chapter 4

1

Chapter 4

Aqueous reactions and solution stoichiometry

2

Aqueous Chemistry

Virtually all chemistry that makes life possible occurs in solution

Common tests for sugar, cholesterol, and iron are all done in solution.

3

Solution Vocabulary

Solution: Homogeneous mix of 2+ substances

Solvent: dissolving medium (water, oil, liquid nitrogen), present in larger qty.

Solute: what is being dissolved (salt, sugar, sodium hydroxide), present in smaller qty.

Aqueous Solution: When water is the dissolving medium (Making Kool-Aid)

Electrolytes: substance that dissolves in water to yield a solution that conducts electricity. (salt water)

44.1

Solution Solvent Solute

Soft drink (l)

Air (g)

Soft Solder (s)

H2O

N2

Pb

Sugar, CO2

O2, Ar, CH4

Sn

5

4.1 General Properties of Solutions

One of the properties of water is its ability to dissolve MANY different substances.

Polar: unequal distribution of charges makes water polar, allowing it to dissolve solutes.

6

Ions dissolving in water

Ionic substances such as salts dissolve in water to release cations (+) and anions

(-).

Ex: NaCl -> Na+ , Cl-

The dissociation of NaCl is called hydration.

7

Solvation- The process in which an ion or a molecule is surrounded by solvent molecules arranged in a specific manner.

Hydration – solvation with water as solvent

8

A Note On Dissociation

You must memorize the charges or all the ions and poly atomic ions in order to predict how molecules will dissociate.

If you don’t know them by now you are in trouble. Make flash cards and study!

9

Solubility

The amount of a substance that dissolves in a given volume of solvent at a given temperature.

If ionic compounds are not greatly attracted to the ions in water then that compound will be less soluble in water.

10

What is not soluble in water?

Like dissolves like

In general polar (unequal distribution of charges) and ionic substances are expected to be more soluble in water than nonpolar substances.

12

Electrolyte Vocabulary

Ionization: the process of adding or removing electrons from an atom or molecule. Which gives the atom a net charge.

Complete Ionization:

substances that only exist as ions in solution

13

Strong and Weak electrolytes

Strong electrolyte: ionize completely in water (single arrow). Are good conductors of electricity Ex: Strong acids, strong bases and salts.

14

Weak Electrolytes

Ionize partially in water (double arrows). Poor conductors of electricity Ex: weak acids (acetic acid shown), weak

bases, partially soluble salts.

And thus The reaction is

reversible

15

Non-electrolytes

Do not IONIZE in water or conduct electricity. They are covalent compounds that are not acids or bases.

Dissolve in water as molecules instead of ions. Sugar (C12H22O11)

16

Identify the strong and weak electrolytes.

18

An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity.

A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity.

nonelectrolyte weak electrolyte strong electrolyte4.1

19

Molecular Compounds

20

Homework

Chang pg 157 # ‘s 1 2 4 7 11

BL Pg 145 #’s 3, 5, 6, 7, 8

21

4.2 Precipitate Reactions

Reactions that result in the formation of an insoluble substance.

(-) anion (+) cation

Ag(NO3)2 (aq) + 2NaI (aq) AgI2 (s) + 2NaNO3 (aq)

23

Question

How would each molecule dissociate? Label cations and anions.

NiSO4 Ca(NO3)2 Na3PO4 Al2(SO4)3

24

Answer

Ni 2+ SO4 2-

Ca2+ (NO3)2 -

Na3+

PO43-

Al23+ (SO4)3

2-

25

Solubility Rules to Memorize

You must memorize the following rules.

You will have a pop quiz on solubility rules in the next week.

If you do not pass with 85% you must write all the rules out 5 times each.

26

Solubility Rules (memorize)Solubility Rules (memorize)1. NH4

+ and alkali metal (group IA) salts are soluble.

2. Nitrate, NO3-, acetate C2H3O2

-, chlorate, ClO3-, perchlorate ClO4 salts

are soluble.

