ch 4. reactions of ions and molecules in aqueous solutions · 4.1. special terminology applies to...

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Ch 4. Reactions of Ions and Molecules in Aqueous Solutions

Brady & Senese 5th Ed.

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Link to Sections

4.1. Special terminology applies to solutions4.2. Ionic compounds conduct electricity when dissolved in water4.3. Acids and bases are classes of compounds with special

properties4.4. Naming acids and bases follows a system4.5. Ionic reactions can often be predicted4.6. The composition of a solution is described by its concentration4.7. Molarity is used for problems in solution stoichiometry4.8. Chemical analysis and titration are applications of solution

stoichiometry

4.1. Special terminology applies to solutions 3

Solutions

• solution–a homogeneous mixture in which the two or more components mix freely

• solvent- the component present in the largest amount

• solute– the substance dissolved in the solvent. The solution is named by the solute.

• concentration- a solute-to-solvent ratio describing the composition of the mixture


The dilute solutionon the left has less solute per unit volume than the (more) concentrated solutionon the right

Relative concentration terms


• saturated–no more solute can be dissolved at the current temperature in the given amount of solvent

• solubility - the amount of solute that can dissolve in the specified amount of solvent at a given temperature (usually g solute/ 100 g solvent or moles solute/L solution)

• unsaturated- contains less solute than the solubility allows

• supersaturated-contains more solute thansolubility predicts

Solubility


Your turn!

The solubility of NaCl is 39.0 g / 100 g water at

100 ºC. If 10.0 g are dissolved in 50.0 g water at this temperature, the solution is:

A. saturated

B. unsaturated

C. supersaturated

D. none of these


• Most solid solutes are more soluble at higher temperatures.

• Careful cooling of saturated solutions may result in a supersaturated solution

• often form a precipitate (ppt.)

Supersaturated solutions are unstable

4.2 Ionic Compounds Conduct Electricity When Dissolved in Water 8

Ionic compounds in water

• Water molecules arrange themselves around the ions and dissociatethem from the lattice.

• The separated ions are “hydrated” and conduct electrical current (act as electrolytes)

• Polyatomic ions remain intact in the dissociation process.

4.2. Ionic compounds conduct electricity when dissolved in water 9

Molecular compounds in water

• The solute particles are surrounded by the water, but the molecules are not dissociated


Electrical conductivity

• Strong electrolyte– aqueous solution that conducts electricity because solute is 100% dissociated into ions

• Weak electrolyte–aqueous solution that weakly conducts electricity due to low ionization

• Non-electrolyte– an aqueous solution that doesn’t conduct electricity because solute does not dissociate into ions


Ionic equations show dissociated ions

• hydrated ions, with the symbol (aq), are written separately

• Na2SO4(s)→ 2Na+(aq) + SO4

2-(aq)

• you might encounter the equation as:

• Na2SO4(s)→ 2Na+ + SO42-

� Accepted because only 2 states allow for dissociated ions (plasma and aqueous). Aqueous is far more common

� It is vague and not preferred


Learning check

Write the equations that illustrate the dissociation of the following:

• Na3PO4(aq)→

• Al 2(SO4)3(aq) →

• CaCl2(aq) →

• Ca(MnO4)2(aq) →

2Al3+(aq) + 3SO4

2-(aq)

3Na+(aq) + PO4

3-(aq)

Ca2+(aq) + 2Cl-(aq)

Ca2+(aq) + 2MnO4

-(aq)


Your turn!

Which of the following would not be expected to produce Cl-(aq) when dissolved?

A. PCl3(aq)

B. NaCl(aq)

C. HCl(aq)

D. none of the above

E. all produce Cl-(aq)


Your turn!

How many ions form on the dissociation of Na3PO4?

A. 1

B. 2

C. 3

D. 4

E. none of the above


Your turn!

How many ions form on the dissociation of Al2(SO4)3?

A. 2

B. 3

C. 4

D. 5

E. none of the above


Writing chemical equations

• Molecular equation: � Balanced, shows states, all substances electrically neutral

� AgNO3(aq)+ KCl(aq)→AgCl(s) + KNO3(aq)

• Ionic equation:� Balanced, shows states, shows strong electrolytesas

dissociatedions, net charges balance

� Ag+(aq) + NO3

-(aq) + K+

(aq) + Cl-(aq)→AgCl(s) + K+(aq) + NO3

-(aq)

• Net ionic equation:� Balanced, shows states, eliminates spectator ionsfrom the

ionic equation, net charges balance

� Ag+(aq) + Cl-(aq)→AgCl(s)


Writing ionic equations

Since strong electrolytes exist as dissociated ions in solution, we can show this in an equation.

