CH131 Note 04 Quantity

Chemical Quantities and Chemical Quantities and Aqueous ReactionsAqueous Reactions

Chapter 4

2

Global WarmingGlobal Warming

• Scientists have measured an average 0.6°C rise in atmospheric temperature since 1860

• During the same period atmospheric CO2levels have risen 25%

• Are the two trends causal?

3

The Sources of Increased COThe Sources of Increased CO22

• One source of CO2 is the combustion reactions of fossil fuels we use to get energy

• Another source of CO2 is volcanic action• How can we judge whether global warming is natural

or due to our use of fossil fuels?

4

Quantities in Chemical ReactionsQuantities in Chemical Reactions

• The amount of every substance used and made in a chemical reaction is related to the amounts of all the other substances in the reaction

Law of Conservation of MassBalancing equations by balancing atoms

• The study of the numerical relationship between chemical quantities in a chemical reaction is called stoichiometry

5

Reaction Reaction StoichiometryStoichiometry

• The coefficients in a balanced chemical equation specify the relative amounts in moles of each of the substances involved in the reaction2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)

2 molecules of C8H18 react with 25 molecules of O2to form 16 molecules of CO2 and 18 molecules of H2O

2 moles of C8H18 react with 25 moles of O2to form 16 moles of CO2 and 18 moles of H2O

2 mol C8H18 : 25 mol O2 : 16 mol CO2 : 18 mol H2O

6

Making PizzaMaking Pizza

• The number of pizzas you can make depends on the amount of the ingredients you use

• This relationship can be expressed mathematically1 crust : 5 oz. sauce : 2 cu cheese : 1 pizza

1 crust + 5 oz. tomato sauce + 2 cu cheese → 1 pizza

• If you want to make more or less than one pizza, you can use the amount of cheese you have to determine the number of pizzas you can make

assuming you have enough crusts and tomato sauce

7

Predicting Amounts from Predicting Amounts from StoichiometryStoichiometry

• The amounts of any other substance in a chemical reaction can be determined from the amount of just one substance

• How much CO2 can be made from 22.0 moles of C8H18 in the combustion of C8H18?

2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)2 moles C8H18 : 16 moles CO2

8

PracticePractice

• According to the following equation, how many moles of water are made in the combustion of 0.10 moles of glucose?

C6H12O6 + 6 O2 → 6 CO2 + 6 H2O

9

Practice Practice −− How many moles of water are made in How many moles of water are made in the combustion of 0.10 moles of glucose?the combustion of 0.10 moles of glucose?

Check:

Solution:

Conceptual Plan:

Relationships:

Given:Find:

Answer: 0.60 moles

10

Example: Estimate the mass of COExample: Estimate the mass of CO22 produced in produced in 2007 by the combustion of 3.5 2007 by the combustion of 3.5 ×× 10101515 g g gasolnegasolne

• Assuming that gasoline is octane, C8H18, the equation for the reaction is

2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)• The equation for the reaction gives the mole

relationship between amount of C8H18 and CO2, but we need to know the mass relationship, so the conceptual plan will be

11

Example: Estimate the mass of COExample: Estimate the mass of CO22 produced in produced in 2007 by the combustion of 3.5 2007 by the combustion of 3.5 ×× 10101515 g gasolineg gasoline

because 8x moles of CO2 as C8H18, but the molar mass of C8H18 is 3x CO2, the number makes sense

Check:

Solution:

Conceptual Plan:

Relationships:

Given:Find:

12

Which Produces More COWhich Produces More CO22; ; Volcanoes or Fossil Fuel Combustion?Volcanoes or Fossil Fuel Combustion?• Our calculation just showed that the world

produced 1.1 × 1016 g of CO2 just from petroleum combustion in 2007

1.1 × 1013 kg CO2

• Estimates of volcanic CO2 production are 2 × 1011 kg/year

• This means that volcanoes produce less than 2% of the CO2 added to the air annually

%.%.

.81100

10111002

yrkg13

yrkg11

=××

×

13

Example 4.1: How many grams of glucose can be Example 4.1: How many grams of glucose can be synthesized from 37.8 g of COsynthesized from 37.8 g of CO22 in photosynthesis?in photosynthesis?

because 6x moles of CO2 as C6H12O6, but the molar mass of C6H12O6 is 4x CO2, the number makes sense

37.8 g CO2, 6 CO2 + 6 H2O → C6H12O6+ 6 O2g C6H12O6

Check:

Solution:

Conceptual Plan:

Relationships:

Given:Find:

14

Practice Practice —— How many grams of OHow many grams of O22 can be made can be made from the decomposition of 100.0 g of PbOfrom the decomposition of 100.0 g of PbO22??

2 PbO2 PbO22((ss)) →→ 2 2 PbO(PbO(ss) + O) + O22((gg))

Check:

Solution:

Conceptual Plan:

Relationships:

Given:Find:

Answer: 6.689 g

15

StoichiometryStoichiometry Road MapRoad Map

a A → b B

Moles A Moles B

mass A mass B

volume A (l) volume B(l)

Pur

eS

ubst

ance

Sol

utio

n

Volume A(g) Volume B(g)

% A(aq)ppm A(aq)

% B(aq)ppm B(aq)

M A(aq) M B(aq)

MM MM

density density

equation

equation

22.4 L 22.4 L

M = molesL

16

More Making PizzasMore Making Pizzas

• We know that

1 crust + 5 oz. tomato sauce + 2 cu cheese → 1 pizza

• But what would happen if we had 4 crusts, 15 oz. tomato sauce, and 10 cu cheese?