3. Chloride, Cl-, bromide, Br-,iodide, I-, salts are soluble. EXCEPT: Ag+, Hg2

2+, Pb2+ (AgBr, Hg2I2, PbCl2)

4. Sulfate, SO42- , salts are soluble.

EXCEPT: PbSO4, HgSO4, CaSO4, BaSO4, AgSO4, SrSO4

5. Most Hydroxide, OH- ,salts are slightly soluble. Hydroxide salts of Group I elements are soluble (Li, Na, K, Rb, Cs, Fr).Hydroxide salts of Group II elements (Ca, Sr, and Ba) are slightly soluble.Hydroxide salts of transition metals and Al3+ are insoluble. Thus, Fe(OH)3,

Al(OH)3, Co(OH)2 are not soluble.

27

Use Solubility Rules to Classify as Soluble or Insoluble

Sodium carbonate Na2CO3

Lead Sulfate PbSO4

Cobalt (II) hydroxide Barium nitrate Ammonium phosphate

28

Answer

Sodium carbonate Na2CO3 Soluble (#1)

Lead Sulfate PbSO4 Insoluble (#4)

Co(OH)2 Insoluble (#5)

Ba(NO3)2 Soluble (#2)

Ammonium phosphate (NH4)3PO4

New Polyatomic: PO4-3 Phosphate

29

Goal

Use solubility rules to predict whether a precipitate will form when electrolytic solutions are mixed.

Hint: find products that have insoluble salt(s). This implies a precipitate reaction.

30

Example

Predict what will happen when the following pairs of solutions are mixed. (Hint: break it into ions)

KNO3(aq) and BaCl2 (aq)

Na2SO4(aq) and Pb(NO3)2(aq)

31

Answer

KNO3(aq) and BaCl2 (aq)

Reactants: K+ + NO3- + Ba2+ + Cl-

Products: KCl + Ba (NO3)2

Both are soluble according to the rules and thus no precipitate forms.

Rule #1 Rule # 2

32

Na2SO4(aq) and Pb(NO3)2(aq)

Reactants: Na+ + SO42- + Pb2+ + NO3

-

Products: NaNO3 + PbSO4

NaNO3 Soluble according to rule #1.

PbSO4 Insoluble according to rule # 4

33

Demonstrations

Aqueous iron (III) nitrate reacts with sodium hydroxide

34

Demonstrations

CoCl2(aq) + Ca(OH)2(aq)

35

Predicting Precipitates

Predict the precipitate that forms when solutions mix and write a balanced chemical equation.

BaCl2 (aq) + K2SO4 (aq)

Fe2(SO4)3 (aq) + LiOH (aq)

36

Answer

BaCl2 (aq)+ K2SO4 (aq) BaSO4 (s) + 2KCl

Fe2(SO4)3 (aq) + 6LiOH (aq) 2Fe(OH)3 (s) + 3Li2SO4 (aq)

37

Describing reactions

In this section we will talk about the types of equations used to represent reactions in solution.

Three types of equations: Molecular Complete ion Net ionic

38

Precipitation Reactions

Precipitate – insoluble solid that separates from solution

molecular equation

ionic equation

net ionic equation

Pb2+ (aq) + 2NO3- (aq) + 2Na+

(aq) + 2I- ( aq) PbI2 (s) + 2Na+ (aq) + 2NO3

- (aq)

Na+ and NO3- are spectator ionsP

bI

2

Pb(NO3)2 (aq) + 2NaI (aq) PbI2 (s) + 2NaNO3 (aq)

precipitate

Pb2+aq + 2I- aq PbI2 (s)

4.2

39

Molecular Equations/Chemical equation Shows the complete chemical formula for

reactants and products

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

Strong acid Strong Base Soluble Salt weak electrolyte

Strong elect. Strong elect.

40

Complete Ionic Equation

Shows the formula of cations and anions for ionic compounds.

H+(aq) + Cl-(aq) + Na+

(aq) + OH-(aq) Na+

(aq) + Cl-(aq) + H2O(l)

Water (l) and solid (s) precipitates do not break down into ions

Solids on reactant side can break down in water (i.e adding salt to water)

41

Net Ionic Equation

Includes only those solutions components directly involved in the reaction.

