.

43

A.Ba2+(aq) + SO42–(aq) → BaSO4(s)

B. 2NH4+(aq) + 2CH3CO2

–(aq) → 2NH4CH3CO2(s)

C.Ba2+(aq) + SO42–(aq) → BaSO4(aq)

D.2NH4+(aq) + Ba2+(aq) + SO4

2–(aq) + 2CH3CO2–(aq) →

2NH4+(aq) + 2CH3CO2

–(aq) + BaSO4(s)

E. 2NH4+(aq) + 2CH3CO2

–(aq) → 2NH4CH3CO2(aq)


Criteria for Balancing Ionic and Net Ionic Equations

1. Material Balance

▪ There must be the same number of atoms of each kind on both sides of the arrow

2. Electrical Balance

▪ The net electrical charge on the left must equal the net electrical charge on the right

▪ Charge does not have to be zero

44


Learning Check: Balancing Equations for Mass & Charge

Balance molecular equation for mass

2Na3PO4(aq) + 3Pb(NO3)2(aq) →

6NaNO3(aq) + Pb3(PO4)2(s)

▪ Can keep polyatomic ions together when counting

Balance ionic equation for charge

6Na+(aq) + 2PO43–(aq) + 3Pb2+(aq) + 6NO3

–(aq) →6Na+(aq) + 6NO3

–(aq) + Pb3(PO4)2(s)

▪ Charge must add up to zero on both sides.

Net ionic equation balanced for mass and charge

3Pb2+(aq) + 2PO43–(aq) ⎯→ Pb3(PO4)2(s)

45


Acids and Bases

▪ Common laboratory reagents▪ Also found in food and household products

▪ vinegar, citrus juice, and cola contain acids▪ drain cleaners and ammonia contain bases

▪ Acids▪ Tart, sour taste

▪ Bases▪ Bitter taste and slippery feel

▪ Caution: Never taste, feel, or smell laboratory chemicals

46


▪ Substance that reacts with water to produce the hydronium ion, H3O

+

Acid + H2O ⎯→ Anion + H3O+

HA + H2O ⎯→ A– + H3O+

Arrhenius Acid

47

HCl(g) + H2O Cl–(aq) + H3O+(aq)⎯→


Ionization reaction definition

▪ Ions form where none have been before

Arrhenius Acid

Another exampleHC2H3O2(aq) + H2O H3O

+(aq) + C2H3O2−(aq)⎯→

48


Arrhenius Base

▪ Substance that produces OH–

▪ Ionic substances containing OH– or O2-

▪ Molecular substances

Ionic compound containing OH–

a. Metal hydroxides

▪ Dissociate into metal and hydroxide ions

NaOH(s) ⎯→ Na+(aq) + OH–(aq)

Mg(OH)2(s) ⎯→ Mg2+(aq) + 2OH–(aq)

49


Strong Acids

▪ Dissociate completely when dissolved in water

e.g., HBr(g) + H2O ⎯→ H3O+(aq) + Br–(aq)

▪ Good electrical conduction (i.e., strong electrolytes)

HClO4(aq) perchloric acid

HClO3(aq) chloric acid

HCl(aq) hydrochloric acid

HBr(aq) hydrobromic acid

HI(aq) hydroiodic acid

HNO3(aq) nitric acid

H2SO4(aq) sulfuric acid

50


Strong Bases

▪ Bases that dissociate completely in water

▪ Soluble metal hydroxides

▪ KOH(aq) ⎯→ K+(aq) + OH–(aq)

▪ Good electrical conductors (i.e., strong electrolytes)

▪ Behave as aqueous ionic compounds

▪ Common strong bases are:

▪ Group 1A metal hydroxides

▪ LiOH, NaOH, KOH, RbOH, CsOH

▪ Group 2A metal hydroxides

▪ Ca(OH)2, Sr(OH)2, Ba(OH)2

51


Weak Acids

▪ Any acid other than seven strong acids

▪ Are also weak electrolytes, i.e, ionize < 100%

Organic acids

HC2H3O2(aq) + H2O ⎯→ H3O+(aq) + C2H3O2

–(aq)

e.g., HCO2H(aq) + H2O → H3O

+(aq) + HCO2–(aq)

52

Only this H comes off as H+

Acetic AcidMolecule,HC2H3O2

Acetate ion, C2H3O2–


Why is Acetic Acid Weak?

53

CH3COO–(aq) + H3O+ (aq) → CH3COOH (aq) + H2O

CH3COOH(aq) + H2O → CH3COO-(aq) + H3O+(aq)


Your Turn

Which of the following is a weak acid?

A. HCl (hydrochloric acid)

B. HNO3(nitric acid)

C. HClO4 (perchloric acid)

D. HC2H3O2 (acetic acid)

E. H2SO4 (sulfuric acid)

54


Your Turn

Which of the following is a strong acid?

A. HF (hydrofluoric acid)

B. HClO3 (chloric acid)

C. H3PO4 (phosphoric acid)

D. HNO2 (nitrous acid)

E. H2SO3 (sulfurous acid)

55


Dynamic Equilibrium

▪ Two opposing reactions occurring at same rate

▪ Also called chemical equilibrium

Equilibrium

▪ Concentrations of substances present in solution do not change with time

Dynamic

▪ Both opposing reactions occur continuously

▪ Represented by double arrow

HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2

–(aq)

Forward reaction – forms ions

Reverse reaction – forms molecules56


Arrhenius Bases

2. Molecular Bases

▪ Undergo ionization (hydrolysis) reaction to form hydroxide ions

Base + H2O ⎯→ BaseH+(aq) + OH–(aq)

B + H2O ⎯→ BH+(aq) + OH–(aq)

NH3(aq) + H2O ⎯→ NH4+(aq) + OH–(aq)

57

NH3 H2O NH4+ OH–⎯→


Weak Bases

▪ Molecular bases

▪ Do not dissociate

▪ Accept H+ from water inefficiently

▪ Are weak electrolytes

e.g.,

NH3(aq) + H2O NH4+(aq) + OH–(aq)

58


Equilibrium for Weak Base

Forward reaction

59

Reverse reaction

Net is dynamic equilibrium

NH3(aq) + H2O NH4+(aq) + OH–(aq)


General Ionization Equations

▪ Strong acid in waterHX(aq) + H2O ⎯→ H3O

+(aq) + X–(aq)

▪ Strong base, M(OH)n

M(OH)n ⎯→ Mn+(aq) + nOH–(aq)

▪ Weak acid in waterHA(aq) + H2O H3O

+(aq) + A–(aq)