1. identify the strong electrolytes

2. distinguish counting subscripts (those present only to make charges cancel) from characteristic subscripts-- counting subscripts become multipliers

3. separate the ions in the strong electrolytes

4. show the states as recorded in the molecular equations


Learning check:

• Write the ionic equations for each:

• BaCl2(aq)+ Pb(NO3)2(aq)→PbCl2(s) + Ba(NO3)2(aq)

• Ba2+(aq) + 2Cl-(aq) + Pb2+

(aq) + 2NO3-(aq)→PbCl2(s) +

Ba2+(aq) + 2NO3

-(aq)

• Na2CO3(aq)+CaCl2(aq) →CaCO3(s) +2NaCl(aq)

• 2Na+(aq) + 2CO3

2-(aq) + Ca2+

(aq) + 2Cl-(aq)→CaCO3(s) + 2Na+

(aq) + 2Cl-(aq)


Writing net ionic equations

• Show only those ions that were changed by the process

• Omits spectator ions:� When we compare the reactant to product, spectator

ions are those ions that are not changed in any way


Learning check:

• Write the following as net ionic equations:

• Pb2+(aq) + 2NO3

-(aq) + 2K+

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

-(aq)

• Ba2+(aq)+ 2Cl-(aq) + 2Na+

(aq) + SO42-

(aq)→ BaSO4(s) + 2Na+(aq) )+ 2Cl-(aq)

• 2Na+(aq) )+ 2Cl-(aq) + Hg2

2+(aq) + 2NO3

-(aq)→ 2Na+

(aq) + 2NO3-(aq) +

Hg2Cl2(s)

2Cl-(aq) + Hg22+

(aq)→ Hg2Cl2(s)

Ba2+(aq)+ SO4

2-(aq)→ BaSO4(s)

Pb2+(aq) + 2I-(aq)→ PbI2(s)


Your turn!

Consider the following reaction :

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

Which is the correct total ionic equation?

A. 2Na+(aq) + SO4

2-(aq) + Ba2+

(aq)+ Cl22-(aq)→

2Na+(aq) +2Cl-(aq) + BaSO4(s)

B. 2Na+(aq) + SO4

2-(aq) + Ba2+

(aq)+ 2Cl-(aq)→ 2Na+(aq)

+2Cl-(aq) + BaSO4(s)

C. 2Na+(aq) + SO42-

(aq) + Ba2+(aq)+ Cl22-

(aq)→2Na+

(aq) +2Cl-(aq) + Ba2+(s) + SO4

2-(s)

D. None of these

4.3.Acids and bases are classes of compounds with special properties 22

• An acid is a substance that ionizes in a reaction with water to form the hydronium ion, H3O

+

• Strong acidsare 100% ionized when dissolved, whereas weak acidsare far less efficiently ionized

The Arrhenius definition of acids

(weak)OHCOHOH OHHC

(strong)Cl OH OH HCl

(aq)-

232)(3 (l)22(aq)32

)(-

)(3(l)2)(

++

+→++

+

aq

aqaqg


• It is common to encounter the hydrogen ion (H+) instead of the hydronium ion

• The previous ionization is, for simplicity, also written as:

H+ does not ever exist in aqueous solution- it is always attached to a water molecule as the hydronium ion

(aq)-

(aq)OH

(g) Cl H HCl 2 + → +

What is H(aq)+?


Nonmetal oxides can be acids

• Nonmetal oxides, or “acidic anhydrides” react with water to form acid solutions

• SO2(g) + H2O(l)→H2SO3(aq)

• CO2(g) + H2O(l)→H2CO3(aq)


Arrhenius bases

• Base- substance that produces hydroxideions in water

• Molecular bases undergo an ionization(hydrolysis) reaction to form the hydroxide ions, and are weak bases

• Many N-compounds are molecular bases� B(aq) + H2O(l) HB+

(aq) + OH-(aq)


Metal oxides and hydroxides are bases

• Metal hydroxide solutions dissociateinto metal and hydroxide ions and are strong bases.

• NaOH(s)→Na+(aq) + OH-

(aq)

• Soluble metal oxides “basic anhydrides” react with water to form metal hydroxides that are strong bases

• CaO(s) +H2O(l) → Ca2+(aq) + 2OH-

(aq)


Strong vs. weak

• Some acids ionize 100% in water, and are termed “strong acids” and are also “strong electrolytes”� HCl, HClO4, HNO3, HBr, HI, H2SO4

• The very soluble metal hydroxides are strong electrolytes and “strong bases”.� Ia hydroxides and Ca, Ba, and Sr hydroxides.