H+(aq) + OH-

(aq) H2O(l)

Spectator ions: ions that appear in identical forms in reactants and product side of a chemical rxn that do not participate in the rxn directly.

NOTE: The net ionic equation of any strong acid-base

neutralization rxn is always like the above rxn.

42

Write the three types of equations for the following rxn.

Pb(NO3)2 (aq) + KI (aq)

43

Molecular

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

Complete Ionic

Pb 2+ (aq) + 2NO3-1

(aq) + 2K+ (aq) + 2I- (aq) PbI2 (s) + 2K+ (aq) + 2NO3- (aq)

• Note subscripts that a re “multiplied” through become coefficients • Coefficients apply to all atoms in a compound

Net Ionic

Pb 2+ (aq) + 2I- (aq) PbI2 (s)

44

Writing Net Ionic Equations1. Write the balanced molecular equation.

2. Write the ionic equation showing the strong electrolytes completely dissociated into cations and anions.

3. Cancel the spectator ions on both sides of the ionic equation

4. Check that charges and number of atoms are balanced in the net ionic equation

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

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

-

Ag+ + Cl- AgCl (s) 4.2

Write the net ionic equation for the reaction of silver nitrate with sodium chloride.

45

4.2 Homework

Chang pg 156-157 #’s 9,10,12,15,18,19,21, 22,

BL: Pg 145 #’s 11, 12, 14, 15, 16, 19

46

4.3 Acids

Substances that are able to donate a hydrogen ion (H+) and increase [H+] in aqueous solutions.

MEMORIZE

Nitric acid HNO3

chloric acid HClO3

perchloric acid HClO4

sulfuric acid H2SO4

hydrochloric acid

HCl

hydrobromic acid

HBr

hydroiodic acid

HI

47

Properties of acids

Sour taste Acids neutralize bases Acids corrode active

metals Acids release a

hydrogen ion into water (aqueous) solution

Strong acids conduct electricity

48

Properties of Bases

Bitter taste Slippery feel Bases denature protein Bases neutralize acids Bases release a

hydroxide ion into water solution

Strong bases conduct electricity

49

Bases

Are soluble ionic compounds containing hydroxide ion (OH-).

When dissolved in water the cations and OH- ions separate and move independently.

MEMORIZE /LEARN

Hydroxides of 1A

NaOH

LiOH

KOH

RbOH

CsOH

FrOH

Hydroxides of 2A

Be (OH)2

Mg(OH)2

Ca(OH)2

Sr(OH)2

Ba(OH)2

Ra(OH)2

50

Strong electrolytes

All of the strong bases and acids that you memorized are strong electrolytes because they ionize completely.

All weak acids and bases are weak electrolytes because they only partially ionize.

51

Salts

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space.

Another name for an ionic compound. When a salt dissolves in water, it breaks up into its ions, which move about independently.

ionic substances are electrolytes

Result from acid base neutralizations

52

Classify the following dissolved substances as strong or weak electrolytes:

CaCl2

HNO3

C2H5OH

HCHO2

53

CaCl2 = ionic = strong electrolyte

HNO3 = strong acid = strong electrolyte

C2H5OH = molecular = nonelectrolyte

HCHO2 = molecular = nonelectrolyte

54

Neutralization Rxn

Neutralization: when an acid and a base mix and their products share no characteristics with their reactants. (i.e acid base qualities)

55

Neutralization Rxn

Neutralization reaction between an acid and a metal hydroxide produce water and a salt.

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

(acid) (base) (water) (salt)

56

Molecular equation

HCl (aq) + NaOH (aq) H2O (l) + NaCl (aq)(acid) (base) (water) (salt)


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

Net Ionic Equation

H+ + OH- H2O

57

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

58

Question

Write a balanced complete chemical equation for the reaction between aqueous solutions of acetic acid and barium hydroxide.