▪ Weak base in waterB(aq) + H2O HB+(aq) + OH–(aq)

60


Learning Check

▪ Write the ionization equation for each of the following with water:

1.Weak base methylamine, CH3NH2

2.Weak acid nitrous acid, HNO2

3.Strong acid chloric acid, HClO3

4.Strong base strontium hydroxide, Sr(OH)2

61

CH3NH2(aq) + H2O CH3NH3+(aq) + OH–(aq)

HClO3(aq) + H2O ⎯→ H3O+(aq) + ClO3

–(aq)

Sr(OH)2(aq) ⎯→ Sr2+(aq) + 2OH–(aq)

HNO2(aq) + H2O H3O+(aq) + NO2

–(aq)


Brief summary

▪ Strong acids and bases are strong electrolytes

▪ Weak acids and bases are weak electrolytes

▪ Strong electrolyte

▪ Completely ionizes

▪ Forward reaction dominates

▪ Mostly products

▪ Strong acids & bases

▪ Little reverse reaction

▪ Write eqn. as ⎯→

62

▪ Weak electrolyte

▪ Small % ionizes

▪ Reverse rxn dominates

▪ Mostly reactants

▪ Weak acids and bases

▪ Lots of reverse reaction

▪ Write eqn. as


Polyprotic Acids

Monoprotic Acids

▪ Furnish only one H+

HNO3(aq) + H2O ⎯→ H3O+(aq) + NO3

–(aq)

HC2H3O2(aq) + H2O → H3O+(aq) + C2H3O2

–(aq)

Diprotic acids — furnish two H+

H2SO3(aq) + H2O ⎯→ H3O+(aq) + HSO3

–(aq)

HSO3–(aq) + H2O ⎯→ H3O

+(aq) + SO32–(aq)

Polyprotic acids

▪ Furnish more than one H+

63


Polyprotic Acids

Polyprotic acids

▪ Triprotic acids — furnish three H+

H3PO4 ⎯→ H2PO4– ⎯→ HPO4

2– ⎯→ PO43–

▪ Stepwise equations

H3PO4(aq) + H2O ⎯→ H3O+(aq) + H2PO4

–(aq)

H2PO4–(aq) + H2O ⎯→ H3O

+(aq) + HPO42–(aq)

HPO42–(aq) + H2O ⎯→ H3O

+(aq) + PO43–(aq)

Net:

H3PO4(aq) + 3H2O ⎯→ 3H3O+(aq) + PO4

3–(aq)

64

– H+ – H+ – H+


Learning Check

▪ Write the stepwise ionization reactions for citric acid, H3C6H5O7, in water.

65

H3C6H5O7(aq) + H2O ⎯→ H3O+(aq) + H2C6H5O7

–(aq)

HC6H5O72-(aq) + H2O ⎯→ H3O

+(aq) + C6H5O73-(aq)

H2C6H5O7–(aq) + H2O ⎯→ H3O

+(aq) + HC6H5O72-(aq)


Acidic Anhydrides

Nonmetal Oxides

▪ Act as Acids

▪ React with water to form molecular acids that contain hydrogen

SO3(g) + H2O ⎯→ H2SO4(aq)

sulfuric acid

N2O5(g) + H2O ⎯→ 2HNO3(aq)

nitric acid

CO2(g) + H2O ⎯→ H2CO3(aq)

carbonic acid66


Ionic Oxidesb. Basic Anhydrides

▪ Soluble metal oxides

▪ Undergo ionization (hydrolysis) reaction to form hydroxide ions

▪ Oxide reacts with water to form metal hydroxide

CaO(s) + H2O ⎯→ Ca(OH)2(aq)

▪ Then metal hydroxide dissociates in water

Ca(OH)2(aq) ⎯→ Ca2+(aq) + 2OH–(aq)

67

2OH–O2– + H2O ⎯→


Your Turn

Which of the following is not a strong base?

A. NaOH

B. CH3NH2

C. Cs2O

D. Ba(OH)2

E. CaO

68


Your Turn

Which of the following is an acid?

A. NaO2

B. SO2

C. CH3NH2

A. Ba(OH)2

B. CaO

69


Your Turn

Which of the following is a base?

A. N(CH3)3

B. SO2

C. CH3COOH

D. HF

E. HNO2

70

N(CH3)3(aq) + H2O ⎯→ HN(CH3)3+(aq) + OH–(aq)

NH3(aq) + H2O ⎯→ NH4+(aq) + OH–(aq)

Just like ammonia, NH3


Acid—Base Nomenclature

▪ System for naming acids and bases

Acids

▪ Binary acid system e.g., HCl(aq), H2S(aq)

▪ Oxoacid system e.g., H2SO4, HClO2

▪ Acid salt system e.g., NaHSO4, NaHCO3

Bases

▪ Metal hydroxide/oxide system e.g., NaOH, CaO

▪ Molecular base system e.g., NH3, (CH3)3N

71


Naming Acids

A. Binary Acids — hydrogen + nonmetal

▪ Take molecular name

▪ Drop –gen from H name

▪ Merge hydro– with nonmetal name

▪ Replace –ide with –ic acid

Name of Molecular Compound

Name of Binary Acid in water

HCl(g) hydrogen chloride HCl(aq) hydrochloric acid

H2S(g) hydrogen sulfide H2S(aq) hydrosulfuricacid

72


Naming Acids

B. Oxo Acids

▪ Acids with hydrogen, oxygen and another nonmetal element

▪ A table of polyatomic ions can be found in the book

▪ To name:

▪ Based on parent oxoanion name

▪ Take parent ion name

▪ Anion ends in –ate change to –ic (more O's)

▪ Anion ends in –ite change to –ous (less O's)

▪ End name with acid to indicate H+

73


Oxoacids (Aqueous)

Named according to the anion suffix

▪ Anion ends in -ite, acid name is -ous acid

▪ Anion ends in -ate, acid name is -ic acid

Name of Parent Oxoanion

Name of Oxoacid

NO3− HNO3

SO42− H2SO4

ClO2− HClO2

PO32− H2PO3

74

sulfate

chlorite

phosphite

sulfuric acid

chlorous acid

phosphorous acid

nitrate nitric acid


Learning Check: Name Each Aqueous Acid

▪ HNO2

▪ HCN

▪ HClO4

▪ HF

▪ H2CO3

▪ nitrous acid

▪ hydrocyanic acid

▪ perchloric acid

▪ hydrofluoric acid

▪ carbonic acid

75


Your Turn

What is the correct name for HClO4 (aq)?