17

More Making Pizzas (contMore Making Pizzas (cont’’d)d)

• Each ingredient could potentially make a different number of pizzas

• But all the ingredients have to work together!• We only have enough tomato sauce to make

three pizzas, so once we make three pizzas, the tomato sauce runs out no matter how much of the other ingredients we have.

18

More Making Pizzas (contMore Making Pizzas (cont’’d)d)

• The tomato sauce limits the amount of pizzas we can make. In chemical reactions we call this the limiting reactant.

also known as the limiting reagent• The maximum number of pizzas we can make

depends on this ingredient. In chemical reactions, we call this the theoretical yield.

it also determines the amounts of the other ingredients we will use!

19

The Limiting ReactantThe Limiting Reactant

• For reactions with multiple reactants, it is likely that one of the reactants will be completely used before the others

• When this reactant is used up, the reaction stops and no more product is made

• The reactant that limits the amount of product is called the limiting reactant

sometimes called the limiting reagentthe limiting reactant gets completely consumed

• Reactants not completely consumed are called excess reactants

• The amount of product that can be made from the limiting reactant is called the theoretical yield

20

Limiting and Excess Reactants in the Limiting and Excess Reactants in the Combustion of MethaneCombustion of Methane

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)• Our balanced equation for the combustion of

methane implies that every one molecule of CH4reacts with two molecules of O2

21

Limiting and Excess Reactants in the Limiting and Excess Reactants in the Combustion of MethaneCombustion of Methane

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)• If we have five molecules of CH4 and eight molecules

of O2, which is the limiting reactant?

22

Practice Practice —— How many moles of SiHow many moles of Si33NN44 can be made from can be made from 1.20 moles of 1.20 moles of SiSi and 1.00 moles of Nand 1.00 moles of N22 in the reaction in the reaction 3 3 SiSi + 2 N+ 2 N22 →→ SiSi33NN44??

Solution:

Conceptual Plan:

Relationships:

Given:Find:

Answer: 0.400 moles

23

More Making PizzasMore Making Pizzas

• Let’s now assume that as we are making pizzas, we burn a pizza, drop one on the floor, or other uncontrollable events happen so that we only make two pizzas. The actual amount of product made in a chemical reaction is called the actual yield.

• We can determine the efficiency of making pizzas by calculating the percentage of the maximum number of pizzas we actually make. In chemical reactions, we call this the percent yield.

24

Theoretical and Actual YieldTheoretical and Actual Yield

• As we did with the pizzas, in order to determine the theoretical yield, we should use reaction stoichiometry to determine the amount of product each of our reactants could make

• The theoretical yield will always be the least possible amount of product

the theoretical yield will always come from the limiting reactant

• Because of both controllable and uncontrollable factors, the actual yield of product will always be less than the theoretical yield

Example 4.4:Example 4.4:Finding limiting reactant, theoretical Finding limiting reactant, theoretical

yield, and percent yieldyield, and percent yield

26

Example 4.4• When 28.6 kg of C are allowed to react with 88.2

kg of TiO2 in the reaction below, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield.

TiO2(s) + 2 C(s) → Ti(s) + 2 CO(g)

27

Example 4.4Example 4.4When 28.6 kg of C reacts with When 28.6 kg of C reacts with 88.2 kg of TiO88.2 kg of TiO22, 42.8 kg of Ti , 42.8 kg of Ti are obtained. Find the limiting are obtained. Find the limiting reactant, theoretical yield, and reactant, theoretical yield, and percent yield percent yield TiOTiO22((ss) + 2 C() + 2 C(ss) ) →→ Ti(Ti(ss) + 2 CO() + 2 CO(gg))

• Write down the given quantity and its units Given: 28.6 kg C

88.2 kg TiO2

42.8 kg Ti produced

28

Example 4.4Example 4.4Find the limiting Find the limiting reactant, theoretical reactant, theoretical yield, and percent yield, and percent yield yield TiOTiO22((ss) + 2 C() + 2 C(ss) ) →→Ti(Ti(ss) + 2 CO() + 2 CO(gg))

• Write down the quantity to find and/or its unitsFind: limiting reactant

theoretical yieldpercent yield

InformationGiven: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti

29


• Write a conceptual plan

InformationGiven: 28.6 kg C, 88.2 kg TiO2, 42.8 kg TiFind: lim. rxt., theor. yld., % yld.

kgTiO2

kgC

smallestamount is

from limitingreactant

smallestmol Ti

30


• Collect needed relationships1000 g = 1 kgMolar Mass TiO2 = 79.87 g/molMolar Mass Ti = 47.87 g/molMolar Mass C = 12.01 g/mol1 mole TiO2 : 1 mol Ti (from the chem. equation)2 mole C : 1 mol Ti (from the chem. equation)

InformationGiven: 28.6 kg C, 88.2 kg TiO2, 42.8 kg TiFind: lim. rxt., theor. yld., % yld.CP: kg rxt → g rxt → mol rxt → mol Tipick smallest mol Ti → TY kg Ti → %Y Ti