Weak acids and bases are weak electrolytes

4.4. Naming acids and bases follows a system 29

Naming binary acids (aqueous)

• prefix hydro-+ nonmetal stem + the suffix –ic, followed by the word acid

• Stem is first syllable of element name. i.e. Chlorine• P and S stems use 2 syllables phosphorus, sulfur• the name of the (aq) form differs from other states

due to the ionization that occurs in water

hydrosulfuric acidH2S(aq)hydrogen sulfideH2S(g)

hydrochloric acidHCl(aq)hydrogen chlorideHCl(g)

Aqueous Binary AcidMolecular compound


Your turn!

Which of the following is not named as an hydro___ic acid?

A. HCl

B. H2S

C. HNO3

D. HF

E. all are named in this way


Oxoacids (aqueous)

• named according to the anion suffix� anion ends in -ite, the acid name is -ous acid� ends in -ate, the acid name is -ic acid

phosphorous acidH2PO3(aq)sulfuric acidH2SO4(aq)

chlorous acidHClO2(aq)nitric acidHNO3(aq)

-ite anion acids-ate anion acids


Learning check:

• HNO2

• HCN

• HClO4

• HF

• HMnO4

• H2CO3

• nitrous acid

• hydrocyanic acid

• perchloric acid

• hydrofluoric acid

• permanganic acid

• carbonic acid

Name each aqueous acid


Your turn!

Which of the following is the correct name for HClO4(aq)?

A. chloric acid

B. hydrochloric acid

C. perchloric acid



Your turn!

Which of the following is the correct name for H2SO3(aq)?

A. sulfuric acid

B. sulfurous acid

C. hydrosulfuric acid



• polyprotic acidscan be partially neutralized to form acid salts

• acidic salt-contains an anion that is capable of furnishing additional hydrogen ions

• The number of hydrogen atoms that can still be neutralized is also indicated in the name

phosphate dihydrogen sodium PONaH

phosphatehydrogen sodium HPONa

sulfatehydrogen sodium NaHSO

42

42

4

Acid salts

4.5. Ionic reactions can often be predicted 36

A reaction will exist if…

• A precipitate (insoluble product) forms from soluble reactants

• An acid reacts with a base

• A weak electrolyte product is formed from strong electrolyte reactants

• A gas is formed from a mixture of reactions


Metathesis (double replacement) reactions

• AB + CD → AD + CB

• Cations change partners

• Charges on each ion don’t change

• Formulas of the products are determined by the charges of the reactant ions

• Metathesis reactions occur only if they form a weak electrolyte or non-electrolyte as a product (otherwise, all ions are spectator ions)


Predicting metathesis reactions

• Identify the ions involved: � Do not confuse counting subscripts (those present only

to make charges cancel) with those that are characteristic of a polyatomic ion

• Swap partners and make neutral with appropriate subscripts

• Assign states using solubility rules• Balance the equation

HCl(aq) + Ca(OH)2(aq) →

ions: H+, Cl- Ca2+ , OH-

counting subscript

CaCl2 + H2O(aq) (l)22


Solubility rules: soluble compounds

A general idea as to whether a fair amount of solid will dissolve is achieved using solubility rules

1. All compounds of the alkali metals (Group IA)

2. All salts containing NH4+, NO3−, ClO4

−, ClO3−, and

C2H3O2−

3. All chlorides, bromides, and iodides (salts containing Cl−, Br−, or I−) except when combined with Ag+, Pb2+, and Hg22+

4. All sulfates (salts containing SO42−) except those of

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


Solubility rules: insoluble 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 of Ca2+, Sr2+, and Ba2+

When metal oxides dissolve, they react with water to form hydroxides. The oxide ion, O2−, does not exist in water. For example, Na2O(s) +H2O(l)

→ 2NaOH(aq)

6. All salts that contain PO43−, CO32−, SO3

2−, and S2−

are insoluble, except those of Group IA and NH4+.


Learning check:

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

Ca(C2H3O2)2

FeCO3

AgCl

Yes

No

No


Learning Check:

• Pb(NO3)2(aq)+ Ca(OH)2(aq) →

• BaCl2(aq)+ Na2CO3(aq) →

• Na3PO4(aq)+ Hg2(NO3)2(aq) →

• NaCl(aq) + Ca(NO3)2(aq)→

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

BaCO3(s) + NaCl(aq)

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

CaCl2(aq)+ NaNO3(aq)

NR (No reaction)

Predict the products of the following:


Your turn!