A. chloric acid

B. hydrochloric acid

C. perchloric acid

D. hypochlorous acid

E. chlorous acid

76


Your Turn

What is the correct name for H2SO3(aq)?

A. sulfuric acid

B. sulfurous acid

C. hydrosulfuric acid

D. hydrosulfurous acid

E. hydrogen sulfite acid

77


C. Naming BasesOxides & Hydroxides

▪ Ionic compounds

▪ Named like ionic compounds

▪ Ca(OH)2 calcium hydroxide

▪ Li2O lithium oxide

Molecular Bases

▪ Named like molecules

▪ NH3 ammonia

▪ CH3NH2 methylamine

▪ (CH3)2NH dimethylamine

▪ (CH3)3N trimethylamine78


1. Predicting Precipitation Reactions

Metathesis Reaction

▪ Reactions where anions and cations exchange partners.

▪ Also called double replacement reaction

▪ Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq)

Precipitation reactions

▪ Metathesis reactions where precipitate forms

How can we predict if compounds are insoluble?

▪ Must know solubility rules

79


Table 4.1 Solubility Rules

Soluble Compounds

1. All salts of the alkali metals (Group 1A) are soluble.

2. All salts containing NH4+, NO3

–, ClO4–, ClO3

–, and C2H3O2

– are soluble.

3. All chlorides, bromides, and iodides (salts containing Cl–, Br–, or I–) are soluble except when combined with Ag+, Pb2+, and Hg2

2+

(note the subscript 2).

4. All salts containing SO42– are soluble except those

of Pb2+, Ca2+, Sr2+, Ba2+, and Hg22+.

80


Table 4.1 Solubility RulesInsoluble Compounds

5. All metal hydroxides (ionic compounds containing OH−) and all metal oxides (ionic compounds containing O2− are insoluble except those of Group 1A and those of Ca2+, Sr2+, and Ba2+.

▪ When metal oxides do dissolve, they react with water to form hydroxides. The oxide ion, O2−, does not exist in water. For example:

Na2O(s) + H2O ⎯→ 2NaOH(aq)

6. All salts containing PO43–, CO3

2–, SO32– and S2–

are insoluble except those of Group 1A and NH4+

81


Learning Check: Solubility Rules

82

Which of the following compounds are

expected to be soluble in water?

Ca(C2H3O2)2

FeCO3

AgCl

Yes

No

No


Metathesis (Double Replacement) Reaction

AB + CD → AD + CB

▪ Cations and anions change partners

▪ Charges on each ion don’t change

▪ Formulas of products are determined by charges of reactants

▪ Occurs only if solid, gas, weak electrolyteor non-electrolyte product forms

▪ Otherwise, all ions are spectator ions

83


Predicting Products of Double Replacement Reactions

1. Identify the ions involved:

▪ Distinguish between subscripts that count ions and those that are characteristic of a polyatomic ion.

2. Swap partners and make neutral with appropriate subscripts

3. Assign states using solubility rules

4. Balance equation

84

HCl(aq)+ Ca(OH)2(aq) ⎯→

ions: H+, Cl– Ca2+, 2OH –

counting subscript

CaCl2 + H2O(aq) 22


Predict if Ionic Reaction Occurs

1. Write molecular equation for metathesis reaction

2. Determine which ion combinations form insoluble salt, water, weak electrolyte, non-electrolyte, or gas.

3. Translate molecular equation into ionic equation

4. Cancel spectator ions, to give net ionic equation

5. Check for driving force: formation of weak electrolyte, solid, gas, water or non-electrolyte

85


Your Turn

What will be the solid product of the reaction of Ca(NO3)2(aq) + Na2CO3(aq) ⎯→ ?

A. CaCO3

B. NaNO3

C. Na(NO3)2

D. Na2(NO3)2

E. H2O

86


Learning Check: Predict Products

Pb(NO3)2(aq) + Ca(OH)2(aq) ⎯→

BaCl2(aq) + Na2CO3(aq) ⎯→

2Na3PO4(aq) + 3Hg2(NO3)2(aq) ⎯→

2 NaCl(aq) + Ca(NO3)2(aq) ⎯→

Pb(OH)2(s) + Ca(NO3)2(aq)

87

BaCO3(s) + 2NaCl(aq)

6NaNO3(aq) + (Hg2)3(PO4)2(s)

NR (No reaction)

CaCl2(aq) + 2NaNO3(aq)


Learning Check▪ Predict the reaction that will occur when aqueous

solutions of Cd(NO3)2 and Na2S are mixed. Write molecular, ionic and net ionic equations.

Molecular equation:

Cd(NO3)2(aq) + Na2S(aq) ⎯→

Ionic equation:

Cd2+(aq) + 2NO3–(aq) + 2Na+(aq) + S2–(aq) ⎯→

Net ionic equation:

Cd2+(aq) + S2–(aq) ⎯→ CdS(s)

88

CdS(s) + 2NaNO3(aq)

CdS(s) + 2NO3–(aq) + 2Na+(aq)


Learning Check▪ Write molecular, ionic and net ionic equations for the

reaction that occurs when Pb(NO3)2 and Fe2(SO4)3

are mixed in solution.

Molecular equation

3Pb(NO3)2(aq) + Fe2(SO4)3(aq) → PbSO4(s) + 2Fe(NO3)3(aq)

Ionic equation

3Pb2+(aq) + 6NO3–(aq) + 2Fe3+(aq) + 6SO4

2–(aq) →

2Fe3+(aq) + 6NO3–(aq) + PbSO4(s)

Net ionic equation

3Pb2+(aq) + 6SO42–(aq) ⎯→ 3PbSO4(s)

Pb2+(aq) + 2SO42–(aq) ⎯→ PbSO4(s)

89

1 12


Your TurnWill mixing aqueous solutions of Mg(C2H3O2)2 and CsCl yield a precipitate?

A. Yes

B. No

Molecular equation:

Mg(C2H3O2)2(aq) + 2CsCl(aq) →

MgCl2(aq) + 2CsC2H3O2(aq)

Ionic equation:

Mg2+(aq) + 2C2H3O2–(aq) + 2Cs+(aq) + 2Cl–(aq) →

Mg2+(aq) + 2Cl–(aq) + 2Cs+(aq) + 2C2H3O2–(aq)

90


Your TurnWill mixing aqueous solutions of K2SO3 and Ba(NO3)2 yield a precipitate?

A. Yes

B. No

Molecular equation:

K2SO3(aq) + Ba(NO3)2(aq) → BaSO3(s) + 2KNO3(aq)

Ionic equation:

2K+(aq) + SO32-(aq) + Ba2+(aq) + 2NO3

–(aq) →

BaSO3(s) + 2K+(aq) + 2NO3–(aq)

91


Your TurnWill mixing aqueous solutions of NH4OH and Zn(ClO3)2 yield a precipitate?