59

Answer

Complete chemical equation

We are given an acid and a base (metal hydroxide) so the result should be water and a salt

2HC2H3O2 (aq) + Ba(OH)2 (aq) 2H2O (l)+ Ba(C2H3O2)2 (aq)

60

Now Write the complete ionic equation

Steps:

Determine if (aq) solutions are strong or weak electrolytes to see how they will dissociate.


(weak elect) (strong Base/elect (strong elect

weak acid Rule 5 exception) ionic salt)

61



2HC2H3O2 (aq) + Ba2+ (aq) + 2OH- (aq) 2H2O (l)+ Ba 2+ (aq) + 2C2H3O2- (aq)

* Note subscripts become coefficients

62

Now write the Net Ionic Equation

2HC2H3O2 (aq) + Ba2+ (aq) + 2OH- (aq) 2H2O (l)+ Ba 2+ (aq) + 2C2H3O2- (aq)

Net Ionic Equation

2HC2H3O2 (aq) + 2OH- (aq) 2H2O (l)+ 2C2H3O2- (aq)

Simplify coefficients for final answer

HC2H3O2 (aq) + OH- (aq) H2O (l)+ C2H3O2- (aq)

63

Acid Base reactions forming Gases

Sulfide ion and carbonate ion react with acids to form gases with low solubility's in water.

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

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

64

Homework

BL Pg 146 #’s : 23, 24, 25, 27, 29, 31

65

4.4 Oxidation –Reduction Reactions Rxn’s in which one or more electrons are

transferred between reactants.

Ex: 2 Na + Cl2 2NaCl

neutral neutral Na+ Cl-

66

Symantics

Charges are written: # sign : 1- 3+

Oxidation numbers are written: sign # : -1 +3

67

Photosynthesis is a redox rxn

Most energy producing rxns are redox rxns. Such as combustion rxn (fuel)

Rusting of iron is a redox rxn

Ca (s) + 2H+ (aq) Ca2+ (aq) + H2 (g)

68

Oxidation

An atom ion, molecule, becomes more positive it has lost electrons.

We say that that atom, ion, molecule has been oxidized.

Ca (s) + 2H+ (aq) Ca2+ (aq) + H2 (g)

69

Reduction

When an atom, ion, molecule has become more negatively charged its has gained electrons.

We say this atom, ion, molecule has been reduced.

Ca (s) + 2H+ (aq) Ca2+ (aq) + H2 (g)

70

Oxidation reduction

When one atom loses electrons it is gained by the other atom involved in the reaction.

Ca (s) + 2H+ (aq) Ca2+ (aq) + H2 (g)

71

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 4.4

72

Oxidation numbers

Helps us keep track of the electrons being gained and lost.

Oxidation number is the actual charge of the atom if it is a mono-atomic ion, other wise it is the hypothetical charged assigned to the atom using a set of rules.

73

The Rules 1. The rule is that the cation is written first in a formula,

followed by the anion. Example: in NaH, the H is H-; in HCl, the H is H+. + - + -2. The oxidation number of a free element is always 0. Example: The atoms in He and N2, for example, have

oxidation numbers of 0.

3. The oxidation number of a monatomic ion equals the charge of the ion.

Example: oxidation number of Na+ is +1; the oxidation number of N3- is -3.

4. The oxidation number of oxygen in compounds is usually -2.

74

5. The oxidation number of a Group 1 element in a compound is +1.



8. The oxidation number of a Group 7 element in a compound is -1, except when that element is combined with one having a higher electronegativity.

75

9. The sum of the oxidation numbers of all of the atoms in a neutral compound is 0.

EX: CO2 = we know O = -2 and there are 2 O’s and CO2 is neutral so

C + 2(-2) = O C = +4

10. The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

EX: the sum of the oxidation numbers for SO4 2- is -2. S + 4(-2) = - 2 S = +6

78

10. The sum of the oxidation numbers of all of the atoms in a neutral compound is 0.

EX: CO2 = we know O = -2 and there are 2 and CO2 is neutral so C + 2(-2) = O C = +4

11. The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

EX: the sum of the oxidation numbers for SO4 2- is -2. S + 4(-2) = - 2 S = +6

79

The oxidation numbers of elements in their compounds

4.4

80

ExamplesHINT: start with what you know for sure

H2S Neutral molecule so all oxidation numbers

must add up to 0. Let X = the oxidation number of S. H has an

oxidation number of (+1)2X +2(+1) = 0 so charge of S = -2

81

S8

In elemental form so oxidation number is 0 (rule 1)

82

SCl2

This is a binary compound. We expect Cl to have an oxidation number or -1.