31

• Apply the conceptual plan to determine the limiting rxt

InformationGiven: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: lim. rxt., theor. yld., % yld.CP: kg rxt → g rxt → mol rxt → mol Tipick smallest mol Ti → TY kg Ti → %Y TiRel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO2 = 79.87g; 1000g = 1 kg; 1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti


32

• Apply the conceptual plan to calculate the theoretical yield



33

• Apply the conceptual plan to calculate the percent yield

InformationGiven: 28.6 kg C, 88.2 kg TiO2, 42.8 kg TiFind: lim. rxt., theor. yld., % yld.CP: kg rxt → g rxt → mol rxt → mol Tipick smallest mol Ti → TY kg Ti → %Y TiRel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO2 = 79.87g; 1000g = 1 kg; 1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti


34

• Check the solutions

limiting reactant = TiO2theoretical yield = 52.9 kgpercent yield = 80.9%

Because Ti has lower molar mass than TiO2, the T.Y. makes sense and the percent yield makes sense as it is less than 100%



35

Practice Practice —— How many grams of NHow many grams of N22((gg) can be made from 9.05 g of NH) can be made from 9.05 g of NH33reacting with 45.2 g of reacting with 45.2 g of CuOCuO??2 NH2 NH33((gg) + 3 ) + 3 CuO(CuO(ss) ) →→ NN22((gg) + 3 Cu() + 3 Cu(ss) + 3 H) + 3 H22O(O(ll) ) If 4.61 g of NIf 4.61 g of N22 are made, what is the percent yield?are made, what is the percent yield?

1 mol NH3 = 17.03g, 1 mol CuO = 79.55g, 1 mol N2 = 28.02 g2 mol NH3 : 1 mol N2, 3 mol CuO : 1 mol N2

Conceptual Plan:

Relationships:

Given:Find:

Choosesmallest

36

Check:

Solution:

Practice Practice —— How many grams of NHow many grams of N22((gg) can be made from 9.05 g of NH) can be made from 9.05 g of NH33reacting with 45.2 g of reacting with 45.2 g of CuOCuO??2 NH2 NH33((gg) + 3 ) + 3 CuO(CuO(ss) ) →→ NN22((gg) + 3 Cu() + 3 Cu(ss) + 3 H) + 3 H22O(O(ll) ) If 4.61 g of NIf 4.61 g of N22 are made, what is the percent yield?are made, what is the percent yield?

Answer: percent yield = 86.8%

37

SolutionsSolutions

• When table salt is mixed with water, it seems to disappear, or become a liquid – the mixture is homogeneous

the salt is still there, as you can tell from the taste, or simply boiling away the water

• Homogeneous mixtures are called solutions• The component of the solution that changes state

is called the solute• The component that keeps its state is called the

solventif both components start in the same state, the major component is the solvent

38

Describing SolutionsDescribing Solutions

• Because solutions are mixtures, the composition can vary from one sample to another

pure substances have constant compositionsaltwater samples from different seas or lakes have different amounts of salt

• So to describe solutions accurately, we must describe how much of each component is present

we saw that with pure substances, we can describe them with a single name because all samples are identical

39

Solution ConcentrationSolution Concentration

• Qualitatively, solutions are often described as dilute or concentrated

• Dilute solutions have a small amount of solute compared to solvent

• Concentrated solutionshave a large amount of solute compared to solvent

40

ConcentrationsConcentrations——Quantitative Quantitative Descriptions of SolutionsDescriptions of Solutions

• A more precise method for describing a solution is to quantify the amount of solute in a given amount of solution

• Concentration = amount of solute in a given amount of solution

occasionally amount of solvent

41

Solution ConcentrationSolution ConcentrationMolarityMolarity

• Moles of solute per 1 liter of solution• Used because it describes how many

molecules of solute in each liter of solution

42

Preparing 1 L of a 1.00 M Preparing 1 L of a 1.00 M NaClNaCl SolutionSolution

43

Example 4.5: Find the Example 4.5: Find the molaritymolarity of a solution that of a solution that has 25.5 g has 25.5 g KBrKBr dissolved in 1.75 L of solutiondissolved in 1.75 L of solution

because most solutions are between 0 and 18 M, the answer makes sense

Check:

Solution:

Conceptual Plan:

Relationships:

Given:Find:

44

Practice Practice —— What Is the What Is the molaritymolarity of a solution containing of a solution containing 3.4 g of NH3.4 g of NH3 3 (MM 17.03) in 200.0 (MM 17.03) in 200.0 mLmL of solution?of solution?

Check:

Solve:

Conceptual Plan:

Relationships:

Given:Find:

Answer: 1.0 M

45

Using Using MolarityMolarity in Calculationsin Calculations

• Molarity shows the relationship between the moles of solute and liters of solution

• If a sugar solution concentration is 2.0 M, then 1 liter of solution contains 2.0 moles of sugar

2 liters = 4.0 moles sugar 0.5 liters = 1.0 mole sugar

• 1 L solution : 2 moles sugar

46

Example 4.6: How many liters of 0.125 M Example 4.6: How many liters of 0.125 M NaOHNaOH contain 0.255 mol contain 0.255 mol NaOHNaOH??

because each L has only 0.125 mol NaOH, it makes sense that 0.255 mol should require a little more than 2 L

Check:

Solution:

Conceptual Plan:

Relationships:

Given:Find:

47

Practice Practice —— Determine the mass of CaClDetermine the mass of CaCl2 2 (MM = 110.98) in 1.75 L of 1.50 M solution(MM = 110.98) in 1.75 L of 1.50 M solution

Check:

Solution:

Conceptual Plan:

Relationships:

Given:Find:

Answer: 291 g CaCl2

48

Example: How would you prepare Example: How would you prepare 250.0 250.0 mLmL of a of a 1.00 M solution 1.00 M solution CuSOCuSO44••5 H5 H22O(MM 249.69)?O(MM 249.69)?

the unit is correct, the magnitude seems reasonable as the volume is ¼ of a liter

Check:

Solution:

Conceptual Plan:

Relationships:

Given:Find:

49

Practice Practice –– How would you prepare 250.0 How would you prepare 250.0 mLmL of of 0.150 M CaCl0.150 M CaCl2 2 (MM = 110.98)?(MM = 110.98)?

Check:

Solution:

Conceptual Plan:

Relationships:

Given:Find:

Answer: Dissolve 4.16 g of CaCl2 in enough water to total 250.0 mL

50

DilutionDilution

• Often, solutions are stored as concentrated stock solutions

• To make solutions of lower concentrations from these stock solutions, more solvent is added

the amount of solute doesn’t change, just the volume of solutionmoles solute in solution 1 = moles solute in solution 2

• The concentrations and volumes of the stock and new solutions are inversely proportional

M1·V1 = M2·V2

51

Example 4.7: To what volume should you dilute Example 4.7: To what volume should you dilute 0.200 L of 15.0 M 0.200 L of 15.0 M NaOHNaOH to make 3.00 M to make 3.00 M NaOHNaOH??

because the solution is diluted by a factor of 5, the volume should increase by a factor of 5, and it does

Check:

Solution:

Conceptual Plan:

Relationships:

Given:Find:

52

Practice Practice –– What is the concentration of a solution prepared What is the concentration of a solution prepared by diluting 45.0 by diluting 45.0 mLmL of 8.25 M HNOof 8.25 M HNO33 to 135.0 to 135.0 mLmL??

Check:

Solution:

Conceptual Plan:

Relationships:

Given:Find:

Answer: 2.75 M

53

Practice Practice –– How would you prepare 200.0 How would you prepare 200.0 mLmL of of 0.25 M 0.25 M NaClNaCl solution from a 2.0 M solution?solution from a 2.0 M solution?

Check:

Solution:

Conceptual Plan:

Relationships:

Given:Find:

Answer: Dilute 25 mL of 2.0 M solution up to 200.0 mL

54

Solution Solution StoichiometryStoichiometry

• Because molarity relates the moles of solute to the liters of solution, it can be used to convert between amount of reactants and/or products in a chemical reaction

55

Example 4.8: What volume of 0.150 M Example 4.8: What volume of 0.150 M KClKCl is required to is required to completely react with 0.150 L of 0.175 M Pb(NOcompletely react with 0.150 L of 0.175 M Pb(NO33))2 2 in the in the reaction 2 reaction 2 KCl(KCl(aqaq) + Pb(NO) + Pb(NO33))22((aqaq) ) →→ PbClPbCl22((ss) + 2 KNO) + 2 KNO33((aqaq)?)?

because you need 2x moles of KCl as Pb(NO3)2, and the molarity of Pb(NO3)2 > KCl, the volume of KCl should be

more than 2x the volume of Pb(NO3)2

1 L Pb(NO3)2 = 0.175 mol, 1 L KCl = 0.150 mol, 1 mol Pb(NO3)2 : 2 mol KCl

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

56

Practice Practice –– 43.8 43.8 mLmL of 0.107 M of 0.107 M HClHCl is needed to neutralize is needed to neutralize 37.6 37.6 mLmL of Ba(OH)of Ba(OH)22 solution. What is the solution. What is the molaritymolarity of the of the base? 2 base? 2 HCl(HCl(aqaq) + Ba(OH)) + Ba(OH)22((aqaq) ) →→ BaClBaCl22((aqaq) + 2 H) + 2 H22O(O(aqaq))

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

Answer: 0.0623 M

57

What Happens When a Solute Dissolves?What Happens When a Solute Dissolves?

• There are attractive forces between the solute particles holding them together

• There are also attractive forces between the solvent molecules

• When we mix the solute with the solvent, there are attractive forces between the solute particles and the solvent molecules

• If the attractions between solute and solvent are strong enough, the solute will dissolve

58

Table Salt Dissolving in WaterTable Salt Dissolving in WaterEach ion is attracted to the surrounding water molecules and pulled off and away from the crystalWhen it enters the solution, the ion is surrounded by water molecules, insulating it from other ions The result is a solution with free moving charged particles able to conduct electricity

59

Electrolytes and Electrolytes and NonelectrolytesNonelectrolytes• Materials that dissolve

in water to form a solution that will conduct electricity are called electrolytes

• Materials that dissolve in water to form a solution that will not conduct electricity are called nonelectrolytes

60

Molecular View of Molecular View of Electrolytes and Electrolytes and NonelectrolytesNonelectrolytes

• To conduct electricity, a material must have charged particles that are able to flow

• Electrolyte solutions all contain ions dissolved in the water

ionic compounds are electrolytes because they dissociate into their ions when they dissolve