Which 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. None of the above


Predicting acid-base reactions

• Neutralization: metathesis reaction in which acid + metal hydroxide or metal oxide forms water and salt � NaOH(aq) + HCl(aq)→H2O(l) + NaCl(aq)

• Acid-base reaction: reaction of weak base and acid transferring a H+ ion, driven by the formation of a weaker acid.� HCl(aq) + NH3(aq)→NH4Cl(aq)


Learning check

Determine the molecular, total ionic and net ionic equations

• Molecular Equation

• Total Ionic Equation (TIE)

• Net Ionic Equation (NIE)

2HCl(aq) + Ca(OH)2(aq) → 2H2O(l) + CaCl2(aq)

H+(aq) + OH-

(aq)→ H2O(l)

2H+(aq)+2Cl-(aq)+ Ca2+

(aq)+2OH-(aq) 2H2O(l)→ + Ca 2+

(aq)+ 2Cl-(aq)


Your turn!

Which of the following is not a product of the reaction: NH3(aq)+HCN(aq) →?

A. NH3CN(aq)

B. NH4+

(aq)

C. CN-(aq)

D. None of the above


Your turn!

Which is the net ionic equation for the reaction:

NaOH(aq) + HF(aq)→?

A. Na+(aq)+ OH-

(aq) + H+(aq) + F-

(aq) →H2O(l) + NaF(aq)

B. OH-(aq) + H+

(aq) →H2O(l)

C. OH-(aq) + HF(aq)→H2O(l) + F-

(aq)

D. Na+(aq)+ OH-

(aq) + HF(aq) →H2O(l) + NaF(aq)



Metathesis and gas formation

• If the product of a metathesis reaction is one of the following, formation of a gas is a driving force.

• Gases formed by metathesis: H2S, HCN

• Unstable compounds that decompose and form a gas:� H2CO3 (H2O & CO2(g))

� NH4OH (H2O & NH3(g))

� H2SO3 (H2O & SO2(g))


Metathesis overview

• Precipitation:2 solutions form solid product

• Neutralization: acid + metal hydroxide or oxide form water and a salt

• Gas-forming: metathesis reaction forms one of these products:

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

• Check for a driving force: formation of weak electrolyte or non-electrolyte


Your turn!

Which of the following combinations will not react?

A. Na2CO3 (aq)+ HCl(aq)

B. Na2CO3(aq)+ CaCl2(aq)

C. NaCl(aq) + H2C3O2(aq)

D. None of these

E. All of these

4.6. The composition of a solution is described by its concentration 51

Molar concentrations

• In solutions, solutes are dispersed in a larger volume

• Molarity expresses the relationship between the moles of solute and the volume of the solution

• Molarity (M)=moles solute/L solution� Hence, a 6.0Msolution of HCl contains 6.0 mole HCl

in a liter of solution


Learning check:

• What is the molarity of a solution created by dissolving 10.2g KNO3 in enough water to make 350 mL solution?

• What mass of KNO3 are found in 25.33 mL of .0500M KNO3 solution?

=

×

×× −

3

33 2

KNOmol

KNOg1033.101

soln L

KNOmol0.050010533.2 L

MM KNO3 = 101.1033 g/mol

0.128 g

33 89

KNO Msoln L 0.350

KNO mol0.100 =

0.29 M


Your turn!

If 10.0 g NaCl (58.443 g/mol) are dissolved in 75.0 mL. What is the molarity?

A. 0.133 M

B. 2.28 M

C. 7.5 M

D. 0.00228M



Dilution

• Adding solvent to a solution creates a less concentrated solution

• moles of solute do not change, hence CstockVstock= CnewVnew� C=concentration� V=volume

• Using volumetric glassware ensures that the volumes are known precisely


Dilution allows molecules more room

• Adding solvent does not change how many moles of solute are present

• The total volume does change• The concentration of the solution is decreased while

the actual amount of solute is unchanged


Dilution of K2Cr2O7

• A volumetric pipette is used to transfer the stock solution

• A volumetric flask is used to receive the final solution


Learning Check

• What volume of 12.1M HCl are needed to create 250. mL of 3.2 M HCl?