The sum of the ox #’s must equal zero because this is a neutral compound.

X + 2(-1) = 0 X = +2

83

Na2SO3

Oxidation numbers of Alkali metals always have an oxidation number of +1 in compounds. Oxygen has a common oxidation state of -2.

Let x = number of S 2(+1) + X + 3(-2) = 0 X = +4

84

Assign Oxidation numbers for

CO2

SF6-

NO3-

85

CO2 C = +4 O= -2(2)

SF6- S = +5 F = -1(6)

NO3- N = +5 O = -2(3)

5+ -6 = -1

86

Oxidations of metals by acids and saltsGeneral pattern

A + BX AX + B

Zn (s) + 2HBr (aq) ZnBr2 (aq) + H2 (g) 0 +1 -1 +2 -1(2) 0Gain e- reduced Lose e- oxidized Stays the same

87

Fe = +3(2) = +6

O = -2(3) = -6

Oxidation for

free elements

is zero

88

L - E -O

Lose electrons Oxidized

The Lion Says

G – E - R

Gain Electrons Reduced

I think you

are pretty

and smart!!!

89

Leo the Lion Says Ger!!!

OXIDATION

REDUCTION

90

In this reaction, oxygen maintains a -2 oxidation number throughout.

Iron becomes reduced from +3 to 0 oxidation number, while carbon becomes oxidized from 0 to +4 state.

The species that becomes reduced is called the oxidizing agent since it is accepting electrons from some other species. Conversely, the species that becomes oxidized is called the reducing agent since it is giving up electrons.

92

Identify the atoms being oxidized and reduced as well as the oxidizing and reducing agents.

2Al + 3I2 2AlI3

93

2Al + 3I2 2AlI3

0 0 +3 -1

* Since each Al atom changes from a 0 +3 Al is Oxidized

* Iodine is reduced 0 -1

* Al donates the electrons so it is the reducing agent

* I2 accepts the electrons and it is the oxidizing agent

94

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

4.4

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

95

Activity Series

We need to know what metals are most likely to oxidize others.

Example: We can’t store nickel nitrate in an iron container because the solution would eat through the container.

96

Activity Series

A list of metals arranged in order of decreasing ease of oxidation.

Page 139 table

97

Using activity series

Any metal on the list can be oxidized by the metal below it.

Give: FeCl2 + Mg Find: will iron oxidize Magnesium metal?

1. I finger on Fe 2. 1 finger on Mg3. Is the bound chemical below4. Yes Fe is below Mg. 5. Then complete the reaction

98

Give: NaCl2 + Mg Find: will sodium oxidize Magnesium metal?

1. I finger on Na 2. 1 finger on Mg3. Is the bound chemical below4. no5. Then the reaction is not

possible

99

The Activity Series for Halogens

Halogen Displacement Reaction

4.4

Cl2 + 2KBr 2KCl + Br2

0 -1 -1 0

F2 > Cl2 > Br2 > I2

I2 + 2KBr 2KI + Br2

101

Homework

Chang Pg 158-159 #’s 46,48, 50, 54,56 (id what is red and ox)

BL146-147

#’s : 35, 39, 41, 42, 44, 45,

102

Concentrations of Solutions 4.5

A solution is a homogeneous mixture of two or more substances. One of these substances is a solvent the other is the solute.

Solvent: component in greater quantity Solute: component in lesser quantity. Solution = solvent + solute Concentration: amount of solute dissolved in a

given amount of solution. Units vary.