A. Yes

B. No

Molecular equation:

2NH4OH(aq) + Zn(ClO3)2(aq) →

Zn(OH)2(s) + 2NH4ClO3(aq)

Ionic equation:

2NH4+(aq) + 2OH–(aq) + Zn2+(aq) + 2ClO3

–(aq) →

Zn(OH)2(s) + 2NH4+(aq) + 2ClO3

–(aq)

92


2. Predicting Acid−Base Reactions

Neutralization reaction

▪ Combining an acid and base to form a salt and water

Salt

▪ Ionic compound formed by a neutralization reaction

▪ Acid + Base ⎯→ Salt + Water

HClO4(aq) + NaOH(aq) → NaClO4(aq) + H2O

93


2. Predicting Acid−Base Reactions

Neutralization reaction

▪ Can be viewed as a metathesis reaction

HClO4(aq) + NaOH(aq) → NaClO4(aq) + H2O

Ionic equationH+(aq) + ClO4

–(aq) + Na+(aq) + OH–(aq) →H2O + Na+(aq) + ClO4

–(aq)

Net ionic equation

H+(aq) + OH–(aq) → H2O

94


Neutralization Between Strong Acid and Strong Base

Molecular equation

2HCl(aq) + Ca(OH)2(aq) →

Ionic equation

2H+(aq) + 2Cl–(aq) + Ca2+(aq) + 2OH–(aq) →2H2O + Ca2+(aq) + 2Cl–(aq)

Net ionic equation

2H+(aq) + 2OH–(aq) → 2H2O

H+(aq) + OH–(aq) → H2O

True for any strong acid and strong base

2H2O + CaCl2(aq)

95


Weak Acid with Strong Base

Molecular Equation:

HC2H3O2(aq) + NaOH(aq) →

Ionic Equation:

HC2H3O2(aq) + Na+(aq) + OH–(aq) →

H2O +Na+(aq) + C2H3O2–(aq)

Net Ionic Equation:

HC2H3O2(aq) + OH–(aq) → H2O + C2H3O2–(aq)

96

H2O + NaC2H3O2(aq)


Neutralization of Strong Acid with Insoluble Base

Insoluble Hydroxides

Molecular Equation

Mg(OH)2(s) + 2HCl(aq) →

Ionic Equation

Mg(OH)2(s) + 2H+(aq) + 2Cl–(aq) →

Mg2+(aq) + 2Cl–(aq) + 2H2O

Net Ionic Equation

Mg(OH)2(s) + 2H+(aq) → Mg2+(aq) + 2H2O 97

MgCl2(aq) + 2H2O


Neutralization of Strong Acid with Insoluble Base

Insoluble Oxides – Basic Anhydrides

Molecular Equation

Al2O3(s) + 6HCl(aq) →

Ionic Equation

Al2O3(s) + 6H+(aq) + 6Cl–(aq) →

2Al3+(aq) + 6Cl–(aq) + 3H2O

Net Ionic Equation

Al2O3(s) + 6H+(aq) ⎯→ 2Al3+(aq) + 3H2O

98

2AlCl3(aq) + 3H2O


Strong Acid with Weak Base

Molecular equation:

NH3(aq) + HCl(aq) ⎯→

Ionic equation :

NH3(aq) + H+(aq) + Cl–(aq) → NH4+(aq) + Cl–(aq)

Net ionic equation :

NH3(aq) + H+(aq) ⎯→ NH4+(aq)

99

NH4Cl(aq)

NH3(aq) + H3O+(aq) → NH4

+(aq) + H2O


Learning Check

100

What is the net ionic equation for the reaction between the following reactants:

1. HNO3(aq) + Ca(OH)2(aq) ⎯→

H+(aq) + OH–(aq) ⎯→ H2O

2. N2H4(aq) + HI(aq) ⎯→

N2H4(aq) + H+(aq) ⎯→ N2H5+(aq)

3. CH3NH2(aq) + HC4H7O2(aq) ⎯→

CH3NH2(aq) + HC4H7O2(aq) →

CH3NH3+(aq) + C4H7O2

–(aq)


Your Turn

What is the net ionic equation for the reaction:

NaOH(aq) + HF(aq) → ?

A. Na+(aq) + OH–(aq) + H+(aq) + F–(aq) →

H2O + NaF(aq)

B. OH–(aq) + H+(aq) → H2O

C. Na+(aq) + OH–(aq) + HF(aq) → H2O + NaF(aq)

D. OH–(aq) + HF(aq) → H2O + F–(aq)

E. No reaction

101


Your Turn

What is the net ionic equation for the reaction:

NH3(aq) + CH3COOH(aq) → ?

102

A. NH2–(aq) +H+(aq) + CH3COOH(aq) → NH3(aq) +

HCH3COOH(aq)

B. NH3(aq) + CH3COO–(aq) + H+(aq) → NH4+(aq) +

CH3COO–(aq)

C. NH3(aq) + CH3COO–(aq) + H+(aq) →

NH4CH3COO(s)

D. NH3(aq) + CH3COOH(aq) → NH4+(aq) + CH3COO– (aq)

E. No reaction


Challenge ProblemWhat is the net ionic equation for reaction of an insoluble hydroxide and a weak acid?

Molecular equation

Mg(OH)2(s) + 2HC2H3O2(aq) ⎯→

Ionic equation

Mg(OH)2(s) + 2HC2H3O2(aq) ⎯→

Mg2+(aq) + 2H2O + 2C2H3O2–(aq)

▪ There are NO spectator ions!