• 25 mL of 6 M HCl are diluted to 500 mL with water. What is the molarity of the resulting solution?

newVmL 500 M6mL 25 ×=× 0.3 M

3.2MmL 250. M12.1Vstock ×=× 66 mL


Your Turn!

When 20.00 mL of 3.11 M HCl are added to 15.00 mL of water, what is the resulting concentration?

A. 1.77 M

B. 4.15 M

C. 1.33 M

D. None of these

4.7. Molarity is used for problems in solution stoichiometry 59

Solution stoichiometry

• A balanced equation is needed to start any stoichiometry problem

• If we are given starting quantities of more than one reactant, must determine the limiting reagent

• The difference arises in how we calculate moles of reacting substance


• What volume of 2M HCl is needed to react 25.2 g Na2CO3

(MM=105.9887) completely?

• How many moles of BaSO4 will form if 20.0 mL of 0.600 M BaCl2 is mixed with 30.0 mL of 0.500 M MgSO4?

• BaCl2(aq)+ MgSO4(aq)→BaSO4(s) + MgCl2(aq)

Solution stoichiometry

L 0.238HCl mol 2

L

CONa mol 1

HCl mol 2

105.9887g

CONa mol 1

1

CONa g 25.2

32

3232 =×××

44

44

42

42

BaSO mol 0150.0MgSO mol 1

BaSO mol 1

L

MgSO mol 0.500

1

L 0.0300

BaSO mol 0.0120BaCl mol 1

BaSO mol 1

L

BaCl mol 0.600

1

L 0.0200

=××

=××0.0120 mol


Your turn!

What mass of Na2CO3 (MM=105.9887) can be neutralized with 25.00 mL of 3.11 M HCl?

A. 53.0 g

B. 1.65(102) g

C. 8.24 g

D. 4.12 g

E. None of these


Ion Concentrations

• The chemical formula for a strong electrolyte relates the moles of ions that will be released on dissociation to the chemical formula

• Thus, the formula can be used to relate the ion concentration to the solution concentration

• Learning check: What is the concentration of Cl-

in 0.600 M BaCl2?

−=−

Cl M 1.20BaCl 1mol

Cl 2molx

soln L

BaCl mol 0.600

2

2


Learning check:

What volume of solution containing 0.5M Ag+ will be required to react 100.0mL of 0.0075M Cl-?

Ag+(aq) + Cl-(aq)→AgCl(s)

moles 0.0075M*0.100L=7.5×10-4 mol Cl-

1.5×10-3 L = 2 mL

stoichiometry: 7.5×10-4mol Cl-×(1molAg+ / 1 mol Cl-)

× +−

Ag mol 0.5

soln L*mol107.5 4

4.8. Chemical analysis and titration are applications of solution stoichiometry 64

.

Titration

• Is the controlled addition of one reactant (titrant) to a known quantity of another (titrate) until the reaction is complete

• Often, an indicator is used to signal the reaction completion

• Endpoint: the volume of titrant required to complete the reaction


Titration in practice:


Solving titration problems

• Write the balanced equation

• Calculate the moles of the known component� M ×L = moles or mass/MM=moles

• Use stoichiometry to determine moles of the unknown

• Convert moles to desired quantity


Path for working titration problems


Learning Check:

• 25.00 mL of HCl are titrated with 75.00 mL of 1.30M Ca(OH)2. What is the concentration of HCl?

• 2HCl(aq) + Ca(OH)2(aq)→CaCl2(aq)+ 2H2O(l)

=×××0.02500L

1

Ca(OH) mol 1

HCl mol 2

L

Ca(OH) mol 1.30

1

L 0.07500

2

2

7.80 M HCl


Learning Check:

A sample of metal ore is reacted according to the following reaction: Fe(s) + 2H+

(aq)→ Fe2+(aq) + H2(g).

If 25.00 mL of 2.3M HCl are used, what mass of Fe was in the ore?

1.6gFe mol

g 55.845

H mol 2

Fe mol 1

HCl mol 1

H mol 1

L

HCl mol 2.3

1

L 0.02500 =×+

×××+


Your Turn!

The CO32- content of rock is determined by titration

with acid according to the reaction :

CO32- + 2H+ →.CO2(g)_+H2O(l)

If 12.50 mL of 3.5 M H2SO4 are required to titrate the carbonate in the rock, what mass of CaCO3

(MM=100.089) is present in the sample?

A. 4.38 g

B. 0.0428 g

C. 9.76 g

D. none of these

ch 4. reactions of ions and molecules in aqueous solutions · 4.1. special terminology applies to...

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