chapter 5 molecular view of reactions in aqueous solutions: part 2 chemistry: the molecular nature...

Chapter 5 Molecular View of

Reactions in Aqueous Solutions: Part 2

Chemistry: The Molecular Nature of Matter, 6E

Jespersen/Brady/Hyslop

Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Predicting Precipitation ReactionsMetathesis 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

2


Table 5.1 Solubility Rules Soluble Compounds1. All salts of the alkali metals (Group IA) 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+.

3


Table 5.1 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 IA 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 IA and NH4+

4


Learning Check: Solubility Rules

5

Which of the following compounds are expected to be soluble in water?

Ca(C2H3O2)2

FeCO3

AgCl

Yes

No

No


Your TurnWhich of the following 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

6


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 Occur only if solid, gas, weak

electrolyte or non-electrolyte product forms Otherwise, all ions are spectator ions

7


Predicting Products of Double Replacement Reactions

1. Identify the ions involved: Distinguish between subscripts that count and

those that are characteristic of a polyatomic ion.

2. Swap partners and make neutral with appropriate subscripts

3. Assign states using solubility rules4. Balance equation

8

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, 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 or non-electrolyte

9


Learning Check: Predict ProductsPb(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)

10

BaCO3(s) + 2 NaCl (aq)

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

NR (No reaction)

CaCl2(aq) + 2 NaNO3(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)

11

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)

12

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)13


2. Predicting Acid−Base ReactionsNeutralization reaction

When mixed in 1:1 molar ratio, acid and base solutions lose their acidic and basic properties

Combination of H+ and OH– to form salt and water

Salt

Ionic compound without H+, OH–, or O2–

Acid + Base Salt + Water

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

14


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

15

2H2O + CaCl2(aq)


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)

16

H2O + NaC2H3O2(aq)


Neutralization of Strong Acid with Insoluble Base

Insoluble HydroxidesMolecular 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 17

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

18

2AlCl3(aq) + 3H2O


Strong Acid with Weak BaseMolecular 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)

19

NH4Cl(aq)

NH3(aq) + H3O+(aq) NH4+ (aq) + H2O


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 20

Mg(C2H3O2)2(aq) + 2H2O


Learning Check

21

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) N2H4+(aq)

3. CH3NH2(aq) + HC4H7O2(aq)

CH3NH2(aq) + HC4H7O2(aq) CH3NH3+(aq)

+ C4H7O2

–(aq)


Your TurnWhich 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

22


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 escaped, can’t go back in, drives

reaction to completion Many anions that give rise to gases are

insoluble Dissolving in acid followed by gas formation

drives reaction to completion Dissolves insoluble salt

23


Metathesis and Gas Formation1. Gases formed 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)

24


Reactions that Release CO2

a) Acid with Bicarbonate (HCO3–)

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

a) Acid with Carbonate (CO32–)

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

25

a) b)


Reactions that Release GasesAcid with Sulfites (SO3

2–) or Bisulfites (HSO3

2–)

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

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

Acid with Sulfides2HCl(aq) + Na2S(aq) 2NaCl(aq) + H2S(g)

Acid with CyanidesHNO3(aq) + CsCN(aq) HCN(g) + CsNO3(aq)

Bases with AmmoniumNaOH(aq) + NH4Cl(aq) NH3(g) + H2O + NaCl(aq)

26


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) + SO3

2–(aq) + 2HCHO2(aq)

2CHO2–(aq) + 2Li+(aq) + SO2(g) + H2O

Net Ionic Equation: SO3

2 –(aq) + 2HCHO2(aq) 2CHO2

–(aq) + SO2(g) + H2O

27


Your TurnWhat 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

28


Metathesis OverviewPrecipitation: 2 solutions form solid product

Neutralization: Acid + metal hydroxide or oxide forms water and

salt

Gas-forming: Metathesis reaction forms one of these products:

HCN, H2S, H2CO3(aq) , H2SO3(aq) , NH4OH(aq)

Formation of Weak Electrolyte: Salt of weak acid reacts with base to form

molecule29


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

30


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

31


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

32


Your TurnWhat 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.