103

Molarity (M)

Number of moles of solute in one liter of solution

Molarity = moles of solute liter of solution

Units = mol/L = M

104

To make 250 mL (0.250 L) of 1.00 M CuSO4

A. Use the formula grams needed = Molecular weight x Volume x Molarity

g = 159.6 x 0.250 x 1 = 39.9 g CUSO4

B. Place chemical into flask and add a small quantity of water to dissolve.

C. Mix solution

D. Bring total volume up to 0.250 L

A B C D

105

Question

Calculate the molarity of a solution made by dissolving 5.00g of C6H12O6 (MW = 180 amu) in sufficient water to form 100 ml solution.

Recall : M = mole/ L

106

Answer

5.0 g 1mol = 0.027 mol

180g

M = 0.027 = 0.27M

0.1

107

Question

How many grams of Na2SO4 are there in 5ml of 0.50 M Na2SO4.

108

Answer

g = M.W x V x L

g = 142 x 5/1000 x .5

109

Finding Concentration of one type of atom We can find the concentration of one type

of atom in a molecule by multiplying the molarity of the solution by that number of atoms.

110

Question

Which of the following solutions of strong electrolytes contains the largest concentration of chloride ions

A.0.30 M AlCl3

B. 0.60M MgCl2 C. 0.40 NaCl

111

Answer

A. 0.30(3 Cl) = 0.90 M Cl

B. 0.60 (2Cl) = 1.2 M Cl

C. 0.40 ( 1 Cl) = 0.4 M Cl

MgCl2 > AlCl3 > NaCl

112

Dilutions

Moles before dilution = moles after dilution

M1V1 = M2V2

How many milliliters of 5.0 M K2Cr2O7 solution must be diluted in order to prepare a 250 mL of 0.10M solution.

NOTE: All Volumes must be in L to satisfy units of Molarity!!!!!!

113

Answer

250 mL = .250 L

5.0 M stock ( V1) stock = (0.10 M)want 0.250 L want

V1 = .005 L stock

To make this solution we will add 0.005 L of stock to 0.245 L of water to make a 0.10M solution.

114

Homework Sections

Chang pg 159 #’s 59,60,61,63,69, 74

Page 147

#’s 49, 51, 52, 54, 56, 59, 60, 61

115

4.7 Stoichiometry of precipitation reactions

2 differences

1. its hard to predict products in solution, so we need to think and remember the rules.

2. To obtain moles of reactants we must use the volume of the solution and its molarity. Recall M = mol

1L

116

Question

How many moles are in 1 L of 0.3M solution?

How many moles are in 6.9L of 0.45M solution?

117

Example:

What is the mass of NaCl solid that must be added to 1.50L of a 0.100M AgNO3 solution to precipitate all the Ag+ ions in the form of AgCl according to the balanced equation below?

NaCl + AgNO3 AgCl + NaNO3

118

Apply solubility rules

Products: AgCl + NaNO3

NaNO3 is soluble (rule 1) AgCl is insoluble (rule 3)

Forming a solid.

SO…

Lets add enough Cl- ions (form NaCl) to react with all of the Ag+ (from AgNO3) to form a precipitate of AgCl. But how many moles of AgNO3 do we have?

119

Given:1.50L 0.100M AgNO3

we know from our homework that that there is 0.1M(1 Ag) = 0.1M Ag 0.1M( 1 NO3) = 0.1M NO3

We can use our molarity (moles per one liter) and how many Liters we are given to find how many moles we have in that volume.

1.5 L AgNO3 0.100 mol AgNO3 = 0.150 mol AgNO3in 1.5L

1L AgNO3

120

Because AgNo3 and NaCl react in a 1:1 ratio (see rxn)

NaCl + AgNO3 AgCl + NaNO3

0.150 mol AgNO3 are present (we just solved for that) thus 0.150 mol NaCl are needed. But we need to know grams not moles.

0.150 mol NaCl 58.45g NaCl = 8.77g NaCl 1 mol NaCl

121

Question

When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed, PbSO4 precipitates. Calculate the mass of PbSO4 formed when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00L of 0.0250 M Na2SO4 are mixed?