• Nonelectrolyte solutions contain whole molecules dissolved in the water

generally, molecular compounds do not ionize when they dissolve in water

the notable exception being molecular acids

61

Salt vs. Sugar Dissolved in WaterSalt vs. Sugar Dissolved in Water

ionic compounds dissociate into ions when

they dissolve

molecular compounds do not dissociate when

they dissolve

62

AcidsAcids• Acids are molecular compounds that ionize when

they dissolve in waterthe molecules are pulled apart by their attraction for the waterwhen acids ionize, they form H+ cations and also anions

• The percentage of molecules that ionize varies from one acid to another

• Acids that ionize virtually 100% are called strong acids

HCl(aq) → H+(aq) + Cl−(aq)• Acids that only ionize a small percentage are

called weak acidsHF(aq) ⇔ H+(aq) + F−(aq)

63

Strong and Weak ElectrolytesStrong and Weak Electrolytes

• Strong electrolytes are materials that dissolve completely as ions

ionic compounds and strong acidstheir solutions conduct electricity well

• Weak electrolytes are materials that dissolve mostly as molecules, but partially as ions

weak acidstheir solutions conduct electricity, but not well

• When compounds containing a polyatomic ion dissolve, the polyatomic ion stays together

HC2H3O2(aq) ⇔ H+(aq) + C2H3O2−(aq)

64

Classes of Dissolved MaterialsClasses of Dissolved Materials

65

Dissociation and IonizationDissociation and Ionization

• When ionic compounds dissolve in water, the anions and cations are separated from each other. This is called dissociation.

Na2S(aq) → 2 Na+(aq) + S2-(aq)• When compounds containing polyatomic ions

dissociate, the polyatomic group stays together as one ion

Na2SO4(aq) → 2 Na+(aq) + SO42−(aq)

• When strong acids dissolve in water, the molecule ionizes into H+ and anions

H2SO4(aq) → 2 H+(aq) + SO42−(aq)

66

Practice Practice –– Write the equation for the process Write the equation for the process that occurs when the following strong that occurs when the following strong

electrolytes dissolve in waterelectrolytes dissolve in water

CaCl2

HNO3

(NH4)2CO3

67

Solubility of Ionic CompoundsSolubility of Ionic Compounds

• Some ionic compounds, such as NaCl, dissolve very well in water at room temperature

• Other ionic compounds, such as AgCl, dissolve hardly at all in water at room temperature

• Compounds that dissolve in a solvent are said to be soluble, where as those that do not are said to be insoluble

NaCl is soluble in water, AgCl is insoluble in waterthe degree of solubility depends on the temperatureeven insoluble compounds dissolve, just not enough to be meaningful

68

When Will a Salt Dissolve?When Will a Salt Dissolve?

• Predicting whether a compound will dissolve in water is not easy

• The best way to do it is to do some experiments to test whether a compound will dissolve in water, then develop some rules based on those experimental results

we call this method the empirical method

69

Solubility RulesSolubility RulesCompounds that Are Generally Compounds that Are Generally SolubleSoluble in Waterin Water

70

Solubility RulesSolubility RulesCompounds that Are Generally Compounds that Are Generally InsolubleInsoluble in Waterin Water

71

Practice Practice –– Determine if each of the Determine if each of the following is soluble in waterfollowing is soluble in water

KOH

AgBr

CaCl2

Pb(NO3)2

PbSO4

72

Precipitation ReactionsPrecipitation Reactions

• Precipitation reactions are reactions in which a solid forms when we mix two solutions

reactions between aqueous solutions of ionic compounds produce an ionic compound that is insoluble in water the insoluble product is called a precipitate

73

2 2 KI(KI(aqaq)) + + Pb(NOPb(NO33))22((aqaq)) →→PbIPbI22((ss) + 2 KNO) + 2 KNO33((aqaq))

74

No Precipitate Formation = No Precipitate Formation = No ReactionNo Reaction

KI(aq) + NaCl(aq) → KCl(aq) + NaI(aq)all ions still present, ∴ no reaction

75

Process for Predicting the Products of a Process for Predicting the Products of a Precipitation ReactionPrecipitation Reaction

1. Determine what ions each aqueous reactant has2. Determine formulas of possible products

exchange ions(+) ion from one reactant with (-) ion from other

balance charges of combined ions to get formula of each product

3. Determine solubility of each product in wateruse the solubility rulesif product is insoluble or slightly soluble, it will precipitate

4. If neither product will precipitate, write no reaction after the arrow

76

Process for Predicting the Products of a Process for Predicting the Products of a Precipitation ReactionPrecipitation Reaction

5. If any of the possible products are insoluble, write their formulas as the products of the reaction using (s) after the formula to indicate solid. Write any soluble products with (aq)after the formula to indicate aqueous.