▪ So net ionic and ionic equations are the same in this case

103

Mg(C2H3O2)2(aq) + 2H2O


Acid Salts

Polyprotic acids can be neutralized stepwise

▪ Can halt neutralization at each step

▪ Name must specify number of hydrogens remaining in the salt

Acid salt

▪ Formula contains a cation, a hydrogen, and an anion

▪ The acid salt can react with a base

H2SO4(aq) + KOH(aq) ⎯→ KHSO4(aq) + H2O(l )

acid salt104


Naming Acid Salts—Polyprotic

▪ Must specify number of hydrogens still attached to the anion

▪ Can be neutralized by additional base

e.g., Na2HPO4

NaH2PO4

KHSO4

▪ Some acid salts have common names

▪ NaHCO3

105

sodium hydrogen carbonate

or sodium bicarbonate

sodium hydrogen phosphate

sodium dihydrogen phosphate

potassium hydrogen sulfate


Metathesis and Gas Formation

▪ Metathesis reactions involving certain ions lead to formation of a gas

▪ Low solubility of gas in solvent (water) leads to escape of gas

▪ Once a gas escapes, it cannot redissolve. This drives the reaction to completion

▪ Many compounds that contain anions that give rise to gases are insoluble

▪ Adding an acid to these anions forms the gas

106


Metathesis and Gas Formation

1. Gases formed CO2, SO2, NH3 by metathesis

▪ H2S, HCN

2. Unstable compounds—decompose and form gas

▪ H2CO3 ⎯→ H2O and CO2(g)

▪ H2SO3 ⎯→ H2O and SO2(g)

▪ NH4OH ⎯→ H2O and NH3(g)

107


Reactions that Release CO2

a) Acid with Bicarbonate (HCO3–)

NaHCO3(aq) + HI(aq) → NaI(aq) + H2O + CO2(g)

b) Acid with Carbonate (CO32–)

CaCO3(aq) + 2HCl(aq) → CaCl2(aq) + H2O + CO2(g)

108

a) b)


Reactions that Release Gases

Acid with Sulfites (SO32–) or Bisulfites (HSO3

2–)

K2SO3(aq) + 2HClO4(aq) → SO2(g) + 2KClO4(aq) + H2O

LiHSO3(aq) + HClO3(aq) → SO2(g) + H2O + LiClO3(aq)

Acid with Sulfides

2HCl(aq) + Na2S(aq) → 2NaCl(aq) + H2S(g)

Acid with Cyanides

HNO3(aq) + CsCN(aq) → HCN(g) + CsNO3(aq)

Bases with Ammonium salts

NaOH(aq) + NH4Cl(aq) →

NH3(g) + H2O + NaCl(aq)

109


Learning Check Write the molecular, ionic and net ionic equations for the reaction of Li2SO3 with formic acid, HCHO2

Molecular equation:

Li2SO3(aq) + 2HCHO2(aq) →

2LiCHO2(aq) + SO2(g) + H2O

Ionic equation:

2Li+(aq) + SO32–(aq) + 2HCHO2(aq) →

2CHO2–(aq) + 2Li+(aq) + SO2(g) + H2O

Net ionic equation:

SO32 –(aq) + 2HCHO2(aq) →

2CHO2–(aq) + SO2(g) + H2O

110


Your Turn

What is the net ionic equation for the reaction of HCl with KHCO3?

A. HCl(aq) + KHCO3(aq) ⎯→ KCl(aq) + H2CO3(aq)

B. H+(aq) + HCO3–(aq) ⎯→ H2CO3(aq)

C. HCl(aq) + KHCO3(aq) → KCl(aq) + CO2(g) + H2O

D. H+(aq) + Cl–(aq) + K+(aq) + HCO3–(aq) ⎯→

K+(aq) + Cl–(aq) + CO2(g) + H2O

E. H+(aq) + HCO3–(aq) ⎯→ CO2(g) + H2O

111


Your Turn

What is the net ionic equation for the reaction of HBr with K2SO3?

A. H+(aq) + SO32-(aq) ⎯→ SO2(g) + OH-(aq)

B. H+(aq) + Br–(aq) + 2K+(aq) + SO32-(aq) ⎯→

2K+(aq) + Br–(aq) + SO2(g) + H2O

C. 2H+(aq) + SO32-(aq) ⎯→ SO2(g) + H2O

D. HBr(aq) + K+(aq) → KBr(s)

E. H+(aq) + SO32-(aq) ⎯→ SO2(g) + H2O

112


Metathesis Overview

Precipitation:

▪ Two solutions form solid product

Neutralization:

▪ acid + base → salt + water

Gas-forming:

▪ Metathesis reaction forms one of these products:

▪ HCN, H2S, H2CO3(aq) , H2SO3(aq) , NH3(aq)

Formation of Weak Electrolyte:

▪ Salt of weak acid reacts with acid to form molecule

113


Predicting Reactions and Writing Their Equations

What reaction, if any, occurs between potassium nitrate and ammonium chloride?

▪ Need to know whether net ionic equation exists.

1.Determine formulas of reactants

▪ KNO3 + NH4Cl ⎯→ ?

2.Write molecular equation

▪ KNO3 + NH4Cl ⎯→ KCl + NH4NO3

3.Check solubilities

▪ All are soluble

114


Predicting Reactions and Writing Their Equations

▪ Predicted molecular equation

▪ KNO3(aq) + NH4Cl(aq) ⎯→ KCl(aq) + NH4NO3(aq)

▪ Write ionic equation

▪ K+(aq) + NO3–(aq) + NH4

+(aq) + Cl–(aq) ⎯→

K+(aq) + Cl–(aq) + NH4+(aq)

+ NO3–(aq)

▪ Same on both sides

▪ All ions cancel out

▪ No gases, solids, water, or weak electrolytes formed

115


Learning Check

Determine the net ionic equation for the following reactions.

1. Co(OH)2 + HNO2

Co(OH)2(s) + 2H+(aq) → Co2+(aq) + 2H2O

2. KCHO2 + HCl

CHO2–(aq) + H+(aq) → HCHO2(aq)

3. CuCO3 + HC2H3O2

CuCO3(s) + 2HC2H3O2(aq) →

Cu2+(aq) + CO2(g) + C2H3O2–(aq) + H2O

116


Your Turn

What is the net ionic reaction when aqueous solutions of NaOH and NiCl2 are mixed?

A. Ni2+(aq) + 2OH–(aq) → Ni(OH)2(s)

B. NaOH(aq) + NiCl2(aq) → NaCl(aq) + Ni(OH)2(s)

C. 2NaOH(aq) + NiCl2(aq) → 2NaCl(aq) + Ni(OH)2(s)

D. 2Na+(aq) + 2OH–(aq) + Ni2+(aq) + 2Cl–(aq) →

2Na+(aq) + 2Cl–(aq) + Ni(OH)2(s)

E. No reaction

117


Your Turn

Which of the following combinations will not react?