33


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) + HClO4(aq)

E. KCN(aq) + H2SO4(aq)

34


Synthesize by Metathesis Reaction Practical use of metathesis reactions

Desired compound should be easily separated from reaction mixture. Two principal approaches1. Desired compound is insoluble in water

Start with 2 soluble reactants Product isolated by filtration

2. Desired compound is soluble in water Acid-base neutralization Reaction of metal carbonate and acid Either way, product isolated by

evaporation of water 35


Learning Check

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

Use solubility rules NiSO4 is soluble in water

So, there are 2 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 36


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

37

Volumemole

solution of literssolute of moles

(M)Molarity


Molarity (M)

38

Number of moles of solute per liter of solution.

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

Hence, 0.100M 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

NaCl 100.0soln NaClL 0.100

NaCl mol 0.0100soln NaClL 1.00

NaCl mol 0.100M


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

39

Molarity

Moles of a Substance

Volume of a solution of Substance

soln NaClL 1.00NaCl mol 0.100

NaCl 100.0 M


Molarity as Conversion Factor Gives equivalence relationship between

“mol NaCl” and “L soln” Forms two conversion factors

Three basic types of calculations:

40



0.100 mol NaCl 1.00 L soln

Vmol

M molV MMmol

V


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

41

NaOH mol 288.0NaOH g 00.40

NaOH mol 1 NaOH g 5.11

solnL 50.1NaOH mol 288.0

lnsoL NaOH moles

M

NaOH 192.0 M


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

42


NaCl 250.0 M

= 400 mL of 0.250 M NaCl solution

soln NaClL 1soln NaClmL 1000


NaCl mol 100.0


Preparing Solution of Known Molarity

a b c d ea) Weigh solid and transfer to volumetric flaskb) Add part of the waterc) Dissolve solute completelyd) Add water to reach etched linee) Stopper flask and invert to mix thoroughly

43

a) b) c) d) e)


Learning Check: Preparing Solution of Known Molarity from

SolidHow many grams of strontium nitrate are required to prepare 250.0 mL of 0.100 M Sr(NO3)2 solution? M × V mol × MM g1. Convert Molarity and Volume to mole

= 0.0250 mol Sr(NO3)2

2. Convert mol to g

44

L 1 100.0

mL 1000L 1

soln )Sr(NOmL 250 23M

= 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 gB. 50.0 gC. 0.316 gD. 7.90 gE. 198 g

45

g 09.7KMnO mol

g 03.158L 1mol 200.0

mL 1000L 1

mL 2504


Preparing a Solution of Known Molarity by Dilution

When making solutions, don’t always begin with solute as pure solid, liquid or gas.

Can take solution of high concentration and dilute to a lower concentration.

Amount of MOLES does NOT change!Remains the same

46

Small Volume

Concentrated Solution

LargeVolume Dilute

Solution

Add lots of water


Diluting Solutions

edconcentratdilute M

usedbeto

M

preparedbeto

solution edconcentrat

of Volume

solution dilute

of Volume

Moles of solute do not change upon dilution Just changing volume# moles in dilute = # moles in concentrated

47

Moles of solute in the dilute solution

Moles of solute in the

concentrated solutionVdil Mdil = Vconc Mconc


Learning Check: DilutionsWhat volume (in mL) of 16.0 M H2SO4 must be used to prepare 1.00 L of 2.00 M H2SO4?

48

MM 0.16V 00.2L 00.1

mol/L 0.16mol 00.2

0.16 2.00L 00.1

V

M

M

= 125 mL

Rearranging gives

L 1.00mL 1000

L 125.0V


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

49


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

50

MM 3.2mL 250. 12.1Vconc

Vconc = 66 mL


Your Turn25 mL of 6.0 M HCl are 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

dilMmL 500 6.0mL 25 M

51

Mdil = 0.30 M


Solution Stoichiometry

Often work with solutions when running 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.

52


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:

53

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

54

Learning Check: Solution Stoichiometry

KOHL 1

KOH mol 100.0KOHmL 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 1mL 1000

POH mol 0.0475 POHL 1

POH mol10 50.143

4343

3

= 31.6 mL H3PO4


Stoichiometry of Ionic Equations Calculating Concentrations of Ions in

Solutions of Electrolytes. When using solutions of electrolytes, need to

know [ions] in solution. Ions don’t stay together, [ions] may not be same Can easily calculate from M electrolyte in

molecular form. Concentration of particular ion equals

concentration of salt multiplied by number of ions of that kind in one formula unit of salt.