6. Balance the equationremember to only change coefficients, not subscripts

77

Example 4.10: Write the equation for the precipitation Example 4.10: Write the equation for the precipitation reaction between an aqueous solution of potassium reaction between an aqueous solution of potassium

carbonate and an aqueous solution of nickel(II) chloridecarbonate and an aqueous solution of nickel(II) chloride

1. Write the formulas of the reactantsK2CO3(aq) + NiCl2(aq) →

2. Determine the possible productsa) determine the ions present

(K+ + CO32−) + (Ni2+ + Cl−) →

b) exchange the Ions(K+ + CO3

2−) + (Ni2+ + Cl−) → (K+ + Cl−) + (Ni2+ + CO32−)

c) write the formulas of the productsbalance charges

K2CO3(aq) + NiCl2(aq) → KCl + NiCO3

78



3. Determine the solubility of each productKCl is soluble

NiCO3 is insoluble4. If both products are soluble, write no reaction

does not apply because NiCO3 is insoluble

79



5. Write (aq) next to soluble products and (s) next to insoluble products

K2CO3(aq) + NiCl2(aq) → KCl(aq) + NiCO3(s)6. Balance the equation

K2CO3(aq) + NiCl2(aq) → 2 KCl(aq) + NiCO3(s)

80

Practice Practice –– Predict the products and Predict the products and balance the equationbalance the equation

KCl(aq) + AgNO3(aq) →

Na2S(aq) + CaCl2(aq) →

81

PracticePractice –– Write an equation for the reaction that Write an equation for the reaction that takes place when an aqueous solution of (NHtakes place when an aqueous solution of (NH44))22SOSO44 is is

mixed with an aqueous solution of Pb(Cmixed with an aqueous solution of Pb(C22HH33OO22))22..

82

Ionic EquationsIonic Equations

• Equations that describe the chemicals put into the water and the product molecules are called molecular equations2 KOH(aq) + Mg(NO3)2(aq) → 2 KNO3(aq) + Mg(OH)2(s)

• Equations that describe the material’s structure when dissolved are called complete ionic equations

aqueous strong electrolytes are written as ionssoluble salts, strong acids, strong bases

insoluble substances, weak electrolytes, and nonelectrolytes are written in molecule form

solids, liquids, and gases are not dissolved, therefore molecule form2K+

(aq) + 2OH−(aq) + Mg2+

(aq) + 2NO3−

(aq) → 2K+(aq) + 2NO3

−(aq) + Mg(OH)2(s)

83

Ionic EquationsIonic Equations

• Ions that are both reactants and products are called spectator ions

2 K+(aq) + 2 OH−

(aq) + Mg2+(aq) + 2 NO3

−(aq) → 2 K+

(aq) + 2 NO3−

(aq) + Mg(OH)2(s)

An ionic equation in which the spectator ions are removed is called a net ionic equation

2 OH−(aq) + Mg2+

(aq) → Mg(OH)2(s)

84

PracticePractice –– Write the ionic and net Write the ionic and net ionic equation for eachionic equation for each

K2SO4(aq) + 2 AgNO3(aq) → 2 KNO3(aq) + Ag2SO4(s)

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

85

AcidAcid--Base ReactionsBase Reactions

• Also called neutralization reactions because the acid and base neutralize each other’s properties

2 HNO3(aq) + Ca(OH)2(aq) → Ca(NO3)2(aq) + 2 H2O(l)• The net ionic equation for an acid-base

reaction isH+(aq) + OH−(aq) → H2O(l)

as long as the salt that forms is soluble in water

86

Acids and Bases in SolutionAcids and Bases in Solution

• Acids ionize in water to form H+ ionsmore precisely, the H from the acid molecule is donated to a water molecule to form hydronium ion, H3O+

most chemists use H+ and H3O+ interchangeably• Bases dissociate in water to form OH− ions

bases, such as NH3, that do not contain OH− ions, produce OH− by pulling H off water molecules

• In the reaction of an acid with a base, the H+ from the acid combines with the OH− from the base to make water

• The cation from the base combines with the anion from the acid to make the salt

acid + base → salt + water

87

Common AcidsCommon Acids

88

Common BasesCommon Bases

89

HCl(HCl(aqaq)) + + NaOH(NaOH(aqaq) ) →→ NaCl(NaCl(aqaq) + H) + H22O(O(ll))

90

Example: Write the molecular, ionic, and netExample: Write the molecular, ionic, and net--ionic ionic equation for the reaction of aqueous nitric acid with equation for the reaction of aqueous nitric acid with aqueous calcium hydroxideaqueous calcium hydroxide

1. Write the formulas of the reactants→

2. Determine the possible productsa) determine the ions present when each reactant

dissociates or ionizes→

b) exchange the ions, H+ combines with OH− to make H2O(l)

→c write the formula of the salt

→

91


3. Determine the solubility of the saltCa(NO3)2 is

4. Write an (s) after the insoluble products and an (aq) after the soluble products

5. Balance the equation

92


6. Dissociate all aqueous strong electrolytes to get complete ionic equation

not H2O

7. Eliminate spectator ions to get net-ionic equation

93


HCl(aq) + Ba(OH)2(aq) →

H2SO4(aq) + Sr(OH)2(aq) →

94

TitrationTitration• Often in the lab, a solution’s concentration is

determined by reacting it with another material and using stoichiometry – this process is called titration

• In the titration, the unknown solution is added to a known amount of another reactant until the reaction is just completed. At this point, called the endpoint, the reactants are in their stoichiometric ratio.

the unknown solution is added slowly from an instrument called a burette

a long glass tube with precise volume markings that allows small additions of solution

95

AcidAcid--Base TitrationsBase Titrations

• The difficulty is determining when there has been just enough titrant added to complete the reaction

the titrant is the solution in the burette• In acid-base titrations, because both the reactant

and product solutions are colorless, a chemical is added that changes color when the solution undergoes large changes in acidity/alkalinity

the chemical is called an indicator• At the endpoint of an acid-base titration, the

number of moles of H+ equals the number of moles of OH−

also known as the equivalence point

96

TitrationTitration

97

TitrationTitrationThe titrant is the base solution in the burette.As the titrant is added tothe flask, the H+ reacts with the OH– to form water. But there is still excess acid present so the color does not change.