A. Na2CO3(aq) + HCl(aq)

B. Na2SO3(aq) + CaCl2(aq)

C. NaCl(aq) + HC2H3O2(aq)

D. NH4Cl(aq) + Ba(OH)2(aq)

E. KCN(aq) + H2SO4(aq)

118


Synthesize Salts via Metathesis Reactions

▪ Practical use of metathesis reactions

▪ Desired compound should be easily separated from reaction mixture. Three principal approaches

1. Desired compound is insoluble in water

▪ Start with two soluble reactants

▪ Product isolated by filtration

2. Desired compound is soluble in water

▪ Acid-base neutralization

▪ Reaction of metal carbonate or other gas forming anion and acid

▪ Product isolated by evaporation of water119


Synthesize Salts via Metathesis Reactions

3. Desired compound is soluble in water

▪ Add acid to supply desired anion (e.g., HCl for Cl–)

▪ Add excess metal carbonate to supply the metal

(e.g., Na2CO3 for Na+)

▪ CO32- reacts with H+ to form CO2(g), which is not

soluble in water and leaves the system.

▪ Metal sulfides and sulfites also work

▪ Avoided because H2S and SO2 are poisonous

▪ Product isolated by evaporation of water

120


Learning Check

What reaction might we use to synthesize nickel sulfate, NiSO4?

Use solubility rules

▪ NiSO4 is soluble in water

So, there are two possible methods

▪ Use acid + base

H2SO4(aq) + Ni(OH)2(s) ⎯→ NiSO4(aq) + 2H2O

▪ Use acid + carbonate

H2SO4(aq) + NiCO3(s) → NiSO4(aq) + CO2(g) + 2H2O

121


Molarity (M)

122

▪ Number of moles of solute per liter of solution.

▪ Allows us to express relationship between moles of solute and volume of solution

▪ Hence, 0.100 M solution of NaCl contains 0.100 mole NaCl in 1.00 liter of solution

▪ Same concentration results if you dissolve 0.0100 mol of NaCl in 0.100 liter of solution

0.100 mol NaCl

1.00 L NaCl soln=

0.0100 mol NaCl

0.100 L NaCl soln= 0.100 M NaCl


Molar Concentration

Dissolve solutes. Make separate solutions

Mix Solutions

Allow Reaction to occur

▪ Need to know quantitatively HOW MUCH of each solute we used.

▪ Define

123

Molarity (M) =moles of solute

liters of solution=

mole

volume


Molarity as Conversion Factor

▪ Often have stoichiometry problems involving amount of chemical and volume of solution

▪ Solve the problem using molarity

▪ Molarity provides conversion factors between moles and volume

▪ M = mole per liter

124

Molarity

Moles of a substance

Volume of a solution of substance

0.100 M NaCl =0.100 mol NaCl

1.00 L NaCl soln


Molarity as Conversion Factor

▪ Gives equivalence relationship between “mol NaCl” and “L soln”

▪ Forms two conversion factors

▪ Three basic types of calculations:

125

soln NaClL 1.00

NaCl mol 0.100

NaCl mol 0.100

soln NaClL 1.00

0.100 mol NaCl 1.00 L soln

M =mol

VM ´V = mol V =

mol

M


Learning Check: Calculating Molarity (from grams and volume)

Calculate the molarity (M) of a solution prepared by dissolving 11.5 g NaOH (40.00 g/mol) solid in enough water to make 1.50 L of solution.

g NaOH ⎯→ mol NaOH ⎯→ M NaOH

126

NaOH mol 288.0NaOH g 00.40

NaOH mol 1 NaOH g 5.11 =

M =moles NaOH

L soln=

0.288 mol NaOH

1.50 L soln

= 0.192 M NaOH


Learning Check: Calculating Volume (from Molarity and moles)

How many mL of 0.250 M NaCl solution are needed to obtain 0.100 mol of NaCl?

▪ Use M definition

▪ Given molarity and moles, need volume

127

soln NaClL 00.1

NaCl mol 250.0 NaCl 250.0 =M

= 400 mL of 0.250 M NaCl solution

soln NaClL 1

soln NaClmL 1000

NaCl mol 250.0

soln NaClL 00.1 NaCl mol 100.0


Learning Check: Calculating Moles (from Molarity and Volume)

How many moles of Ca(NO3)2 are in 250 mL of 0.150 M of NaCl?

▪ Use M definition

▪ Given molarity and volume, find moles

128

soln )Ca(NO L 00.1

)Ca(NO mol 150.0 )Ca(NO 150.0

23

2323 =M

= 3.75 x 10-2 moles of Ca(NO3)2

=soln )Ca(NO L 1

)Ca(NO mol 150.0 soln )Ca(NO L .2500

23

2323

Convert mL to Liters


Preparing Solution of Known Molarity

a b c d e f

a) Weigh solid and transfer to volumetric flask

b) Add part of the water

c) Dissolve solute completely

d) Add water to reach etched line

e) Stopper flask and invert to mix thoroughly

129

a) b) c) d) e)


Learning Check: Preparing Solution of Known Molarity from Solid

How many grams of strontium nitrate are required to prepare 250.0 mL of 0.100 M Sr(NO3)2 solution?

M × V ⎯→ mol × MM ⎯→ g

1. Convert molarity and volume to mole

= 0.0250 mol Sr(NO3)2

2. Convert mol to g

130

L 1

100.0

mL 1000

L 1soln )Sr(NOmL 250 23

M

= 5.29 g Sr(NO3)2 mol 1

g 11.622 mol 0250.0


Your TurnHow many grams of KMnO4 must you weigh out if you want to make 250.mL of a 0.200 M KMnO4

solution?

A. 7900 g

B. 50.0 g

C. 0.316 g

D. 7.90 g

E. 198 g

131

g 09.7KMnO mol

g 03.158

L 1

mol 200.0

mL 1000

L 1mL 250

4

=


Preparing a Solution of Known Molarity by Dilution

▪ Can take solution of higher concentration and dilute it to a lower concentration.

▪ Amount of MOLES does NOT changeRemains the same

132

Small

Volume

Concentrated

Solution

Large

Volume

Dilute

Solution

Add solvent


Diluting Solutions

Volume of

dilute solution

to be prepared

æ

è

ççç

ö

ø

÷÷÷´M

dilute=

Volume of

concentrated solution

to be used

æ

è

ççç

ö

ø

÷÷÷´M

concentrated

▪ Moles of solute do not change upon dilution

▪ Just changing volume

Number of moles in dilute = number of moles in concentrated

133

Moles of solute in the dilute solution

Moles of solute in the concentrated solution

Vdil Mdil = Vconc Mconc


Learning Check: Dilutions

What volume (in mL) of 16.0 M H2SO4 must be used to prepare 1.00 L of 2.00 M H2SO4?

134

1.00 L ´2.00 M =V ´16.0 M

V =1.00 L ´ 2.00 M

16.0 M =

2.00 mol

16.0 mol/L

= 125 mL

Rearranging gives

V = 0.125 L ´1000 mL

1.00 L


Preparing Solution of Known M▪ Using volumetric glassware ensures that the

volumes are known precisely

▪ Use a volumetric pipette to transfer the stock solution

▪ Use a volumetric flask to receive the final solution

135


Your TurnWhat volume of 12.1 M HCl is needed to create 250. mL of 3.2 M HCl?