55

Note: [XX] = concentration of whatever species is in square bracket


Learning Check: Ion ConcentrationsIf 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:

56

Na 300.0CONa mol 1

Na mol 2CONaL 1

CONa mol 150.0

3232

32 M

23

32

23

32 CO 150.0CONa mol 1

CO mol 1CONa 150.0 MM


Your TurnIf 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

57

--

mol

molM

24

34224 SO 3

)(SO Al1SO 750.0

= 0.250 M Al2(SO4)3


Learning Check: Net Ionic Eqns in Solution Stoichiometry

CalculationsWhat 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) + 2 OH–(aq) Fe(OH)2(s)

2. Determine the game plan

58

M FeCl2

M Fe2+

mol Fe2+ mol OH–

V OH–

V KOH


Learning Check: Net Ionic Eqns in Solution Stoichiometry

Calculations3. Convert M FeCl2 M Fe2+ mol Fe2+

4. Convert mol Fe2+ mol OH–

5. Convert mol OH– V OH– V KOH

59

mL 0100L 1

mL 0.60FeCl mol 1Fe mol 1

FeCl 250.02

2

2

M

OH mol 0030.0

Fe mol 1

OH mol 2Fe mol 0015.0

22

= 0.0150 mol Fe2+

= 60.0 mL KOH L 1mL 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) will form if 20.0 mL of 0.800 M FeI3 (436.5) is mixed with 50.0 mL of 0.300 M Pb(NO3)2 (269.2)?

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

Net Ionic Eqn: Pb2+(aq) + 2I(aq) PbI2(s)

StrategyVol Pb(NO3)2 mol Pb(NO3)2 mol Pb2+ mol PbI2 g PbI2

Vol FeI3 mol FeI3 mol I mol PbI2 g PbI2

Whichever is less determines how much is formed and which reagent is limiting

60


Limiting Reagent ProblemStarting with Pb(NO3)2

Starting with FeI3

61

2

222

23

2

23

23

23

2323

PbI mol 1PbI 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 1PbI g 0.461

I mol 2

PbI mol 1

FeI mol 1I mol 3

FeIL 1FeI mol 080.0

FeImL 1000FeIL 1

FeImL .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 AnalysisQualitative Analysis

What substances are present in a sample

Quantitative Analysis Measure the amounts of various substances

in a sample Chemical reactions useful

strategy 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 orginial sample (unknown)

62


Learning Check: Chemical AnalysisInsecticide contains H, C, Cl. Carry out

reactions on 1.000 g sample to convert all chlorine to Cl– in H2O. Resulting solution treated with AgNO3 solution in excess. AgCl ppts and is collected, dried, and weighed. Mass AgCl = 2.022 g. What is the percentage by mass of Cl in original sample?

Strategy:

g AgCl mol AgCl mol Cl g Cl

63


How Much Cl in 2.022 g of AgCl?

64

Cl mol 1Cl g 35.45

AgClmol 1Cl mol 1

AgClg 3.143 AgClmol 1

AgClg 202.2

= 0.5002 g Cl

% Cl in original Sample?

100sample of mass

Cl of mass%Cl

100sample g 1.000

Cl g 0.5002%Cl = 50.02% Cl

g AgCl mol AgCl mol Cl g Cl


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 precise 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

65


Titration Controlled addition of 1 reactant to known

quantity of another until reaction is complete

Acid-Base Titration Very common type of titration

Ex. 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 acid 66


Titration in practice:

67

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 exactly equal68

.


TitrationIndicator

Dye that is 1 color in acid and 2nd color in base

Ex. 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 same69


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

70


Learning Check: Oxalic acid + NaOH

Step 2: Calculate moles of Base used

71

NaOH mmol 5718.soln NaOHmL 1NaOH mmol 0.500

soln NaOHmL 37.5

Step 3: Calculate moles of Oxalic Acid

422422 OCH mmol 579.3

NaOH mmol 2OCH mmol 1

NaOH mmol 578.1

Step 4: Calculate M H2C2O4

solnmL 25OCH mmol 579.3 422 = 0.375 M H2C2O4


Summary of Stoichiometry Calculations

72

M A

NA


Your Turn25.00 mL of HNO3 are titrated with 75.00 mL of 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

73

3

2

32

HNOmL 00.25Ca(OH) mmol 1

HNO mmol 2Ca(OH) 30.1mL 00.75 M

= 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 g74

Fe mol 1Fe g 85.55

HCl mol 2Fe mol 1

HCl 3.2mL 1000

L 1HClmL 00.25

M

= 1.61 g

chapter 5 molecular view of reactions in aqueous solutions: part 2 chemistry: the molecular nature...

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