At the titration’s endpoint,just enough base has been added to neutralize all the acid. At this point the indicator changes color.

98

Example 4.14:Example 4.14:The titration of 10.00 The titration of 10.00 mLmL of of HClHCl solution of unknown solution of unknown concentration requires 12.54 concentration requires 12.54 mLmL of 0.100 M of 0.100 M NaOHNaOH solution solution to reach the end point. What to reach the end point. What is the concentration of the is the concentration of the unknown unknown HClHCl solution?solution?

• Write down the given quantity and its units Given: 10.00 mL HCl

12.54 mL of 0.100 M NaOH

99

• Write down the quantity to find, and/or its units Find: concentration HCl, M


InformationGiven: 10.00 mL HCl

12.54 mL of 0.100 M NaOH

100

• Collect needed equations and conversion factors HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

∴ 1 mole HCl = 1 mole NaOH0.100 M NaOH ∴0.100 mol NaOH ≡ 1 L sol’n



12.54 mL of 0.100 M NaOHFind: M HCl

101

• Write a conceptual plan

mLNaOH

LNaOH

molNaOH

molHCl


12.54 mL of 0.100 M NaOHFind: M HClRel: 1 mol HCl = 1 mol NaOH

0.100 mol NaOH = 1 LM = mol/L


mLHCl

LHCl

102

• Apply the conceptual plan


12.54 mL of 0.100 M NaOHFind: M HClRel: 1 mol HCl = 1 mol NaOH


CP: mL NaOH → L NaOH →

mol NaOH → mol HCl;mL HCl → L HCl & mol ⇒ M

Example 4.14:Example 4.14:The titration of 10.00 The titration of 10.00 mLmLof of HClHCl solution of unknown solution of unknown concentration requires 12.54 concentration requires 12.54 mLmL of 0.100 M of 0.100 M NaOHNaOHsolution to reach the end solution to reach the end point. What is the point. What is the concentration of the concentration of the unknown unknown HClHCl solution?solution?

103

• Apply the conceptual plan


12.54 mL NaOHFind: M HClRel: 1 mol HCl = 1 mol NaOH





104



12.54 mL NaOHFind: M HClRel: 1 mol HCl = 1 mol NaOH




• Check the solutionHCl solution = 0.125 M

The units of the answer, M, are correct.The magnitude of the answer makes sense because

the neutralization takes less HCl solution thanNaOH solution, so the HCl should be more concentrated.

105

Practice Practice —— What is the concentration of What is the concentration of NaOHNaOH solution that solution that requires 27.5 requires 27.5 mLmL to titrate 50.0 to titrate 50.0 mLmL of 0.1015 M Hof 0.1015 M H22SOSO44? ? 2 2 NaOH(NaOH(aqaq) + H) + H22SOSO44((aqaq) ) →→ NaNa22SOSO44((aqaq) + 2 H) + 2 H22O(O(aqaq))

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

Answer: 0.369 M

106

GasGas--Evolving ReactionsEvolving Reactions

• Some reactions form a gas directly from the ion exchange

K2S(aq) + H2SO4(aq) → K2SO4(aq) + H2S(g)• Other reactions form a gas by the decomposition

of one of the ion exchange products into a gas and water

K2SO3(aq) + H2SO4(aq) → K2SO4(aq) + H2SO3(aq)H2SO3 → H2O(l) + SO2(g)

107

NaHCONaHCO33((aqaq) + ) + HCl(HCl(aqaq) ) →→NaCl(NaCl(aqaq) + CO) + CO22((gg) + H) + H22O(O(ll))

108

Compounds that UndergoCompounds that UndergoGasGas--Evolving ReactionsEvolving Reactions

109

Example 4.14: When an aqueous solution of sodium Example 4.14: When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, carbonate is added to an aqueous solution of nitric acid, a gas evolvesa gas evolves

1. Write the formulas of the reactantsNa2CO3(aq) + HNO3(aq) →

2. Determine the possible productsa) determine the ions present when each reactant

dissociates or ionizes(Na+ + CO3

2−) + (H+ + NO3−) →

b) exchange the anions(Na+ + CO3

2−) + (H+ + NO3−) → (Na+ + NO3

−) + (H+ + CO32−)

c) write the formula of compoundsbalance the charges

Na2CO3(aq) + HNO3(aq) → NaNO3 + H2CO3

110


3. Check to see if either product is H2S - No4. Check to see if either product decomposes –

YesH2CO3 decomposes into CO2(g) + H2O(l)

Na2CO3(aq) + HNO3(aq) → NaNO3 + CO2(g) + H2O(l)

111


5. Determine the solubility of other productNaNO3 is soluble

6. Write an (s) after the insoluble products and an (aq) after the soluble products

Na2CO3(aq) + 2 HNO3(aq) → 2 NaNO3(aq) + CO2(g) + H2O(l)