122

Step 1:

Write a balanced equation and apply solubility rules:

Na2SO4 (aq) + Pb(NO3)2 (aq) PbSO4 (s)+ 2NaNO3 (aq)

We need to find the mass of PbSO4 (s) formed. Where is the Pb and SO4 coming from?

Which one is going to limit us?

How do I find this answer?

123

Step 2: Find limiting reactant g mol mol g

Given: 1.25 L of 0.0500 M Pb(NO3)2

2.00L of 0.0250 M Na2SO4

Pb2+(aq) + SO4

2-(aq) PbSO4 (s)

1.25 L Pb(NO3) x 0.0500 mol Pb(NO3) = 0.0625 mol Pb2+ formed 1L

2.00 L Na2SO4 x 0.0250 mol Na2SO4 = 0.0500 mol SO4

2- formed 1 L

124

Step 3:

calculate moles of product PbSO4 that can be formed according to how much LR we have been given. ( g-mol-mol-g)

.05 molNa2SO4 x 1 mol PbSO4 = 0.05mol PbSO4

1 mol Na2SO4

only 0.0500 mol of solid PbSO4 will be formed since we only have 0.0500 mole of S04 available to us because it is the limiting recatant.

125

Step 4:

convert moles to grams because that is the units the question wants us to report our answers in:

The mass of PbSO4 formed can be calculated using the molar mass of PbSO4 (303.3g/mol)

0.0500 mol PbSO4 303.3g PbSO4 = 15.2 gPbSO4

1mol PbSO4

126

With your partner solve:

What mass of NaCl is needed to precipitate all of the silver ions from 20.0 mL of 0.100 M AgNO3 solution?

127

Example: neutralization rxn

What volume of 0.100 M HCl solution is needed to neutralize 25.0 ml of 0.350 M NaOH?

Step 1: list reactants

HCl + NaOH

H+ + Cl- + Na+ + OH-

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Step 2:

What possible rxn’s will occur

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

Na+ + Cl- NaCl (soluble and cant neutralize)

H+ + OH- H2O (insoluble and can neutralize)

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Step 3: write a balanced net ionic equation and calculate moles of reactant needed.

H+ + OH- H2O

0.025 L NaOH 0.35 mol OH = 8.75 x 10-3 mol OH-

1 L NaOH

We do not need to determine limiting reactant since the addition of H+ ions react exactly with the OH-

present in a1:1 ratio. Thus 8.75 x 10-1 mol H+ is required to neutralize the solution.

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Convert to volume required to neutralize the rxn.

Volume X 0.100 mol H+ = 8.75 x 10-3 mol H+

1L

= 8.75 x 10-2 L of .100M HCL is required to neutralize 25.0 mL of NaOH

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Titrations

A procedure used for determining the concentration of an acid or a base in a solution by addition of a base or an acid of a known concentration.

We know the solution is at its “end point” or stoichiometric point when the indicator changes color.

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Titration vocabulary

Standard solution: Solution with a known concentration.

Equivalence point: = when unknown solution and standard solution are at the same concentration.

Indicators: help establish equivalence point by a color change.

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Equivalence point: the point in the titration when exactly enough base is added to neutralize the acid.

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Indicators

Organic dyes that change color as they go from an acidic solution to basic solution.

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Titration Question

What volume of 0.25M HNO3 is required to titrate (neutralize) a solution containing 0.200 g of KOH.

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Given:

0.200 g of KOH.

0.25M HNO3

Find: volume of 0.25M HNO3 required to neutralize 0.200 g KOH.

KOH + HNO3 KNO3 + H2O

Grams to moles to moles to liters

0.2 g KOH 1 mol KOH x 1 mol HNO3 x 1 L HNO3 = 0.014 L HNO3

56.1 g KOH 1 mol KOH 0.25 mol HNO3

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Homework

Pg 148 #’s 65, 67, 69, 73

Bonus 2 pts number 76 must be turned in to the box tomorrow first thing. On a piece of a paper with all units clearly worked out.

S.O.S

chapter 4

Documents

water polar

medium water

properties of water

solution strong

solution vocabulary

ionic solution

solution of nacl

aqueous solution