A. 66 mL

B. 800 mL

C. 3025 mL

D. 945 mL

E. 9680 mL

136

250. mL ´ 3.2 M =Vconc

´12.1 M

Vconc = 66 mL


Your Turn

A 25 mL of 6.0 M HCl is diluted to 500 mL with water. What is the molarity of the resulting solution?

A. 150 M

B. 3.0 M

C. 0.120 M

D. 120 M

E. 0.30 M

500 mL ´Mdil

= 25 mL ´ 6.0 M

137

Mdil = 0.30 M


Solution Stoichiometry

▪ Often work with solutions when conducting reactions

▪ How do we determine amounts needed to completely react one compound?

▪ Like any other stoichiometry problem

▪ Now use volume and molarity to obtain moles

of each substance.

138


Solution Stoichiometry

Reactantmolarity

Productmolarity

mole-to-mole ratio

Volume of reactant Moles of reactant

Moles of product Volume of product

▪ General scheme

139


Learning Check: Solution Stoichiometry

How many milliliters of 0.0475 M H3PO4 could be completely neutralized by 45.0 mL of 0.100 M KOH? The balanced equation for the reaction is

H3PO4(aq) + 3KOH(aq) ⎯→ K3PO4(aq) + 3H2O

Strategy:

140

KOH solution

Vol and M ofKOH soln

mol KOH mol H3PO4

H3PO4 soln

mol and M ofH3PO4 soln

Coefficients of

Balanced equation


1. Calculate moles of KOH

2. Use coefficients to calculate the moles H3PO4

required

3. Calculate volume of H3PO4 needed

141

Learning Check: Solution Stoichiometry

KOHL 1

KOH mol 100.0

KOHmL 1000

KOHL 1KOHmL 0.45

= 4.50 × 10–3 mol KOH

KOH mol 3

POH mol 1KOH mol10 50.4 433 −

= 1.50 × 10–3 mol H3PO4

L 1

mL 1000

POH mol 0.0475

POHL 1POH mol10 50.1

43

4343

3 −

= 31.6 mL H3PO4


Stoichiometry of Ionic Equations

▪ Sometimes we need to know concentrations of ions

▪ Important for net ionic reaction stoichiometry

▪ Molar concentration of particular ion equals molar concentration of salt multiplied by number of ions of that kind in one formula unit of salt.

142


Learning Check: Ion Concentrations

If you have 0.150 M Na2CO3 (aq), what is the concentration of each type of ion in solution?

Means Na2CO3(aq) → 2 Na+(aq) + CO32–(aq)

Concentration of Na+ ions is:

Concentration of CO32– ions is:

143

0.150 mol Na2CO

3

1 L Na2CO

3

´2 mol Na+

1 mol Na2CO

3

= 0.300 M Na+

0.150 M Na2CO

3´

1 mol CO3

2-

1 mol Na2CO

3

= 0.150 M CO3

2-


Your Turn

If the solution concentration of sulfate ion is 0.750 M, what is the concentration of Al2(SO4)3,

assuming that all of the sulfate ion comes aluminum sulfate?

A. 0.750 M

B. 2.25 M

C. 0.250 M

D. 1.50 M

E. 0.500 M

144

0.750 M SO4

2– ´1 mol Al

2(SO

4)

3

3 mol SO4

2–

= 0.250 M Al2(SO4)3


Learning Check: Net Ionic Eqns in Solution Stoichiometry Calculations

What volume, in mL, of 0.500 M KOH is needed to react completely with 60.0 mL of 0.250 M FeCl2 to form Fe(OH)2 solid?

1. Write Balanced Net Ionic Equation

Fe2+(aq) + 2OH–(aq) ⎯→ Fe(OH)2(s)

2. Determine the game plan

145

M FeCl2

M Fe2+

mol Fe2+ mol OH–

V OH–

V KOH


Learning Check: Net Ionic Eqns in Solution Stoichiometry Calculations

3. Convert M FeCl2 → M Fe2+ → mol Fe2+

4. Convert mol Fe2+ → mol OH–

5. Convert mol OH– → V OH– → V KOH

146

0.250 M FeCl2

´1 mol Fe2+

1 mol FeCl2

´ 60.0 mL ´1 L

1000 mL

−

+

−+ = OH mol 0030.0

Fe mol 1

OH mol 2Fe mol 0015.0

2

2

= 0.0150 mol Fe2+

= 60.0 mL KOH

L 1

mL 1000

OH mol 1

KOH mol 1

OH mol 050.0

soln OHL 1OH mol 0030.0

−−

−−


Learning Check: Solution Limiting Reagent Problem

How many grams of PbI2 (461.0 g/mol) will form if 20.0 mL of 0.800 M FeI3 (436.5 g/mol) is mixed with 50.0 mL of 0.300 M Pb(NO3)2 (269.2 g/mol)?

3Pb(NO3)2(aq) + 2FeI3(aq) → 3PbI2(s) + 2Fe(NO3)3(aq)

Net ionic equation: Pb2+(aq) + 2I−(aq) → PbI2(s)

Strategy

vol Pb(NO3)2 → mol Pb(NO3)2 → mol Pb2+ → mol PbI2 → g PbI2

vol FeI3 → mol FeI3 → mol I− → mol PbI2 → g PbI2

▪ The calculation that gives the least PbI2 determines how much is formed and which reagent is limiting

147


Limiting Reagent Problem

Starting with Pb(NO3)2

Starting with FeI3

148

2

222

23

2

23

23

23

2323

PbI mol 1

PbI g 0.461

Pb mol 1

PbI mol 1

)Pb(NO mol 1

Pb mol 1

)Pb(NOL 1

)Pb(NO mol .3000

)Pb(NOmL 1000

)Pb(NOL 1)Pb(NOmL 0.50

+

+

= 6.92 g PbI2

PbI mol 1

PbI g 0.461

I mol 2

PbI mol 1

FeI mol 1

I mol 3

FeIL 1

FeI mol 080.0

FeImL 1000

FeIL 1FeImL .020

2

22

33

3

3

33

−

−

= 11.06 g PbI2

Pb(NO3)2 is limiting and only 6.92 g of PbI2 can be made.