7. Balance the equationNa2CO3(aq) + 2 HNO3(aq) → 2 NaNO3(aq) + CO2(g) + H2O(l)

112


HCl(aq) + Na2SO3(aq) →

H2SO4(aq) + CaS(aq) →

113

Other Patterns in ReactionsOther Patterns in Reactions

• The precipitation, acid-base, and gas-evolving reactions all involve exchanging the ions in the solution

• Other kinds of reactions involve transferring electrons from one atom to another – these are called oxidation-reduction reactions

also known as redox reactionsmany involve the reaction of a substance with O2(g)

4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)

114

Combustion as Combustion as RedoxRedox2 H2 H22((gg) + O) + O22((gg)) →→ 2 2 HH22O(O(gg))

115

RedoxRedox without Combustionwithout Combustion2 Na(2 Na(ss)) + Cl+ Cl22((gg) ) →→ 2 2 NaCl(NaCl(ss))

2 Na → 2 Na+ + 2 e−

Cl2 + 2 e− → 2 Cl−

116

Reactions of Metals with NonmetalsReactions of Metals with Nonmetals

• Consider the following reactions:4 Na(s) + O2(g) → 2 Na2O(s)2 Na(s) + Cl2(g) → 2 NaCl(s)

• The reactions involve a metal reacting with a nonmetal

• In addition, both reactions involve the conversion of free elements into ions

4 Na(s) + O2(g) → 2 Na+2O2– (s)

2 Na(s) + Cl2(g) → 2 Na+Cl–(s)

117

Oxidation and ReductionOxidation and Reduction• To convert a free element into an ion, the atoms

must gain or lose electronsof course, if one atom loses electrons, another must accept them

• Reactions where electrons are transferred from one atom to another are redox reactions

• Atoms that lose electrons are being oxidized, atoms that gain electrons are being reduced

2 Na(s) + Cl2(g) → 2 Na+Cl–(s)Na → Na+ + 1 e– oxidation

Cl2 + 2 e– → 2 Cl– reductionLeo

Ger

118

Electron BookkeepingElectron Bookkeeping

• For reactions that are not metal + nonmetal, or do not involve O2, we need a method for determining how the electrons are transferred

• Chemists assign a number to each element in a reaction called an oxidation state that allows them to determine the electron flow in the reaction

even though they look like them, oxidation states are not ion charges!

oxidation states are imaginary charges assigned based on a set of rules ion charges are real, measurable charges

119

Rules for Assigning Oxidation StatesRules for Assigning Oxidation States

• Rules are in order of priority1. free elements have an oxidation state = 0

Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g)2. monatomic ions have an oxidation state equal

to their chargeNa = +1 and Cl = −1 in NaCl

3. (a) the sum of the oxidation states of all the atoms in a compound is 0

Na = +1 and Cl = −1 in NaCl, (+1) + (−1) = 0

120


3. (b) the sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion

N = +5 and O = −2 in NO3–, (+5) + 3(−2) = −1

4. (a) Group I metals have an oxidation state of +1 in all their compounds

Na = +1 in NaCl4. (b) Group II metals have an oxidation state of +2

in all their compoundsMg = +2 in MgCl2

121


5. in their compounds, nonmetals have oxidation states according to the table below

nonmetals higher on the table take priority

122

Example: Determine the oxidation states of all Example: Determine the oxidation states of all the atoms in a the atoms in a propanoatepropanoate ion, Cion, C33HH55OO22

––

• There are no free elements or free ions in propanoate, so the first rule that applies is Rule 3b

• Because all the atoms are nonmetals, the next rule we use is Rule 5, following the elements in order:

H = +1O = −2 Note: unlike charges,

oxidation states can be fractions!

123

Practice Practice –– Assign an oxidation state Assign an oxidation state to each element in the followingto each element in the following

• Br2

• K+

• LiF

• CO2

• SO42−

• Na2O2

124

Oxidation and ReductionOxidation and ReductionAnother DefinitionAnother Definition

• Oxidation occurs when an atom’s oxidation state increases during a reaction

• Reduction occurs when an atom’s oxidation state decreases during a reaction

CH4 + 2 O2 → CO2 + 2 H2O

125

OxidationOxidation––ReductionReduction• Oxidation and reduction must occur simultaneously

if an atom loses electrons another atom must take them

• The reactant that reduces an element in another reactant is called the reducing agent

the reducing agent contains the element that is oxidized

• The reactant that oxidizes an element in another reactant is called the oxidizing agent

the oxidizing agent contains the element that is reduced

2 Na(s) + Cl2(g) → 2 Na+Cl–(s)is oxidized, is reduced

is the reducing agent, is the oxidizing agent

126

Fe + MnO4− + 4 H+ → Fe3+ + MnO2 + 2 H2O

Example: Assign oxidation states, determine the element Example: Assign oxidation states, determine the element oxidized and reduced, and determine the oxidizing agent oxidized and reduced, and determine the oxidizing agent and reducing agent in the following reactions:and reducing agent in the following reactions:

127

Practice Practice –– Assign oxidation states, determine the element Assign oxidation states, determine the element oxidized and reduced, and determine the oxidizing agent oxidized and reduced, and determine the oxidizing agent and reducing agent in the following reactions:and reducing agent in the following reactions:

Sn4+ + Ca → Sn2+ + Ca2+

F2 + S → SF4

is oxidized, is reducedis the reducing agent, is the oxidizing agent

is oxidized, is reducedis the reducing agent, is the oxidizing agent

Documents

CH131 Note 04 Quantity