Chemical Analysis

Qualitative analysis

▪ What substances are present in a sample

Quantitative analysis

▪ Measure the amounts of various substances in a sample

▪ Convert all of an element present in a sample into a substance of known formula

▪ Use the amount of this known to determine amount of element present in the original sample (unknown or analyte)

149


Learning Check: Chemical Analysis

A 1.000 g sample of insecticide is decomposed so that all the chlorine is converted to Cl–(aq). Silver nitrate is added to precipitate all the chloride as AgCl. The solid after filtering and drying is found to weigh 2.022 g. What is the percentage, by mass, of the chloride in the insecticide?

Strategy:

g AgCl → mol AgCl → mol Cl → g Cl

150


How Much Cl in 2.022 g of AgCl?

151

Cl mol 1

Cl g 35.45

AgClmol 1

Cl mol 1

AgClg 3.143

AgClmol 1 AgClg 202.2

= 0.5002 g Cl

Percentage Cl in original sample?

%Cl =mass of Cl

mass of sample´100%

%Cl =0.5002 g Cl

1.000 g sample´100% = 50.02% Cl

g AgCl → mol AgCl → mol Cl → g Cl


Your Turn

A 4.000 g sample that contains sodium carbonate and an inert, soluble salt is dissolved in 100 g of water. Hydrochloric acid is added to react with the carbonate ions. If 33.5 mL of 0.525 M HCl was required to just completely react with all of the sodium carbonate, what was the mass percent of sodium carbonate in the original sample?

Strategy:

1. Write the balance chemical equation

2. M HCl → mol HCl → mol Na2CO3 → g Na2CO3

152


Your TurnA 4.000 g sample that contains sodium carbonate and an inert, soluble salt…

A. 0.466%

B. 23.3%

C. 23,300%

D. 46.6%

E. 0.233%153

Brady/Jespersen/Hyslop, Chemistry7E, Copyright © 2015 John Wiley & Sons, Inc. All Rights Reserved 154

soln HClL

HClmol 0.525soln HClL1033.5 3-

= 1.76 10-2 moles HCl

HClmoles 2

CONa mole 1 HClmoles1076.1 322 −

1. Write the balanced chemical equation

Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O



= 8.80 10-3 moles Na2CO3


100%sample g 4.00

CONa g 0.933CONa % 32

32 =

%100sample of mass

CONa of massCONa % 32

32 =

= 23.3% Na2CO3

32

3232

3

CONa mole 1

CONa g 106 CONa moles1080.8 −


= 0.933 g Na2CO3

Percentage Na2CO3 in original sample?

155


Titrations

Widely used analytical technique

▪ Used to determine concentration of solute

▪ Used daily to monitor:

▪ Water purity

▪ Quality control in food industry

How it works:

▪ Must know reaction that occurs

▪ Reaction must be rapid and complete

▪ Must know exact quantity of one reactant

▪ Use stoichiometry to find exact amount of any other substance in solution

156


Titration▪ Controlled addition of one reactant to known

quantity of another until reaction is complete

Acid-Base Titration

▪ Very common type of titration

e.g., Analysis of citric acid in orange juice by neutralization with NaOH

▪ Know MNaOH and measure exact VNaOH needed to completely neutralize citric acid

▪ MNaOH × VNaOH = mol NaOH

▪ mol NaOH mol citric acid

▪ mol citric acid × MM citric acid = g citric acid157


Titration in practice:

158

Buret

▪ Volumetric measuring device with 0.10 mL markings

Stopcock

▪ Permits flow of titrant to stop when reaction is complete

Volume titrant

used = Vf – Vi


Titration: DefinitionsTitrant

▪ Solution in the buret

▪ Known concentration

▪ Can be either acid or base depending on nature of the analyte

Analyte

▪ Solution being analyzed

▪ Solution in flask

▪ Solution of unknown concentration

Equivalence point

▪ Volume of titrant where moles of titrant and moles of analyte are stoichiometrically equal

159


Titration: Definitions, continuedIndicator

▪ Dye that is one color in acid and 2nd color in base

e.g., phenolphthalein

▪ Colorless in acid and bright pink in base

▪ Color change signals end point of titration

Endpoint:

▪ Volume of titrant required to complete reaction monitored by color change of indicator

▪ Choose indicator so endpoint and equivalence point are

the same

160


Millimoles

▪ We can use metric prefixes with molarity units.

For example

We can use the last ratio in problems with mL units.

161

2.83 M =2.83 mol

1 L=

2.83 mmol

1 mL


Learning Check

Suppose that 25.00 mL of a solution of oxalic acid, H2C2O4, extracted from rhubarb leaves, is titrated with 0.500 M NaOH(aq) and that the stoichiometric point is reached when 37.5 mL of the solution of base is added. What is the molarity of the oxalic solution?

Step 1: Write the balanced equation.

H2C2O4(aq) + 2NaOH(aq) → Na2C2O4(aq) + 2H2O

162


Learning Check: Oxalic acid + NaOH

Step 2: Calculate millimoles of base used

163

NaOH mmol 5718.soln NaOHmL 1

NaOH mmol 0.500soln NaOHmL 37.5 =

Step 3: Calculate millimoles of oxalic acid

422422 OCH mmol 579.3

NaOH mmol 2

OCH mmol 1NaOH mmol 578.1 =

Step 4: Calculate M H2C2O4

9.375 mmol H2C

2O

4

25.00 mL soln= 0.375 M H2C2O4


Summary of Stoichiometry Calculations

164


Your TurnA 25.00 mL sample of HNO3 is titrated with 75.00 mLof 1.30 M Ca(OH)2. What is the concentration of HNO3

in the initial sample?

2HNO3(aq) + Ca(OH)2(aq) → 2AgBr(s) + Ca(NO3)2(aq)

A. 0.433 M

B. 1.95 M

C. 0.867 M

D. 3.90 M

E. 7.80 M

165

75.00 mL Ca(OH)2

´1.30 mmol Ca(OH)

2

1 mL Ca(OH)2

´2 mmol HNO

3

1 mmol Ca(OH)2

25.00 mL HNO3

= 7.80 M HNO3


Your TurnA sample of metal ore is reacted according to the following reaction:

Fe(s) + 2H+(aq) → Fe2+(aq) + H2(g)

If 25.00 mL of 2.3 M HCl are used, what mass of Fe was in the ore? (Atomic mass of Fe is 55.85 g/mol)

A. 0.515 g

B. 1.03 g

C. 1.21 g

D. 1.61 g

E. 3.20 g

166

0.02500 L HCl ´2.3 mol HCl

1 L HCl ´

1 mol Fe

2 mol HCl´

55.85 g Fe

1 mol Fe

= 1.61 g

Documents

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